1、工程热力学计算题工程热力学计算题1、已知某柴油机混合加热理想循环,p1=0.17MPa、t1=60,压缩比为 v1 / v2 =14.5, 定容升压比 p3 / p2 = 1.43,定压预胀比 v4 / v3 = 1.42,设工质比热容为定值,CP=1.004 kJ/(K)、Cv=0.718 kJ/(K),求:(1)试画出该循环的p-v图和T-s图;(2)循环中各点的温度、压力;(3)循环热效率,并于同温度限的卡诺循环热效率作比较。解:(1)试画出该循环的p-v图和T-s图;p-v图; 3分T-s图; 3分(2)循环中各点的温度、压力;点1:k = CP / Cv =1.004 kJ/(K)/
2、0.718 kJ/(K) = 1.4p1 =0.17Mpa=1.7105Pa; t1=60 = 333 K 2分点2:1-2为绝热压缩过程T2 = T1v1/ v2(k-1) = T1v1/ v2(k-1) = 333K14.50.4 = 970.5K P2 = P1(v1/v2)k = P1k = 170kPa14.51.4 = 7.184106Pa 2分点3:2-3为等容过程 T3 = T2 P3 / P2 = T2 = 970.5K1.43 = 1387.8K P3 / P2 = ; P3 = P2 = 7.184106Pa1.43 = 1.027107Pa2分点4:3-4为等压过程 T
3、4 = T3v4/ v3 = T3 = 1387.8K1.42 = 1970.7K P4 = P3 = 1.027107Pa 2分点5:4-5为绝热膨胀过程 v4 = RgT4 /P4 = CP - Cv T4 / P4 = 1004 J/(K)- 718 J/(K) 1970.7K /1.027107Pa= 0.0548 m3/kg v5 = v1 = RgT1/P1 = CP - Cv T1 / P1 = 1004 J/(K)- 718 J/(K) 333K /1.7105Pa = 0.5602 m3/kgT5 = T4v4/ v5(k-1) = T4v4/ v1(k-1) = 1970.
4、7K0.0548 /0.56020.4 = 777.7K P5 = RgT5/v5 = 1004 J/(K)- 718 J/(K) 777.7K /0.5602 m3/kg = 3.97105Pa 2分(3)循环热效率,并于同温度限的卡诺循环热效率作比较。t = wnet / q1 = 1- q2 / q1 = 1- Cv(T5-T1)/ Cv(T3-T2) + Cp(T4-T3)= 1- 319.3 / (299.6 + 585.2) = 63.9 2分 c = 1- T1 / T4 = 1- 333K/1970.7K = 83.1 t 2分2、已知某定压燃气轮机装置理想循环,参数如下:p1
5、 = 100kPa;T1 300K;循环增温比为:= 5.24 ;循环增压比为: p2 / p1 = 12。假设工质为空气且比热容为定值,k=1.4,CP = 1.005 kJ/(kg.K)。求:(1)试画出该循环的p-v图和T-s图;(2)循环中吸收和放出的热q1和q2和循环功wnet;(3)循环热效率,并于同温度限的卡诺循环热效率作比较。解:、试画出该循环的p-v图和T-s图; p-v图; 3分T-s图; 3分、循环中吸收和放出的热q1和q2和循环功wnet;1-2、34是绝热过程T2 = T1p2 / p1(k-1)/k = T1(k-1)/k = 300K120.4/1.4= 610.
6、2K 2分= 5.24 = T3 /T1 ; T3 = 5.24T1 = 1572 K 1分T4= T3P4/P3(k-1)/k= 1572K(1/) (k-1)/k= 772.9K 1分 2-3、4-1为等压过程吸收的热:q1 = q2-3 = CP(T3-T2) = 1.005kJ/(kg.K)(1572-610.2)K=966.6 kJ/kg 2分放出的热:q2 = q4-1 = CP(T1-T4)= 1.005kJ/(kg.K)(300-772.9)K = -475.3 kJ/kg 2分循环净功wnet = q1 q2 = 966.6 kJ/kg - 475.3 kJ/kg = 491
7、.3kJ/kg 2分、循环热效率,并于同温度限的卡诺循环热效率作比较。 t = wnet / q1 = 491.3 (kJ/kg )/ 966.6 (kJ/kg) = 50.8 2分 c = 1- T1 / T3 = 1- 300/1572 = 80.9 t 2分 3、已知某活塞式内燃机定容加热理想循环,工质为空气,压缩比为 6, 气体压缩过程起点状态是p1=98.1kPa、t1=60,加热过程中气体吸热879kJ/。假定比热容为定值且CP=1.005 kJ/(K)、k=1.4,求:(1)试画出该循环的p-v图和T-s图;(2)循环中各点的温度和压力;(3)循环热效率,并于同温度限的卡诺循环热
8、效率作比较。解:(1)试画出该循环的p-v图和T-s图; p-v图 2分T-s图 2分(2)循环中各点的温度和压力; 点1: k = 1.4 = CP / CV ; CP =1.005 kJ/(K) CV = CP /1.4 = 1.005 kJ/(K)/1.4 = 0.718 kJ/(K) CP - CV = Rg ; Rg = 0.287 kJ/(K) T1 = (60+273.15)K = 333.15K p1 = 98.1kPa 3分点2:P2 = P1(v1/v2)k = P1k = 98.1kPa61.4 = 1205.3kPa T2 = T1(v1/v2)k-1 = T1k-1
