JP4421735B2

JP4421735B2 - Heat storage control building air conditioning system

Info

Publication number: JP4421735B2
Application number: JP2000122283A
Authority: JP
Inventors: 泰鍋島; 寧村田
Original assignee: Sinko Industries Ltd
Current assignee: Sinko Industries Ltd
Priority date: 2000-04-24
Filing date: 2000-04-24
Publication date: 2010-02-24
Anticipated expiration: 2020-04-24
Also published as: JP2001304659A

Description

【０００１】
【発明の属する技術分野】
本発明は、冷媒が循環する二次側が重力式ヒートパイプに接続する個別分散するビル空調システムの躯体蓄熱運転の技術分野に属する。
【０００２】
【従来の技術】
近時、夜間電力を利用して熱を蓄えておき昼間に放熱する様々な蓄熱システムがこれまで研究開発されてきた。その中でも空調機で躯体への夜間蓄熱し昼間放熱を行う躯体蓄熱システムは、蓄熱槽構築費が不要である等で低コストな蓄熱システムとして普及が期待されており、躯体蓄熱空調システムの利点は、負荷平準化による熱源効率の向上と夜間電力利用による経済性の2項目に集約される。
躯体蓄熱空調システムは、予めの実験データにより夜間蓄熱量を設定して定性的に蓄熱していた。
なお、空気調和システムにおけるビル空気調和に費やした消費エネルギーの量は、従来は吹出温度と室温との温度差、及び、風量により算出していた。
【０００３】
【発明が解決しようとする課題】
ところで、ビル空調システムの躯体蓄熱システムの問題点は、中央式熱源を用いた空調機等と異なり、躯体への蓄熱温度や設定室温との温度差に対して、蓄放熱制御、特に負荷に応じた素早い放熱の制御が難しいことにある。
したがって、１日のうち冷房負荷と暖房負荷の発生する中間期あるいは中間期に近い冷暖房期には“冷やしすぎ”“暖めすぎ”が発生するが、これは室内環境上の問題であると同時に、室内機が冷暖自動切替えの場合、冷房（暖房）躯体蓄熱が暖房（冷房）負荷を発生させてしまうというシステム上の問題でもある。
しかも、蓄熱対象が建物躯体であるため、従来の蓄熱槽のように蓄熱量の定量的な把握が困難であり、そのため、ビル空調システムの躯体蓄熱システムの夜間蓄熱運転方法もいまだ定性的である。
【０００４】
本発明は、上記の問題点に鑑みてなされたもので、その課題は、ビル空調システムの躯体蓄熱システムを定量的に制御して、最適な空気調和状態を維持して無駄なエネルギーの損失を防ぎ、かつ、自動的に運転が可能な蓄熱量制御ビル空気調和システムを提供することにある。
【０００５】
【課題を解決するための手段】
上記の課題を解決するために、請求項１に記載の発明は、冷媒が循環する二次側が重力式ヒートパイプに接続して個別分散するビル空気調和システムであって、前記空気調和システムからの送風を切り替えダンパーによって通常運転時には居室内を冷房あるいは暖房し、躯体蓄熱運転時には建物躯体に向けて送風して躯体に蓄熱する蓄熱量制御ビル空気調和システムにおいて、通常運転時間における消費エネルギーの算出を冷房時には膨張弁の開度と時間の積算値から、暖房時には電磁弁の開時間の積算値から求め、算出された消費エネルギーから躯体運転時間を求めてビル空気調和装置の躯体蓄熱運転する蓄熱量制御ビル空気調和システムである。
上記の課題を解決するために、請求項２に記載の発明は、冷媒が循環する二次側が重力式ヒートパイプに接続して個別分散するビル空気調和システムであって、前記空気調和システムからの送風を切り替えダンパーによって通常運転時には居室内を冷房あるいは暖房し、躯体蓄熱運転時には建物躯体に向けて送風して躯体に蓄熱する蓄熱量制御ビル空気調和システムにおいて、基準躯体運転時間と基準通常運転時間と通常運転１時間あたりに必要な躯体単位時間を経験値より求め、下記の式によりその日の躯体運転時間を算出し、算出された躯体運転時間に基づいてビル空気調和装置の躯体蓄熱運転する蓄熱量制御ビル空気調和システムである。
躯体運転時間＝
基準躯体運転時間−(基準通常運転時間−通常運転時間)×躯体単位時間
上記の課題を解決するために、請求項３に記載の発明は、前記基準躯体運転時間は、前日の躯体運転時間とする請求項２に記載の蓄熱量制御ビル空気調和システムである。
【０００６】
【発明の実施の形態】
本発明の本質は、ビル空調システムの躯体蓄熱システムを定量的に運転制御することであり、躯体蓄熱システムで夜間蓄熱するエネルギーを昼間使用時の消費エネルギーよりも少な目にして、昼間時には各部位ごとに必要熱量を演算して空調システムを稼働することである。
【０００７】
[実施例]
ここで、本発明の好適な蓄熱量制御ビル空調システムの概略構成の実施例を図面に沿って説明する。
図１において、本実施例に用いた空調システムは重力式空調システムで、各構成はそれぞれの設置位置が高さに関して特定され、冷熱源である第１蓄熱槽としての氷蓄熱槽１は建物の例えば屋上のような高所に設置され、一方、温熱源である第２蓄熱槽としての温水蓄熱槽２は建物の地下のような低所に設置されている。氷蓄熱槽１は顕熱に比べて約80倍もの潜熱も含めて蓄熱するので、顕熱のみによって蓄熱する冷水蓄熱槽よりも大きな蓄熱容量を有しており、したがって同等容量の蓄熱槽を得るには蓄熱槽をはるかに小さくすることができ、屋上設置も可能となる。また、室内ユニットとしての空調機３は氷蓄熱槽１と温水蓄熱槽２との間の高さ位置となる各階の居室に設置されている。
