CN106777675B - The hot relevant parameter design method of graphite resistance heater - Google Patents
The hot relevant parameter design method of graphite resistance heater Download PDFInfo
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- 229910002804 graphite Inorganic materials 0.000 title claims abstract description 42
- 239000010439 graphite Substances 0.000 title claims abstract description 42
- 238000013461 design Methods 0.000 title claims abstract description 41
- 238000010438 heat treatment Methods 0.000 claims abstract description 122
- 238000009413 insulation Methods 0.000 claims abstract description 97
- 230000005855 radiation Effects 0.000 claims description 31
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- 239000007789 gas Substances 0.000 description 26
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- G—PHYSICS
- G06—COMPUTING; CALCULATING OR COUNTING
- G06F—ELECTRIC DIGITAL DATA PROCESSING
- G06F30/00—Computer-aided design [CAD]
- G06F30/30—Circuit design
- G06F30/36—Circuit design at the analogue level
- G06F30/367—Design verification, e.g. using simulation, simulation program with integrated circuit emphasis [SPICE], direct methods or relaxation methods
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- H—ELECTRICITY
- H05—ELECTRIC TECHNIQUES NOT OTHERWISE PROVIDED FOR
- H05B—ELECTRIC HEATING; ELECTRIC LIGHT SOURCES NOT OTHERWISE PROVIDED FOR; CIRCUIT ARRANGEMENTS FOR ELECTRIC LIGHT SOURCES, IN GENERAL
- H05B3/00—Ohmic-resistance heating
- H05B3/10—Heating elements characterised by the composition or nature of the materials or by the arrangement of the conductor
- H05B3/12—Heating elements characterised by the composition or nature of the materials or by the arrangement of the conductor characterised by the composition or nature of the conductive material
- H05B3/14—Heating elements characterised by the composition or nature of the materials or by the arrangement of the conductor characterised by the composition or nature of the conductive material the material being non-metallic
- H05B3/145—Carbon only, e.g. carbon black, graphite
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- H—ELECTRICITY
- H05—ELECTRIC TECHNIQUES NOT OTHERWISE PROVIDED FOR
- H05B—ELECTRIC HEATING; ELECTRIC LIGHT SOURCES NOT OTHERWISE PROVIDED FOR; CIRCUIT ARRANGEMENTS FOR ELECTRIC LIGHT SOURCES, IN GENERAL
- H05B3/00—Ohmic-resistance heating
- H05B3/40—Heating elements having the shape of rods or tubes
- H05B3/42—Heating elements having the shape of rods or tubes non-flexible
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- G—PHYSICS
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Abstract
The invention discloses a kind of hot relevant parameter design methods of graphite resistance heater, graphite resistance heater is made of components such as rectangular spiral passage, heating tube, the thermally conductive, thermal insulation layers in end, the present invention passes through selected heater structure parameter and given heater power, simplified mathematical model is combined using a kind of air-flow-heating tube-thermal insulation layer temperature, the Axial Temperature Distribution of the air-flow of heater, heating element, thermal insulation layer is calculated, required by exit flow temperature index, efficiency index etc. it is up to standard, complete graphite resistance heater hot relevant parameter design calculate.
Description
Technical field
The present invention relates to the hot phases of a kind of superhigh temperature continous way gas heater more particularly to a kind of graphite resistance heater
Close Parameters design.
Background technique
Graphite resistance heater is a kind of superhigh temperature continous way gas heater, can be used for the heating of the inert gases such as nitrogen,
Heating temperature is up to 2500K, can be applied to the gas medium heating of the ground simulation tests equipment such as superb wind-tunnel.
The hot relevant parameter design method of graphite resistance heater is a kind of continuous heaters thermal design side of engineering
Method.Existing continuous heaters Thermal design is broadly divided into two kinds: one is the heater designs for being engineered simple fuel factor
Method only considers the simple fuel factors such as convection current, the energy；Another kind is the heater design method for refining analogue simulation, usually
Minutia for Accurate Analysis heater fuel factor.
(1) the heater Thermal design of simple fuel factor: this is a kind of heater engineering Thermal design, only considers pipe
The simple fuel factor such as interior air-flow convection current, the energy, governing equation successively indicate that air-flow increases energy, round tube and air-flow pair from left to right
Stream heat exchange energy, power supply loaded energy, wherein the heat convection of round tube and air-flow is engineering heat treatment method (please see Figure 1).
Governing equation indicates are as follows:
(2) the heater design method of analogue simulation is refined: right in the document that Chen Huajun, Shen Xinrong et al. are delivered
Convective heat transfer characteristic is studied in detail in the helix tube of rectangular section, analyzes under rotation, curvature and torsion joint effect
Secondary flow in pipeline, axial mainstream distribution, coefficient of friction than the situation of change between, pipeline Nusselt number and each parameter,
The fluid interchange mechanism in rectangular spiral passage is studied emphatically.In the document that Wu Shuanying, Chen Sujun et al. are delivered, annular is cut
Convective heat transfer characteristic is studied in detail in surface helix pipe, analyzes Reynolds number, song in the case where inner ring heats outer ring adiabatic condition
The influence of rate, screw pitch to pipeline Nusselt number, coefficient of friction.
