CN112066594A - An orderly control method for large-scale electric heat pump based on μPMU device - Google Patents

An orderly control method for large-scale electric heat pump based on μPMU device Download PDF

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CN112066594A
CN112066594A CN202010893913.6A CN202010893913A CN112066594A CN 112066594 A CN112066594 A CN 112066594A CN 202010893913 A CN202010893913 A CN 202010893913A CN 112066594 A CN112066594 A CN 112066594A
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余昆
陈星莺
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    • FMECHANICAL ENGINEERING; LIGHTING; HEATING; WEAPONS; BLASTING
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Abstract

本发明公开了一种基于μPMU装置的大规模电热泵有序用电控制方法,利用μPMU的快速、准确的测量数据,结合电热泵模型及启动特性选择特征量并制定电热泵启动状态的特征判据,然后按照考虑集群电热泵的配电网电压约束模型所耦合的功率约束对三种工作状态下的电热泵负荷群进行有序用电控制方法。该方法控制电热泵有序停机,削减稳态功率以及控制启动序列,避免了电热泵启动冲击叠加越限。

Figure 202010893913

The invention discloses a method for controlling orderly power consumption of a large-scale electric heat pump based on a μPMU device, which utilizes the fast and accurate measurement data of the μPMU, selects characteristic quantities in combination with an electric heat pump model and start-up characteristics, and formulates a characteristic judgment of the start-up state of the electric heat pump. Then, according to the power constraint coupled by the distribution network voltage constraint model considering the cluster electric heat pump, the orderly power consumption control method is carried out for the electric heat pump load group under the three working states. The method controls the electric heat pump to stop in an orderly manner, reduces the steady-state power and controls the starting sequence, and avoids the superposition of the electric heat pump starting shock and exceeding the limit.

Figure 202010893913

Description

一种基于μPMU装置的大规模电热泵有序用电控制方法An orderly control method for large-scale electric heat pump based on μPMU device

技术领域technical field

本发明涉及一种电热泵用电控制方法,尤其涉及一种基于μPMU装置的大 规模电热泵有序用电控制方法。The invention relates to a method for controlling electricity consumption of an electric heat pump, in particular to a method for controlling electricity consumption in an orderly manner for a large-scale electric heat pump based on a μPMU device.

背景技术Background technique

电热泵是以电能驱动热泵机械运转的热泵系统,具有适用范围广,运行成本 低,性能稳定和环保的优点,目前已广泛用于住房的采暖和热水供应,但电热泵 具有单机功率大、用户用电集中的特点,对负荷日峰谷差、季节峰谷差造成了显 著影响,尤其在负荷量急剧增加的供暖季,加上居民供暖、出行的不规律性,使 得配电网负荷峰谷差变大、负荷三相不平衡问题显著。在低压配电网中,电热泵 短暂的启动过程造成的启动冲击会引起明显的电流涌流的突增型脉冲和电压跌 落型脉冲。Electric heat pump is a heat pump system that drives heat pump mechanical operation by electric energy. It has the advantages of wide application range, low operating cost, stable performance and environmental protection. At present, it has been widely used for heating and hot water supply in housing. The characteristics of centralized power consumption by users have a significant impact on the daily peak-to-valley difference and seasonal peak-to-valley difference, especially in the heating season when the load increases sharply, coupled with the irregularity of residents' heating and travel, the peak load of the distribution network. The valley difference becomes larger and the load three-phase imbalance problem is obvious. In the low-voltage distribution network, the start-up shock caused by the short-term start-up process of the electric heat pump will cause obvious current inrush pulses and voltage-drop pulses.

受电热泵的运行不规律和启动冲击影响,三相电压的瞬时波动极易超出电能 质量的规定,严重的电压降低会造成电热泵的启动失败,启动冲击不仅影响低压 配电网的电能质量、可能使得电热泵自身无法启动,还可能影响到其他电热泵及 其他用电设备运行和启动。现有技术通过对配电网进行广泛地建设和改造来保障 用电,但这种方法不仅经济性差,且建设周期长不适于实际应用。Affected by the irregular operation of the electric heat pump and the start-up shock, the instantaneous fluctuation of the three-phase voltage can easily exceed the power quality regulations. A serious voltage drop will cause the start-up failure of the electric heat pump. The start-up shock not only affects the power quality of the low-voltage distribution network. It may make the electric heat pump itself unable to start, and may also affect the operation and start of other electric heat pumps and other electrical equipment. The existing technology guarantees electricity consumption by extensively constructing and transforming the distribution network, but this method is not only economical, but also has a long construction period and is not suitable for practical application.

发明内容SUMMARY OF THE INVENTION

发明目的:本发明提出一种基于μPMU设备的大规模电热泵有序用电控制 方法,减小电热泵大规模使用时的启动冲击。Purpose of the invention: The present invention proposes a method for controlling the orderly power consumption of a large-scale electric heat pump based on μPMU equipment, which reduces the start-up impact when the electric heat pump is used on a large scale.

技术方案:本发明采用的技术方案为一种基于μPMU装置的大规模电热泵 有序用电控制方法,包括以下步骤:Technical scheme: the technical scheme adopted in the present invention is a large-scale electric heat pump based on μPMU device orderly electricity control method, comprising the following steps:

1)电热泵制热系统建模;1) Modeling of electric heat pump heating system;

2)基于μPMU量测辨识电热泵状态;2) Identify the state of the electric heat pump based on μPMU measurement;

3)基于量测辨识的电热泵有序用电策略。3) Orderly electricity consumption strategy of electric heat pump based on measurement and identification.

所述步骤1)中电热泵制热系统中Said step 1) in the electric heat pump heating system

压缩机电功率Pcmp关于转速N、制冷剂流量mr的函数关系式:The functional relationship between the electric power P cmp of the compressor and the rotational speed N and the refrigerant flow rate m r :

Figure BDA0002657814280000011
Figure BDA0002657814280000011

上式中c1为压缩机吸气比容,n为压缩过程多变指数,ηcmp为压缩机指示功 率和耗电功率的比值,

Figure BDA0002657814280000021
表示排气压力与吸气压力之比;In the above formula, c 1 is the suction specific volume of the compressor, n is the variable index of the compression process, η cmp is the ratio of the indicated power of the compressor to the power consumption,
Figure BDA0002657814280000021
Represents the ratio of exhaust pressure to suction pressure;

冷凝器的换热模型如下:The heat transfer model of the condenser is as follows:

Figure BDA0002657814280000022
Figure BDA0002657814280000022

式中,Cr,sh,cond为制冷剂比热系数;T2为流入冷凝器的制冷剂温度;Tcond为 冷凝器的过热温度;T3为流出冷凝器的制冷剂温度;rcond为板式冷凝器的结构半 径;Ai,cond为板式冷凝器的换热面积;Tw是冷凝器热水侧的出水口温度;Usc,cond、 Ush,cond、Utp,cond分别为冷凝器冷水测、过热区、两相区换热综合系数,Tw为热 水温度;In the formula, C r, sh, cond is the specific heat coefficient of the refrigerant; T 2 is the temperature of the refrigerant flowing into the condenser; T cond is the superheat temperature of the condenser; T 3 is the temperature of the refrigerant flowing out of the condenser; r cond is The structural radius of the plate condenser; A i,cond is the heat exchange area of the plate condenser; Tw is the outlet temperature of the hot water side of the condenser; U sc,cond , U sh,cond , U tp,cond are the condensation is the comprehensive coefficient of heat transfer in cold water measurement, superheated zone and two-phase zone of the boiler, and Tw is the temperature of hot water;

蒸发器的换热量满足:The heat exchange of the evaporator satisfies:

Qeva=Qtp+Qsh=Qa (3)Q eva =Q tp +Q sh =Q a (3)

上式中Qeva为蒸发器中转移给制冷剂的热量,Qa为蒸发器吸收自热源介质的 热量。In the above formula, Q eva is the heat transferred to the refrigerant in the evaporator, and Q a is the heat absorbed by the evaporator from the heat source medium.

异步电机的d-q坐标系复变量模型描述为:The complex variable model of the d-q coordinate system of the asynchronous motor is described as:

Figure BDA0002657814280000023
Figure BDA0002657814280000023

上式中,

Figure BDA0002657814280000024
分别定子电压、定子电流、转 子电流、定子磁链和转子磁链的复数向量;Rs、Ls、M、Rr、Lr分别为定子 电阻、定子电感、定转子互感、转子电阻和转子电感;ls=Ls-M和lr=Lr-M分 别为定子和转子漏电感;ωr为转子电角速度,单位为rad/s;j是复数单位。In the above formula,
Figure BDA0002657814280000024
Complex vectors of stator voltage, stator current, rotor current, stator flux linkage and rotor flux linkage, respectively; R s , L s , M, R r , L r are stator resistance, stator inductance, stator-rotor mutual inductance, rotor resistance and rotor, respectively Inductance; l s =L s -M and l r =L r -M are the stator and rotor leakage inductances, respectively; ω r is the rotor electrical angular velocity, in rad/s; j is a complex unit.

所述步骤1)中电热泵启动过程有功功率Ps关于时间t的闭区间函数关系为:In the step 1), the closed interval functional relationship of the active power P s with respect to the time t in the starting process of the electric heat pump is:

Ps(t)=-1160(t-0.2)2+Pmax t∈[t0,t1] (5)P s (t)=-1160(t-0.2) 2 +P max t∈[t 0 ,t 1 ] (5)

上式中t0为启动初始时刻,t1为启动完成电热泵进入稳态的时刻,Pmax为电 热泵启动功率最大值。In the above formula, t 0 is the initial start time, t 1 is the time when the electric heat pump enters a steady state after starting up, and P max is the maximum starting power of the electric heat pump.

所述步骤2)中首先建立首端节点电压Ui约束,台区首端节点电压Ui需满 足:In the step 2), the constraint of the voltage U i of the head-end node is first established, and the voltage U i of the head-end node of the station area needs to satisfy:

Ui≥{max(Ulim.k)|Tekmaxrk,Ulim.k)=Tmmaxrk),k∈[1,n]} (6)U i ≥{max(U lim.k )|T ekmaxrk ,U lim.k )=T mmaxrk ),k∈[1,n]} (6)

上式中Ulim.k为节点k处电热泵负荷能正常启动时临界值,Tekmaxrk,Ulim.k)为 第k台电热泵负荷中作用在电机转子上的电磁转矩,ωmaxrk为Tek取最大值时所对 应的最大转子转速,Ulim.k为节点k处临界启动电压,Tmmaxrk)为异步电机转速 标幺值ωmaxrk下的电热泵机械转矩。In the above formula, U lim.k is the critical value when the electric heat pump load at node k can start normally, T ekmaxrk , U lim.k ) is the electromagnetic torque acting on the motor rotor in the kth electric heat pump load, ω maxrk is the maximum rotor speed when T ek takes the maximum value, U lim.k is the critical starting voltage at node k, and T mmaxrk ) is the mechanical torque of the electric heat pump at the per-unit value ω maxrk of the asynchronous motor speed.

接着根据功率冲击和电压暂降判断电热泵启动状态,其中功率冲击判据如 下:Then judge the start-up state of the electric heat pump according to the power shock and voltage sag. The power shock criterion is as follows:

Figure BDA0002657814280000031
Figure BDA0002657814280000031

上式中P(t)为当前时刻t的负荷功率;P(t-1)为前一量测时刻的负荷功率; N为时刻t的量测频数,功率变化量取n个量测数据依次作差并求和得到;P′(t) 为当前时刻t的负荷功率变化速率;λ1、λ2为判据对应的阈值。In the above formula, P(t) is the load power at the current time t; P(t-1) is the load power at the previous measurement time; N is the measurement frequency at time t, and the power change takes n measurement data in turn Differences and summations are obtained; P'(t) is the rate of change of the load power at the current time t; λ 1 and λ 2 are the thresholds corresponding to the criteria.

电压暂降判据为:The voltage sag criterion is:

Figure BDA0002657814280000032
Figure BDA0002657814280000032

式中,λ3和λ4为用于描述Ut-1与Ut-2的接近程度阈值,N为时刻t的量测频数。In the formula, λ 3 and λ 4 are the thresholds used to describe the proximity of U t-1 and U t-2 , and N is the measurement frequency at time t.

所述步骤2)中利用电热泵启动冲击功率中二次函数的数量,辨识出处于启 动状态的电热泵台数。In described step 2), utilize the quantity of quadratic function in electric heat pump start-up impact power, identify the number of electric heat pumps in start-up state.

所述步骤3)中将负荷电压转化为负荷功率,具体如下:In the step 3), the load voltage is converted into load power, as follows:

首端节点n的电压和注入有功功率满足:The voltage and injected active power of the head-end node n satisfy:

Un=kP·[P-P(t0)]+Un(t0) (9)U n =k P ·[PP(t 0 )]+U n (t 0 ) (9)

式中,Un为首端节点n所求的电压;P为首端节点n所求的注入功率, Un(t0)、P(t0)为t0时刻测出的当前节点电压和功率,kp为斜率,在其中一个 量已知的情况下可以计算出另外一个量的值。In the formula, U n is the voltage required by the head-end node n; P is the injected power required by the head-end node n , Un (t 0 ), P (t 0 ) are the current node voltage and power measured at time t 0 , k p is the slope, and when one of the quantities is known, the value of the other quantity can be calculated.

所述有序用电策略实现步骤如下:The steps of implementing the orderly power consumption strategy are as follows:

1)根据量测、辨识方法得到负荷信息并确定处于不同状态的负荷群;1) Obtain load information according to measurement and identification methods and determine load groups in different states;

2)对启动和运行状态的两个设备群的电热泵负荷进行优先级排序;2) Prioritize the electric heat pump loads of the two equipment groups in startup and running states;

3)确定受控制的电热泵设备顺序及数量。3) Determine the order and quantity of controlled electric heat pump equipment.

所述步骤2)中针对同一时刻下启动的电热泵启动冲击功率的叠加峰值超出 限制,将稳态电压优先级、不同温度优先级以及当前序列位置号相加得到启动序 列编号。In the described step 2), the superimposed peak value of the electric heat pump startup impulse power started at the same moment exceeds the limit, and the steady-state voltage priority, the different temperature priority and the current sequence position number are added to obtain the startup sequence number.