これら各蓄熱槽1,2と各被空調室に設けられた空調機３との間は重力式のヒートパイプ4,5によって接続されている。ヒートパイプ４は冷房用回路を形成し、ヒートパイプ５は暖房用回路を形成する。
冷房用回路を形成するヒートパイプ４は、その凝縮部として、氷蓄熱槽１内に形成された冷房用凝縮部６を備えており、冷房用凝縮部６が氷蓄熱槽１により冷却される。
【０００８】
これらヒートパイプ4,5内の冷媒（図中実線矢印は冷媒液流、破線矢印は冷媒ガス流を示す）は、各ブロックに配置された複数の空調機３といずれか一方の蓄熱槽１もしくは２内で、あるいは凝縮部６で熱交換することによって相変化し、これらの間を往復するようにヒートパイプ4,5内を循環して流れる。氷蓄熱槽１および凝縮部６を経由するヒートパイプ４の冷媒液流部分には、各空調機３へ液体の冷媒を分配する前の位置に受液器61が介設され、さらにその下流側に各空調機３に対応して膨張弁62が設けられている。したがって、各被空調室における冷房負荷に応じて受液器61から対応空調機３の蒸発器31へ必要量の冷媒液を供給することができる。
【０００９】
また、温水蓄熱槽２を経由するヒートパイプ５の冷媒液流部分には、各空調機３の凝縮器32から温水蓄熱槽２へ還流するが、凝縮器32からの冷媒液は電磁弁21で流量を調節して受液器22を介して環流する。したがって、熱搬送媒体として被空調室側へ導かれるのは水ではなく冷媒であり、たとえこれが配管から漏れたとしてもすぐに気化するので居室を汚すようなことはない。
各蓄熱槽1,2に蓄冷ないし蓄熱するための熱源装置は製氷器を内蔵したヒートポンプチラー７であり、氷蓄熱槽１と製氷器との間にスラリーポンプ71が介設され、製氷器で作られた氷がスラリーポンプ71によって氷蓄熱槽１へ圧送され、氷蓄熱槽１の冷水はポンプ11によって冷房用凝縮部６の熱交換器に圧送される。温水蓄熱槽２は、ヒートポンプチラー７の凝縮器に形成された熱交換器との間に温水熱回収管72が設けられ、これに接続するポンプ73によって温水が温水蓄熱槽２へ圧送される。
各空調機３では、居室内から集められた空気は蒸発器31あるいは凝縮器32によって冷やされ、あるいは暖められたて送風機33よって、送風ダクト部８に送られる。送風ダクト部８は切り替えダンパー81によって、昼間時には居室天井の送風口82に居室内に送風され居室内を冷房あるいは暖房し、夜間時には空調機３の近傍の蓄熱用開口83から躯体９に向けて送風して躯体９に蓄熱する。
【００１０】
本実施例では、ヒートポンプチラー７は夜間の低電力料金で運転され、その熱エネルギを蓄熱槽１または２に蓄えるので、基本的には各蓄熱槽１または２の蓄熱量によって空調負荷に対応し、この躯体蓄熱システムは複数の区画の空調機３が躯体９の近傍に配置される構成であるから、空調部位と蓄熱部位が近傍にあると同時に、蓄熱部位の分散化が可能である。
そして、通常運転時での室内空調機３の処理能力の調整は、冷房時においては膨張弁62の開度を変更することにより行い、暖房時には電磁弁21の開閉によって行なっている。すなわち、通常運転において、躯体蓄熱利用後の昼間エネルギーの消費量は、冷房時においては膨張弁62の開度積算値の関数であり、暖房時には電磁弁21の開時間の関数である。なお、躯体蓄熱利用時には膨張弁62および開閉弁21は閉状態である。
したがって、本実施例における躯体蓄熱システムは、負荷特性の異なる部位別に蓄熱量（蓄熱時間）を制御することが可能であり、基本的には昼間時に使用した熱エネルギーを、夜間時に蓄積し補充すれば無駄のない運転が可能なる。
本発明者らは、冷房蓄熱運転時間および暖房蓄熱運転時間を経験的に演算する方法を鋭意研究した結果、以下に述べるように、冷房蓄熱運転時間および暖房蓄熱運転時間を経験的に演算する方法を見出した。
【００１１】
まず、躯体蓄熱を行わない場合のファン運転と膨張弁または電磁弁の動作を図２に示し、躯体蓄熱を行なった場合のファン運転と膨張弁または電磁弁の動作を図３に示す。この場合に、昼間での通常運転では、室温が設定範囲内になるように制御されている。
ここで、従来より昼間の稼働時の空気調和システムにおけるビル空気調和の消費エネルギーの量は、室温と風量から算出しており、非常に不安定で計測もしづらく正確に算出来ないが、本発明者らは、消費エネルギーが、冷房時には膨張弁62の開度と時間の積算値の関数となること、暖房時には電磁弁21の開時間の積算値の関数となることを実験値より見出し、この数値を利用することにより、冷媒が循環する二次側が重力式ヒートパイプに接続して個別分散するビル空気調和システムにおいて、躯体蓄熱運転が容易となることを見出した。
このことを用いて、基準躯体運転時間と基準通常運転時間と通常運転1時間あたりに必要な躯体単位時間を求めるが、図２および図３において、ファンは昼間の8:00から18:00までの10時間連続して稼働しているが、通常運転時の室内機の処理能力の調整は、冷房時には、図１における膨張弁62の開度を変更することにより調整され、昼間時のエネルギーの消費量は、膨張弁62の開度と時間の積算値となる。
一方、暖房時には、電磁弁21の開閉により調整され、昼間時のエネルギーの消費量は、電磁弁21の開時間の積算値となる。なお、躯体蓄熱利用時には、膨張弁62および電磁弁21は閉状態である。
【００１２】
