工程热力学(第三版)习题答案全解可打印第八章

Acr = A喉 =
qm vcr 1.5kg/s × 0.3287m3 /kg = = 1.27 ×10−3 m 2 387.21m/s ccr
cf = 2 ×1005J/(kg ⋅ K) × (350 − 326.5)K = 217.34m/s Acf 2.6 ×10−3 m 2 × 217.34m/s qm = = = 3.07kg/s v 0.1839m3 / kg
pcr = ν cr p* = 0.528 × 0.65MPa = 0.3432MPa > pb
8-3 燃气经过燃气轮机中渐缩喷管形的通道绝热膨胀，燃气的初参数为 p1=0.7MPa、t1=75℃， 燃气在通道出口截面上的压力 p2 = 0.5MPa ，经过通道的流量 qm = 0.6kg/s ，若通道进口处 流速及通道中的磨擦损失均可忽略不计，求燃气外射速度及通道出口截面积。 （燃气比热容按 变值计算，设燃气的热力性质近似地和空气相同，其热焓可查附表） 解： 查附表
p Tcr = T1 cr p1 vcr = RgTcr pcr =
κ −1 κ
= T1ν cr
κ −1 κ
= (43.3 + 273.15)K × 0.528 1.4 = 263.67K
1.4 −1
287J/(kg ⋅ K) × 263.67K = 0.0104m3 / kg 6 7.2864 × 10 Pa
κ −1 κ
p2 = pcr
= 823.15K × 0.528 1.4 = 685.86K
1.4 −1
′ = 1.4 × 287J/(kg ⋅ K) × 685.86K = 524.96m/s cf′ 2 = κ RgTcr
cf 2 − cf′ 2 526.87m/s − 524.96m/s = = 0.36% cf 2 526.87m/s
cf 2 = 2(h * −h2 ) = 2c p (T * −T2 ) = 2 ×1005J/(kg ⋅ K)(350 − 291.62)K = 342.55m/s A2 = qm v2 3.07kg/s × 0.2439m3 / kg = = 2.19 ×10−3 m 2 342.55m/s cf 2
ccr = c = κ RgTc = 1.4 × 287J/(kg ⋅ K) × 263.67K = 325.49m/s
103
第八章 气体和蒸气的流动
qm =
Acr ccr 1× 10−6 m 2 × 325.49m/s = = 0.0313kg/s 0.0104m3 / kg vcr
pcr : p2 = 10 :1，p2 = 0.72864MPa p T2 = T1 2 p1
cf 2 = 2(h1 − h2 ) = 2 × (1074.28kJ/kgK − 976.56kJ/kgK) ×103 = 435.25m/s v2 = RgT2 p2 = 287J/(kg ⋅ K) × 939.73K = 0.5394m 3 / kg 0.5 × 106 Pa
qm v2 0.6kg/s × 0.5394m3 / kg = = 7.44 ×10−4 m 2 A2 = 435.25m/s cf 2
o 8-4 有一玩火箭装满空气，其参数为： p = 13.8MPa、t = 43.3 C ℃。空气经缩放喷量排向
2 大气产生推力。已知：喷管喉部截面积为 1mm ，出口上截面压力与喉部压力之比为 1︰10，
试求稳定情况下火箭的净推力。 （ p0 = 0.1MPa ） 解： pcr = ν cr p1 = 0.528 ×13.8MPa = 7.2864MPa
κ −1 κ
0.72864MPa 1.4 = 316.45K × = 136.57K 13.8MPa
1.4 −1
v2 =
RgT 2 p2
=ห้องสมุดไป่ตู้
287J/(kg ⋅ K) ×136.57K = 0.0538m3 / kg 6 0.72864 × 10 Pa
cf 2 = 2c p (T1 − T2 ) = 2 ×1004J/(kg ⋅ K) × (316.45 − 136.57)K = 601.0m/s
=
102
第八章 气体和蒸气的流动
cf 2 = 2(h1 − h2 ) = 2cp (T1 − T2 ) = 2 × 1005J/(kg ⋅ K) × (300.15 − 250.09)K = 317.2m/s qm = A2 cf 2 10 ×10−4 m 2 × 317.2m/s = = 4.66kg/s v2 0.0680m 3 / kg
速度很小可忽略不计。求空气经喷管射出时的速度，流量以及出口截面处空气的状态参数
v2、t2 。 设空气取定值比热容， c p = 1005J/(kg ⋅ K)、κ = 1.4 ， 喷管的背压力 pb 分别为 1.5MPa
和 1MPa。 解： pcr = ν cr p1 = 0.528 × 2MPa = 1.056MPa 当背压 pb = 1.5MPa 时， p2 = pb = 1.5MPa
cf 2 = 2(h1 − h2 ) = 2cp (T1 − T2 ) = 2 × 1005J/(kg ⋅ K) × (300.15 − 276.47)K = 218.2m/s A2 cf 2 10 × 10−4 m 2 × 218.2m/s qm = = = 4.12kg/s v2 0.0529m3 / kg
当 背压 pb = 1MPa 时
p2 = pcr = 1.056MPa
1.4 −1
T2 = T1ν cr v2 = RgT2 p2
κ −1 κ
= 300.15K × 0.528 1.4 = 250.09K ； t2 = −23.06 o C 287J/(kg ⋅ K) × 250.09K = 0.0680m 3 / kg 1.05 ×106 Pa
第八章 气体和蒸气的流动
第八章 气体和蒸汽的流动