所述步骤2)中针对配电变压器稳态运行的负荷较大时叠加部分启动功率导 致的功率越限,将当前序列位置号、电热泵按照外加电压均值排序的优先级,以 及不同温度优先级相加得到关闭序列编号。In the step 2), for the power exceeding the limit caused by the superimposed partial starting power when the load of the steady-state operation of the distribution transformer is relatively large, the current sequence position number, the priority of the electric heat pump sorted according to the average value of the applied voltage, and the priority of different temperatures Add up to get the closing sequence number.

有益效果:本发明有序用电控制方法基于电热泵状态识别以及μPMU快速 量测,提出一种优化启动冲击的电热泵有序用电控制方法,利用μPMU的快速、 准确的测量数据,结合电热泵模型及启动特性选择特征量并制定电热泵启动状态 的特征判据,然后按照考虑集群电热泵的配电网电压约束模型所耦合的功率约束 对三种工作状态下的电热泵负荷群进行有序用电控制方法。该方法控制电热泵有 序停机,削减稳态功率以及控制启动序列,避免了电热泵启动冲击叠加越限。Beneficial effects: The orderly power consumption control method of the present invention is based on the state identification of the electric heat pump and the fast measurement of the μPMU, and proposes an orderly power consumption control method of the electric heat pump that optimizes the start-up impact, using the fast and accurate measurement data of the μPMU, combined with the electric heat pump. Heat pump model and start-up characteristics Select the characteristic quantities and formulate the characteristic criteria of the start-up state of the electric heat pump. Then, according to the power constraints coupled by the distribution network voltage constraint model considering the cluster electric heat pump, the load groups of the electric heat pump under the three working states are evaluated. Sequence electricity control method. The method controls the electric heat pump to stop in an orderly manner, reduces the steady-state power and controls the starting sequence, and avoids the superposition of the electric heat pump starting shocks exceeding the limit.

附图说明Description of drawings

图1为电热泵机组制热循环图;Fig. 1 is the heating cycle diagram of the electric heat pump unit;

图2为蒸发器传热示意图;Figure 2 is a schematic diagram of heat transfer in an evaporator;

图3为异步电机T型等值电路图;Figure 3 is an equivalent circuit diagram of the T-type asynchronous motor;

图4为异步电机的机械特性图;Fig. 4 is the mechanical characteristic diagram of the asynchronous motor;

图5为辐射状低压配电网等值电路图;Figure 5 is an equivalent circuit diagram of a radial low-voltage distribution network;

图6为节点k的电热泵端口等效电路图;Fig. 6 is the electric heat pump port equivalent circuit diagram of node k;

图7为简化后的戴维南等值电路图;Figure 7 is a simplified Thevenin equivalent circuit diagram;

图8为低压配电网的线路等效图;Figure 8 is a line equivalent diagram of a low-voltage distribution network;

图9为电热泵有序用电控制方法流程图;Fig. 9 is the flow chart of the control method of orderly electricity consumption of electric heat pump;

图10为电热泵负荷的热行为和功率特性图。Figure 10 is a graph of thermal behavior and power characteristics of the electric heat pump load.

具体实施方式Detailed ways

一、电热泵制热系统建模1. Modeling of electric heat pump heating system

电热泵机组的主要内部元件有蒸发器、压缩机、四通阀、冷凝器、通膨阀。 其中压缩机、冷凝器、蒸发器是构成电热泵机组的主要换热、耗能部件:The main internal components of the electric heat pump unit are evaporator, compressor, four-way valve, condenser, and expansion valve. Among them, the compressor, condenser and evaporator are the main heat exchange and energy consumption components that constitute the electric heat pump unit:

A、压缩机:通过不断旋转地做功将机械能转换成流体的内能,被称为电热 泵机组的心脏。压缩机是制冷剂循环中的动力来源,压缩机的转子旋转做功将来 自蒸发器的制冷剂蒸汽进行绝热压缩,并将具有更高压强的蒸汽不断地推入冷凝 器中。A. Compressor: It converts mechanical energy into internal energy of fluid by continuously rotating and doing work, which is called the heart of the electric heat pump unit. The compressor is the power source in the refrigerant cycle. The rotor of the compressor performs adiabatic compression of the refrigerant vapor from the evaporator and continuously pushes the vapor with higher pressure into the condenser.

B、蒸发器、冷凝器:是电热泵机组中的重要换热部件。蒸发器是制冷剂和 热源介质对流蒸发吸热的过程,使得制冷剂获得热能的部件。冷凝器则是制冷剂 经过冷凝放热过程向冷却水中释放热量,实现制冷剂热量转移的部件。B. Evaporator and condenser: It is an important heat exchange component in the electric heat pump unit. The evaporator is a process in which the refrigerant and the heat source medium convectively evaporate and absorb heat, so that the refrigerant obtains heat energy. The condenser is the part that the refrigerant releases heat to the cooling water through the process of condensation and heat release, and realizes the heat transfer of the refrigerant.

图1中元件1为蒸发器;2为压缩机;3为四通阀;4为冷凝器;5为通膨阀;6 为关断水阀;7为室内散热器;8为循环水泵。Ta为环境温度,Tw为热水温度;Qo为输出制热量;Teva,i、Teva,o为蒸发器流入、流出的制冷剂温度;Tcond,i、Tcond,o为 冷凝器流入、流出的制冷剂温度;Tw,i、Tw,o为冷凝器流入、流出的冷却水温度; peva,i、peva,o为蒸发器流入、流出介质的压力;pcond,i、pcond,o为冷凝器流入、 流出介质的压力。1 is an evaporator; 2 is a compressor; 3 is a four-way valve; 4 is a condenser; 5 is an expansion valve; 6 is a shut-off valve; 7 is an indoor radiator; T a is the ambient temperature, Tw is the hot water temperature; Q o is the output heating capacity; T eva,i , T eva,o are the refrigerant temperatures flowing into and out of the evaporator; T cond,i , T cond,o are The temperature of the refrigerant flowing into and out of the condenser; Tw, i and Tw, o are the temperatures of the cooling water flowing into and out of the condenser; p eva,i and p eva,o are the pressures of the medium flowing into and out of the evaporator; p cond,i and p cond,o are the pressures of the medium flowing into and out of the condenser.

如图2所示,一般采用分区模型描述蒸发器内的换热过程:分为两相区和过 热区。蒸发管内制冷剂在两相区吸热,吸收热量后制冷剂沸腾汽化成制冷剂蒸汽。 由于过热区的制冷剂换热系数低,且过热区单位面积的换热量也比两相区的小很 多。因此进入蒸发器的热源介质主要是在两相区向制冷剂完成热量的交换。As shown in Figure 2, a partition model is generally used to describe the heat exchange process in the evaporator: it is divided into two-phase zone and superheat zone. The refrigerant in the evaporating tube absorbs heat in the two-phase region, and after absorbing the heat, the refrigerant boils and vaporizes into refrigerant vapor. Because the heat transfer coefficient of the refrigerant in the superheated zone is low, and the heat exchange per unit area of the superheated zone is also much smaller than that of the two-phase zone. Therefore, the heat source medium entering the evaporator mainly completes the heat exchange to the refrigerant in the two-phase region.

蒸发器热源介质侧的换热量:Heat exchange on the heat source medium side of the evaporator:

Qa=ζ·Cp,a·ma·(Ta-T0) (10)Q a =ζ·C p,a ·m a ·(T a -T 0 ) (10)

式中,Ta为流入的热源介质温度,℃;ζ为析湿系数;Cp,a为介质的比热 容,可取值为1.003kJ/(kg*K);ma为流入蒸发器的低位热源介质质量,且ma=ρ·V; T0为流出蒸发器的介质温度,单位℃。In the formula, Ta is the temperature of the incoming heat source medium, ℃; ζ is the moisture separation coefficient; C p,a is the specific heat capacity of the medium, which can be 1.003kJ /(kg*K); The mass of the heat source medium, and m a =ρ·V; T 0 is the temperature of the medium flowing out of the evaporator, in °C.

蒸发器制冷剂侧的换热量:Heat exchange on the refrigerant side of the evaporator:

蒸发器两相区制冷剂侧换热方程:The heat transfer equation on the refrigerant side of the evaporator in the two-phase zone:

Figure BDA0002657814280000051
Figure BDA0002657814280000051

上式中,γi为管内污垢系数;atp为制冷剂两相换热系数;Din为换热管截面 积;TW2为两相区壁面温度;Teva.i为进入蒸发器的液态制冷剂温度;ltp为两相区 长度。In the above formula, γ i is the fouling coefficient in the tube; a tp is the two-phase heat transfer coefficient of the refrigerant; D in is the cross-sectional area of the heat exchange tube; T W2 is the wall temperature of the two-phase zone; T eva.i is the liquid entering the evaporator. Refrigerant temperature; l tp is the length of the two-phase region.

蒸发器过热区制冷剂侧换热方程:The heat transfer equation of the refrigerant side in the superheated zone of the evaporator:

Figure BDA0002657814280000052
Figure BDA0002657814280000052

式中,aV为制冷剂单相换热系数;TW1为过热区壁面温度;Teva.o为制冷剂蒸 汽温度;lsh为过热区长度。In the formula, a V is the single-phase heat transfer coefficient of the refrigerant; T W1 is the wall temperature of the superheated zone; T eva.o is the refrigerant vapor temperature; l sh is the length of the superheated zone.

可见,蒸发器吸收自热源介质的热量Qa主要由热源介质的温度及流量决定。 由式(11)至式(12)说明蒸发器所能传递至制冷剂的热量主要是由蒸发器两相 区、过热区的结构参数决定的。It can be seen that the heat Q a absorbed by the evaporator from the heat source medium is mainly determined by the temperature and flow rate of the heat source medium. Equations (11) to (12) show that the heat that the evaporator can transfer to the refrigerant is mainly determined by the structural parameters of the two-phase zone and the superheat zone of the evaporator.

压缩机的数学模型Mathematical model of compressor

为了保证制冷剂循环的正常进行,压缩机消耗电功率并将机械能以绝热压缩 输入能量至制冷剂,并不断地推动蒸发器中形成的制冷剂蒸汽经过绝热压缩后送 入冷凝器中。压缩机推动了整个制冷剂循环的进行。In order to ensure the normal operation of the refrigerant cycle, the compressor consumes electrical power and inputs mechanical energy to the refrigerant in adiabatic compression, and continuously pushes the refrigerant vapor formed in the evaporator to be sent to the condenser after adiabatic compression. The compressor drives the entire refrigerant cycle.

压缩机的输入功率与制冷剂循环中的压强关系如下:The relationship between the input power of the compressor and the pressure in the refrigerant cycle is as follows:

Figure BDA0002657814280000061
Figure BDA0002657814280000061

上式中,Pcmp为压缩机消耗的电功率,kW;频率f取50HZ,s为异步电机 的转差率;Vd为压缩机理论输气量(m3);ηV为容积系数,取值为

Figure BDA0002657814280000062
ηcmp为压缩机指示功率和耗电功率的比值,取为0.64;
Figure BDA0002657814280000063
表示排气压力与吸气压力之比,即压缩比;n为压缩过程多变指数,绝热压缩过 程中n为常数。In the above formula, P cmp is the electric power consumed by the compressor, kW; the frequency f is 50HZ, s is the slip rate of the asynchronous motor; V d is the theoretical gas delivery volume of the compressor (m 3 ); value is
Figure BDA0002657814280000062
η cmp is the ratio of the indicated power of the compressor to the power consumption, which is taken as 0.64;
Figure BDA0002657814280000063
Represents the ratio of the exhaust pressure to the suction pressure, namely the compression ratio; n is the variable index of the compression process, and n is a constant in the adiabatic compression process.

制冷剂质量流量mr满足:The refrigerant mass flow m r satisfies:

Figure BDA0002657814280000064
Figure BDA0002657814280000064

式中,N=60f(1-s)为压缩机转速(rpm);c1为压缩机吸气比容(m3/kg)。In the formula, N=60f(1-s) is the compressor rotational speed (rpm); c 1 is the compressor suction specific volume (m 3 /kg).

由上式(13)、式(14)可以建立压缩机电功率Pcmp关于转速N、制冷剂流 量mr的函数关系式:From the above equations (13) and (14), the functional relationship of the compressor electric power P cmp with respect to the rotational speed N and the refrigerant flow rate m r can be established:

Figure BDA0002657814280000065
Figure BDA0002657814280000065

可见,压缩机消耗电能参与制冷剂循环的过程中,其运行功率和转速直接影 响到制冷剂的流量和制冷剂蒸汽的压强(亦是内能)。It can be seen that in the process of the compressor consuming electric energy and participating in the refrigerant cycle, its operating power and speed directly affect the flow rate of the refrigerant and the pressure (also internal energy) of the refrigerant vapor.

冷凝器的数学模型Mathematical model of condenser

压缩机绝热压缩后的高温高压制冷剂蒸汽被推入冷凝器,冷凝器是一种可供 制冷剂向冷却水中释放热量实现制冷剂热能转移的换热部件。冷凝器常采用板式 换热器,根据能量守恒,冷凝器的换热模型如下:The high-temperature and high-pressure refrigerant vapor after adiabatic compression by the compressor is pushed into the condenser, which is a heat exchange component that can release heat from the refrigerant to the cooling water to transfer the heat energy of the refrigerant. The condenser often uses a plate heat exchanger. According to the conservation of energy, the heat exchange model of the condenser is as follows:

Figure BDA0002657814280000071
Figure BDA0002657814280000071

式中,Cr,sh,cond为制冷剂比热系数;T2为流入冷凝器的制冷剂温度;Tcond为 冷凝器的过热温度;T3为流出冷凝器的制冷剂温度;rcond为板式冷凝器的结构半 径;Ai,cond为板式冷凝器的换热面积;Tw是冷凝器热水侧的出水口温度;Usc,cond、 Ush,cond、Utp,cond分别为冷凝器冷水测、过热区、两相区换热综合系数。In the formula, C r, sh, cond is the specific heat coefficient of the refrigerant; T 2 is the temperature of the refrigerant flowing into the condenser; T cond is the superheat temperature of the condenser; T 3 is the temperature of the refrigerant flowing out of the condenser; r cond is The structural radius of the plate condenser; A i,cond is the heat exchange area of the plate condenser; Tw is the outlet temperature of the hot water side of the condenser; U sc,cond , U sh,cond , U tp,cond are the condensation Comprehensive coefficient of heat transfer in cold water measurement, superheat zone and two-phase zone.