In both of the above method, the heater Thermal design of simple fuel factor does not consider conductive heat loss, radiation damage
The fuel factors such as mistake, only adapt to the design of low temperature continuous heaters.And refine the heater design method of analogue simulation by
It is extremely complex in governing equation, it is simply possible to use in research local heat effect, engineer application can not be adapted to.
Summary of the invention
In view of the above technical problems, the present invention has designed and developed a kind of hot relevant parameter design side of graphite resistance heater
Method can pass through theory deduction completely, without passing through experience curve, to obtain the hot relevant parameter of heater；This method can be with
Adapt to the design of the engineering continuous heaters to superhigh temperature, low temperature.
Technical solution provided by the invention are as follows:
A kind of hot relevant parameter design method of graphite resistance heater, the graphite resistance heater includes rectangular coil
Channel, heating tube, cold end heat-conducting piece, hot end heat-conducting piece and thermal insulation layer, the graphite resistance heating tube are sheathed on the rectangle
The outside of helical duct, the cold end heat-conducting piece are connected to the cold end of the heating tube, and the hot end heat-conducting piece is connected to described
The hot end of heating tube, the thermal insulation layer are set in the outside of the graphite resistance heating tube, and a heating power supply is electrically connected to described
Cold end heat-conducting piece and the hot end heat-conducting piece, to start the heating tube fever；
It the described method comprises the following steps:
Step 1: being set in the following way to the parameter of rectangular spiral passage, heating tube and hot end heat-conducting piece
And calculating:
The height of the rectangular spiral passageWherein, G is air flow rate, and P is gas
Flowing pressure, R are thermodynamic equilibrium constant, ugLFor the outlet air flow velocity of the rectangular spiral passage；By the rectangular spiral passage edge
Axially discrete is M infinitesimal, then is t in i-th of infinitesimal overdraught temperaturegi, tg1It is micro- at the 1st for the rectangular spiral passage
The gas flow temperature of member；
The width b=3.5a of the rectangular spiral passage；
Screw pitch s=a+ the Δ a, Δ a=3~5mm of the rectangular spiral passage；
The outside diameter d of the rectangular spiral passagew=1.8s；
The thickness δ of the heating tubew=5~10mm；
The length of the heating tubeWherein, CpFor gas level pressure
Heat capacity ratio, α are convection transfer rate, tg0For the end air flow temperature of the rectangular spiral passage, Δ tgw=250~350K is
The designed temperature difference of heating tube and air-flow, tgMFor the hot end gas flow temperature of the rectangular spiral passage；
The thermal coefficient λ of the heating tubew=50~100W/ (m*K)；
Step 2: the parameter of thermal insulation layer is set and is calculated in the following manner:
The thickness δ of the thermal insulation layerf=60~100mm；
Spacing h between the thermal insulation layer and the outer wall of the heating tube is selected at the beginning of one in the range of 20~30mm
Initial value；
Step 3: the parameter of cold end heat-conducting piece and hot end heat-conducting piece is set and is calculated in the following manner:
The outer end temperature t of the cold end heat-conducting pieceOutside w0=300K, the outer end temperature t of the hot end heat-conducting pieceOutside w1=300K,
Then have:
Wherein, q0For cold end heat conduction heat flux, λw0Equivalent for cold end heat-conducting piece is led
Hot coefficient, L0For the equivalent passage of heat length of cold end heat-conducting piece, S0For the equivalent heat-conducting section area of cold end heat-conducting piece, tw0For
The cold junction temperature of the heating tube；
Wherein, q1For hot end heat conduction heat flux, λw1Equivalent for hot end heat-conducting piece is led
Hot coefficient, L1For the equivalent passage of heat length of hot end heat-conducting piece, S1For the equivalent heat-conducting section area of hot end heat-conducting piece, tw1For
The hot-side temperature of the heating tube；
It is calculated Step 4: air-flow-heating tube-thermal insulation layer temperature joint simplifies:
By rectangular spiral passage along it is axial it is discrete be M infinitesimal, then be t in i-th of infinitesimal overdraught temperaturegi, will be described
It is M infinitesimal that heating tube edge is axially discrete, then heating tube temperature is t on i-th of infinitesimalwi, by the thermal insulation layer along axial direction from
It dissipates for M separate cylinder radial direction conduction model；
(1) gas flow optimized equation is established:
Wherein, tg0=273K, n are the rectangular spiral passage in unit length
Wetted perimeter is long；
(2) heating tube governing equation is established:
Wherein, the cross-sectional area of the heating tube
Unit
The exterior surface area A of the heating tube of length2=π dw, α is convection transfer rate, σbFor Stefan-bohr hereby climing constant, ρwFor
The density of the heating tube, CwFor the material specific heat capacity of heating tube, λwFor the material conducts heat rate of the heating tube, τ is time, xi
For i-th of infinitesimal of heating tube；
Separately have, the radiation energy that the heating tube is receivedWherein, described
RADIATION ANGLE COEFFICIENT of j-th of the infinitesimal of thermal insulation layer inner surface to i-th of infinitesimal of the heating tube Radiation energy E (the x of described heat exchange j-th of infinitesimal of pipe internal surfacej)=σb·taj 4, the thermal insulation layer inner surface