仮に、躯体運転により、昼間での(１)冷房時には膨張弁運転時間を４時間から２時間に変更でき、(２)暖房時には電磁弁運転時間を４時間から１時間に変更できたとすれば、躯体蓄熱を100％消費して冷房時には運転２時間分を、暖房時には運転３時間分を畜熱したこと、即ち、通常運転において、冷房時には躯体運転により２時間分軽減され、暖房時には躯体運転により３時間分軽減されたことになる。
【００１３】
この状態であるならば、冷房(暖房)運転２(３)時間分に相当する蓄熱時間との関連を定義することで、必要躯体蓄熱時間を演算することが可能となる。
即ち、媒体が循環する二次側が重力式ヒートパイプに接続して個別分散するビル空調システムにおいて、基準躯体運転時間と基準通常運転時間とを経験値より求め、通常運転1時間あたりに必要な躯体単位時間を経験値を加味し以下に述べる所定の計算式より求め、これらの数値を下記の式に当てはめてその日の躯体運転時間を算出して自動運転することができる。
躯体運転時間＝
基準躯体運転時間−(基準通常運転時間−通常運転時間)×躯体単位時間
【００１４】
そこで、実際に躯体エリアのユニットでの能力と躯体側の蓄熱量を考えると、空調ユニット側(熱量を供給する側)は、次の計算により見出すことができる。
(1)…必要熱量(供給側)＝ユニット能力×通常運転時間×要求効率※¹
ここで、※１の「要求効率」とは、暖房運転時の室内機に対する負荷要求の比率である。
【００１５】
次に、躯体側で保有できる躯体保有可能熱量は、蓄熱量と蓄熱面積から算出される。
(2)…躯体保有可能熱量＝蓄熱量×蓄熱面積
今、処理能力3890kcal/hのユニットを40m²に1台設置した場合を考えると
冷房時必要熱量＝ 3890×2×0.6 ＝4668 kcal
暖房時必要熱量＝ 3890×3×0.6 ＝7002 kcal
冷房時躯体保有可能熱量＝ 12.8※²×40 ＝512 kcal/h
暖房時躯体保有可能熱量＝ 35※³×40 ＝1400 kcal/h
となる。
ここで※２の冷房蓄熱量「12.8 kcal/h.m²」、および、※３暖房蓄熱量「35 kcal/h.m²」は実験により求められた経験値である。
【００１６】
これらの条件から、蓄熱に必要な時間は次式より求められる。
(3)…必要躯体蓄熱時間＝必要熱量／躯体熱量
必要躯体冷房畜熱時間＝ 4668／512 ＝9時間
必要躯体暖房畜熱時間＝ 7002／1400 ＝5時間
この必要躯体蓄熱時間が、外気の気温変動が大きくない場合には、基礎躯体運転時間となる。即ち、この場合に、冷房時においては９時間が基礎躯体運転時間となり、暖房時においては５時間が基礎躯体運転時間となる。
【００１７】
これにより、通常運転1時間あたりに必要な躯体運転時間、即ち、躯体基準時間次のようになる。
▲４▼…通常運転1時間あたりの躯体単位時間＝
必要躯体蓄熱時間／躯体運転により軽減された冷暖房運転時間
冷房時躯体運転時間＝ 9／2 ＝4.5 時間(4時間30分) [変更可能係数］
暖房時躯体運転時間＝ 5／3 ＝1.67 時間(約1時間40分)[変更可能係数]
【００１８】
以上のように、おおよその躯体運転時間が算出される。そして実際には躯体運転後に、通常運転での実運転をモニターすることで微調整すればよく、次の躯体運転に必要な時間を導き出せる。
【００１９】
[運転例１]
ここで、前記▲３▼で述べたように、このビルと空気調和機の能力から、必要躯体蓄熱時間を基礎躯体運転時間として５時間と設定した場合、自動運転の運転例１を図４を参照して説明する。
先ず、基礎躯体運転時間を５時間と設定したが、空調するビルと空調装置の能力から、外気温度が５℃程度で室温20℃を必要としたときの効率のよい基礎躯体運転時間で、実験から得られる経験値であるが、運転例２で述べるように外気温度の変化が大きいときは補正しうる値である。また、昼間の基準通常運転を１時間とするが、基準通常運転を大きくすると節約効果がなくなり、また零時間とすると通常運転での修正ができなくなるので、好ましくは、30分から4時間の範囲内に設定すればよく、さらに好ましくは、30分から2時間の範囲内に設定すればよく、ここでは、基準通常運転は１時間に設定する。なお、ビルを閉じる休祭日には空気調和機による躯体蓄熱運転および通常運転は稼働しない。
また、前記▲３▼で算出したように、対象のビルの躯体単位時間が1.67であるので、これを採用する。
まず、５時間の基礎運転時間の躯体暖房運転後、通常運転を開始し設定された室温範囲での暖房運転が上記の想定とおりに延べの暖房運転の基準通常運転時間と同じ１時間となった場合は、想定どおりなので、その次の躯体運転も同様に５時間行えばよいことになる(補正0時間)。
【００２０】
次に、想定した時間以外の暖房運転となった場合は、躯体運転の補正が必要となる。
ケース1：暖房運転が1時間以下の場合（躯体過多）
これは、前回の躯体運転が多すぎた場合で、1時間以下の暖房運転時間より躯体運転を補正することが必要である。例えば、躯体過多運転時間は30分の場合
（30分の通常運転)には、
躯体過多運転時間＝基準通常運転時間−30分の通常運転時間 1-0.5＝0.5h）
実運転結果による補正1
躯体運転時間＝
基準躯体運転時間−(基準通常運転時間−通常運転時間)×躯体単位時間
＝ 5−0.5×1.67 ＝ 4.17 時間（約4時間10分）
【００２１】
ケース2：暖房運転が1時間以上の場合（躯体不足）
これは、前回の躯体運転が不足した場合で、1時間以上の暖房運転時間より躯体運転を補正することが必要である。例えば、躯体不足運転時間は１時間の場合（2時間の通常運転 1−2＝-1h）には、