上述电热泵机组内各个部件的数学模型分析得到了两种循环介质在各部件 所转换的热量。各个环节通过介质的流量、温度将电热泵的完整运行情况关联在 一起,而在运行控制方面着重关心的参数往往是电热泵的制热量、耗电量以及能 效比。The mathematical model analysis of each component in the above electric heat pump unit has obtained the heat converted by the two circulating media in each component. Each link is related to the complete operation of the electric heat pump through the flow and temperature of the medium, and the parameters that focus on operation control are often the heating capacity, power consumption and energy efficiency ratio of the electric heat pump.

由式(2-2)可知集热器聚集的低位热能Qc主要由热源介质的外界温度To、 流量和集热效率决定。假设这部分热量全部进入蒸发器,则蒸发器热源介质的换 热量等于集热器汇聚的热能,即Qa=Qc。而蒸发器中转移给制冷剂的热量Qeva等 于蒸发器内两相区换热量Qtp和过热区换热量Qsh之和,由式(10)至式(12), 当忽略过热区、两相区换热损失时,蒸发器的换热量满足:From the formula (2-2), it can be known that the low-level heat energy Q c collected by the heat collector is mainly determined by the external temperature To , flow rate and heat collection efficiency of the heat source medium. Assuming that all this heat enters the evaporator, the heat exchange of the heat source medium of the evaporator is equal to the heat energy collected by the heat collector, that is, Q a =Q c . The heat Q eva transferred to the refrigerant in the evaporator is equal to the sum of the heat exchange Q tp in the two-phase area and the heat exchange Q sh in the superheated area. , When the heat exchange loss in the two-phase area, the heat exchange of the evaporator satisfies:

Qeva=Qtp+Qsh=Qa (3)Q eva =Q tp +Q sh =Q a (3)

除了低位热源的热能输入,电热泵机组借助电能做功也为制冷剂循环提供了 能量。由前文的分析,电热泵系统的耗电元件有压缩机、风机、水泵。而其中压 缩机做功对电热泵的制热影响最大。由式(1)可知电热泵的压缩机电功率Pcmp会 直接影响到制冷剂循环中制冷剂的流量mr和压强,式(2)说明制冷剂的流量mr和制冷剂的温度Tcond决定了冷凝器的换热量Qcond。因此,电热泵最终输出的制热 量应当是由两部分构成:低位热源的热能及电热泵压缩机的电能做功。可以得到 电热泵输出的制热量Qo与蒸发器换热量Qeva、压缩机消耗的电功率Pcmp存在如下 等式关系:In addition to the thermal energy input of the low-level heat source, the electric heat pump unit also provides energy for the refrigerant circulation by means of electrical energy. From the previous analysis, the power-consuming components of the electric heat pump system include compressors, fans, and water pumps. Among them, the compressor work has the greatest influence on the heating of the electric heat pump. It can be seen from equation (1) that the electric power P cmp of the compressor of the electric heat pump will directly affect the flow rate m r and pressure of the refrigerant in the refrigerant cycle. Equation (2) shows that the flow rate m r of the refrigerant and the temperature T cond of the refrigerant are determined by the heat exchange Q cond of the condenser. Therefore, the final output heat of the electric heat pump should be composed of two parts: the heat energy of the low-level heat source and the electric energy of the electric heat pump compressor. It can be obtained that the heating quantity Q o output by the electric heat pump, the heat exchange quantity Q eva of the evaporator and the electric power P cmp consumed by the compressor have the following equation relationship:

Qo=Qeva+Pcmp·ηcmp (15)Q o =Q eva +P cmp ·η cmp (15)

能量守恒等式描述了电热泵制热量Qo包含从外界吸收的热量和电功率做功 这两部分能量。工程上描述温控类设备的节能性能时常常用到能效比的概念,能 效比COP就是设备输出的制热量与设备消耗的电功率间的比值:The energy conservation equation describes that the heating capacity Q o of the electric heat pump contains two parts of energy, the heat absorbed from the outside and the work done by the electric power. In engineering, the concept of energy efficiency ratio is often used when describing the energy-saving performance of temperature control equipment.

Figure BDA0002657814280000081
Figure BDA0002657814280000081

COP反应了电热泵的节能制热能力,COP值越大则单位耗电量下电热泵产 生的热量越多。由电热泵供热系统的模型分析可知:电热泵的COP与外界热源 介质温度To、换热器(蒸发器和冷凝器)的换热效率以及主要耗电元件(压缩机) 的运行功率Pcmp有关。COP reflects the energy-saving heating capacity of the electric heat pump. The larger the COP value, the more heat the electric heat pump generates per unit power consumption. From the model analysis of the electric heat pump heating system, it can be known that the COP of the electric heat pump is related to the temperature T o of the external heat source medium, the heat exchange efficiency of the heat exchanger (evaporator and condenser), and the operating power P of the main power-consuming component (compressor). cmp related.

由前文电热泵供热系统的机组各部件模型分析,电热泵供热系统的耗电元件 有压缩机、风机、水泵,其中压缩机是电热泵中耗电较大的元器件。电热泵启动 时的功率需求主要来自压缩机。另外,电热泵的压缩机几乎都是采用具备良好调 速特性的异步电机,因此可采用异步电机的模型来表征电热泵的耗电特性。According to the model analysis of the units of the electric heat pump heating system, the power-consuming components of the electric heat pump heating system include compressors, fans, and water pumps, among which the compressor is the most power-consuming component in the electric heat pump. The power demand when the electric heat pump starts is mainly from the compressor. In addition, almost all compressors of electric heat pumps use asynchronous motors with good speed regulation characteristics, so the model of asynchronous motors can be used to characterize the power consumption characteristics of electric heat pumps.

由于存在高阶的非线性磁链耦合,异步电机在静止坐标系中的数学模型较为 复杂,常常变换到d-q两相旋转坐标系中。异步电机的d-q坐标系复变量模型描 述为:Due to the existence of high-order nonlinear flux coupling, the mathematical model of the asynchronous motor in the static coordinate system is relatively complex, and it is often transformed into the d-q two-phase rotating coordinate system. The complex variable model of the d-q coordinate system of the asynchronous motor is described as:

Figure BDA0002657814280000082
Figure BDA0002657814280000082

上式中,

Figure BDA0002657814280000083
分别定子电压、定子电流、转 子电流、定子磁链和转子磁链的复数向量;Rs、Ls、M、Rr、Lr分别为定子 电阻、定子电感、定转子互感、转子电阻和转子电感;ls=Ls-M和lr=Lr-M分 别为定子和转子漏电感;ωr为转子电角速度,单位为rad/s;j是复数单位。In the above formula,
Figure BDA0002657814280000083
Complex vectors of stator voltage, stator current, rotor current, stator flux linkage and rotor flux linkage, respectively; R s , L s , M, R r , L r are stator resistance, stator inductance, stator-rotor mutual inductance, rotor resistance and rotor, respectively Inductance; l s =L s -M and l r =L r -M are the stator and rotor leakage inductances, respectively; ω r is the rotor electrical angular velocity, in rad/s; j is a complex unit.

由上式(4)可得感应电动机的T型暂态等值电路的矩阵方程式:From the above formula (4), the matrix equation of the T-type transient equivalent circuit of the induction motor can be obtained:

Figure BDA0002657814280000084
Figure BDA0002657814280000084

式中,p为微分算子;

Figure BDA0002657814280000085
为感应电动机的速度感应电势。In the formula, p is the differential operator;
Figure BDA0002657814280000085
Induces an electric potential for the speed of an induction motor.

因为T型暂态等值模型中es的存在,使得模型成为复杂的复数微分方程, 使求解和推导的计算量较大、较困难。考虑到笼型转子异步电机的速度感应电势 为零[40],即es=0,可进一步简化,选择无速度感应电动势的感应电动机进行分 析。其复变量模型可描述为:Because of the existence of es in the T-type transient equivalent model, the model becomes a complex complex differential equation, which makes the calculation amount of solution and derivation larger and more difficult. Considering that the speed induced electromotive force of the squirrel-cage rotor asynchronous motor is zero [40] , that is, es = 0, it can be further simplified, and an induction motor without speed induced electromotive force is selected for analysis. Its complex variable model can be described as:

Figure BDA0002657814280000091
Figure BDA0002657814280000091

如图3所示,在化简的T型暂态等值电路中。忽略了转子回路中的速度感 应电势es,而采用转子等效电阻进行代替。当稳态运行时p=jω,因子

Figure BDA0002657814280000092
T型暂态等值电路都退化为T型稳态等值电路,As shown in Figure 3, in the simplified T-type transient equivalent circuit. The speed induced potential es in the rotor circuit is ignored, and the rotor equivalent resistance is used instead. When p=jω in steady state operation, the factor
Figure BDA0002657814280000092
T-type transient equivalent circuits are degenerated into T-type steady-state equivalent circuits.

可见,暂态T型等值模型中定子电阻、电抗与稳态模型一致,只是暂态等 效模型的转子电阻是稳态转子等效电阻与微分因子

Figure BDA0002657814280000093
的乘积构成。然而, 上述电机动态数学模型具有非线性、强耦合的特点,加上微分算子p及启动过程 中转差率s这类不可观测量的存在使得模型求解和参数辨识较为困难且不便于 研究使用。It can be seen that the stator resistance and reactance in the transient T-type equivalent model are consistent with the steady-state model, but the rotor resistance of the transient equivalent model is the steady-state rotor equivalent resistance and differential factor.
Figure BDA0002657814280000093
The product of . However, the above-mentioned dynamic mathematical model of the motor has the characteristics of nonlinearity and strong coupling, and the existence of unobservables such as differential operator p and slip s in the starting process makes the model solution and parameter identification difficult and inconvenient for research and use.

在异步电机启动的瞬时,因为电机转子还未转动。故异步电机的转差率s=1, 此时电机结构可以等效为次级短路的变压器,所以有

Figure BDA0002657814280000094
将上式代入式 (18)可以得到:At the moment when the asynchronous motor starts, because the motor rotor has not yet turned. Therefore, the slip ratio of the asynchronous motor is s=1, and the motor structure can be equivalent to a secondary short-circuit transformer at this time, so there are
Figure BDA0002657814280000094
Substituting the above formula into formula (18) can get:

Figure BDA0002657814280000095
Figure BDA0002657814280000095

Figure BDA0002657814280000096
Figure BDA0002657814280000096

上式中,L为定子漏感系数。In the above formula, L is the stator leakage inductance coefficient.

在异步机处于空载稳定运行时,转子转速与电磁转速接近。此时转差s很小 (约为0.03),近似s=0。此时电机的等值模型与二次侧开路的变压器模型类似, 存在

Figure BDA0002657814280000097
将上式代入式(18)可以得到:When the asynchronous machine is running stably with no load, the rotor speed is close to the electromagnetic speed. At this time, the slip s is very small (about 0.03), approximately s=0. At this time, the equivalent model of the motor is similar to the transformer model with an open circuit on the secondary side, there are
Figure BDA0002657814280000097
Substituting the above formula into formula (18) can get:

Figure BDA0002657814280000101
Figure BDA0002657814280000101

Figure BDA0002657814280000102
Figure BDA0002657814280000102

化简后的式(20)与式(22)是异步电机启动瞬时以及空载稳定运行两种状 态下的端口特性,而从启动开始至稳定运行这个暂态过程中转差率s由1至近似 为0的变化过程难以描述。The simplified equations (20) and (22) are the port characteristics of the asynchronous motor under the two states of instantaneous startup and no-load stable operation, and the slip s in the transient process from startup to stable operation is from 1 to approximately The change process of 0 is difficult to describe.

基于上述的分析,接下来对电热泵的启动过程作以下两方面的研究:基于试 验的启动功率函数拟合,基于模型参数辨识的临界启动电压分析。Based on the above analysis, the following two aspects of the start-up process of the electric heat pump are studied: the start-up power function fitting based on the test, and the critical start-up voltage analysis based on the model parameter identification.

为解决启动过程中转差率s难以描述而无法建立启动功率模型的问题,试验 测试平台使电热泵多次在额定电压下启动,并记录了电流、电压量测值。由启动 电流、电压量测值可以计算得到电热泵启动过程的若干启动功率数据并绘制成曲 线。可选取具有较高试验精度的试验数据作为模型辨识的源数据,并在源数据的 基础上按照电热泵启动冲击倍数(5.9~7.1)、启动时长(0.22~0.36s)符合均匀分 布模拟得到丰富的试验模拟数据。最后采用曲线拟合的方法辨识出电热泵的启动 功率模型。In order to solve the problem that the slip rate s is difficult to describe and the starting power model cannot be established during the starting process, the test platform makes the electric heat pump start at the rated voltage for many times, and records the current and voltage measurement values. From the measured values of starting current and voltage, some starting power data of the starting process of the electric heat pump can be calculated and drawn into a curve. The test data with higher test accuracy can be selected as the source data for model identification, and based on the source data, the electric heat pump start-up shock multiple (5.9-7.1) and start-up duration (0.22-0.36s) conform to the uniform distribution simulation to enrich the simulation experimental simulation data. Finally, the method of curve fitting is used to identify the starting power model of the electric heat pump.