9、= 333.15K60.4 = 682.2K 3分点3:q1 = Cv (T3 - T2 ) T3 = T2 + q1 /Cv = 682.2K + 879kJ/kg /0.718 kJ/(K) = 1906K P3 = P2(T3 / T2 ) = 1205.3kPa 1906 K /682.2K = 3367.5kPa 3分点4:P4 = P3(v3/v4)k = P3(1/)k = 3367.5kPa(!/6)1.4 = 274.1 kPa T4 = T1 P4 / P1= 333.15K274.1 kPa/ 98.1kPa = 930.1K 3分(3)循环热效率,并于同温度限的卡诺循环
10、热效率作比较。t =1 - q2/ q1 = 1 (T4-T1)/( T3-T2)= 1-(930.1K -333.15K)/( 1906K -682.2K)= 1- 596.95/1223.8= 0.512 = 51.2 2分 c = 1- T1 /T3 = 1- 333.15K /1906K = 82.5 t 2分4、已知某活塞式内燃机定压加热理想循环,压缩比为 17, 输入1空气的热量q1750kJ/。若气体压缩过程起点状态是p1=100kPa、t1=25,假定比热容为定值且CP=1.005 kJ/(K)、Cv=0.718 kJ/(K),求:(1)试画出该循环的p-v图和T-s图;(2)
11、循环中各点的温度和压力和比体积;(3)循环热效率,并于同温度限的卡诺循环热效率作比较。解:(1)试画出该循环的p-v图和T-s图; p-v图; 2分 T-s图; 2分(2)循环中各点的温度和压力和比体积;点1:p1 = 100kPa; T1 = 298K ;Rg = CP - Cv = 1.005 kJ/(K) - 0.718 kJ/(K) = 0.287 kJ/(K)v1 = RgT1 / P1 = 0.287 kJ/(K)298K /100kPa = 0.8553m3/kg 3分点2:v2 = v1/ = 0.8553m3/kg / 17 = 0.0503 m3/kg k = CP / C
12、v = 1.005 kJ/(K) / 0.718 kJ/(K) = 1.4 P2 = P1 v1/v2k = 100kPa 171.4 = 5279.9 kPa T2 = P2v2 / Rg = 5279.9 kPa 0.0503 m3/kg / 0.287 kJ/(K) = 925.4K 另一方法; T2 = T1(v1/v2) k-1 =298K170.4 = 925.5K 3分 点4: q1 = Cp ( T4- T2)T4 = T2 + q1/Cp = 925.4 K + 750kJ/ / 1.005 kJ/(K) = 1496.3KP4 = P2 = 5279.9 kPa v4=Rg
13、T4/P4=0.287 kJ/(K)1496.3K / 5279.9 kPa = 0.0813 m3/kg 3分点5:v5 = v1 = 0.8553m3/kg P5 = P4v4/v5k = P4v4/v1k = 5279.9 kPa0.0813 m3/kg / 0.8553m3/kg1.4 = 195.8 kPaT5 = P5v5/ Rg = 195.8 kPa0.8553m3/kg / 0.287 kJ/(K) = 583.5K (另一方法:T5 = P5T1/ P1 =195.8 kPa298K / 100kPa = 583.5K ) 3分(3)循环热效率,并于同温度限的卡诺循环热效率
14、作比较。t = 1 q2 / q1 = 1 Cv (T5 T1) / CP(T4 T2) = 1 (T5 T1) /k(T4 T2) = 1- 285.5/(1.4570.9) = 64.3 2分c = 1 TL / TH = 1 - 298K / 1496.3K = 80.1 2分5、设气缸中有1 kg氧气,其压力为0.1MPa、温度为70。如进行一个定压过程,气体对外作功8 kJ。试求过程中气体(1)热力学能的变化;(2)熵的变化;(3)气体的吸收热量。设比热容为定值,已知:氧气的气体常数为0.2598kJ/(kg.K),比定容热容为 0.625 kJ/(kg.K)。解: m = 1 kg
15、氧气 ; P1 = P2 = 0.1Mpa ; T1 = 343K ;Rg = 0.2598kJ/(kg.K) ; Cv = 0.625 kJ/(kg.K) ;Cp = Cv + Rg = 0.625 kJ/(kg.K) + 0.2598kJ/(kg.K) = 0.8848 kJ/(kg.K) ;3分(1)热力学能的变化 P1 = P2 w = P v = P (v2 v1)= 8 kJ = P2v2 P1v1 = Rg(T2 T1 )T2 T1 = 8 kJ/ kg / 0.2598kJ/(kg.K) = 30.79 K3分T2 = 30.79 K + 343K = 373.79K2分u =