The area dA of j-th of infinitesimalj=π (dw+2h)·dxj, tajFor the temperature of the jth infinitesimal of the inner surface of the thermal insulation layer, Rij
For the distance between the i-th infinitesimal of j-th infinitesimal and the heating tube of the thermal insulation layer inner surface, xjFor the thermal insulation layer
J-th of infinitesimal；
(3) thermal insulation layer outer surface governing equation and thermal insulation layer inner surface governing equation are established:
Thermal insulation layer inner surface governing equation are as follows: Wherein, tbjFor the temperature of the outer surface jth infinitesimal of the thermal insulation layer, tajFor the thermal insulation layer inner surface jth infinitesimal
Temperature, λfFor the material conducts heat rate of the thermal insulation layer, da=dw+2δw+ 2h, db=dw+2δw+2h+δf；
Separately have, the radiation energy that the thermal insulation layer inner surface is receivedIts
In, RADIATION ANGLE COEFFICIENT of described j-th of the infinitesimal of thermal insulation layer inner surface to i-th of infinitesimal of the heating tubeRadiation energy E (the x of i-th of infinitesimal of the heating tubei)=σb·twi 4,
The area dA of i-th of infinitesimal of the heating tubei=π dw·dxi；
Thermal insulation layer outer surface governing equation are as follows:
Wherein, αsFor NATURAL CONVECTION COEFFICIENT OF HEAT,
t0=303K；
(4) by q obtained in the step 30、q1、tOutside w1And tOutside w0As boundary condition, by gas flow optimized equation, heating
Pipe governing equation, thermal insulation layer inner surface governing equation and thermal insulation layer outer surface governing equation joint are calculated in conjunction with boundary condition
M t outgiValue, M twiValue, M tbjValue and M tajValue；
(5) a heater power N is given, (1)~(4) is utilized to calculate tgi、twi、tbj、tajValue passes through given difference
N value, until tgM≥tgL, tgLFor gas flow temperature threshold value；
Step 5: heater efficiency calculates
Calculate heater efficiencyWherein, the value of N is by the number finally determined in step 4
Value, when the calculated heater efficiency of institute is less than η0, then change the value of h, and re-execute step 4, until η >=η0, wherein
η0For heater efficiency index.
Preferably, in the hot relevant parameter design method of the graphite resistance heater, the rectangular spiral passage
Outlet air flow velocity ugL=30m/s.
Preferably, in the hot relevant parameter design method of the graphite resistance heater, η0Value is 50%.
Preferably, in the hot relevant parameter design method of the graphite resistance heater, αsValue is 8.7W/ (m*
K)。
Preferably, in the hot relevant parameter design method of the graphite resistance heater, in the step 5, change
When the value of h, value in the range of except 20~30mm.
The hot relevant parameter design method of graphite resistance heater of the present invention is by selecting heater structure parameter
With given heater power, simplified mathematical model is combined using air-flow-heating tube-thermal insulation layer temperature, heater is calculated
Air-flow, heating tube, thermal insulation layer Axial Temperature Distribution, required by exit flow temperature index, efficiency index etc. it is up to standard, it is complete
The design of hot relevant parameter and calculating of pairs of graphite resistance heater.
Detailed description of the invention
Fig. 1 is the schematic diagram of the heater Thermal design of simple fuel factor in the prior art.
Fig. 2 is the schematic diagram of graphite resistance heater of the present invention.
Fig. 3 is the schematic cross-section of rectangular spiral passage of the present invention.
Fig. 4 is heating tube of the present invention and heat-insulated lamellar spacing radiation loss simplified mathematical model.
Fig. 5 is that function is given in the embodiment one of the hot relevant parameter design method of graphite resistance heater of the present invention
Rate N is 1300kw, heter temperature distribution curve when spacing h is 100mm.
Fig. 6 is that function is given in the embodiment one of the hot relevant parameter design method of graphite resistance heater of the present invention
Rate N is 1200kw, heter temperature distribution curve when spacing h is 50mm.
Fig. 7 is that function is given in the embodiment two of the hot relevant parameter design method of graphite resistance heater of the present invention
Rate N is 150kw, heter temperature distribution curve when spacing h is 30mm.
Fig. 8 is that function is given in the embodiment two of the hot relevant parameter design method of graphite resistance heater of the present invention
Rate N is 1200kw, heter temperature distribution curve when spacing h is 10mm.
Specific embodiment
Present invention will be described in further detail below with reference to the accompanying drawings, to enable those skilled in the art referring to specification text
Word can be implemented accordingly.
As shown in Figures 2 and 3, the present invention provides a kind of hot relevant parameter design method of graphite resistance heater, described
Graphite resistance heater includes rectangular spiral passage 2, heating tube 1, cold end heat-conducting piece 5, hot end heat-conducting piece 6 and thermal insulation layer 3, institute
The outside that graphite resistance heating tube is sheathed on the rectangular spiral passage is stated, the cold end heat-conducting piece is connected to the heating tube
Cold end, the hot end heat-conducting piece are connected to the hot end of the heating tube, and the thermal insulation layer is set in the graphite resistance heating tube
Outside, a heating power supply 4 is electrically connected to the cold end heat-conducting piece and the hot end heat-conducting piece, to start the heating tube hair
Heat.