実運転結果による補正2
躯体運転時間＝
基準躯体運転時間−(基準通常運転時間−通常運転時間)×躯体単位時間
＝ 5＋1.0×1.67 ＝6.67 時間（約6時間40分）
以上の演算により躯体運転時間を算出し、その後は自動的に躯体運転時間を設定して自動的に運転する。そして、図４の表に見られるように、昼間の通常運転時間は２時間以下に抑えられ、経費が節約される躯体蓄熱運転のビル空気調和システムとなる。
【００２２】
[運転例２]
運転例１において、基準通常運転時間を１時間としたが、基準躯体運転時間を当初、前記▲３▼で述べたように、基礎躯体運転時間を必要躯体蓄熱時間の５時間としたために、外気が大きく変動して暖かくなった場合に基準通常運転時間を０時間、即ち、通常運転で暖房運転をまったく行わなくとも躯体運転を過度にしてしまう。
具体的には、躯体運転時間＝
基準躯体運転時間−(基準通常運転時間−通常運転時間)×躯体単位時間
＝ 5−1×1.67 ＝3.33 （約3時間20分）
となり、通常運転で暖房運転をまったく行わなくとも、本来必要のない3時間20分間の躯体運転を行うことになる。
【００２３】
言い換えれば、外気の気温変動が少ない場合には、運転例１のように基準躯体運転時間を固定してもこのような不都合は生じないが、気温変動が大きい場合には、基準躯体運転時間を補正する必要が生じる。
運転例２は、基準躯体運転時間を前日の躯体運転時間とすることで補正をかけて自動運転することによって、適切な躯体蓄熱運転が可能となる。即ち、次式より算出された躯体運転時間に基づいて、自動的に躯体運転時間を設定して運転する。
躯体運転時間＝基準躯体運転時間(前日の躯体運転時間)
−(基準通常運転時間−通常運転時間)×躯体単位時間
ただし、躯体運転が10時間を越えないように最大時間を設定することで、躯体運転時間を制限する。
この運転例が図５の表Ａ，Ｂ，Ｃであって、大きな躯体運転時間および通常運転時間の変動もなく、スムースに効率よく躯体蓄熱運転が行われている。そして、昼間の通常運転時間は３．５時間以下に抑えられ、経費が節約される躯体蓄熱運転のビル空気調和システムとなる。特に、表Ｂの運転においては、運転例１で運転した場合と異なり、月曜日の基準躯体運転時間の５時間が、火曜日には約3時間20分に補正され、水曜日には約1時間40分に、以後は躯体運転も通常運転も行われなくても適切な空気調和がなされ、かつ、運転による経費も少なくてすむビル空気調和システムとなる。
【００２４】
なお、本発明の特徴を損うものでなければ、上記の実施例に限定されるものでないことは勿論である。例えば、請求項２および３において、昼間の稼働時の空気調和システムにおけるビル空気調和の消費エネルギーの量を、正確で簡易に算出できる手段を用いることを妨げるものではない。
【００２５】
【発明の効果】
以上説明したように、請求項１の発明によれば、冷媒が循環する二次側が重力式ヒートパイプに接続して個別分散するビル空気調和システムにおいて、通常運転時間における消費エネルギーの算出を冷房時には膨張弁の開度と時間の積算値から、暖房時には電磁弁の開時間の積算値から求め、算出された消費エネルギーから躯体運転時間を求めてビル空気調和装置の躯体蓄熱運転をするので、適格で無駄が少なく、かつ、比較的容易に躯体蓄熱運転が可能となるという効果が得られる。
【００２６】
また、請求項２および請求項３の発明によれば、冷媒が循環する二次側が重力式ヒートパイプに接続して個別分散するビル空気調和システムにおいて、基準躯体運転時間と基準通常運転時間と通常運転1時間あたりに必要な躯体単位時間を経験値より求め、所定の式によりその日の躯体運転時間を算出し、算出された躯体運転時間に基づいてビル空気調和装置の躯体蓄熱運転する蓄熱量制御ビル空気調和システムであるから、最適な空気調和状態を維持して無駄なエネルギーの損失を防ぎ、かつ、自動的にビル空気調和システムにおける躯体蓄熱運転が可能となるという効果が得られる。
【図面の簡単な説明】
【図１】本発明の好適な実施例を蓄熱量制御ビル空気調和システムの概略を示す説明図
【図２】躯体蓄熱を行わない場合のファン運転と膨張弁または電磁弁の動作状態に示した説明図
【図３】躯体蓄熱を行なった場合のファン運転と膨張弁または電磁弁の動作状態を示した説明図
【図４】本発明の実施例における暖房時における躯体蓄熱運転の第１の運転例の表を示した図
【図５】本発明の実施例における暖房時における躯体蓄熱運転の第２の運転例の表を示した図である。
【符号の説明】
１…氷蓄熱槽
11,73…ポンプ
２… 温水蓄熱槽(暖房用蒸発器)
21…電磁弁
22…受液器
３…空調機
31…蒸発器
32…凝縮器
33…送風機
4,5…ヒートパイプ
６…冷房用凝縮部
61…冷受液器
62…膨張弁
７…ヒートポンプチラー
71…スラリーポンプ
72…温水熱回収管
８…送風ダクト部
81…切り替えダンパー
82…送風口
83…蓄熱用開口
９…躯体[0001]
BACKGROUND OF THE INVENTION
The present invention belongs to the technical field of housing heat storage operation of individually distributed building air conditioning systems in which the secondary side through which the refrigerant circulates is connected to a gravity heat pipe.