选用二次函数拟合试验数据,对比观测有功功率曲线后发现:电热泵的启动 功率与时间的曲线呈现较明显的二次函数特点。因此可以构建有功功率Ps关于时 间t的二次函数关系如下:The quadratic function is used to fit the test data, and after comparing and observing the active power curve, it is found that the curve of the starting power and time of the electric heat pump presents obvious quadratic function characteristics. Therefore, the quadratic function relationship of active power P s with respect to time t can be constructed as follows:

Ps(t)=a(t-b)2+c (23)P s (t)=a(tb) 2 +c (23)

上式中,a、b、c为参数待辨识。结合电热泵启动特点和二次函数顶点式 特性可得:Ps(t)曲线的顶点坐标(tmax,Pmax)对应于电热泵启动功率达到最大值 的时间tmax和峰值Pmax。利用试验数据辨识模型参数得到:In the above formula, a, b, and c are parameters to be identified. Combining the start-up characteristics of the electric heat pump and the apex characteristic of the quadratic function, the vertex coordinates (t max , P max ) of the P s (t) curve correspond to the time t max and the peak value P max when the electric heat pump start-up power reaches the maximum value. Using the experimental data to identify the model parameters, we get:

Figure BDA0002657814280000103
Figure BDA0002657814280000103

值得说明的是:式中的t定义域应当是启动初始时刻t0至启动完成电热泵进 入稳态的时刻t1,b近似于半个启动周期。所以电热泵启动过程有功功率Ps关于 时间t的闭区间函数关系为:It is worth noting that: the definition domain of t in the formula should be from the initial start time t 0 to the time t 1 when the electric heat pump enters a steady state after the start is completed, and b is approximately half a start cycle. Therefore, the closed-interval functional relationship between the active power P s and the time t during the starting process of the electric heat pump is:

Ps(t)=-1160(t-0.2)2+Pmax t∈[t0,t1] (5)P s (t)=-1160(t-0.2) 2 +P max t∈[t 0 ,t 1 ] (5)

表1给出了二次函数在各个拟合点的相似度以及均值,二次函数对电热泵的 试验功率曲线拟合的平均相似度达到0.931。Table 1 shows the similarity and mean of the quadratic function at each fitting point. The mean similarity of quadratic function to the test power curve fitting of the electric heat pump reaches 0.931.

表1Table 1

Figure BDA0002657814280000111
Figure BDA0002657814280000111

前文中式(20)、式(22)中定子漏感L、定子电感Ls、定子电阻Rs、转 子电阻Rr是有待确定的未知参数。可以利用异步电机启动的瞬时以及异步电机 进入空载稳定运行两个时段的试验量测数据对参数进行辨识。The stator leakage inductance L , the stator inductance L s , the stator resistance R s , and the rotor resistance R r in the foregoing equations (20) and (22) are unknown parameters to be determined. The parameters can be identified by using the test measurement data of the instant when the asynchronous motor starts and the two periods when the asynchronous motor enters the no-load stable operation.

为了简化辨识的复杂程度,对式(20)、式(22)的微分方程设计一阶滤波 器,将微分方程的辨识转化为利用滤波数据的最小二乘法辨识。并借助试验数据 辨识出T型等值模型的定转子电阻和电抗参数,如表2。In order to simplify the complexity of identification, a first-order filter is designed for the differential equations of equations (20) and (22), and the identification of the differential equations is transformed into the least squares identification using the filtered data. And with the help of the test data, the stator and rotor resistance and reactance parameters of the T-type equivalent model are identified, as shown in Table 2.

表2Table 2

Figure BDA0002657814280000112
Figure BDA0002657814280000112

辨识得到异步电机的阻抗参数后,便可以利用T型等值电路对异步电机的 启动过程作电磁转矩、机械转矩分析。当异步电机转子转动时,旋转的转子从定 子侧吸收的电磁功率PE。电磁功率在磁场当中通过电磁转矩的方式施加到转子 上并使其转动,由此完成了电能向转子机械能的转化过程。电磁转矩Te与PE存在 以下关系:After identifying the impedance parameters of the asynchronous motor, the T-type equivalent circuit can be used to analyze the electromagnetic torque and mechanical torque of the asynchronous motor's starting process. When the rotor of the asynchronous motor rotates, the electromagnetic power PE absorbed by the rotating rotor from the stator side. Electromagnetic power is applied to the rotor through electromagnetic torque in the magnetic field and makes it rotate, thus completing the conversion process of electrical energy to mechanical energy of the rotor. The electromagnetic torque T e and P E have the following relationship:

PE=△P+Pmec=f(Us 2,Z) (25)P E =ΔP+P mec =f(U s 2 ,Z) (25)

Figure BDA0002657814280000113
Figure BDA0002657814280000113

式中,PE为异步电机的电磁功率;△P为转子电阻铜耗;Pmec为转轴上产生的 总机械功率;Us为异步电机的端电压;Z为异步电机的端口等值阻抗;ω为异 步电机的转速标幺值。In the formula, PE is the electromagnetic power of the asynchronous motor; △P is the copper loss of the rotor resistance; P mec is the total mechanical power generated on the rotating shaft; U s is the terminal voltage of the asynchronous motor; Z is the port equivalent impedance of the asynchronous motor; ω is the per-unit value of the speed of the asynchronous motor.

异步电机的电磁转矩Te与转速和端电压相关,而电热泵负荷的机械转矩Tm与 转速平方成正比。电热泵负荷的机械转矩由与转速平方成正比的变化部分、摩擦 损耗决定的固定部分组成。电热泵的机械转矩Tm为:The electromagnetic torque T e of the asynchronous motor is related to the speed and terminal voltage, while the mechanical torque T m of the electric heat pump load is proportional to the square of the speed. The mechanical torque of the electric heat pump load consists of a variable part proportional to the square of the rotational speed and a fixed part determined by friction loss. The mechanical torque T m of the electric heat pump is:

Tm(ω)=T0+△Tω2 (27)T m (ω)=T 0 +ΔTω 2 (27)

上式中T0为机械转矩固定部分的常系数;△T为机械转矩关于ω2正比变化的 常系数,取T0=0.8471,△T=0.1875[44]In the above formula, T 0 is the constant coefficient of the fixed part of the mechanical torque; ΔT is the constant coefficient of the proportional change of the mechanical torque about ω 2 , take T 0 =0.8471, ΔT=0.1875 [44] .

当不计及网络阻抗时,基于上述的模型参数取值可以得到不同端电压(取电 机端电压分别为1.0pu、0.98pu、0.95pu)下异步电机的机械特性(异步电动机的 电磁转矩与转速之间的关系曲线)如图4。When the network impedance is not taken into account, the mechanical characteristics of the asynchronous motor (the electromagnetic torque and rotational speed of the asynchronous motor) under different terminal voltages (taking the motor terminal voltages as 1.0pu, 0.98pu, and 0.95pu, respectively) can be obtained based on the values of the above model parameters. The relationship between the curves) is shown in Figure 4.

由异步电机的电磁转矩、机械转矩曲线可知:当电机端口的启动电压足够高 (U≥0.95pu)时,电磁转矩Te与机械转矩Tm存在交点(平衡运行点),在这种 情况下异步电机能够启动且加速至稳定运行状态;当启动电压低于某个阈值 (U<0.95pu)时,Te与Tm没有交点,机械转矩始终高于启动电压对应的最大 电磁转矩,电机无法加速至稳定运行的状态。所以电磁转矩Te与机械转矩Tm是否 存在平衡运行点是电机成功启动与否的条件。而电磁转矩的峰值大小由启动电压 决定,所以Te与Tm是否存在平衡运行点取决于电机启动时端口电压值的大小。From the electromagnetic torque and mechanical torque curves of the asynchronous motor, it can be known that when the starting voltage of the motor port is high enough (U ≥ 0.95pu), the electromagnetic torque T e and the mechanical torque T m have an intersection point (balanced operating point), at In this case, the asynchronous motor can start and accelerate to a stable operating state; when the starting voltage is lower than a certain threshold (U<0.95pu), there is no intersection between T e and T m , and the mechanical torque is always higher than the maximum corresponding to the starting voltage. Electromagnetic torque, the motor cannot be accelerated to a state of stable operation. Therefore, whether there is a balanced operating point between the electromagnetic torque T e and the mechanical torque T m is the condition for the successful start of the motor. The peak value of electromagnetic torque is determined by the starting voltage, so whether there is a balanced operating point between T e and T m depends on the value of the port voltage when the motor starts.

定义使得异步电机的电磁转矩Te与机械转矩Tm存在唯一交点的电压值为临 界启动电压Ulim。由式(26)及图4可见异步电机的最大电磁转矩Te,max及最大转 速ωmax只与异步电机自身阻抗有关,而与端口电压无关。因此只要确定最大转速 ωmax就可通过式(26)电磁转矩Te与式(27)机械转矩Tm相等计算得到电热泵 负荷能正常启动的临界电压UlimThe voltage value at which there is a unique intersection point between the electromagnetic torque Te and the mechanical torque T m of the asynchronous motor is defined as the critical starting voltage U lim . It can be seen from the formula (26) and Fig. 4 that the maximum electromagnetic torque T e,max and the maximum rotational speed ω max of the asynchronous motor are only related to the impedance of the asynchronous motor itself, but have nothing to do with the port voltage. Therefore, as long as the maximum rotational speed ω max is determined, the critical voltage U lim at which the electric heat pump load can start normally can be calculated by equation (26) equal to the electromagnetic torque T e and equation (27) mechanical torque T m .

Temax,Ulim)=Tmmax) (28)T emax ,U lim )=T mmax ) (28)

图4中,当端口电压为0.95pu时,电热泵的机械转矩和电磁转矩存在唯一 交点(Te,maxmax),即其临界启动电压为0.95pu。当考虑配电网阻抗及电机启动 的电压波动情况下,异步电动机能否正常运转取决于所在节点电压值是否低于其 能正常启动的临界电压UlimIn Figure 4, when the port voltage is 0.95pu, there is a unique intersection (T e,max , ω max ) between the mechanical torque and the electromagnetic torque of the electric heat pump, that is, its critical starting voltage is 0.95pu. When considering the impedance of the distribution network and the voltage fluctuation of the motor starting, whether the asynchronous motor can run normally depends on whether the voltage value of the node where it is located is lower than the critical voltage U lim that it can start normally.

二、基于μPMU量测辨识电热泵状态2. Identify the state of the electric heat pump based on μPMU measurement

首先确定首端节点电压Ui的约束。按照现有理论,电热泵启动是否成功主 要取决于压缩机是否成功启动。对于采用单相异步电机的定频压缩机来说,其成 功启动与否取决于端口电压是否达到异步电机临界启动电压UlimFirst determine the constraints of the head-end node voltage U i . According to the existing theory, whether the electric heat pump starts successfully depends mainly on whether the compressor starts successfully. For a fixed-frequency compressor using a single-phase asynchronous motor, the successful start-up depends on whether the port voltage reaches the critical start-up voltage U lim of the asynchronous motor.

实际低压配电网中,电热泵分散在电网各处,由于配电线路长存在压降而导 致各个电热泵的电压不同。因此必须使首端节点电压足够高才能保证经线路降压 后的各电热泵电压不低于其临界启动电压UlimIn the actual low-voltage distribution network, the electric heat pumps are scattered all over the power grid, and the voltage of each electric heat pump is different due to the long voltage drop in the distribution line. Therefore, the voltage of the head-end node must be high enough to ensure that the voltage of each electric heat pump after being stepped down by the line is not lower than its critical starting voltage U lim .

运用电热泵的T型等值模型和含阻抗的配电线路并行联结的结构特点可得含 n台电热泵的网络等值电路如图5所示,其中Ui为电源接入点电压(取低压配电 变压器出口电压),Rk和Xk(k=1,2,...,n)为第k段线路的电阻和电抗。Rsk和Rrk分 别是第k台电热泵负荷的定子电阻和转子电阻,Xsk和Xrk分别为第k台电热泵负 荷的定子和转子的电抗;Xmk为励磁电抗;转差sk=(ωsrk)/ωs,ωs为同步角 速度,ωrk为转子角速度。Using the T-type equivalent model of the electric heat pump and the structural characteristics of the parallel connection of the distribution lines with impedance, the equivalent circuit of the network containing n electric heat pumps can be obtained as shown in Figure 5, where U i is the power supply access point voltage (take the low voltage. Distribution transformer outlet voltage), R k and X k (k=1,2,...,n) are the resistance and reactance of the k-th line. R sk and R rk are the stator resistance and rotor resistance of the k-th electric heat pump load respectively, X sk and X rk are the reactances of the stator and rotor of the k-th electric heat pump load, respectively; X mk is the excitation reactance; slip sk = ( ω srk )/ω s , ω s is the synchronous angular velocity, and ω rk is the rotor angular velocity.

为了得到第k台电热泵负荷接入点的电压,可以对上述网络等值电路采用节 点电压法进行计算。图中含有n个节点,由节点导纳矩阵定义和等值电路的特点 可以得到网络的节点导纳矩阵Y:In order to obtain the voltage of the kth electric heat pump load access point, the node voltage method can be used to calculate the above-mentioned network equivalent circuit. The figure contains n nodes, which are defined by the node admittance matrix and the characteristics of the equivalent circuit. The node admittance matrix Y of the network can be obtained:

Figure BDA0002657814280000131
Figure BDA0002657814280000131

上式中,Ykk为节点k的自导纳,包含线路导纳和位于节点k的电热泵等值导 纳;Yk+1,k为节点k+1与节点k的互导纳,即是线路k+1的导纳。因为只有相邻节 点之间存在互导纳,所以节点导纳矩阵Y是一个稀疏的n×n矩阵。由此可以建 立矩阵形式的节点电压方程Y·U=I如下:In the above formula, Y kk is the self-admittance of node k, including the line admittance and the equivalent admittance of the electric heat pump located at node k; Y k+1, k is the mutual admittance between node k+1 and node k, that is is the admittance of line k+1. Since there is only mutual admittance between adjacent nodes, the node admittance matrix Y is a sparse n×n matrix. From this, the node voltage equation Y·U=I in matrix form can be established as follows:

Figure BDA0002657814280000132
Figure BDA0002657814280000132

上式中,

Figure BDA0002657814280000141
为节点k的节点电压;
Figure BDA0002657814280000142
为注入节点1的电流且
Figure BDA0002657814280000143
Z1为节点1至电源接入点的线路阻抗且Z1=R1+jX1。在线路阻抗 和电热泵T型等值电路阻抗已知的情况下,根据节点电压方程(30)可以求解 得到节点k的电热泵端口电压
Figure BDA0002657814280000144
是关于
Figure BDA0002657814280000145
的线性函数:
Figure BDA0002657814280000146
In the above formula,
Figure BDA0002657814280000141
is the node voltage of node k;
Figure BDA0002657814280000142
is the current injected into node 1 and
Figure BDA0002657814280000143
Z 1 is the line impedance from node 1 to the power access point and Z 1 =R 1 +jX 1 . When the line impedance and the T-type equivalent circuit impedance of the electric heat pump are known, the electric heat pump port voltage of node k can be obtained by solving the node voltage equation (30).
Figure BDA0002657814280000144
its about
Figure BDA0002657814280000145
The linear function of :
Figure BDA0002657814280000146

图6为端口电压为Uk的电热泵等值电路。为了便于计算其关于端口电压及自 身阻抗的电磁转矩,将转子阻抗等效支路以外的电路部分作简化等效。可得到简 化后的端口等值电路如图7所示。Fig. 6 is the equivalent circuit of the electric heat pump whose terminal voltage is U k . In order to facilitate the calculation of its electromagnetic torque related to the port voltage and its own impedance, the circuit parts other than the equivalent branch of the rotor impedance are simplified and equivalent. The simplified port equivalent circuit can be obtained as shown in Figure 7.

