# 东北大学2003级《工程热力学》期末试题(2005年7月15日星期五)

Engineering Thermodynamics

1. Write out the Following Concepts and Laws:

(1) The first law of thermodynamics; (5%) (2) Dew point; (5%)

(3) The Kelvin-Planck statement of the second law of thermodynamics; (5%) (4) The Clausius statement of the second law of thermodynamics; (5%) (5) The distinguish between ideal and real gas; (5%)

(6) The COP (coefficient of performance) of a heat pump. (5%) 2. The piston of a vertical piston-cylinder device

containing a gas has a mass of 60kg and a cross-sectional area of 0.04 m 2, as shown in Fig. 1-66. The local atmosphere pressure is 0.97bar (0.097MPa), and the gravitational acceleration is 9.81 m/s 2. (a) Determine the pressure inside the cylinder. (b) If some heat is transferred to the gas and its volume is doubled, do you expect the pressure inside the cylinder to change? (10%)

4. An ideal Diesel cycle with air as the working fluid (figure 8-24), the volume of fresh air is

compressed to 1/18, then constant-pressure heat make volume double. At the beginning of the compression process, the working fluid is at 101325Pa, 27℃, and 1600 cm 3. Assumed the specific heats of air are constant. Determine (a) the temperature and pressure of the air at the end of each process, (b) the net work output and the thermal efficiency. (15%) 5. (Fig. 7-3-7) Fresh steam parameters of steam

power set operated with reheat cycle are p 1=16.50MPa, T 1=808.15K. Reheat pressure and temperature are p 3=3.84MPa, T 3

=808.15K.

P atm Figure 1-66 Schematic for Example 1-10, and the free-body

diagram of the piston.

Excluded steam pressure is p=0.004MPa. Output power of set is 300MW. The set operate 8000 hours annually. Burning one-kilogram coal can release 23000kJ heat. Please calculate annual power generation and coal consumption. (Efficiency of boiler and efficiency changing mechanical energy to electricity equal 1.)(15%)

Known:

When p=4000Pa, t s =28.981℃,

h'=121.41kJ/kg, h"=2554.1kJ/kg, s'=0.4224kJ/(kg ?K), s"=8.4747kJ/(kg ?K); When p=165bar and t=535℃, h=3390.625kJ/kg, s=6.4109kJ/(kg ?K); When p=165bar and t=29.354℃, h=137.98kJ/kg, s=0.4224kJ/(kg ?K); When p=38.4bar and t=535℃, h=3526.07kJ/kg, s=7.2118kJ/(kg ?K); When p=3.84MPa and t=304.17℃, h=2979.2kJ/kg, s=6.4109kJ/(kg ?K). 6.

Experimentation of engineering thermodynamics. You have got it from Ms. Li Juan.(10%)

Fig. 7-3-7

1. (1) open system; (5%)

(2) The temperature at which air becomes saturated and produces dew.

(3) The Kelvin-Planck statement of the second law of thermodynamics; (5%) (4) The Laval nozzle; (5%) (5) Ideal gas; (5%) (6)

2. A gas is contained in a vertical cylinder with a heavy piston. The pressure inside the cylinder and the effect of volume change on pressure are to be determined.

Assumptions: Friction between the piston and the cylinder is negligible.

Analysis: (a) The gas pressure in the piston-cylinder device depends on the atmospheric pressure and the weight of the piston. Drawing the free-body diagram of the piston an shown in Fig. 1-66 and balancing the vertical forces yield

pA =p atm A +W Solving for P and substituting, p =p atm +

A

mg

=0.97?105+

04

.081

.960?=111715Pa =1.12bar

(b) The volume change will have no effect on the free-body diagram draw in part (a), and therefore the pressure inside the cylinder will remain the same. 3. The mass of air in a room is to be determined.