[0002]
[Prior art]
Recently, various heat storage systems that use nighttime power to store heat and dissipate it during the day have been researched and developed. Among them, the frame heat storage system that stores the heat in the frame at night by using an air conditioner and dissipates it during the day is expected to spread as a low-cost heat storage system because it does not require a heat storage tank construction cost. It can be summarized into two items: improvement of heat source efficiency by load leveling and economic efficiency by using nighttime power.
The frame heat storage air-conditioning system qualitatively stores heat at night by setting the amount of heat storage based on previous experimental data.
Note that the amount of energy consumed for building air conditioning in the air conditioning system is conventionally calculated based on the temperature difference between the blowing temperature and room temperature and the air volume.
[0003]
[Problems to be solved by the invention]
By the way, the problem of the building heat storage system of the building air conditioning system is different from the air conditioner etc. using a central heat source, etc., depending on the heat storage temperature to the housing and the temperature difference from the set room temperature, especially according to the load. It is difficult to control heat dissipation quickly.
Therefore, in the middle or near the middle period when the cooling load and heating load occur in the day, “too cold” and “too warm” occur, but this is a problem in the indoor environment, When the indoor unit is automatically switched between cooling and heating, it is also a system problem that cooling (heating) housing heat storage generates a heating (cooling) load.
Moreover, since the heat storage target is a building frame, it is difficult to quantitatively grasp the amount of heat storage as in a conventional heat storage tank, so the night heat storage operation method of the building heat storage system of the building air conditioning system is still qualitative. .
[0004]
The present invention has been made in view of the above-mentioned problems, and its problem is to quantitatively control the building heat storage system of the building air conditioning system to maintain an optimal air-conditioning state and reduce wasteful energy loss. The object is to provide a heat storage control building air conditioning system which can be prevented and automatically operated.
[0005]
[Means for Solving the Problems]
In order to solve the above-mentioned problem, the invention according to claim 1 is a building air conditioning system in which a secondary side through which a refrigerant circulates is connected to a gravity heat pipe and dispersed individually , and the air conditioning system includes: In a heat storage control building air conditioning system that cools or heats the room during normal operation by switching dampers and heats the room during normal operation, and blows air toward the building body during heat storage operation to store heat in the building , calculation of energy consumption during normal operation time The amount of heat stored in the building heat storage operation of the building air conditioner using the integrated value of the opening and time of the expansion valve during cooling, the integrated value of the open time of the solenoid valve during heating, and the building operating time from the calculated energy consumption Control building air conditioning system.
In order to solve the above-mentioned problem, the invention according to claim 2 is a building air conditioning system in which a secondary side through which a refrigerant circulates is connected to a gravity heat pipe and individually dispersed . In a heat storage control building air conditioning system that cools or heats the room during normal operation with a damper and cools or heats during normal operation, and blows air toward the building body and stores heat in the building body during heat storage operation, the standard body operation time and the standard normal operation time And the unit time required for normal operation per hour is calculated from experience values, the body operation time of the day is calculated by the following formula, and the heat storage operation of the building air conditioner in the building heat storage operation based on the calculated body operation time It is a quantity control building air conditioning system.
Housing operation time =
Reference chassis operation time-(reference normal operation time-normal operation time) x chassis unit time In order to solve the above problem, the invention according to claim 3 is characterized in that the reference chassis operation time is calculated as follows: The heat storage amount control building air conditioning system according to claim 2.
[0006]
DETAILED DESCRIPTION OF THE INVENTION
The essence of the present invention is to quantitatively control the building heat storage system of the building air conditioning system. The energy stored in the building heat storage system at night is less than the energy consumed during daytime use. The necessary heat amount is calculated and the air conditioning system is operated.
[0007]
[Example]
Here, the Example of schematic structure of the suitable heat storage amount control building air-conditioning system of this invention is described along drawing.
In FIG. 1, the air conditioning system used in the present embodiment is a gravity air conditioning system, and each configuration is identified with respect to height, and the ice thermal storage tank 1 as the first thermal storage tank, which is a cold heat source, is For example, the hot water storage tank 2 as a second heat storage tank that is a heat source is installed in a low place such as the basement of a building. The ice heat storage tank 1 stores about 80 times as much latent heat as sensible heat, so it has a larger heat storage capacity than a cold water storage tank that stores heat only by sensible heat, and thus obtains a heat storage tank of the same capacity. The heat storage tank can be made much smaller and can be installed on the rooftop. Moreover, the air conditioner 3 as an indoor unit is installed in the living room of each floor used as the height position between the ice thermal storage tank 1 and the warm water thermal storage tank 2. FIG.
The heat storage tanks 1 and 2 and the air conditioner 3 provided in each air-conditioned room are connected by gravity heat pipes 4 and 5. The heat pipe 4 forms a cooling circuit, and the heat pipe 5 forms a heating circuit.
The heat pipe 4 forming the cooling circuit includes a cooling condensing unit 6 formed in the ice heat storage tank 1 as the condensing unit, and the cooling condensing unit 6 is cooled by the ice heat storage tank 1.
[0008]
The refrigerant in these heat pipes 4 and 5 (solid line arrows in the figure indicates a refrigerant liquid flow, and broken line arrows indicate a refrigerant gas flow) includes a plurality of air conditioners 3 arranged in each block and one of the heat storage tanks 1 or 2 or by heat exchange in the condensing unit 6, and the phase changes, and flows through the heat pipes 4 and 5 so as to reciprocate between them. In the refrigerant liquid flow portion of the heat pipe 4 that passes through the ice heat storage tank 1 and the condensing unit 6, a liquid receiver 61 is provided at a position before the liquid refrigerant is distributed to each air conditioner 3, and further downstream side thereof. An expansion valve 62 is provided corresponding to each air conditioner 3. Therefore, a required amount of refrigerant liquid can be supplied from the liquid receiver 61 to the evaporator 31 of the corresponding air conditioner 3 in accordance with the cooling load in each air-conditioned room.