由戴维南电路等效原理,计算得到简化后的串联等值电势Uek和等值阻抗 Rek+jXek满足以下等式:According to the equivalent principle of Thevenin circuit, the simplified series equivalent potential U ek and equivalent impedance Re +jX ek are calculated to satisfy the following equations:

Figure BDA0002657814280000147
Figure BDA0002657814280000147

Figure BDA0002657814280000148
Figure BDA0002657814280000148

由简化电路及新的等值参数可得第k台电热泵负荷中作用在电机转子上的 电磁转矩Tekrk,Ui)为:From the simplified circuit and the new equivalent parameters, the electromagnetic torque T ekrk ,U i ) acting on the motor rotor in the kth electric heat pump load can be obtained as:

Figure BDA0002657814280000149
Figure BDA0002657814280000149

上式中,sk和Uek分别为转子转速ωrk和台区首端节点电压Ui的线性函数。 令式(33)关于转子转速ωrk的一阶导数为0,便可以得到Tek取最大值时所对应 的最大转子转速ωmaxrkIn the above formula, sk and U ek are the linear functions of the rotor speed ω rk and the voltage U i at the head-end node of the platform, respectively. Let the first derivative of equation (33) with respect to the rotor speed ω rk be 0, then the maximum rotor speed ω maxrk corresponding to the maximum value of T ek can be obtained.

Figure BDA00026578142800001410
Figure BDA00026578142800001410

由式(34)可以看出:最大转子转速ωmaxrk仅受到电热泵负荷转子阻抗、定 子阻抗和励磁阻抗以及外部网络的等值阻抗影响,而与端口电压无关。因此,在 各电热泵负荷阻抗、网络阻抗参数确定后,即可通过式(34)计算各电热泵负荷 的最大转子转速ωmaxrk。得到最大电磁转矩对应的最大转子转速ωmaxrk后,根据机 械转矩与转速的关系式(27)可得到第k台电热泵的机械转矩如下式:It can be seen from equation (34) that the maximum rotor speed ω maxrk is only affected by the electric heat pump load rotor impedance, stator impedance and excitation impedance and the equivalent impedance of the external network, and has nothing to do with the port voltage. Therefore, after the load impedance and network impedance parameters of each electric heat pump are determined, the maximum rotor speed ω maxrk of each electric heat pump load can be calculated by formula (34). After obtaining the maximum rotor speed ω maxrk corresponding to the maximum electromagnetic torque, the mechanical torque of the kth electric heat pump can be obtained according to the relationship between mechanical torque and speed (27) as follows:

Figure BDA00026578142800001411
Figure BDA00026578142800001411

上式中Tm(ω)为异步电机转速标幺值ω下的电热泵机械转矩,T0为机械转矩 固定部分的常系数;△T为机械转矩关于ω2正比变化的常系数,通常取 T0=0.8471,△T=0.1875。Tk0和△Tk为第k台电热泵的机械转矩固定和变化部分的 常系数,其取值同式(27)。再由临界启动电压的方程式(28),结合式(33) 与式(35)即有Tekmaxrk,Ui)=Tmmaxrk)。由此等式便可以计算得到低压配电网中 位于节点k的电热泵能正常启动时台区首端节点电压Ui的临界值Ulim.kIn the above formula, T m (ω) is the mechanical torque of the electric heat pump at the per-unit value ω of the asynchronous motor speed, T 0 is the constant coefficient of the fixed part of the mechanical torque; △T is the constant coefficient of the mechanical torque proportional to the change of ω 2 , usually take T 0 =0.8471, ΔT = 0.1875. T k0 and ΔT k are the constant coefficients of the fixed and variable parts of the mechanical torque of the kth electric heat pump, and their values are the same as those of formula (27). Then, from the equation (28) of the critical startup voltage, combined with the equation (33) and the equation (35), there is T ekmaxrk , U i )=T mmaxrk ). From this equation, the critical value U lim.k of the voltage U i at the head-end node of the station area when the electric heat pump located at node k in the low-voltage distribution network can be normally started can be calculated.

综上,当低压配电变压器的台区内接有n台电热泵时,台区首端节点电压Ui需满足:To sum up, when there are n electric heat pumps connected to the station area of the low-voltage distribution transformer, the voltage U i of the head-end node of the station area must satisfy:

Ui≥{max(Ulim.k)|Tekmaxrk,Ulim.k)=Tmmaxrk),k∈[1,n]} (6)U i ≥{max(U lim.k )|T ekmaxrk ,U lim.k )=T mmaxrk ),k∈[1,n]} (6)

式中,Ui取位于节点k(k=1,2,...,n)处电热泵负荷能正常启动时临界值Ulim.k的最大值。当Ui满足上式时,即使因为线路阻抗和负载阻抗的存在使得网络中 各节点电压发生降落也不会低至电热泵的临界启动电压Ulim,故能保证网络中各 电热泵成功启动。In the formula, U i takes the maximum value of the critical value U lim.k when the electric heat pump load at node k (k=1,2,...,n) can start normally. When U i satisfies the above formula, even if the voltage of each node in the network drops due to the existence of line impedance and load impedance, it will not be lower than the critical start-up voltage U lim of the electric heat pump, so it can ensure that each electric heat pump in the network starts successfully.

微型相量量测单元(μPMU)具有精准、快速的量测优点。本实施例基于以下 两种判据判断电热泵启动状态:Micro Phasor Measurement Unit (μPMU) has the advantages of accurate and fast measurement. The present embodiment judges the start-up state of the electric heat pump based on the following two criteria:

判据一、功率冲击Criterion 1. Power Shock

电热泵启动时刻会伴随有较大的功率冲击,因为压缩机在通电瞬间转速为 零,为了获得足够的电磁转矩,定频电热泵的启动功率峰值高达6倍额定功率, 冲击过程持续时间约0.3s。When the electric heat pump is started, there will be a large power shock, because the compressor rotates at zero at the moment of power-on. In order to obtain sufficient electromagnetic torque, the peak value of the starting power of the fixed-frequency electric heat pump is as high as 6 times the rated power, and the duration of the shock process is about 0.3s.

由电热泵启动功率的二次函数模型也可以看出:启动功率的数值较大且增长 速度较快,对Ps(t)函数(式5)求导可以的到各时刻的曲线斜率,即是启动功率 的变化速率:It can also be seen from the quadratic function model of the start-up power of the electric heat pump: the value of the start-up power is large and the growth rate is fast, and the slope of the curve at each moment can be derived from the P s (t) function (Equation 5), that is, is the rate of change of starting power:

Ps′(t)=-2320(t-0.2) (36)P s '(t)=-2320(t-0.2) (36)

可以发现:当t=0.2时,启动功率的增长速度为0,启动功率达到最大值。 由启动功率二次函数模型可知整个启动过程内Ps′(t)的值域为[-464,464]。据此, 可以根据功率变化量及Ps′(t)的值辨识出当前时刻t是否存在电热泵启动功率:It can be found that: when t=0.2, the growth rate of the starting power is 0, and the starting power reaches the maximum value. From the quadratic function model of starting power, it can be known that the value range of P s '(t) in the whole starting process is [-464,464]. Accordingly, it can be identified whether the electric heat pump starting power exists at the current time t according to the power variation and the value of P s '(t):

Figure BDA0002657814280000151
Figure BDA0002657814280000151

上式中,P(t)为当前时刻t的负荷功率;P(t-1)为前一量测时刻的负荷功 率;N为时刻t的量测频数,功率变化量取n个量测数据依次作差并求和得到; P′(t)为当前时刻t的负荷功率变化速率;λ1、λ2为判据对应的阈值。当存在μPMU 量测数据支撑的情况下可以简便地近似计算|P′(t)|:In the above formula, P(t) is the load power at the current time t; P(t-1) is the load power at the previous measurement time; N is the measurement frequency at time t, and the power change takes n measurement data The difference is made in turn and the sum is obtained; P'(t) is the rate of change of the load power at the current time t; λ 1 and λ 2 are the thresholds corresponding to the criteria. In the presence of μPMU measurement data support, |P′(t)| can be easily approximated:

Figure BDA0002657814280000161
Figure BDA0002657814280000161

式中,P(t2)、P(t1)分别为t2、t1时刻由μPMU装置的电压、电流量测值计 算得到的负荷功率。In the formula, P(t 2 ) and P(t 1 ) are the load powers calculated from the voltage and current measurement values of the μPMU device at time t 2 and t 1 respectively.

值得说明的是:阈值λ1、λ2的取值对辨识结果影响较大,取值太小可能错 误辨识到常规负荷波动,取值太大则可能遗漏某些冲击功率。所以合理的取值应 当既可以过滤掉大部分常规负荷波动,防止误判,也能够准确无遗漏地辨识出电 热泵的冲击功率。It is worth noting that the values of the thresholds λ 1 and λ 2 have a great influence on the identification results. If the values are too small, conventional load fluctuations may be incorrectly identified, and if the values are too large, some impulse power may be missed. Therefore, a reasonable value should not only filter out most of the conventional load fluctuations, prevent misjudgment, but also accurately identify the impact power of the electric heat pump.

判据二、电压暂降Criterion 2. Voltage sag

基于启动试验和启动功率模型分析已知电热泵的启动过程消耗大量的功率, 伴随有电流涌流,大电流会引起明显的电压暂降。而电压暂降△U具有明显的峰 值和瞬变的特征。Based on the start-up test and start-up power model analysis, it is known that the start-up process of the electric heat pump consumes a large amount of power, accompanied by a current inrush, and the large current will cause a significant voltage sag. The voltage sag ΔU has obvious peak and transient characteristics.

μPMU装置可以实时地监测配电网潮流情况,当t时刻的电压量测值Ut远小 于t-1时刻的电压量测值Ut-1,且Ut-1与t-2时刻的电压量测值Ut-2很接近,则可 以判定在t时刻电压发生了暂降,因此电压暂降判据如下式:The μPMU device can monitor the power flow of the distribution network in real time. When the voltage measurement value U t at time t is much smaller than the voltage measurement value U t-1 at time t-1 , and the voltages at time U t-1 and t-2 are If the measured value U t-2 is very close, it can be determined that the voltage sags at time t, so the voltage sag criterion is as follows:

Figure BDA0002657814280000162
Figure BDA0002657814280000162

式中,阈值λ3为一个较大的正数,用于描述Ut远小于Ut-1的程度;阈值λ4为 一个很小的正数,用于描述Ut-1与Ut-2的接近程度;N为时刻t的量测频数。In the formula, the threshold λ 3 is a large positive number, which is used to describe the degree that U t is much smaller than U t-1 ; the threshold λ 4 is a small positive number, which is used to describe U t-1 and U t- 2 closeness; N is the measurement frequency at time t.

值得说明的是:在网络中包含有波动较小且变化缓慢的基础居民负荷PL和 电热泵负荷时,单台电热泵启动首先影响到网络功率迅速爬升,接着引起电压明 显的降落:基于启动试验可知异步电机的启动过程消耗大量的功率(△P=η·PN, η=5~7),伴随有电流涌流,大电流引起电压降落严重。由电压暂降公式It is worth noting that: when the basic residential load PL and electric heat pump load with small fluctuation and slow change are included in the network, the startup of a single electric heat pump first affects the rapid rise of the network power, and then causes the voltage to drop significantly: based on the startup test It can be seen that the starting process of the asynchronous motor consumes a lot of power (ΔP=η· PN , η=5~7), accompanied by current inrush, and the high current causes serious voltage drop. by the voltage sag formula

Figure BDA0002657814280000163
Figure BDA0002657814280000163

由于低压配电网的线路电阻R较大,有功功率P对电压影响较大。节点注 入功率包含负荷稳态功率于电热泵启动冲击功率,即P=PL+△P。所以电压暂 降满足:Due to the large line resistance R of the low-voltage distribution network, the active power P has a great influence on the voltage. The node injection power includes the load steady state power and the electric heat pump startup impulse power, that is, P= PL +△P. So the voltage sag satisfies:

Figure BDA0002657814280000171
Figure BDA0002657814280000171

当有k台电热泵同时启动,则P=PL+△P·k,引起电压降落为:When k sets of electric heat pumps are started at the same time, then P= PL + △P·k, resulting in a voltage drop as:

Figure BDA0002657814280000172
Figure BDA0002657814280000172

启动功率发生叠加使得网络功率爬升的幅度更大,同时引起的电压降落也会 更加严重。λ1、λ2、λ3、λ4的取值是状态判据辨识结果准确与否的关键,合理 选取阈值可使判据准确辨识单台热泵启动和多台热泵同时启动的冲击影响。所以 判据一、二中阈值的取值可以按照单台热泵启动时网络功率、电压变化值进行选 取,并且以此选取的阈值能更加容易地判断多台热泵同时启动的冲击情况。但在 实际电网中,网络功率、电压变化值受到网络负荷P=PL+△P情况的影响,所 以阈值的选取需要根据实际量测数据进行分析并采用枚举的方法进行确定。The superposition of the starting power makes the network power climb larger, and at the same time, the voltage drop will be more serious. The value of λ 1 , λ 2 , λ 3 , and λ 4 is the key to the accuracy of the state criterion identification result. Reasonable selection of the threshold value can accurately identify the impact of the single heat pump startup and the simultaneous startup of multiple heat pumps. Therefore, the threshold values in criteria 1 and 2 can be selected according to the network power and voltage changes when a single heat pump is started, and the selected threshold can more easily judge the impact of multiple heat pumps starting at the same time. However, in the actual power grid, the network power and voltage changes are affected by the network load P= PL + △P, so the selection of the threshold needs to be analyzed according to the actual measurement data and determined by the method of enumeration.