Graphite resistance heater is a kind of continuous heaters, mainly thermally conductive by rectangular spiral passage, heating tube, end
The components such as (i.e. cold end heat-conducting piece and hot end heat-conducting piece), thermal insulation layer composition passes through selected heater structure parameter and given heating
Device power combines simplified mathematical model using a kind of air-flow-heating tube-thermal insulation layer temperature, and the air-flow of heater is calculated, adds
The Axial Temperature Distribution of thermal element, thermal insulation layer requires up to standard, completion graphite by exit flow temperature index, efficiency index etc.
The hot relevant parameter of resistance heater, which designs, to be calculated.
The method specifically includes the following steps:
Graphite resistance heater is designed with this method, it is known that air flow rate G, exit flow temperature requirement tgL, stream pressure
P, heater efficiency requires η0, follow these steps the hot relevant parameter for calculating graphite resistance heater.
Step 1: being set in the following way to the parameter of rectangular spiral passage, heating tube and hot end heat-conducting piece
And calculating:
The height of the rectangular spiral passageWherein, G is air flow rate, and P is
Stream pressure, R are thermodynamic equilibrium constant, ugLFor the outlet air flow velocity of the rectangular spiral passage；By the rectangular spiral passage
It is M infinitesimal that edge is axially discrete, then is t in i-th of infinitesimal overdraught temperaturegi, tg1It is the rectangular spiral passage at the 1st
The gas flow temperature of infinitesimal；
The width b=3.5a of the rectangular spiral passage；
Screw pitch s=a+ the Δ a, Δ a=3~5mm of the rectangular spiral passage；
The outside diameter d of the rectangular spiral passagew=1.8s；
The thickness δ of the heating tubew=5~10mm；
The length of the heating tubeWherein, CpFor gas level pressure
Heat capacity ratio, α are convection transfer rate, tg0For the end air flow temperature of the rectangular spiral passage, Δ tgwFor=250~350K,
For the designed temperature difference of heating tube and air-flow, tgMFor the hot end gas flow temperature of the rectangular spiral passage；
The thermal coefficient λ of the heating tubew=50~100W/ (m*K).
Step 2: the parameter of thermal insulation layer is set and is calculated in the following manner:
The thickness δ of the thermal insulation layerf=60~100mm；
Spacing h between the thermal insulation layer and the outer wall of the heating tube is selected at the beginning of one in the range of 20~30mm
Initial value.
Step 3: the parameter of cold end heat-conducting piece and hot end heat-conducting piece is set and is calculated in the following manner:
The outer end temperature t of the cold end heat-conducting pieceOutside w0=300K, the outer end temperature t of the hot end heat-conducting pieceOutside w1=300K,
Then have:
Wherein, q0For cold end heat conduction heat flux, λw0Equivalent for cold end heat-conducting piece is led
Hot coefficient, L0For the equivalent passage of heat length of cold end heat-conducting piece, S0For the equivalent heat-conducting section area of cold end heat-conducting piece, tw0For
The cold junction temperature of the heating tube；
Wherein, q1For cold end heat conduction heat flux, λw1Equivalent for hot end heat-conducting piece is led
Hot coefficient, L1For the equivalent passage of heat length of hot end heat-conducting piece, S1For the equivalent heat-conducting section area of hot end heat-conducting piece, tw1For
The hot-side temperature of the heating tube.
According to the structure of cold end heat-conducting piece and hot end heat-conducting piece, the equivalent passage of heat length L of cold end heat-conducting piece is determined0、
Equivalent heat-conducting section S0, Equivalent Thermal Conductivities λw0, the equivalent passage of heat length L of hot end heat-conducting piece1, equivalent heat-conducting section S1, etc.
Imitate thermal coefficient λw1This six parameters, keep end conductive heat loss suitable with end practical structures heat loss.It is with cold end heat-conducting piece
Example, equivalent passage of heat length L0It is determined according to the thermally conductive pathways length of cold end heat-conducting piece, equivalent heat-conducting section S0It is led according to cold end
The thermally conductive pathways sectional area of warmware is determining, Equivalent Thermal Conductivities λw0It is determined according to the selected material of cold end heat-conducting piece；Correspondingly, hot
Hold the equivalent passage of heat length L of heat-conducting piece1It is determined according to the thermally conductive pathways length of hot end heat-conducting piece, equivalent heat-conducting section S1Root
It is determined according to the thermally conductive pathways sectional area of hot end heat-conducting piece, Equivalent Thermal Conductivities λw1It is determined according to the selected material of hot end heat-conducting piece.