[0009]
Further, in the refrigerant liquid flow portion of the heat pipe 5 passing through the hot water heat storage tank 2, the refrigerant 32 from the condenser 32 of each air conditioner 3 is returned to the hot water heat storage tank 2. The flow rate is adjusted and refluxed via the receiver 22. Therefore, it is not water but the refrigerant that is guided to the air-conditioned room as the heat transfer medium, and even if this leaks from the pipe, it is vaporized immediately, so that the room is not soiled.
The heat source device for storing or storing heat in each of the heat storage tanks 1 and 2 is a heat pump chiller 7 incorporating an ice maker, and a slurry pump 71 is interposed between the ice heat storage tank 1 and the ice maker, was ice is pumped by slurry pump 71 to the ice thermal storage tank 1, cold water of the ice thermal storage tank 1 is pumped to the heat exchanger of the cooling condensing unit 6 by a pump 11. A hot water heat recovery pipe 72 is provided between the hot water heat storage tank 2 and a heat exchanger formed in the condenser of the heat pump chiller 7, and hot water is pumped to the hot water heat storage tank 2 by a pump 73 connected thereto.
In each air conditioner 3, the air collected from the living room is cooled by the evaporator 31 or the condenser 32, or is warmed and sent to the blower duct unit 8 by the blower 33. The air duct portion 8 is blown into the room by a switching damper 81 during the daytime through the ceiling air outlet 82 to cool or heat the room, and at night, from the heat storage opening 83 near the air conditioner 3 toward the housing 9. It blows and stores heat in the housing 9.
[0010]
In the present embodiment, the heat pump chiller 7 is operated at a low power charge at night, and the thermal energy is stored in the heat storage tank 1 or 2, so basically it corresponds to the air conditioning load by the heat storage amount of each heat storage tank 1 or 2. Since the housing heat storage system has a configuration in which the air conditioners 3 of a plurality of sections are arranged in the vicinity of the housing 9, the air storage portion and the heat storage portion are in the vicinity, and at the same time, the heat storage portion can be dispersed.
The processing capacity of the indoor air conditioner 3 during normal operation is adjusted by changing the opening of the expansion valve 62 during cooling, and by opening and closing the electromagnetic valve 21 during heating. That is, during normal operation, the amount of daytime energy consumption after the use of the housing heat storage is a function of the integrated opening value of the expansion valve 62 during cooling, and a function of the open time of the electromagnetic valve 21 during heating. Note that the expansion valve 62 and the on-off valve 21 are in a closed state when the housing heat storage is used.
Therefore, the frame heat storage system in the present embodiment can control the heat storage amount (heat storage time) for each part having different load characteristics. Basically, the heat energy used in the daytime is accumulated and replenished at nighttime. This makes it possible to drive without waste.
As a result of earnestly researching a method for empirically calculating the cooling heat storage operation time and the heating heat storage operation time, the present inventors have empirically calculated the cooling heat storage operation time and the heating heat storage operation time as described below. I found.
[0011]
First, FIG. 2 shows the fan operation and the operation of the expansion valve or solenoid valve when the housing heat storage is not performed, and FIG. 3 shows the fan operation and the operation of the expansion valve or solenoid valve when the housing heat storage is performed. In this case, during normal daytime operation, the room temperature is controlled to be within the set range.
Here, the amount of energy consumed for building air conditioning in an air conditioning system during daytime operation is calculated from room temperature and air volume, which is very unstable and difficult to measure, but cannot be accurately calculated. mono are energy consumption, be a function of the integrated value of the opening degree of the expansion valve 6 2 times during cooling, heading from the experimental values to be a function of the integrated value of the opening time of the solenoid valve 21 during heating, By using this numerical value, it was found that the building heat storage operation is facilitated in the building air conditioning system in which the secondary side where the refrigerant circulates is connected to the gravity heat pipe and dispersed individually.
By using this, the standard chassis operation time, the standard normal operation time, and the unit time required for the normal operation per hour are obtained. In FIGS. 2 and 3, the fan operates from 8:00 to 18:00 in the daytime. While running 10 hours continuously, the adjustment of the processing capabilities of a normal indoor unit during operation, at the time of cooling is adjusted by changing the opening of the expansion valve 6 2 in FIG. 1, energy daytime consumption becomes the integrated value of the opening degree and time of the expansion valve 6 2.
On the other hand, during heating, adjustment is made by opening and closing the electromagnetic valve 21, and the energy consumption during the daytime is an integrated value of the opening time of the electromagnetic valve 21. At the time of building frame heat storage utilization, expansion valve 6 2 and the electromagnetic valve 21 is closed.
[0012]
Assuming that the expansion valve operation time can be changed from 4 hours to 2 hours during the daytime (1) during cooling, and (2) the solenoid valve operation time can be changed from 4 hours to 1 hour during heating. 100% of the housing heat storage was consumed, and 2 hours of operation during cooling and 3 hours of operation during heating were reduced. In other words, during the normal operation, it was reduced by 2 hours during cooling and by 2 hours during heating. It is reduced for 3 hours.
[0013]
If it is in this state, it becomes possible to calculate the required housing heat storage time by defining the relationship with the heat storage time corresponding to the cooling (heating) operation 2 (3) hours.
That is, in a building air-conditioning system in which the secondary side where the medium circulates is connected to a gravity heat pipe and dispersed individually, the standard housing operation time and the standard normal operation time are obtained from empirical values, and the housing required per hour of normal operation The unit time can be obtained from a predetermined calculation formula described below in consideration of an empirical value, and these numerical values are applied to the following formula to calculate the body operation time of the day and perform automatic driving.