8-1 空气以 cf = 180m/s 的流速在风洞中流动，用水银温度计测量空气的温度，温度计上的读 数定是 70℃，假定气流通在温度计周围得到完全滞止，求空气的实际温度（即所谓热力学温 度） 。 解： T * = T1 +
cf2 2c p
T1 = T * −
8-5 内燃机排出的废气压力为 0.2MPa ，温度为 550℃，流速为 110m / s ，若将之引入渐缩喷 管， 试确定当背压为 0.1MPa 时废气通过喷管出口截面的流速并分析若忽略进口流速时引起的 误差。 解：
T * p* = p1 T1
κ κ −1
829.17K 1.4 −1 = 0.2MPa = 0.205MPa 823.15K
T * κ −1 447.85K 1.4 −1 p* = p1 = 0.617MPa = 0.58MPa × 440K T1 pcr = p喉 = ν cr p* = 0.528 × 0.617MPa = 0.3258MPa Tcr = T喉 = T *ν cr vcr = v喉 = RgTcr pcr
p T2 = T1 2 p1 t2 = 3.32°C v2 = RgT2 p2 =
κ −1 κ
1.5MPa 1.4 = (27 + 273.15)K × = 276.47K 2MPa
1.4 −1
287J/(kg ⋅ K) × 276.47K = 0.0529m3 / kg 6 1.5 × 10 Pa
1.4
2 cf1 (110m/s) 2 T * = T1 + = (550 + 273.15)K + = 829.17K 2cp 2 × 1005J/(kg ⋅ K)
pcr = ν cr p* = 0.528 × 0.205MPa = 0.108MPa > pb、 T2 = Tcr = T *ν cr
cf2 (180m/s)2 = (70 + 273.15)K − = 327.03K 2c p 2 × 1005J/(kg ⋅ K) × 103
t1 = 53.88 o C
8-2 进入出口截面积 A2 = 10cm 的渐缩喷管的空气初参数为 p1 = 2 × 10 Pa、t1 = 27 C ，初
2 6 o
104
第八章 气体和蒸气的流动
8-6 滞止压力为 0.65MPa ，滞止温度为 350K 的空气可逆绝热流经收缩喷管，在截面积为
2.6 × 10−3 m 2 处气流马赫数为 0.6。若喷管背压力为 0.28MPa ，试求喷管出口截面积。
解：
Ma =
c p (T * −T ) cf = c κ RgT
T=
2cpT * 2 ×1005J/(kg ⋅ K) × 350K = = 326.50K 2 κ Rg Ma + 2c p 1.4 × 287J/(kg ⋅ K) × 0.62 + 2 ×1005J/(kg ⋅ K)
κ
1.4
T κ −1 326.50K 1.4 −1 p = p * = 0.510MPa = 0.65MPa × T * 350K v= RgT p = 287J/(kg ⋅ K) × 326.50K = 0.1839m3 / kg 0.51× 106 Pa
v1 =
RgT1 p1
=
287J/(kg ⋅ K) × 440K = 0.2177m3 / kg 0.58 ×106 Pa
105
第八章 气体和蒸气的流动
cf 1 =
qm v1 1.5kg/s × 0.2177m3 / kg = = 125.61m/s 2.6 ×10−3 m 2 A1
cf21 125.61(m/s) 2 T * = T1 + = 440K + = 447.85K 2c p 2 × 1005J/(kg ⋅ K)
κ −1 κ
κ
1.4
= 447.85K × 0.528 1.4 = 373.15K
1.4 −1
=
287J/(kg ⋅ K) × 373.15K = 0.3287m3 / kg 6 0.3258 ×10 Pa
ccr = c = κ RgTcr 1.4 × 287J/(kg ⋅ K) × 373.15K = 387.21m/s
t1 = 750°C
pr1 = 126.984
pr 2 = pr1
h1 = 1074.28kJ/kg
p1 p = r1 p2 pr 2
p2 0.5MPa = 126.984 × = 90.703 p1 0.7MPa
查附表 T2 = 939.73K ，
t2 = 666.58°C
h2 = 979.56kJ/kg
−3
440K ，出口截面上压 8-7 空气等熵流经缩放喷管，进口截面上压力和温度分别为 0.58MPa、
力 p2 = 0.14MPa 。已知喷管进口截面积为 2.6 × 10 m ，空气质量流量为 1.5kg/s ，试求喷
2
管喉部及出口截面和出口流速。空气取定值比热容， c p = 1.005J/(kg ⋅ K) 。 解 ：
A2 =
qm v2 313 × 10−4 kg/s × 0.0538m3 /kg = = 2.8 ×10−6 m 2 601.0m/s cf 2
F = qm cf2 + ∆pA2 = qm cf 2 + ( p2 − p0 ) A2 = 0.0313kg/s × 601m/s + (0.72864 − 0.1) ×106 Pa × 2.8 ×10−6 m 2 = 20.6N
κ −1 κ
p2 = pcr
= 829.17K × 0.528 1.4 = 690.88K
1.4 −1
cf 2 = κ RgTcr = 1.4 × 287J/(kg ⋅ K) × 690.88K = 526.87m/s
若忽略初流速，则
′ = ν cr p1 = 0.528 × 0.2MPa = 0.105MPa > pb pcr ′ = T1ν cr T2′ = Tcr
p2 = pcr = 0.3432MPa ，出口截面即为临界截面。 T2 = Tcr = T *ν cr
κ −1 κ
= 350K × 0.528
1.4 −1 1.4
= 291.62K
v2 =
RgT2 p2
=
287J/(kg ⋅ K) × 291.62K = 0.2439m3 / kg 6 0.3432 × 10 Pa