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工程热力学高教第三版课后习题第十二章答案

κ −1 κ
1.4 −1
T2,a
κ −1 κ
= 290K × 3 1.4 = 396.93K
1.4 −1
T2,b = T1π b
= 290K × 6 1.4 = 483.86K
1.4 −1
qc ,a = c p (T1 − T4,a ) = 1.005kJ/(kg ⋅ K)(290 − 219.18)K = 71.2kJ/kg qc ,b = c p (T1 − T4,b ) = 1.005kJ/(kg ⋅ K)(290 − 179.8)K = 110.7kJ/kg
Pnet =
ψ c 10kW = = 2.63kW 3.80 ε act
10kW = 2.59" 冷吨 " 3.86
(3) 折合冷吨
ψc
3.86
=
12-2 一逆向卡诺制冷循环,其性能系数为 4,问高温热源与低温热源温度之比多少?若输入 功率为 1.5kW。试问制冷量为多少“冷吨”?如果将此系统改作热泵循环,高、低温热源温度 及输入功率维持不变。试求循环的性能系数及能提供的热量。 解: (1) ε c =
h1 = 380kJ/kg
h2 = 406kJ/kg s1 = 1.69kJ/(kg ⋅ K)
h4 = 230kJ/kg s4 = 1.12kJ/(kg ⋅ K) qc = h1 − h5 = h1 − h4 = 380kJ/kg − 230kJ/kg = 150kJ/kg
ε=
qc 1060.7kJ/kg = = 5.77 wnet 184.0kJ/kg
5
(3)
由ψ c = 4.2 ×10 kJ/h , qm =
ψc
qc
=
4.2 × 105 kJ/h = 396.0kg/h=0.11kg/s 1060.7kJ/kg
(4)冷量佣
T (25 + 273.15)K − 1 × 1060.7kJ/kg = 164.35kJ/kg ex ,Q = 0 − 1 qc = (−15 + 273.15)K Tc
第十二章 制冷循环
Байду номын сангаас
第十二章
制冷循环
12-1 一制冷机在-20℃和 30℃的热源间工作,若其吸热为 10kW,循环制冷系数是同温度限 间逆向卡诺循环的 75%,试计算: (1)散热量; (2)循环净耗功量; (3)循环制冷量折合多 少“冷吨”? 解: 在-20℃和 30℃间逆向卡诺循环制冷系数 TL (273.15 − 20)K = = = 5.06 c Th − TL [30 − (−20)]K
4 设该装置的制冷量ψ c = 4.2 ×10 kJ ,求氨的流量 qm ; (4)求该装置的佣效率。
h
解: (1)查 NH3 表, p1 = 0.236MPa;p2 = 1.003MPa (2)查 NH3 表
h1′ = 111.66kJ/kg h1′′ = 1424.6kJ/kg
h1 = xh1 + (1 − x)h1′ = 0.95 × 1424.6kJ/kg + (1 − 0.95) × 111.66kJ/kg = 1358.95kJ/kg ′ = 0.4538kJ/(kg ⋅ K) s1 ′ + x( s1 ′′ − s1 ′) s1 = s1 = 0.4538kJ/(kg ⋅ K) + 0.95 × (5.5397 − 0.4538)kJ/(kg ⋅ K) = 5.2854kJ/(kg ⋅ K)
ψ ' = ε c′ Pnet = 5 × 1.5kW = 6kW
12-3
压缩空气制冷循环运行温度 TC = 290K ,T0 = 300K ,如果循环增压此分别为 3 和
6,分别计算它们的循环性能系数和每 kg,工质的制冷量。假 定空气为理想气体,此热容取定值 c p = 1.005kJ/(kg ⋅ K) 、
12-8
若 12-6 题中制冷剂改为氟里昂 134a(HFC134a), 求 (1) 蒸发压力 p1 和冷凝压力 p 2 ; (2)
qc、Wnet、ε ; (3)HFC134a 的流量; (4)装置佣效率ηex 。
解: (1)据工作温度,查附表得 (2)查 HFC134a 的压焓图得:
p1 = 164.36kPa 、 p2 = 665.49kPa
ε
实际循环制冷系数
ε act = 0.75ε c = 0.75 × 5.06 = 3.80
(1) 散热量
ε act =
ψc
Pnet
=
ψc ψ −ψ c
1 1 = 10kW 1 + = 12.63kW ε act 3.80