It is calculated Step 4: air-flow-heating tube-thermal insulation layer temperature joint simplifies:
By rectangular spiral passage along it is axial it is discrete be M infinitesimal, then be t in i-th of infinitesimal overdraught temperaturegi, will be described
It is M infinitesimal that heating tube edge is axially discrete, then heating tube temperature is t on i-th of infinitesimalwi, by the thermal insulation layer along axial direction from
It dissipates for M separate cylinder radial direction conduction model；
(1) gas flow optimized equation is established:
Wherein, tg0=273K, n are the rectangular spiral passage in unit length
Wetted perimeter is long.
Specifically, the method calculated using numerical value, by airflow channel (i.e. rectangular spiral passage) along it is axial it is discrete be M
Grid node, each grid node overdraught temperature are a constant value tgi, it is discrete using difference method, then it can acquire gas flow temperature
Distribution.
Discrete to gas flow optimized equation progress implicit difference, discrete equation is as follows:
agi、bgi、cgiRespectively represent gas flow optimized generation
Number i-th of equation coefficient of equation group, Δ x --- discrete infinitesimal length,
Arrangement obtains gas flow optimized Algebraic Equation set:
agitgi-1 n+1+bgitgi n+1+cgitwi n+1=0 (i=1 ... ... M),
Wherein:
Application boundary condition:
tg0 n+1=273K.
Gas flow optimized Algebraic Equation set contains M twiUnknown parameter, M tgiUnknown parameter, step (2) and step (3)
Governing equation can then acquire M t using iterative solutiongiValue is to get gas flow temperature distribution.
(2) heating tube governing equation is established:
Heating tube thermal effect be considered as axial thermal conductivity, it is interior can increase, power load, heat convection, the fuel factors such as radiation, control
It successively can increase, the load of axial thermal conductivity, power, element itself radiation loss, receive in thermal insulation layer in representation element from left to right
Layer radiation energy.
Heating tube governing equation are as follows:
Wherein, the cross-sectional area of the heating tube
Unit
The exterior surface area A of the heating tube of length2=π dw, α is convection transfer rate, σbFor Stefan-bohr hereby climing constant, ρwFor
The density of the heating tube, CwFor the material specific heat capacity of heating tube, λwFor the material conducts heat rate of the heating tube, τ is time, xi
For i-th of infinitesimal of heating tube.
It please see Figure 4, heating tube and heat-insulated lamellar spacing radiation loss computation model are reduced to a kind of line gap radiation patterns.Then
Have, the radiation energy that the heating tube is receivedWherein, table in the thermal insulation layer
RADIATION ANGLE COEFFICIENT of j-th of the face infinitesimal to i-th of infinitesimal of the heating tube
Radiation energy E (the x of described heat exchange j-th of infinitesimal of pipe internal surfacej)=σb·taj 4, the face of described j-th of infinitesimal of thermal insulation layer inner surface
Product dAj=π (dw+2h)·dxj, tajFor the temperature of the jth infinitesimal of the inner surface of the thermal insulation layer, RijFor in the thermal insulation layer
The distance between the i-th infinitesimal of j-th infinitesimal and the heating tube on surface, xjFor j-th of infinitesimal of the thermal insulation layer.
Specifically, the method calculated using numerical value, by heating tube along axial discrete for M grid node, each grid
Node temperature is a constant value twi, it is discrete using difference method, then temp of heating element distribution can be obtained.
It is discrete using implicit difference to heating element governing equation, and linearization process is carried out, discrete equation is as follows:
Gap radiation patterns are discrete using explicit difference, and discrete equation is as follows:
Arrangement obtains heating tube control Algebraic Equation set:
awi、bwi、cwi、dwi、Respectively represent heating tube control i-th of system of equations of Algebraic Equation set
Number,
Wherein:
A2=π dwΔ x,
Application boundary condition:
Heating tube control Algebraic Equation set contains M twiUnknown parameter, M tgiUnknown parameter, M tajUnknown parameter, connection
3 × M the algebraic equation closed in step (1) and step (3) can then acquire M t using iterative solutionwiValue is to get heating tube
Temperature Distribution.
(3) thermal insulation layer outer surface governing equation and thermal insulation layer inner surface governing equation are established:
Thermal insulation layer only considers radial conductive heat loss, radiation loss, external heat-exchanging boundary condition, by thermal insulation layer along axial discrete
For M grid node (i.e. M infinitesimal), then thermal insulation layer temperature computation can be reduced to M separate cylinder radial direction conduction model.J
The place's of setting (i.e. j-th of infinitesimal) thermal insulation layer inner surface governing equation successively represent from left to right the thermal insulation layer radial direction conductive heat loss of the position j,
The radiation loss that thermal insulation layer inner surface radiation loss, inner surface receive.
Thermal insulation layer inner surface governing equation are as follows: Wherein, tbjFor the temperature of the outer surface jth infinitesimal of the thermal insulation layer, tajFor the thermal insulation layer inner surface jth infinitesimal
Temperature, λfFor the material conducts heat rate of the thermal insulation layer, da=dw+2δw+ 2h, db=dw+2δw+2h+δf。
At the position j thermal insulation layer outer surface governing equation successively represent from left to right the thermal insulation layer radial direction conductive heat loss of the position j, every
Thermosphere outer surface external ambient atmosphere heat convection.
