作者hanwang ()
看板Physics
标题[题目] 热力学 绝热排气
时间Wed Feb 20 02:17:44 2013
[领域] 热力学 (题目相关领域)
[来源] 考古题 (课本习题、考古题、参考书...)
[题目] 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.)
[瓶颈] (写写自己的想法，方便大家为你解答)
想法一: 能量守恒+质量守恒
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===>带回(1)试得
0=(9.327-P2V2/RT2)*Cp(T2+T1)/2+(P2V2/RT2)*CvT2-m1CvT1 把值带入得
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
问题:将出口焓以平均焓计算，由於压差及温差不算小(P2/P1=1/7)是否使的误差过大?
想法二:起初将绝热排气是为等熵，後觉得不能轻易下此推测故由
(dQ-dW)=h(exit)dm(exit) +(dU)c.v----(1)
依题意Q=W=0===>假设任意时刻之h=h(exit)
(1)试=>0=-h(exit)dm+mdu+udm =>(h-u)dm=mdu =>dm/m=CvdT/RT 其中Cv=R/(k-1)
上试变为 dm/m=1/(k-1)*(dT/T)
两端积分後得 m2/m1=(T2/T1)^[1/(k-1)] 再将m2与m1以P2V2/RT2及P1V1/RT1带入
由於V1=V2故T2/T1=(P2/P1)^[(k-1)/k] 等熵过程成立了@@
故T2带入上试得(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
请问，这样的推导过程哪里出了问题吗?怎麽跟第一种想法的答案有些落差QQ?
最後，两中方法感觉上算出来的值都与实际的值存在误差，想请教各位大大哪种方法较正
确?是否还有较精确的算法?还请各位不吝指教，先感谢各位大大了(鞠躬)
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※ 编辑: hanwang 来自: 140.112.211.216 (02/20 03:02)