ψ = ψ c 1 +
(2) 净功
TL 1 = T TH − TL H −1 TL
TH 1 1 = 1 + = 1 + = 1.25 4 εc TL 1.875kW = 0.486 冷吨 3.86
(2) ε c =
ψc
Pnet
ψ c = ε c Pnet = 1.25 ×1.5kW = 1.875kW
′= (3) ε c
TH 1 1 = = =5 TH − TL 1 − TL 1 − 1 1.25 TH
κ = 1.4 。
解:
εa =
1
π1
κ −1/ κ
−1
=
1 3
(1.4 −1) /1.4
−1
= 2.712
156
第十二章 制冷循环
εb =
1
π2
κ −1/ κ
−1
=
1 6
(1.4 −1) −1
−1
= 1.496
κ −1 κ
1.4 −1
T4,a
p = T3 4 p3
κ −1 κ
循环净功
Wnet ,a = (h1 − h2',a ) + (h3 − h4′,a ) = c p (T1 − T2′,a + T3 − T4′a )
= 1.005kJ/(kg ⋅ K)(290 − 420.40 + 300 − 235.34)K = −66.1kJ/kg
157
第十二章 制冷循环
Wnet ,b = c p (T1 − T2,b + T3 − T4′b )
κ −1 κ
1.3−1
p T3′ = T1 3′ p1
κ
T κ −1 434.93K 1.3−1 p = 14.9 π ′ = 3′ = 3′ = p1 T1 233.15K
1.3
ε1234561 = ε13′5′6 = 1 −
1
π′
h2 − h1 h2′ − h1
h2′ = h1 +
h2 − h1
ηCS
= 1358.95kJ/kg + h2 − h1 wnet
184.0kJ/kg = 1588.9kJ/kg 0.8
′ = h2' − h1 = wnet
ηCS
=
ηCS
=
184.0kJ/kg = 230kJ/kg 0.8
ε=
qc 1060.7kJ/kg = = 4.61 230kJ/kg wnet
12-4 若题 12-3 中压气机绝热效率ηCS = 0.82 ,膨胀机相对内效率ηT = 0.85 ,分别计算 1 kg
工质的制冷量,循环净功及循环性能系数。 解:
ηT =
h3 − h4′ T3 − T4′ = h3 − h4 T3 − T4
′ a = T3 − ηT (T3 − T4 ) T4,
据 s2 = s1 及 p2 = 1.003MPa , 查 NH3 过热蒸汽表, 得 t2 = 52.4 C、h2 = 1542.9kJ/kg ,
o
′′ = 5.539kJ/(kg ⋅ K) s1
查 NH3 饱和蒸汽表, t4 = 25 C、h4 = 298.25kJ/kg
o
qc = h1 − h5 = h1 − h4 = 1358.95kJ/kg − 298.25kJ/kg = 1060.7kJ/kg qc 可用 T-s 图上面积,155’1’1 表示 wnet = h2 − h1 = 1542.9kJ/kg − 1358.95kJ/kg = 184.0kJ/kg
循环制冷量中的冷量佣
T (25 + 273.15)K − 1 × 1060.7kJ/kg = 164.35kJ/kg ex ,Q = 0 − 1 qc = (−15 + 273.15)K Tc
循环佣效率
ηe =
x
ex ,Q 164.35kJ/kg = = 0.715 ′ 230kJ/kg wnet
T4,a = 300K − 0.8 × (300 − 219.18)K = 235.34K T4,b = 300K − 0.8 × (300 − 179.81)K = 203.85K
ηCS =
h2 − h1 T2 − T1 = h2′ − h1 T2′ − T1
T2′ = T1 + (T2 − T1 ) / ηcs
κ −1 κ
= 1−
1 14.9
1− 3−1 1.3
= 0.464
qc ,134561 = c p (T1 − Tb ) = 0.815kJ/(kg ⋅ K)(233.15 − 160.82)K = 58.95kJ/kg
ε 1234561 =
ψc
Pnet
ψ c = ε1234561Pnet = 0.464 × 3kW = 1.39kW
p = T0 1 p2
κ −1 κ
1 = T0 πa
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