Heating tube and heat-insulated lamellar spacing radiation loss computation model are reduced to line gap radiation patterns.Then have, it is described heat-insulated
The radiation energy that layer inner surface is receivedWherein, the thermal insulation layer inner surface
RADIATION ANGLE COEFFICIENT of the j infinitesimal to i-th of infinitesimal of the heating tubeInstitute
State the radiation energy E (x of i-th of infinitesimal of heating tubei)=σb·twi 4, the area of i-th of infinitesimal of the heating tube, dAi=π
dw·dxi。
Thermal insulation layer outer surface governing equation are as follows:Its
In, αsFor NATURAL CONVECTION COEFFICIENT OF HEAT, t0=303K.
Specifically, the method calculated using numerical value, thermal insulation layer internal layer and outer layer equation is discrete, then it can be reduced to 2 × M
A Algebraic Equation set contains M twiUnknown parameter, M tbjUnknown parameter, M tajUnknown parameter combines 2 × M in a and c
A Algebraic Equation set can then acquire M tbjValue and M tajIt is worth to get the Temperature Distribution of thermal insulation layer surfaces externally and internally is arrived.
It is implicit to the progress of thermal insulation layer inner surface governing equation discrete, and linear simplifiation processing is carried out, discrete equation is as follows:
Gap radiation patterns are discrete using explicit difference, and discrete equation is as follows:
Arrangement obtains thermal insulation layer internal layer control Algebraic Equation set:
aajtaj n+1+bajtbj n+1=Raj n(j=1 ... ..., M), aaj, baj, Raj nRespectively represent thermal insulation layer internal layer control algebra
J-th of equation coefficient of equation group,
Wherein:
Implicit to the progress of thermal insulation layer outer surface governing equation discrete, discrete equation is as follows:
Arrangement obtains thermal insulation layer outer layer control Algebraic Equation set:
abjtaj n+1+bbjtbj n+1=Rbj n(j=1 ... ..., M), abj, bbj, Rbj nRespectively represent thermal insulation layer outer layer control algebra
J-th of equation coefficient of equation group,
Wherein:
Rbj n=-αs·t0。
Heat exchange layer internal layer control Algebraic Equation set and thermal insulation layer outer layer control Algebraic Equation set contain M twiUnknown parameter, M
A tajUnknown parameter, M tbjUnknown parameter, joint step (1) and with 2 × M algebraic equation in step (2), using iteration
It solves, then can acquire M tajValue and M tbj, i.e. the Temperature Distribution of thermal insulation layer surfaces externally and internally.
(4) by q obtained in the step 30、q1、tOutside w1And tOutside w0As boundary condition, by gas flow optimized equation, heating
Pipe governing equation, thermal insulation layer inner surface governing equation and thermal insulation layer outer surface governing equation joint are calculated in conjunction with boundary condition
M t outgiValue, M twiValue, M tbjValue and M tajValue.
Specifically, discrete by finite difference method to aforementioned four governing equation, boundary condition is applied to controlling party
Cheng Zhong can then be reduced to 4 × M algebraic equation, contain M twiUnknown parameter, M tgiUnknown parameter, M tbjUnknown parameter
And M tajUnknown parameter can then acquire M tgiValue (obtains gas flow temperature distribution), M twiValue (obtains heating tube temperature
Degree distribution), M tbjValue (obtaining the hull-skin temperature distribution of heat exchanger tube) and M tajValue (obtains the interior table of heat exchanger tube
Face Temperature Distribution).
(5) a heater power N is given, (1)~(4) is utilized to calculate tgi、twi、tbj、tajValue passes through given difference
N value, until tgM≥tgL, tgLFor gas flow temperature threshold value.
Step 5: heater efficiency calculates
Calculate heater efficiencyWherein, the value of N is by the number finally determined in step 4
Value, when the calculated heater efficiency of institute is less than η0, then change the value of h, and re-execute step 4, until η >=η0, wherein
η0For heater efficiency index.
In one embodiment, can be first 20mm by h value, calculate a heater efficiency value, and by the value with
Heater efficiency Indexes Comparison；It is again 30mm by h value, then calculates a heater efficiency value, and by the value and heater
Efficiency index compares；Later according to above-mentioned the case where comparing, relatively accurate selection is carried out to h.
In a preferred embodiment, described in the hot relevant parameter design method of the graphite resistance heater
The outlet air flow velocity u of rectangular spiral passagegL=30m/s.
In a preferred embodiment, in the hot relevant parameter design method of the graphite resistance heater, η0It takes
Value is 50%.
In a preferred embodiment, in the hot relevant parameter design method of the graphite resistance heater, αsIt takes
Value is 8.7W/ (m*K).
In a preferred embodiment, described in the hot relevant parameter design method of the graphite resistance heater
In step 5, when changing the value of h, value in the range of except 20~30mm.During adjusting the value of h, final h
Value may fall outside the above range.
Embodiment one
A) example one
It is known: known throughput G=0.05kg/s, exit flow temperature requirement tgL=1700K, stream pressure P=
1.0MPa, heater efficiency require η0=70%.