Housing operation time =
Standard chassis operation time-(standard normal operation time-normal operation time) x chassis unit time [0014]
Therefore, considering the actual capacity of the unit in the enclosure area and the amount of heat stored on the enclosure side, the air conditioning unit side (the side that supplies the amount of heat) can be found by the following calculation.
(1) … Required heat (supply side) = Unit capacity × Normal operation time × Required efficiency * ¹
Here, “Required efficiency” of * 1 is a ratio of load demand to the indoor unit during the heating operation.
[0015]
Next, the amount of heat that can be held by the housing can be calculated from the amount of heat stored and the heat storage area.
(2) … The amount of heat that can be stored in the housing = heat storage × heat storage area Considering the case where one unit with a processing capacity of 3890 kcal / h is installed in 40 m ² , the required heat during cooling = 3890 × 2 × 0.6 = 4668 kcal
Heat required during heating = 3890 x 3 x 0.6 = 7002 kcal
The amount of heat that can be held in the housing during cooling = 12.8 * ² × 40 = 512 kcal / h
The amount of heat that can be held by the housing during heating = 35 * ³ × 40 = 1400 kcal / h
It becomes.
Here, the cooling heat storage amount of * 2 “12.8 kcal / hm ² ” And * 3 Heating heat storage “35 kcal / hm ² ” is an empirical value obtained through experiments.
[0016]
From these conditions, the time required for heat storage is obtained from the following equation.
(3) ... Necessary body heat storage time = Necessary heat amount / Body heat amount Necessary body cooling heat storage time = 4668/512 = 9 hours Necessary body heating livestock heat time = 7002/1400 = 5 hours This necessary body heat storage time is the temperature of the outside air When the fluctuation is not large, it is the basic chassis operation time. That is, in this case, 9 hours is the basic chassis operation time during cooling, and 5 hours is the basic chassis operation time during heating.
[0017]
As a result, the chassis operation time required per one hour of normal operation, that is, the chassis reference time is as follows.
▲ 4 ▼… Frame unit time per hour of normal operation =
Necessary body heat storage time / Air-conditioning operation time reduced by body operation Cooling body operation time = 9/2 = 4.5 hours (4 hours 30 minutes) [Changeable coefficient]
Heating enclosure operation time = 5/3 = 1.67 hours (approximately 1 hour and 40 minutes) [changeable factor]
[0018]
As described above, the approximate housing operation time is calculated. Actually, after the chassis operation, the actual operation in the normal operation may be monitored for fine adjustment, and the time required for the next chassis operation can be derived.
[0019]
[Operation example 1]
Here, as described in (3) above, when the necessary housing heat storage time is set to 5 hours as the basic housing operation time from the capacity of this building and the air conditioner, FIG. The description will be given with reference.
First, the basic body operation time was set to 5 hours. From the ability of the building to be air-conditioned and the capacity of the air conditioner, the experiment was conducted with an efficient basic body operation time when the outside air temperature was about 5 ° C and room temperature 20 ° C was required. Is an empirical value that can be corrected when the change in the outside air temperature is large, as described in Operation Example 2. The standard normal operation during the day is set to 1 hour. However, if the standard normal operation is increased, the saving effect will be lost, and if it is set to zero time, the normal operation cannot be corrected. Therefore, it is preferably within the range of 30 minutes to 4 hours. More preferably, it may be set within a range of 30 minutes to 2 hours, and here, the standard normal operation is set to 1 hour. Note that the building heat storage operation and normal operation by the air conditioner do not operate on holidays when the building is closed.
Further, as calculated in (3) above, since the unit unit time of the target building is 1.67, this is adopted.
First, after the chassis heating operation of the basic operation time of 5 hours, the normal operation was started and the heating operation in the set room temperature range became the same 1 hour as the standard normal operation time of the total heating operation as assumed above In this case, since it is as expected, the next chassis operation may be similarly performed for 5 hours (correction 0 hours).
[0020]
Next, when it becomes heating operation other than the assumed time, the correction | amendment of a housing operation is needed.
Case 1: When heating operation is less than 1 hour (excessive housing)
This is a case where there was too much previous chassis operation, and it is necessary to correct the chassis operation from the heating operation time of 1 hour or less. For example, if the chassis excessive operation time is 30 minutes (30 minutes normal operation),
Excessive housing operation time = standard normal operation time-30 minutes normal operation time 1-0.5 = 0.5h)
Correction by actual operation result 1
Housing operation time =
Standard chassis operation time-(standard normal operation time-normal operation time) x chassis unit time = 5-0.5 x 1.67 = 4.17 hours (approximately 4 hours and 10 minutes)
[0021]
Case 2: When heating operation is over 1 hour (insufficient body)
This is a case where the previous chassis operation is insufficient, and it is necessary to correct the chassis operation from the heating operation time of 1 hour or more. For example, if the chassis shortage operation time is 1 hour (2 hours normal operation 1-2 = -1h)
Correction based on actual operation results 2
Housing operation time =
Standard chassis operation time-(standard normal operation time-normal operation time) x chassis unit time = 5 + 1.0 x 1.67 = 6.67 hours (approximately 6 hours 40 minutes)
The frame operation time is calculated by the above calculation, and thereafter the frame operation time is automatically set and the vehicle is automatically operated. As shown in the table of FIG. 4, the daytime normal operation time is suppressed to 2 hours or less, and the building heat storage operation building air conditioning system is saved.
[0022]
[Operation example 2]
In the operation example 1, the standard normal operation time is set to 1 hour. However, as described in the above (3), the basic body operation time is set to 5 hours, which is the required body heat storage time. When the temperature fluctuates greatly and becomes warm, the standard normal operation time becomes 0 hours, that is, the housing operation becomes excessive even if the heating operation is not performed at all in the normal operation.
Specifically, chassis operation time =
Standard chassis operation time-(standard normal operation time-normal operation time) x chassis unit time = 5-1 x 1.67 = 3.33 (approximately 3 hours and 20 minutes)
Therefore, even if heating operation is not performed at all in normal operation, the housing operation for 3 hours and 20 minutes, which is not necessary, is performed.