16、 Cv (T2 T1) = 0.625 kJ/(kg.K)30.79 K = 19.24 kJ/kg4分(2)熵的变化 s = Cp ln(T2/T1) = 0.8848/(kg.K)ln (373.79/343) = 0.07606 kJ/(kg.K)4分(3)气体的吸收热量 u = q w q = u + w = 19.24 kJ/kg + 8 kJ/kg = 27.24 kJ/kg 4分6、已知乙醇在48.4时的饱和蒸气压为26.66 kPa,乙醇的正常沸点为78;试求: (a)表示蒸气压与温度关系的方程式 ln ( P/ Pa) = A / T B 中的常数A和B;(b)乙醇的摩尔蒸发
17、焓;(c)在多大压力下乙醇的沸点为100解: (a)方程式 ln( P/ Pa) = A / T B 中的常数A和B;ln ( 101325Pa/ Pa) = A / 351K B 3分ln ( 26660 Pa / Pa) = A / 321.6K B 3分A = 5126.5K 3分B = 26.1 3分(b)乙醇的摩尔蒸发焓;把 lnP = (-vapHm /R )/T+ C 与 ln(P/kPa) = 26.1 5126.5/( T/K )比较vapHm /R 5126.5K ; 2分vapHm 5126.5K8.314 JK-1mol-1 42.62 kJmol-1 2分(c)在多大
18、压力下乙醇的沸点为100100 = 373K 带入ln(P/Pa) = 26.1 5126.5/( T/K )ln(P/Pa) = 26.1 5126.5 / 373 P = 232.36 kPa 4分7、有一台压缩机用于压缩氮气,使其压力由0.3MPa提高到0.9MPa。设比热容为定值及进口温度为500K, ,试求压缩过程中消耗的体积功和技术功:(1)压缩过程为绝热可逆过程;(2)压缩过程为定温过程。对于氮Rg=0.2968kJ/(kg.K);CV =0.741 kJ/(kg.K);1.4。解: (1)压缩过程为绝热可逆过程;初态 P10.3Mpa ; T1=500K; 终态 P20.9Mp
19、aRg = 0.2968kJ/(kg.K);CV =0.741 kJ/(kg.K);1.4。 过程为绝热可逆过程 q = 0 ; T2 = T1P2/P1(k-1)/k = 500K0.9Mpa/0.3Mpa0.2857= 684.4K 4分 体积功: u = q w w = Pdv = -u = CV (T1 T2) = 0.741 kJ/(kg.K)(500K - 684.4K ) = - 136.6 kJ/kg 4分 另一方法:w = 1/(k-1) RgT11- (p2 / p1 )(k-1)/k= 2.50.2968kJ/(kg.K)500K1-1.369 = - 136.8 kJ/
20、kg 技术功:Wt = - vdP = kw = 1.4(- 136.6 kJ/kg) = - 191.2 kJ/kg 另一方法:wt = k/(k-1) RgT11- (p2 / p1 )(k-1)/k= 3.50.2968kJ/(kg.K)500K1-1.369 = - 191.5 kJ/kg 4分(2)压缩过程为定温过程。初态 P10.3Mpa ; T1=500K; 终态 P20.9MpaRg = 0.2968kJ/(kg.K);CV =0.741 kJ/(kg.K);1.4。 过程为定温过程 T2 = T1 ;T = 0 ; u = q w = 0 体积功:w = Pdv = = Rg
21、T1ln( p1 / p2) = 0.2968kJ/(kg.K)500Kln(0.3 Mpa /0.9 Mpa)= - 163.0 kJ/kg 4分技术功:Wt = - vdP = = RgT1ln( p1 / p2) = - 163.0 kJ/kg4分8、固态苯的饱和蒸气压与绝对温度的函数关系为: ln (P/Pa) = 27.57 5320/( T/K )液态苯的饱和蒸气压与绝对温度的函数关系为为: ln (P/Pa) = 23.33 4109/( T/K )试求:(1)苯的三相点的温度、压力; (2)三相点的蒸发焓、升华焓和熔化焓; (3)三相点的熔化熵。解: (1)苯的三相点的温度、压
22、力; ln (P/Pa) = 27.57 5320/( T/K )ln (P/Pa) = 23.33 4109/( T/K ) 27.57 5320/( T/K ) = 23.33 4109/( T/K ) T = 285.6 K 4分 ln(P/Pa) = 27.57 5320/307.4 = 10.56 P = 7650.7 Pa 4分 (2)三相点的蒸发焓、升华焓和熔化焓;把 ln P = (-vapHm /R) / T+ C 与 ln(P/Pa) = 23.33 4109/( T/K )比较vapHm /R 4109K ;vapHm 4109K8.314 JK-1mol-1 34.16 kJmol-1 3分 把 ln P = (-subHm /R)/ T + C 与 ln(P/Pa) = 27.57 5320/( T/K )比较subHm /R 5320K ;subHm 5320K8.314 JK-1mol-1 44.23 kJmol-1 3分 subHm = fusHm + vapHm fusHm = subHm - vapHm = 44.23 kJmol1 - 34.16 kJmol1 = 10.07 kJmol1 3分(3)三相点的熔化熵。 fusSm = fusHm / T = 10070 Jmol1 / 285.6 K = 35.26 J mol1 K1 3分
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