According to step (1)~(3), following parameter is obtained:
ugL=30m/s
A=29mm
B=102mm
δw=5mm
dw=189mm
S=105mm
Δ a=3mm
L=3.5m
λw=100W/ (m*K)
δf=100mm
It is assumed that the equivalent conduction model in boundary is as follows:
Firstly, given power N=1300kw, spacing h=100mm, are calculated heter temperature distribution curve such as Fig. 5,
Exit flow temperature tgM=1800K > tgL=1800K.Heater efficiency η=65% is unsatisfactory for requiring.Fig. 5 into Fig. 8,
Fluid_T indicates gas flow temperature, and elem_T indicates that heating tube temperature, sleev_in_T indicate thermal insulation layer internal surface temperature,
Sleev_out_T indicates thermal insulation layer hull-skin temperature.
Then, power N=1200kw, spacing h=50mm are given, heter temperature distribution curve such as Fig. 6 is calculated, out
Mouth gas flow temperature tgM=1800K > tgL=1800K.Heater efficiency η=70%, meets the requirements.
Embodiment two
It is known: known throughput G=0.05kg/s, exit flow temperature requirement tgL=1700K, stream pressure P=
1.0MPa, heater efficiency require η0=70%.
According to step (1)~(3), following parameter is obtained:
ugL=30m/s
A=16mm
B=54mm
δw=5mm
dw=103mm
S=57mm
Δ a=3mm
L=1.66m
λw=100W/ (m*K)
δf=100mm
It is assumed that the equivalent conduction model in boundary is as follows:
Firstly, given power N=150kw, spacing h=30mm, are calculated heter temperature distribution curve such as Fig. 7, out
Mouth gas flow temperature tgM=1706K > tgL=1700K.Heater efficiency η=52.6% is unsatisfactory for requiring.
It is repeatedly adjusted, gives power N=1200kw, heter temperature distribution curve is calculated such as in spacing h=10mm
Fig. 8, exit flow temperature tgM=1782K > tgL=1700K.Heater efficiency η=75%, meets the requirements.
Although the embodiments of the present invention have been disclosed as above, but its is not only in the description and the implementation listed
With it can be fully applied to various fields suitable for the present invention, for those skilled in the art, can be easily
Realize other modification, therefore without departing from the general concept defined in the claims and the equivalent scope, the present invention is simultaneously unlimited
In specific details and legend shown and described herein.
Claims (5)
1. a kind of hot relevant parameter design method of graphite resistance heater, which is characterized in that the graphite resistance heater packet
Rectangular spiral passage, heating tube, cold end heat-conducting piece, hot end heat-conducting piece and thermal insulation layer are included, the heating tube is sheathed on the square
The outside of shape helical duct, the cold end heat-conducting piece are connected to the cold end of the heating tube, and the hot end heat-conducting piece is connected to institute
The hot end of heating tube is stated, the thermal insulation layer is set in the outside of the heating tube, and a heating power supply is electrically connected to the cold end and leads
Warmware and the hot end heat-conducting piece, to start the heating tube fever；
It the described method comprises the following steps:
Step 1: the parameter of rectangular spiral passage, heating tube and hot end heat-conducting piece is set and is counted in the following way
It calculates:
The height of the rectangular spiral passageWherein, G is air flow rate, and P is air-flow pressure
Power, R are thermodynamic equilibrium constant, ugLFor the outlet air flow velocity of the rectangular spiral passage；By the rectangular spiral passage along axial
Discrete is M infinitesimal, then is t in i-th of infinitesimal overdraught temperaturegi, tg1It is the rectangular spiral passage in the 1st infinitesimal
Gas flow temperature；
The width b=3.5a of the rectangular spiral passage；
Screw pitch s=a+ the Δ a, Δ a=3~5mm of the rectangular spiral passage；
The outside diameter d of the rectangular spiral passagew=1.8s；
The thickness δ of the heating tubew=5~10mm；
The length of the heating tubeWherein, CpFor gas level pressure thermal capacitance
Than α is convection transfer rate, tg0For the end air flow temperature of the rectangular spiral passage, Δ tgw=250~350K, for heating
The designed temperature difference of pipe and air-flow, tgMFor the hot end gas flow temperature of the rectangular spiral passage；
The thermal coefficient λ of the heating tubew=50~100W/ (m*K)；
Step 2: the parameter of thermal insulation layer is set and is calculated in the following manner:
The thickness δ of the thermal insulation layerf=60~100mm；
Spacing h between the thermal insulation layer and the outer wall of the heating tube selectes one initially in the range of 20~30mm
Value；
Step 3: the parameter of cold end heat-conducting piece and hot end heat-conducting piece is set and is calculated in the following manner:
The outer end temperature t of the cold end heat-conducting pieceOutside w0=300K, the outer end temperature t of the hot end heat-conducting pieceOutside w1=300K, then have:
Wherein, q0For cold end heat conduction heat flux, λw0For the equivalent thermally conductive system of cold end heat-conducting piece
Number, L0For the equivalent passage of heat length of cold end heat-conducting piece, S0For the equivalent heat-conducting section area of cold end heat-conducting piece, tw0It is described
The cold junction temperature of heating tube；
Wherein, q1For hot end heat conduction heat flux, λw1For the equivalent thermally conductive system of hot end heat-conducting piece