接下来进一步辨识当前时刻处于启动状态的电热泵台数。某tn时刻的量测值 为P(tn),若有判据确定在tn时刻存在电热泵启动冲击功率,就是说此时负荷功率 中包含有瞬时的突增的脉冲波动。通过比较电热泵启动前的负荷功率就可以将当 前时刻的电热泵启动冲击功率Pimp(tn)(脉冲波动)可以通过下式分离出来:Next, further identify the number of electric heat pumps that are in the starting state at the current moment. The measured value at a certain time t n is P(t n ). If there is a criterion to determine that there is an electric heat pump start-up impulse power at the time t n , that is to say, the load power contains an instantaneous sudden increase in pulse fluctuation. By comparing the load power before the electric heat pump is started, the electric heat pump startup impulse power P imp (t n ) (pulse fluctuation) at the current moment can be separated by the following formula:

Pimp(tn)=P(tn)-P(tm) (41)P imp (t n )=P(t n )-P(t m ) (41)

式中,P(tn)为tn时刻的功率量测值;P(tm)为距离tn时刻很近的无冲击功率 的tm时刻的功率量测值。量测结果不满足启动状态特征判据便视为无冲击功率的 时刻,选择距离tn时刻较近的时刻tm是为了避免负荷的正常起伏波动对式(41) 的计算结果造成误差。因此tm取为由电热泵启动状态特征判所判断出的无冲击功 率的前一时刻。In the formula, P(t n ) is the measured power value at time t n ; P(t m ) is the measured value of power at time t m when there is no impact power very close to time t n . When the measurement result does not satisfy the characteristic criterion of the starting state, it is regarded as the moment when there is no impact power. The moment tm which is closer to the moment tn is selected to avoid the error caused by the normal fluctuation of the load on the calculation result of Equation (41). Therefore, t m is taken as the previous moment of the non-impulse power judged by the characteristic judgment of the starting state of the electric heat pump.

电热泵的启动常常具有随机性,则在tn时刻可能存在若干台电热泵同时处于 启动过程,那么当前时刻的启动冲击功率便是若干台电热泵启动功率叠加而来 的,基于启动冲击功率的二次函数模型可以得出Pimp(tn)是由若干个二次函数 叠加之和:The startup of electric heat pumps is often random, so there may be several electric heat pumps in the startup process at the time t n , then the startup impulse power at the current moment is the superposition of the startup power of several electric heat pumps. From the functional model, it can be concluded that P imp (t n ) is the sum of several quadratic functions:

Pimp(tn)=Ps.1(tn)+Ps.2(tn)+…+Ps.i(tn) (42)P imp (t n )=P s.1 (t n )+P s.2 (t n )+…+P si (t n ) (42)

上式中,Ps.i(tn)为处于启动状态的第i台电热泵的启动功率。In the above formula, P si (t n ) is the starting power of the ith electric heat pump in the starting state.

因此,辨识当前时刻处于启动状态的电热泵台数即是要辨识出Pimp(tn)是由 多少个二次函数叠加而成的。设置启动状态系数K(n)=(k1,k2,...,kN)为和启动延 时系数△t(n)=(t1,t2,...,tN)T对前述电热泵的启动功率函数Ps(t)进行离散取值,得 到处于不同时刻的启动功率函数值Ps(ti),再根据K(n)、△t(n)和Ps(ti)进行组合 便可以描述所有电热泵冲击功率的叠加情况:Therefore, to identify the number of electric heat pumps that are in the starting state at the current moment is to identify how many quadratic functions are superimposed on P imp (t n ). Set the start-up state coefficient K(n)=(k 1 ,k 2 ,...,k N ) as and start-up delay coefficient Δt(n)=(t 1 ,t 2 ,...,t N ) T The starting power function P s (t) of the aforementioned electric heat pump is discretely valued to obtain the starting power function value P s (t i ) at different times, and then according to K(n), Δt(n) and P s ( t i ) can be combined to describe the superposition of all electric heat pump impulse powers:

S(n)=K(n)·Ps[△t(n)]=k1·Ps(t1)+k2·Ps(t2)+…+kN·Ps(tN) (43)S(n)=K(n)·P s [Δt(n)]=k 1 ·P s (t 1 )+k 2 ·P s (t 2 )+…+k N ·P s (t N ) (43)

式中,S(n)为二次函数Ps(t)叠加得到的电热泵冲击功率;N为电热泵启动功 率函数Ps(t)的采样点个数,可结合μPMU的数据采集频率来确定;tN为启动功 率Ps(t)采样点对应的启动时间;Ps(tN)为处于启动时间tN的启动功率函数值;kN为位于启动时间tN的电热泵台数。当kN=0时,表示不存在处于启动时间tN的电 热泵,当kN=k时,表示有k台电热泵正处于启动时间tN,其启动功率为k·Ps(tN)。In the formula, S(n) is the impulse power of the electric heat pump obtained by the superposition of the quadratic function P s (t); N is the number of sampling points of the electric heat pump startup power function P s (t), which can be calculated by combining the data acquisition frequency of the μPMU. Determine; t N is the starting time corresponding to the starting power P s (t) sampling point; P s (t N ) is the starting power function value at the starting time t N ; k N is the number of electric heat pumps at the starting time t N. When k N = 0, it means that there is no electric heat pump at the start-up time t N ; when k N = k, it means that there are k electric heat pumps at the start-up time t N , and the start-up power is k·P s (t N ) .

要辨识出电热泵启动台数便是要求解拟合的叠加功率S(n)与冲击功率量测 值Pimp(tn)相等时对应的启动状态系数K(n)=(k1,k2,...,kN)T。为描述量测功率 Pimp(tn)和叠加功率S(n)的接近程度,可采用Pimp(tn)和S(n)的皮尔森相关系数γ 作为辨识的目标函数,其定义方程式为:To identify the starting number of electric heat pumps, it is necessary to obtain the corresponding starting state coefficient K(n)=(k 1 , k 2 when the superimposed power S(n) of the fitting is equal to the impulse power measurement value P imp (t n ) ,...,k N ) T . In order to describe the closeness of the measured power P imp (t n ) and the superimposed power S(n), the Pearson correlation coefficient γ of P imp (t n ) and S(n) can be used as the identification objective function, which defines the equation for:

γ=cov[Pimp(tn),S(n)]/[μPimp(tn)·μS(n)] (43)γ=cov[P imp (t n ),S(n)]/[μP imp (t n )·μS(n)] (43)

式中,cov[Pimp(tn),S(n)]为电热泵冲击功率Pimp(tn)和拟合的叠加功率S(n)的 协方差,μPimp(tn)、μS(n)是量测功率与叠加功率对应的标准差。where cov[P imp (t n ), S(n)] is the covariance of the electric heat pump impulse power P imp (t n ) and the fitted superimposed power S(n), μP imp (t n ), μS (n) is the standard deviation corresponding to the measured power and the superimposed power.

γ的值越接近1时表明两者的值越相似,即是量测功率Pimp(tn)和叠加功率 S(n)具有较好的拟合,近似Pimp(tn)=S(n)。此时由辨识出的开启状态系数A(n)及 启动延时系数△t(n)就可确定处于启动中电热泵的台数及对应所处的启动时间。The closer the value of γ is to 1, the more similar the two values are, that is, the measured power P imp (t n ) and the superimposed power S(n) have a better fit, and the approximate P imp (t n )=S( n). At this time, the number of electric heat pumps in the process of starting and the corresponding starting time can be determined by the identified starting state coefficient A(n) and starting delay coefficient Δt(n).

上述以皮尔森相关系数γ作为目标函数的辨识过程中,需要对开启状态系数 K(n)=(k1,k2,...,kN)进行寻优,而因为μPMU的量测数据具有较高密度使得其中 N的值较大,同时元素kN是离散的非负整数。所以上述函数问题是一种离散的 组合优化问题,可以借助数学优化软件(如CPLEX)快捷地完成求解。In the above identification process using the Pearson correlation coefficient γ as the objective function, it is necessary to optimize the open state coefficient K(n)=(k 1 , k 2 ,...,k N ), because the measurement data of μPMU Having a higher density makes where the value of N is larger, while the element k N is a discrete non-negative integer. Therefore, the above function problem is a discrete combinatorial optimization problem, which can be solved quickly with the help of mathematical optimization software (such as CPLEX).

三、基于量测辨识的电热泵有序用电方法3. Orderly electricity consumption method of electric heat pump based on measurement and identification

从直接负荷控制的角度来看,往往需要将电压问题转化为负荷功率问题。因 此将台区内接有电热泵负荷的所有节点等效为一个广义节点m,如图8。辐射状 的低压配电网经过等效后可看作两节点网络,首节点连接电网,次节点连接台区 负荷。From the point of view of direct load control, it is often necessary to convert the voltage problem into a load power problem. Therefore, all nodes connected to the electric heat pump load in the station area are equivalent to a generalized node m, as shown in Figure 8. The radial low-voltage distribution network can be regarded as a two-node network after being equivalent, the first node is connected to the power grid, and the second node is connected to the load in the station area.

分析台区首端节点n与负荷广义节点m的电流关系,其值满足:Analyze the current relationship between the head-end node n of the station area and the load generalized node m, and its value satisfies:

Figure BDA0002657814280000181
Figure BDA0002657814280000181

式中,δ为m节点电压与n节点电压之间的相角差。节点n的复功率可以表 示为:In the formula, δ is the phase angle difference between the voltage of the m node and the voltage of the n node. The complex power of node n can be expressed as:

Figure BDA0002657814280000191
Figure BDA0002657814280000191

式中,

Figure BDA0002657814280000192
为电流相量
Figure BDA0002657814280000196
的共轭;
Figure BDA0002657814280000193
为阻抗角。由于低压配电网的线路较短、 线路阻抗R较大且相角差δ较小,所以由式(44)、(45)可以化简有功功率P和 电压Un之间的关系如下:In the formula,
Figure BDA0002657814280000192
is the current phasor
Figure BDA0002657814280000196
the conjugation of ;
Figure BDA0002657814280000193
is the impedance angle. Since the line of the low-voltage distribution network is short, the line impedance R is large, and the phase angle difference δ is small, the relationship between the active power P and the voltage Un can be simplified by equations (44) and (45) as follows:

Figure BDA0002657814280000194
Figure BDA0002657814280000194

可见,首端节点n的注入功率P与其节点电压Un之间近似地满足线性变化关 系。所以当电压越限时可以采取有功功率调节电压。通过μPMU的量测数据可 以便捷地计算得到线性变化系数kpIt can be seen that the injected power P of the head-end node n and the node voltage U n approximately satisfy a linear variation relationship. Therefore, when the voltage exceeds the limit, active power can be used to adjust the voltage. The linear variation coefficient k p can be easily calculated from the measurement data of the μPMU:

Figure BDA0002657814280000195
Figure BDA0002657814280000195

式中,Un(t0)、P(t0)为t0时刻测出的当前节点电压和功率;P(t1)、Un(t1) 为t1时刻测出的当节点功率和电压。与辨识判据不同的是:t0时刻与t1时刻应当 有较长的时间间隔。量测值作差后的比值便可近似为线性关系的斜率kp,因此 得到首端节点n的电压和注入有功功率满足:In the formula, Un (t 0 ) and P(t 0 ) are the current node voltage and power measured at time t 0 ; P(t 1 ) and Un (t 1 ) are the current node power measured at time t 1 and voltage. The difference from the identification criterion is that there should be a longer time interval between time t 0 and time t 1 . The ratio of the difference between the measured values can be approximated as the slope k p of the linear relationship, so the voltage and injected active power of the head-end node n are obtained as follows:

Un=kP·[P-P(t0)]+Un(t0) (9)U n =k P ·[PP(t 0 )]+U n (t 0 ) (9)

式中,Un为首端节点n所求的电压;P为首端节点n所求的注入功率。在 其中一个量已知的情况下可以计算出另外一个量的值。而对于电热泵接入的节 点,为满足电热泵的启动电压要求,节点电压Un的值是由临界启动电压为线的, 即是由确定的Un求解P的过程。In the formula, U n is the voltage required by the head-end node n; P is the injected power required by the head-end node n. When one of the quantities is known, the value of the other can be calculated. For the node connected to the electric heat pump, in order to meet the starting voltage requirement of the electric heat pump, the value of the node voltage Un is linear from the critical starting voltage, that is, the process of solving P from the determined Un .

为了保证台区首端节点的电压约束Ulim,即是要求注入功率不超过 Plim=f(Ulim)。因此电热泵的运行功率与功率限制的缺额就是为满足电热泵启动 所需调节的功率削减量,进而将电热泵启动电压的问题转化为了台区负荷功率约 束的问题。In order to ensure the voltage constraint U lim of the head-end node of the station area, it is required that the injected power does not exceed P lim =f(U lim ). Therefore, the shortage of the operating power and power limit of the electric heat pump is the amount of power reduction required to meet the start-up of the electric heat pump, and then the problem of the starting voltage of the electric heat pump is transformed into the problem of the load power constraint in the platform area.

综合上述步骤,考虑电热泵启动冲击功率下,利用μPMU量测辨识电热泵启 动状态并通过限制台区负荷来确保各节点临界电压的策略流程如图9。在实际应 用中存在两种不同的台区负荷功率越限情形:(1)同一时刻下启动的电热泵启动 冲击功率的叠加峰值超出限制;(2)配电变压器稳态运行的负荷较大时叠加部分 启动功率导致的功率越限。应对这两种不同情形下的电热泵有序用电方法如下:Combining the above steps, considering the impact power of electric heat pump startup, the strategy flow of using μPMU measurement to identify the startup state of electric heat pump and ensure the critical voltage of each node by limiting the load of the station area is shown in Figure 9. In practical applications, there are two different situations where the load power exceeds the limit in the station area: (1) the superimposed peak value of the start-up impulse power of the electric heat pump started at the same time exceeds the limit; (2) the load of the steady-state operation of the distribution transformer is large. The power limit is exceeded due to the superposition of partial starting power. The methods for orderly use of electric heat pumps in response to these two different situations are as follows:

由量测判据及临界电压的限制求解得到所需的功率削减量Pcat后,便可制定 相应的控制策略进行响应。电热泵的运行特性主要取决于压缩机,在启动过程中 的功率消耗是正常运行时的数倍,因此与电热泵的稳态运行策略不同的是,下文 的启动策略将电热泵的运行全过程划分为启动、运行、关闭三种状态。所以,考 虑启动冲击功率时电热泵运行过程的功率特性与室温的关系如图10。After the required power reduction amount P cat is obtained from the measurement criterion and the limit of the threshold voltage, the corresponding control strategy can be formulated to respond. The operating characteristics of the electric heat pump mainly depend on the compressor, and the power consumption during the start-up process is several times that of the normal operation. Therefore, different from the steady-state operation strategy of the electric heat pump, the following startup strategy will be the whole process of the electric heat pump operation. It is divided into three states: startup, running, and shutdown. Therefore, the relationship between the power characteristics of the electric heat pump and the room temperature during the operation of the electric heat pump when the impulse power is considered is shown in Figure 10.