[0023]
In other words, when the temperature fluctuation of the outside air is small, such inconvenience does not occur even if the reference chassis operation time is fixed as in Operation Example 1, but when the temperature fluctuation is large, the reference chassis operation time is set to It is necessary to correct.
In the operation example 2, an appropriate housing heat storage operation can be performed by performing the automatic operation with correction by setting the reference housing operation time as the previous body operation time. That is, based on the chassis operation time calculated from the following equation, the chassis operation time is automatically set for operation.
Chassis operation time = standard chassis operation time (frame operation time of the previous day)
-(Standard normal operation time-normal operation time) x chassis unit time However, the chassis operation time is limited by setting the maximum time so that the chassis operation does not exceed 10 hours.
Examples of this operation are Tables A, B, and C of FIG. 5, and the housing heat storage operation is smoothly and efficiently performed without fluctuations in the large housing operation time and the normal operation time. And the normal operation time of daytime is restrained to 3.5 hours or less, and it becomes the building air conditioning system of the frame heat storage operation which saves cost. In particular, in the operation shown in Table B, unlike the case of driving in Operation Example 1, 5 hours of the standard chassis operation time on Monday is corrected to approximately 3 hours and 20 minutes on Tuesday, and approximately 1 hour and 40 minutes on Wednesday. In addition, a building air conditioning system is obtained in which appropriate air conditioning is performed without the need for frame operation or normal operation, and the operating expenses can be reduced.
[0024]
Needless to say, the present invention is not limited to the above-described embodiments as long as the features of the present invention are not impaired. For example, claims 2 and 3 do not preclude the use of means that can accurately and easily calculate the amount of energy consumed for building air conditioning in an air conditioning system during daytime operation.
[0025]
【The invention's effect】
As described above, according to the first aspect of the present invention, in the building air conditioning system in which the secondary side through which the refrigerant circulates is connected to the gravity heat pipe and dispersed individually, the calculation of the energy consumption during the normal operation time is performed during cooling. It is obtained from the integrated value of the opening and time of the expansion valve, from the integrated value of the open time of the solenoid valve during heating, and the building operation time is obtained from the calculated energy consumption, so that the building air conditioner heat storage operation is qualified. Thus, there is little waste, and the effect that the housing heat storage operation can be performed relatively easily is obtained.
[0026]
Further, according to the invention of claim 2 and claim 3, in the building air conditioning system in which the secondary side through which the refrigerant circulates is connected to the gravity heat pipe and dispersed individually, the reference housing operation time, the reference normal operation time, and the normal Heat storage amount control to calculate the building unit operating time of the building air conditioner based on the calculated building operating time based on the calculated building operating time by calculating the building operating time of the day using a predetermined formula. Since it is a building air-conditioning system, it is possible to obtain an effect that an optimum air-conditioning state is maintained to prevent useless energy loss, and that the building heat storage operation in the building air-conditioning system can be automatically performed.
[Brief description of the drawings]
FIG. 1 is an explanatory view showing an outline of a heat storage amount control building air conditioning system according to a preferred embodiment of the present invention. FIG. 2 shows a fan operation and an operation state of an expansion valve or a solenoid valve when no housing heat storage is performed. Explanatory drawing [FIG. 3] Explanatory drawing which showed the fan operation | movement at the time of frame heat storage, and the operation state of an expansion valve or an electromagnetic valve. [FIG. 4] The 1st driving | operation of the frame heat storage operation at the time of heating in the Example of this invention The figure which showed the table | surface of the example. FIG. 5: The figure which showed the table | surface of the 2nd operation example of the frame heat storage operation at the time of heating in the Example of this invention.
[Explanation of symbols]
1 ... Ice storage tank
11,73 ... Pump 2 ... Hot water storage tank (evaporator for heating)
21 ... Solenoid valve
22 ... Liquid receiver 3 ... Air conditioner
31 ... Evaporator
32 ... Condenser
33 ... Blower
4, 5 ... Heat pipe 6 ... Condensing part for cooling
61 ... Cold receiver
62 ... Expansion valve 7 ... Heat pump chiller
71 ... Slurry pump
72 ... Hot water heat recovery pipe 8 ... Air duct part
81 ... Switching damper
82… Blower
83 ... Opening for heat storage 9 ... Housing

Claims

A building air-conditioning system in which the secondary side where the refrigerant circulates is connected to a gravity heat pipe and dispersed individually , and the air from the air-conditioning system is switched to cool or heat the room during normal operation by a damper to store the housing heat In a heat storage control building air conditioning system that blows air toward the housing during operation and stores heat in the building housing ,
The energy consumption during normal operation is calculated from the integrated value of the opening and time of the expansion valve during cooling, and from the integrated value of the open time of the solenoid valve during heating. A heat storage amount control building air conditioning system characterized by a housing heat storage operation of a harmony device.

A building air-conditioning system in which the secondary side where the refrigerant circulates is connected to a gravity heat pipe and dispersed individually , and the air from the air-conditioning system is switched to cool or heat the room during normal operation by a damper to store the housing heat In a heat storage control building air conditioning system that blows air toward the housing during operation and stores heat in the building housing ,
The standard chassis operation time, the standard normal operation time, and the unit time required for normal operation per hour are obtained from experience values. The chassis operation time of the day is calculated by the following formula, and the building operation time is calculated based on the calculated chassis operation time. A heat storage amount control building air conditioning system characterized in that a housing heat storage operation of an air conditioner is performed.
Housing operation time =
Standard chassis operation time-(standard normal operation time-normal operation time) x chassis unit time

The heat storage amount control building air conditioning system according to claim 2, wherein the reference housing operation time is the housing operation time of the previous day.