Number, L1For the equivalent passage of heat length of hot end heat-conducting piece, S1For the equivalent heat-conducting section area of hot end heat-conducting piece, tw1It is described
The hot-side temperature of heating tube；
It is calculated Step 4: air-flow-heating tube-thermal insulation layer temperature joint simplifies:
By rectangular spiral passage along it is axial it is discrete be M infinitesimal, then be t in i-th of infinitesimal overdraught temperaturegi, by the heating
It is M infinitesimal that pipe edge is axially discrete, then heating tube temperature is t on i-th of infinitesimalwi, by the thermal insulation layer along axial discrete for M
A separate cylinder radial direction conduction model；
(1) gas flow optimized equation is established:
Wherein, tg0=273K, n are the rectangular spiral passage wetted perimeter in unit length
It is long；
(2) heating tube governing equation is established:
Wherein, the cross-sectional area of the heating tube
The exterior surface area A of the heating tube of unit length2=π dw, α is convection transfer rate, σbFor Stefan-bohr hereby climing constant,
ρwFor the density of the heating tube, CwFor the material specific heat capacity of heating tube, λwFor the material conducts heat rate of the heating tube, when τ is
Between, xiFor i-th of infinitesimal of heating tube；
Separately have, the radiation energy that the heating tube is receivedWherein, the thermal insulation layer
RADIATION ANGLE COEFFICIENT of j-th of the infinitesimal of inner surface to i-th of infinitesimal of the heating tube
Radiation energy E (the x of described heat exchange j-th of infinitesimal of pipe internal surfacej)=σb·taj 4, the face of described j-th of infinitesimal of thermal insulation layer inner surface
Product dAj=π (dw+2h)·dxj, tajFor the temperature of the jth infinitesimal of the inner surface of the thermal insulation layer, RijFor in the thermal insulation layer
The distance between the i-th infinitesimal of j-th infinitesimal and the heating tube on surface, xjFor j-th of infinitesimal of the thermal insulation layer；
(3) thermal insulation layer outer surface governing equation and thermal insulation layer inner surface governing equation are established:
Thermal insulation layer inner surface governing equation are as follows:
Wherein, tbjFor the outer of the thermal insulation layer
The temperature of surface jth infinitesimal, tajFor the temperature of the thermal insulation layer inner surface jth infinitesimal, λfFor the material conducts heat of the thermal insulation layer
Rate, da=dw+2δw+ 2h, db=dw+2δw+2h+δf；
Separately have, the radiation energy that the thermal insulation layer inner surface is receivedWherein, institute
J-th of infinitesimal of thermal insulation layer inner surface is stated to the RADIATION ANGLE COEFFICIENT of i-th of infinitesimal of the heating tube Radiation energy E (the x of i-th of infinitesimal of the heating tubei)=σb·twi 4, i-th of infinitesimal of the heating tube
Area dAi=π dw·dxi；
Thermal insulation layer outer surface governing equation are as follows:
Wherein, αsFor NATURAL CONVECTION COEFFICIENT OF HEAT, t0=
303K；
(4) by q obtained in the step 30、q1、tOutside w1And tOutside w0As boundary condition, gas flow optimized equation, heating are managed
Equation, thermal insulation layer inner surface governing equation and thermal insulation layer outer surface governing equation joint processed, in conjunction with boundary condition, calculate M
A tgiValue, M twiValue, M tbjValue and M tajValue；
(5) a heater power N is given, (1)~(4) is utilized to calculate tgi、twi、tbj、tajValue, by giving different N
Value, until tgM≥tgL, tgLFor gas flow temperature threshold value；
Step 5: heater efficiency calculates
Calculate heater efficiencyWherein, the value of N is by the numerical value finally determined in step 4,
When the calculated heater efficiency of institute is less than η0, then change the value of h, and re-execute step 4, until η >=η0, wherein η0
For heater efficiency index.
2. the hot relevant parameter design method of graphite resistance heater as described in claim 1, which is characterized in that the rectangle
The outlet air flow velocity u of helical ductgL=30m/s.
3. the hot relevant parameter design method of graphite resistance heater as claimed in claim 2, which is characterized in that η0Value is
50%.
4. the hot relevant parameter design method of graphite resistance heater as claimed in claim 3, which is characterized in that αsValue is
8.7W/(m*K)。
5. the hot relevant parameter design method of graphite resistance heater as described in claim 1, which is characterized in that the step
In five, when changing the value of h, value in the range of except 20~30mm.
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| CN1475026A (en) * | 2000-11-16 | 2004-02-11 | 麦特逊技术股份有限公司 | Apparatuses and methods for resistively heating thermal processing system |
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| CN1475026A (en) * | 2000-11-16 | 2004-02-11 | 麦特逊技术股份有限公司 | Apparatuses and methods for resistively heating thermal processing system |
| US7230212B1 (en) * | 2006-06-27 | 2007-06-12 | Rocky Research | Golf ball heater |
| CN102573148A (en) * | 2010-11-05 | 2012-07-11 | 图特科有限公司 | Foldable electric resistance heater and method of use |
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