图10中Tmax和Tmin分别为室内温度区间的温度上、下限值,只要能够保证对 电热泵进行控制的策略使其所在的室温不超出该温度限值,就能够满足用户舒适 度需求。在电热泵负荷占比大的条件下,存在大量用户随机开启使用,并且各电 热泵的运行情况是随外界环境和用户需求变化的,如果将图6中的温度变化曲线 细分为若干运行区段,则在任意一个时刻,各电热泵可能处于控制周期的任何一 个区段,消耗的功率包括稳态运行功率和启动冲击功率两种情况。因此,电热泵 有序用电策略不仅要考虑稳态运行功率也要考虑启动冲击功率。即是要同时对处 于启动状态、运行状态的电热泵进行调控。In Figure 10, T max and T min are the upper and lower temperature limits of the indoor temperature range, respectively. As long as the strategy of controlling the electric heat pump can ensure that the room temperature where it is located does not exceed the temperature limit, the user's comfort requirement can be satisfied. . Under the condition that the electric heat pump load accounts for a large proportion, there are a large number of users who randomly turn on and use the electric heat pump, and the operation of each electric heat pump changes with the external environment and user needs. If the temperature change curve in Figure 6 is subdivided into several operating areas section, then at any time, each electric heat pump may be in any section of the control cycle, and the power consumed includes both steady-state operation power and start-up impact power. Therefore, the orderly power consumption strategy of the electric heat pump should consider not only the steady-state operating power but also the start-up impulse power. That is to control the electric heat pump in the starting state and the running state at the same time.

基于以上考虑,按工作状态将电热泵分组为相应状态的负荷群。通过制定控 制策略使不同工作状态下的电热泵相互转化可以改变负荷状态群的整体运行功 率。配电网中任何时刻都存在着处于启动、运行与关闭三种不同状态的电热泵, 使用下列负荷状态群对电热泵进行分组:Based on the above considerations, the electric heat pumps are grouped into load groups of corresponding states according to their working states. The overall operating power of the load state group can be changed by formulating a control strategy to make the electric heat pumps in different working states convert each other. At any time in the distribution network, there are three different states of electric heat pumps: startup, operation and shutdown. The electric heat pumps are grouped using the following load state groups:

Figure BDA0002657814280000201
Figure BDA0002657814280000201

Figure BDA0002657814280000202
Figure BDA0002657814280000202

Figure BDA0002657814280000203
Figure BDA0002657814280000203

式中,t为策略执行的当前时刻;St、Rt和Bt为当前工作于启动、运行和 关闭的电热泵负荷群,对应负荷数量为n1、n2和n3,电热泵负荷总数 n=n1+n2+n3。可见随着电热泵负荷工作的进行,电热泵负荷的工作状态以及St、Rt和Bt中的数目n1、n2和n3都会改变。因此系统中所有状态下的电热泵集合可 表示为:In the formula, t is the current moment of strategy execution; S t , R t and B t are the load groups of the electric heat pump currently working on startup, operation and shutdown, and the corresponding loads are n 1 , n 2 and n 3 , the electric heat pump load The total number n=n 1 +n 2 +n 3 . It can be seen that with the working of the electric heat pump load, the working state of the electric heat pump load and the numbers n 1 , n 2 and n 3 in S t , R t and B t will change. Therefore, the set of electric heat pumps in all states in the system can be expressed as:

Figure BDA0002657814280000211
Figure BDA0002657814280000211

不同工作状态下的不同电热泵负荷在当前时刻t消耗的电功率各不相同,同 样按照负荷群集合的方法定义设备群功率集合:Different electric heat pump loads in different working states consume different electric power at the current time t. Similarly, the power set of equipment group is defined according to the method of load group set:

Figure BDA0002657814280000212
Figure BDA0002657814280000212

式中,

Figure BDA0002657814280000213
为电热泵在时刻t所消耗的电功率,不同工作状态下的电热泵对应 于不同负荷功率(启动功率Ps(t)、额定功率PN和零);Dt内设备在t时刻从电网 中汲取的功率可表示为:In the formula,
Figure BDA0002657814280000213
is the electric power consumed by the electric heat pump at time t, and the electric heat pump in different working states corresponds to different load powers (starting power P s (t), rated power P N and zero); the equipment in D t is removed from the power grid at time t The power drawn can be expressed as:

Figure BDA0002657814280000214
Figure BDA0002657814280000214

式中,

Figure BDA0002657814280000215
为在t时刻处于启动状态的第k台电热泵启动功率;
Figure BDA0002657814280000216
为在t时刻 处于运行状态的第j台电热泵额定功率。因此有序用电控制策略对响应群体中的 电热泵进行温度设定值调整或开关操作,进而改变电热泵负荷运行功率,通过响 应群体中各个电热泵的功率改变,可以影响电热泵负荷群的整体运行功率。In the formula,
Figure BDA0002657814280000215
is the starting power of the kth electric heat pump in the starting state at time t;
Figure BDA0002657814280000216
is the rated power of the jth electric heat pump in operation at time t. Therefore, the orderly power consumption control strategy adjusts the temperature setting value or switches the electric heat pump in the response group, and then changes the load operating power of the electric heat pump. By changing the power of each electric heat pump in the response group, it can affect the electric heat pump load group. overall operating power.

在策略流程图9中得到电热泵的运行功率情况与负荷功率限制Plim的缺额作 为功率削减量Pcat之后,制定控制策略的关键是选择多少数目、哪种工作状态下 的电热泵转化为另一种工作状态来改变负荷状态群整体运行功率。而电热泵工作 状态改变后会引起功率变化的过程有:由运行变为关闭状态、由关闭变为启动状 态。状态改变对应的参数变化如表3。After obtaining the operating power of the electric heat pump and the shortfall of the load power limit P lim as the power reduction amount P cat in the strategy flow chart 9, the key to formulating the control strategy is to choose how many electric heat pumps in which working state are converted into other electric heat pumps. A working state to change the overall operating power of the load state group. After the working state of the electric heat pump changes, the process that will cause the power to change includes: from running to off state, from off to on state. The parameter changes corresponding to the state change are shown in Table 3.

表3table 3

Figure BDA0002657814280000221
Figure BDA0002657814280000221

在有序用电策略的制定时,按照上述两个状态转换过程使不同工作状态下的 电热泵相互转化来改变负荷状态群的整体运行功率即可响应系统的功率调节。有 序用电策略的实现步骤如下:In the formulation of the orderly power consumption strategy, according to the above two state conversion processes, the electric heat pumps in different working states are transformed into each other to change the overall operating power of the load state group to respond to the power regulation of the system. The implementation steps of the orderly electricity consumption strategy are as follows:

1)根据量测、辨识方法得到负荷信息并确定处于不同状态的负荷群。1) According to the measurement and identification methods, the load information is obtained and the load groups in different states are determined.

2)对启动和运行状态的两个设备群的电热泵负荷进行优先级排序。2) Prioritize the electric heat pump loads of the two equipment groups in startup and operation states.

针对前述两种不同的台区负荷功率越限情形,具体采用不同的电热泵有序用 电方法:In view of the above-mentioned two different situations of load power exceeding the limit, different methods of orderly use of electric heat pumps are specifically adopted:

情形一:控制启动序列,限制启动功率Scenario 1: Controlling the startup sequence, limiting startup power

当多台电热泵同时启动时,其叠加的启动功率冲击可能超出台区负荷功率限 制Plim。而在配网负荷不重的情况下,部分热泵的启动冲击功率不会使台区负荷 越限,因此通过人为地控制电热泵的开启序列是避免其同时启动是回避严重功率 冲击的一种方式。为了确定电热泵的启动序列,根据电热泵临界启动电压及其接 入节点的电压水平进行优先级排列,并根据优先级和搜索条件确定启动台数。When multiple electric heat pumps are started at the same time, the superimposed start-up power impact may exceed the load power limit P lim of the station area. When the load of the distribution network is not heavy, the start-up impact power of some heat pumps will not cause the load of the station area to exceed the limit. Therefore, artificially controlling the start-up sequence of the electric heat pumps is a way to avoid their simultaneous start-up and avoid serious power shocks. . In order to determine the starting sequence of the electric heat pump, the priority is arranged according to the critical starting voltage of the electric heat pump and the voltage level of the access node, and the number of starting units is determined according to the priority and search conditions.

首先按照电热泵所在的节点在历史运行时间内的电压均值Uk由小到大的关 系将处于启动以及即将启动的电热泵进行排序,按照“难启动,先启动”(即外 加电压低于启动临界电压的电热泵优先被启动)的原则使所接入电压Uk较大的 电热泵具有更靠后的顺序。然后各电热泵按照Uk与额定电压UN和其临界启动电 压Ulim的偏差生成稳态电压优先级Fu如下。Fu的值越大说明优先级越高,下同。First, according to the relationship between the average voltage U k of the node where the electric heat pump is located in the historical running time from small to large, the electric heat pumps that are in startup and about to be started are sorted. The principle that the electric heat pump with the critical voltage is started first) makes the electric heat pump with a larger connected voltage U k have a later order. Then each electric heat pump generates a steady-state voltage priority F u according to the deviation of U k from the rated voltage U N and its critical starting voltage U lim as follows. The larger the value of F u , the higher the priority, the same below.

Figure BDA0002657814280000222
Figure BDA0002657814280000222

再根据电热泵Nk所在的室内温度Tk,对不同等级室温的电热泵按照“温度低, 先启动”(即室温接近温度区间下限的电热泵优先被启动)的原则生成不同温度 优先级FT如下:Then, according to the indoor temperature T k where the electric heat pump N k is located, the electric heat pumps with different grades of room temperature are generated according to the principle of "low temperature, start first" (that is, the electric heat pumps with room temperature close to the lower limit of the temperature range are preferentially started) to generate different temperature priorities F. T is as follows:

Figure BDA0002657814280000231
Figure BDA0002657814280000231

综合上述的排序及优先级计算方法,将已按历史运行时间内的电压均值Uk排序的电热泵启动序列进行更新。更新方法为当前序列位置号加上相应的优先 级,若电热泵Nk按照历史运行时间内的电压均值Uk排序位于k位,则考虑优先级 后的启动序列编号为:Combining the above sorting and priority calculation methods, the electric heat pump start-up sequence that has been sorted according to the voltage average value U k in the historical running time is updated. The update method is to add the corresponding priority to the current sequence position number. If the electric heat pump N k is ranked in the k position according to the average voltage U k in the historical operation time, the starting sequence number after considering the priority is:

K′=k+Fu+FT (56)K′=k+F u + FT (56)

情形二:控制有序停机,削减稳态功率Scenario 2: Controlled orderly shutdown, reducing steady-state power

当配电变压器稳态运行的负荷较大时,叠加部分启动功率便可能导致台区负 荷功率越限。与控制电热泵的开启序列类似,为了合理削减稳态运行的负荷来降 低台区负荷总量需要合理地安排关闭台数及序列。根据电热泵临界启动电压及其 所在的室内温度水平进行优先级排列,并根据优先级和搜索条件确定关闭台数。When the load of the distribution transformer in steady-state operation is large, the superimposed part of the starting power may cause the load power to exceed the limit in the station area. Similar to controlling the start-up sequence of the electric heat pump, in order to reasonably reduce the steady-state operation load and reduce the total load of the station area, it is necessary to reasonably arrange the number and sequence of shutdown units. Priority is arranged according to the critical starting voltage of the electric heat pump and its indoor temperature level, and the number of units to be turned off is determined according to the priority and search conditions.

首先按照电热泵所在的室内温度Tin由大到小的关系,将处于稳定运行的电 热泵按照“温度高,先关闭”(即室温接近温度区间上限的电热泵优先被关闭) 的原则进行排序,使所在室温接近温度区间下限的电热泵具有更靠后的顺序。然 后根据电热泵Nj所在的室内温度Tj,对不同等级室温的电热泵生成不同温度优先 级F′T如下:First, according to the relationship between the indoor temperature T in where the electric heat pump is located from large to small, the electric heat pumps in stable operation are sorted according to the principle of "high temperature, turn off first" (that is, the electric heat pump whose room temperature is close to the upper limit of the temperature range is preferentially turned off). , so that the electric heat pump whose room temperature is close to the lower limit of the temperature range has a later order. Then, according to the indoor temperature T j where the electric heat pump N j is located, different temperature priorities F′ T are generated for the electric heat pumps with different grades of room temperature as follows:

Figure BDA0002657814280000241
Figure BDA0002657814280000241

再按照“易启动,先关闭”(即外加电压高于启动临界电压的电热泵优先被 关闭)的原则,根据电热泵Nj长时间运行状态下外加电压均值Uj与其临界启动 电压Ulim的偏差生成电压优先级F′U如下:Then according to the principle of "easy to start, first to close" (that is, the electric heat pump with the applied voltage higher than the starting critical voltage is preferentially turned off), according to the difference between the average applied voltage U j and its critical starting voltage U lim of the electric heat pump N j in the long-term operation state. The deviation generation voltage priority F' U is as follows:

Figure BDA0002657814280000242
Figure BDA0002657814280000242

电热泵的关闭序列更新方法与电热泵启动序列更新方式类似,当前序列位置 号加上相应的优先级,若电热泵Nj按Uj排序位于j位,则考虑优先级后的关闭序 列编号为:j′=j+F′T+F′UThe update method of the shutdown sequence of the electric heat pump is similar to the update method of the startup sequence of the electric heat pump. The current sequence position number is added with the corresponding priority. If the electric heat pump N j is located in the j position according to the U j order, the shutdown sequence number after considering the priority is : j'= j +F'T+ F'U .

