[領域] 熱力學 (題目相關領域)
[來源] 考古題 (課本習題、考古題、參考書...)
[題目] A tank with a volume of 2m^3 is filled with air at a pressure of 7
bars and a temperature of 250度C.Determine (a)the final temperature (b)the
percent of the mass left in the tank (c)the quantity of mass,in kilograms,
that left the tank if the mass is permitted to leave the tank under
adiabatic condition until the pressuer dropto 1.0 bar.(k of air is roughly
equal to 1.4, Cp is 1.005KJ/Kg*K and R is0.287KJ/Kg*K.)
[瓶頸] (寫寫自己的想法，方便大家為你解答)
想法一: 能量守恆+質量守恆
m(exit)=m1-m2
Q-W = m(exit)h(exit)+m2u2-m1u1 = m(exit)Cp(T1+T2)/2+m2CvT2-m1CvT1------>(1)
Q=W=0 =>m1=P1V1/RT1=(700*2)/(0.287*(250+273))=9.327 m2=P2V2/RT2
m(exit)=m1-m2=9.327-P2V2/RT2===>帶回(1)試得
0=(9.327-P2V2/RT2)*Cp(T2+T1)/2+(P2V2/RT2)*CvT2-m1CvT1 把值帶入得
0=(9.327-(100*2/0.287*T2))*1.005(T2+523)/2+(100*2)/(0.287*T2)*0.718*T2
-9.327*0.718*523
得(a)T2=315K，m2=P2V2/RT2=2.2123 Kg (b)m2/m1=23.72% (c)m1-m2=9.327-2.2123
=7.1147 Kg
問題:將出口焓以平均焓計算，由於壓差及溫差不算小(P2/P1=1/7)是否使的誤差過大?
想法二:起初將絕熱排氣是為等熵，後覺得不能輕易下此推測故由
(dQ-dW)=h(exit)dm(exit) +(dU)c.v----(1)
依題意Q=W=0===>假設任意時刻之h=h(exit)
(1)試=>0=-h(exit)dm+mdu+udm =>(h-u)dm=mdu =>dm/m=CvdT/RT 其中Cv=R/(k-1)
上試變為 dm/m=1/(k-1)*(dT/T)
兩端積分後得 m2/m1=(T2/T1)^[1/(k-1)] 再將m2與m1以P2V2/RT2及P1V1/RT1帶入
由於V1=V2故T2/T1=(P2/P1)^[(k-1)/k] 等熵過程成立了@@
故T2帶入上試得(a)T2=523*(100/700)^0.4/1.4=300K (b)m2/m1=(300/523)^[1/(1.4-1)]
=0.2492 (c)m2=9.327*0.2492=2.324 => m(exit)=9.327-2.324=7.003 Kg
請問，這樣的推導過程哪裡出了問題嗎?怎麼跟第一種想法的答案有些落差QQ?
最後，兩中方法感覺上算出來的值都與實際的值存在誤差，想請教各位大大哪種方法較正
確?是否還有較精確的算法?還請各位不吝指教，先感謝各位大大了(鞠躬)
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◆ From: 140.112.211.216
※ 編輯: hanwang 來自: 140.112.211.216 (02/20 03:02)