3)确定受控制的电热泵设备顺序及数量。当系统要限制负荷启动功率时, 对负荷群St进行控制来限制启动设备的数量;当系统要降低负荷量以保证启动完 成时,对负荷群Rt进行控制来削减稳态负荷。由系统所需的功率削减量Pcut确定 需要控制的电热泵台数,响应设备台数的确定方法如表4所示。3) Determine the order and quantity of controlled electric heat pump equipment. When the system wants to limit the load starting power, the load group S t is controlled to limit the number of starting equipment; when the system wants to reduce the load to ensure the start-up is completed, the load group R t is controlled to reduce the steady-state load. The number of electric heat pumps to be controlled is determined by the power reduction amount P cut required by the system, and the method for determining the number of response equipment is shown in Table 4.

表4Table 4

Figure BDA0002657814280000251
Figure BDA0002657814280000251

对于启动功率冲击问题的两种不同的情形下,基于此表规则在启动负荷群St和运行负荷群Rt进行条件搜索。由上述搜索方向以及搜索条件可以得到t时刻最 终参与负荷控制的群体R′t及S′t如下。For two different cases of the startup power shock problem, a conditional search is performed on the startup load group S t and the operating load group R t based on this table rule. From the above search direction and search conditions, the groups R' t and S' t that finally participate in load control at time t can be obtained as follows.

Figure BDA0002657814280000252
Figure BDA0002657814280000252

Figure BDA0002657814280000253
Figure BDA0002657814280000253

根据需要依次关闭R′t内设备或开启S′t内设备,其中对启动序列S′t中电热泵 机组的启动时间进行延时干预,确保启动时间尽量错开,避免冲击叠加。其中第 i台将启动的热泵延迟时间为△t=i·t′(i=1,2,...,k1),t′取电热泵最大启动时长。 通过电热泵的启停控制直接改变这些设备的运行状态。达到限制启动台数避免启 动功率叠加越限和削减稳态负荷来优先满足启动需求的目的。Turn off the equipment in R' t or turn on the equipment in S' t in sequence according to the needs, in which delay intervention is performed on the start-up time of the electric heat pump unit in the start-up sequence S' t to ensure that the start-up time is staggered as much as possible to avoid shock superposition. The delay time of the heat pump to be started by the ith unit is Δt=i·t' (i=1,2,...,k 1 ), and t' is the maximum start-up time of the electric heat pump. The operating state of these devices can be directly changed through the start-stop control of the electric heat pump. The purpose of limiting the number of starting units to avoid the superposition of starting power exceeding the limit and reducing the steady-state load to preferentially meet the starting demand.

Claims (10)

1.一种基于μPMU装置的大规模电热泵有序用电控制方法,其特征在于,包括以下步骤:1. a large-scale electric heat pump orderly control method based on μPMU device, is characterized in that, comprises the following steps: 1)电热泵制热系统建模;1) Modeling of electric heat pump heating system; 2)基于μPMU量测辨识电热泵状态;2) Identify the state of the electric heat pump based on μPMU measurement; 3)基于量测辨识的电热泵有序用电策略。3) Orderly electricity consumption strategy of electric heat pump based on measurement and identification. 2.根据权利要求1所述的基于μPMU装置的大规模电热泵有序用电控制方法,其特征在于,所述步骤1)中电热泵制热系统中2. The method for controlling orderly electricity consumption of a large-scale electric heat pump based on a μPMU device according to claim 1, wherein in the step 1) in the electric heat pump heating system 压缩机电功率Pcmp关于转速N、制冷剂流量mr的函数关系式:The functional relationship between the electric power P cmp of the compressor and the rotational speed N and the refrigerant flow rate m r :
Figure FDA0002657814270000011
Figure FDA0002657814270000011
上式中c1为压缩机吸气比容,n为压缩过程多变指数,ηcmp为压缩机指示功率和耗电功率的比值,
Figure FDA0002657814270000012
表示排气压力与吸气压力之比;
In the above formula, c 1 is the suction specific volume of the compressor, n is the variable index of the compression process, η cmp is the ratio of the indicated power of the compressor to the power consumption,
Figure FDA0002657814270000012
Represents the ratio of exhaust pressure to suction pressure;
冷凝器的换热模型如下:The heat transfer model of the condenser is as follows:
Figure FDA0002657814270000013
Figure FDA0002657814270000013
式中,Cr,sh,cond为制冷剂比热系数;T2为流入冷凝器的制冷剂温度;Tcond为冷凝器的过热温度;T3为流出冷凝器的制冷剂温度;rcond为板式冷凝器的结构半径;Ai,cond为板式冷凝器的换热面积;Tw是冷凝器热水侧的出水口温度;Usc,cond、Ush,cond、Utp,cond分别为冷凝器冷水测、过热区、两相区换热综合系数,Tw为热水温度;In the formula, C r, sh, cond is the specific heat coefficient of the refrigerant; T 2 is the temperature of the refrigerant flowing into the condenser; T cond is the superheat temperature of the condenser; T 3 is the temperature of the refrigerant flowing out of the condenser; r cond is The structural radius of the plate condenser; A i,cond is the heat exchange area of the plate condenser; Tw is the outlet temperature of the hot water side of the condenser; U sc,cond , U sh,cond , U tp,cond are the condensation is the comprehensive coefficient of heat transfer in cold water measurement, superheated zone and two-phase zone of the boiler, and Tw is the temperature of hot water; 蒸发器的换热量满足:The heat exchange of the evaporator satisfies: Qeva=Qtp+Qsh=Qa Q eva =Q tp +Q sh =Q a 上式中Qeva为蒸发器中转移给制冷剂的热量,Qa为蒸发器吸收自热源介质的热量。In the above formula, Q eva is the heat transferred to the refrigerant in the evaporator, and Q a is the heat absorbed by the evaporator from the heat source medium. 异步电机的d-q坐标系复变量模型描述为:The complex variable model of the d-q coordinate system of the asynchronous motor is described as:
Figure FDA0002657814270000021
Figure FDA0002657814270000021
上式中,
Figure FDA0002657814270000022
Figure FDA0002657814270000023
分别定子电压、定子电流、转子电流、定子磁链和转子磁链的复数向量;Rs、Ls、M、Rr、Lr分别为定子电阻、定子电感、定转子互感、转子电阻和转子电感;ls=Ls-M和lr=Lr-M分别为定子和转子漏电感;ωr为转子电角速度,单位为rad/s;j是复数单位。
In the above formula,
Figure FDA0002657814270000022
Figure FDA0002657814270000023
Complex vectors of stator voltage, stator current, rotor current, stator flux linkage and rotor flux linkage, respectively; R s , L s , M, R r , L r are stator resistance, stator inductance, stator-rotor mutual inductance, rotor resistance and rotor, respectively Inductance; l s =L s -M and l r =L r -M are the stator and rotor leakage inductances, respectively; ω r is the rotor electrical angular velocity, in rad/s; j is a complex unit.
3.根据权利要求1所述的基于μPMU装置的大规模电热泵有序用电控制方法,其特征在于,所述步骤1)中电热泵制热系统的电热泵启动过程有功功率Ps关于时间t的闭区间函数关系为:3. the orderly control method of large-scale electric heat pump based on μPMU device according to claim 1, is characterized in that, in described step 1), the electric heat pump starting process active power P s of electric heat pump heating system is about time The closed interval functional relationship of t is: Ps(t)=-1160(t-0.2)2+Pmax t∈[t0,t1]P s (t)=-1160(t-0.2) 2 +P max t∈[t 0 ,t 1 ] 上式中t0为启动初始时刻,t1为启动完成电热泵进入稳态的时刻,Pmax为电热泵启动功率最大值。In the above formula, t 0 is the initial start time, t 1 is the time when the electric heat pump enters a steady state after starting up, and P max is the maximum starting power of the electric heat pump. 4.根据权利要求1所述的基于μPMU装置的大规模电热泵有序用电控制方法,其特征在于,所述步骤2)中首先建立首端节点电压Ui约束,台区首端节点电压Ui需满足:4. the method for orderly control of large-scale electric heat pump based on μPMU device according to claim 1, it is characterized in that, in described step 2), first establish head-end node voltage U i constraint, the station head-end node voltage U i must meet: Ui≥{max(Ulim.k)|Tekmaxrk,Ulim.k)=Tmmaxrk),k∈[1,n]}U i ≥{max(U lim.k )|T ekmaxrk ,U lim.k )=T mmaxrk ),k∈[1,n]} 上式中Ulim.k为节点k处电热泵负荷能正常启动时临界值,Tekmaxrk,Ulim.k)为第k台电热泵负荷中作用在电机转子上的电磁转矩,ωmaxrk为Tek取最大值时所对应的最大转子转速,Ulim.k为节点k处临界启动电压,Tmmaxrk)为异步电机转速标幺值ωmaxrk下的电热泵机械转矩。In the above formula, U lim.k is the critical value when the electric heat pump load at node k can start normally, T ekmaxrk , U lim.k ) is the electromagnetic torque acting on the motor rotor in the kth electric heat pump load, ω maxrk is the maximum rotor speed when T ek takes the maximum value, U lim.k is the critical starting voltage at node k, and T mmaxrk ) is the mechanical torque of the electric heat pump at the per-unit value ω maxrk of the asynchronous motor speed. 5.根据权利要求4所述的基于μPMU装置的大规模电热泵有序用电控制方法,其特征在于,根据功率冲击和电压暂降判断电热泵启动状态,其中功率冲击判据如下:5. the orderly control method of large-scale electric heat pump based on μPMU device according to claim 4, is characterized in that, according to power shock and voltage sag to judge electric heat pump start-up state, and wherein power shock criterion is as follows:
Figure FDA0002657814270000024
Figure FDA0002657814270000024
上式中P(t)为当前时刻t的负荷功率;P(t-1)为前一量测时刻的负荷功率;N为时刻t的量测频数,功率变化量取n个量测数据依次作差并求和得到;P′(t)为当前时刻t的负荷功率变化速率;λ1、λ2为判据对应的阈值。In the above formula, P(t) is the load power at the current time t; P(t-1) is the load power at the previous measurement time; N is the measurement frequency at time t, and the power change takes n measurement data in turn Differences and summations are obtained; P'(t) is the rate of change of load power at the current time t; λ 1 and λ 2 are the thresholds corresponding to the criteria. 电压暂降判据为:The voltage sag criterion is:
Figure FDA0002657814270000031
Figure FDA0002657814270000031
式中,λ3和λ4为用于描述Ut-1与Ut-2的接近程度阈值,N为时刻t的量测频数。In the formula, λ 3 and λ 4 are the thresholds used to describe the proximity of U t-1 and U t-2 , and N is the measurement frequency at time t.
6.根据权利要求1所述的基于μPMU装置的大规模电热泵有序用电控制方法,其特征在于,所述步骤2)中利用电热泵启动冲击功率中二次函数的数量,辨识出处于启动状态的电热泵台数。6. The orderly control method for large-scale electric heat pump based on μPMU device according to claim 1, characterized in that, in the step 2), the number of quadratic functions in the electric heat pump startup impulse power is utilized to identify The number of electric heat pumps in the activated state. 7.根据权利要求1所述的基于μPMU装置的大规模电热泵有序用电控制方法,其特征在于,所述步骤3)中将负荷电压转化为负荷功率,具体如下:7. The orderly control method of large-scale electric heat pump based on μPMU device according to claim 1, is characterized in that, in described step 3), load voltage is converted into load power, and is specifically as follows: 首端节点n的电压和注入有功功率满足:The voltage and injected active power of the head-end node n satisfy: Un=kP·[P-P(t0)]+Un(t0)U n =k P ·[PP(t 0 )]+U n (t 0 ) 式中,Un为首端节点n所求的电压;P为首端节点n所求的注入功率,Un(t0)、P(t0)为t0时刻测出的当前节点电压和功率,kp为斜率,在其中一个量已知的情况下可以计算出另外一个量的值。In the formula, U n is the voltage required by the head-end node n; P is the injected power required by the head-end node n , Un (t 0 ), P (t 0 ) are the current node voltage and power measured at time t 0 k p is the slope, and when one of the quantities is known, the value of the other quantity can be calculated. 8.根据权利要求7所述的基于μPMU装置的大规模电热泵有序用电控制方法,其特征在于,所述有序用电策略实现步骤如下:8. The method for controlling orderly power consumption of a large-scale electric heat pump based on a μPMU device according to claim 7, wherein the orderly power consumption strategy implementation steps are as follows: 1)根据量测、辨识方法得到负荷信息并确定处于不同状态的负荷群;1) Obtain load information according to measurement and identification methods and determine load groups in different states; 2)对启动和运行状态的两个设备群的电热泵负荷进行优先级排序;2) Prioritize the electric heat pump loads of the two equipment groups in startup and running states; 3)确定受控制的电热泵设备顺序及数量。3) Determine the order and quantity of controlled electric heat pump equipment. 9.根据权利要求8所述的基于μPMU装置的大规模电热泵有序用电控制方法,其特征在于,所述步骤2)中针对同一时刻下启动的电热泵启动冲击功率的叠加峰值超出限制,将稳态电压优先级、不同温度优先级以及当前序列位置号相加得到启动序列编号。9. The method for controlling the orderly power consumption of a large-scale electric heat pump based on a μPMU device according to claim 8, wherein in the step 2), the superimposed peak value of the start-up impulse power of the electric heat pump started at the same moment exceeds the limit , add the steady-state voltage priority, different temperature priority and the current sequence position number to get the startup sequence number. 10.根据权利要求8所述的基于μPMU装置的大规模电热泵有序用电控制方法,其特征在于,所述步骤2)中针对配电变压器稳态运行的负荷较大时叠加部分启动功率导致的功率越限,将当前序列位置号、电热泵按照外加电压均值排序的优先级,以及不同温度优先级相加得到关闭序列编号。10. The method for orderly power consumption control of a large-scale electric heat pump based on a μPMU device according to claim 8, wherein in the step 2), when the load for the steady-state operation of the distribution transformer is large, the superimposed part of the starting power The resulting power exceeds the limit, the current sequence position number, the priority of the electric heat pump sorted by the average value of the applied voltage, and the priority of different temperatures are added to obtain the shutdown sequence number.
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