Cogeneration plants use the same fuel source to produce both electricity and heat (steam and/or hot water). The steam and/or hot water can be used for heating or for use in a process facility adjacent to cogeneration facility, with the excess electricity sold to a local utility for it use. The utility benefits from cogeneration by gaining an additional generating source to help it meet its demand and also by purchasing the power at a rate lower than the rate at which it sells power. Whether the system is based on steam turbines, gas turbines, or a combination of both, cogeneration can achieve a significant increase in overall fuel efficiency over conventional power plants and steam generators. The advantage of cogeneration lies in the fact that both power and a transfer of heat for an intended use can be accomplished with a total fuel expenditure that is less than would be required to produce them individually. The advantages of cogeneration:

1. The big boiler has more efficiency than small one. Generally, the boiler that output power is less 5MW has efficiency 70%, and the boiler of electricity plant has 95%.

2. Energy utilized efficiency of cogeneration plants is larger than the individual sets,

so energy sources are saving.

3. 由于减少燃料消耗，从而减少污染物排放。

4. 由于大型电厂环保设施完善，所以减少污染物排放。

5. 由于大型电厂远离居住区，所以排放的污染物对人们身体健康影响小。

6. 减少小锅炉、煤贮存、灰贮存占用的土地，尤其是中心城区的土地。

7. 减少人工费用。

8. 减少居民区布置锅炉和煤、灰运输带来的安全隐患。

4. m=

001883.01

1

1=T R V p g kg p 1=101325Pa; T 1=300K; V 1=1600cm 3=0.0016m 3 p 2=p 1k

V

V ???

? ??2

1

=101325?181.4=57955596Pa; T 2= T 11

21-???

?

??k V

V =300?181.4-1=953.30K

p 3=p 2; T 3=

g

R V p 33=28790016

.0579*******??=g R V p =1906.53K p 4=p 3k

V

V ???

? ??43

=5795596?9-1.4=267398Pa; T 4= T 31

43-???

? ??k V

V =1906.53?9-(1.4-1)=791.67K

Q 23=mc p (T 3-T 2)=1.8039kJ

Q 41=mc v (T 1-T 4)=-0.6638kJ

W net =∑Q=1.8039-0.6638=1.1401kJ

ηt =8039

.11401

.123=Q W net =63.20%

5. ηt =2

361541211h h h h h h Q Q Q W

-+---=-= h 1=3390.625kJ/kg h 2=2979.2kJ/kg h 3=3526.07kJ/kg x 4=

'4

"4'

43'4

"4'

44s s s s s s s s --=

--

4224

.04747.84224

.02118.7--=

=0.8432

h 4=x 4(h 4"–h 4')+ h 4'

=0.8432(2554.1-121.41)+121.41

=2172.56kJ/kg

Fig. 7-3-7

h 5=h 4'=121.41kJ/kg h 6=137.98kJ/kg

ηt =2

.297907.352698.137625.339041

.12156.21721-+---=46.02%

Annual power generation: W total =Time ?W=8000?300?106=2.4?109kWh = 24?108kWh Annual heat consumption: Q total =W total /ηt =2.4?109/0.4602=5.215?109kWh=1.877?1013kJ Annual coal consumption: M=Q total /Q dw =1.877?1013/23000=816.28?106kg=816280Ton 6.

（2005年7月15日星期五，小语种）

1.有一个密封的圆筒形塑料瓶，内部几乎充满了水，水

2.写出热力学第二定律的克劳修斯说法和开尔文说法。

3.用公式表示过程功（膨胀功）、技术功、流动功、内能

（热力学能）和焓之间的关系。

4.湿球温度、干球温度、露点温度哪一个最高？哪一个

5.三级压缩中间冷却的中间压力p A和p B如何确定？

6.已知：水的p1=22.120MPa，T1=64

7.3K，v1=0.00317m3/kg。试求：水的范德瓦尔

7.在可逆绝热过程中，过程的膨胀功与技术功有什么关系？内能（热力学能）变化

8.渐缩形喷管的进口参数不变时，逐渐降低出口外的背压，试分析出口压力，出口

9.什么是理想气体？

10.解释概念：热源、工质、准静态过程。

1.如图所示可逆的热力循环，空气在1-2的绝热压缩过程中的压缩比为20，定容过

250kJ/kg，4-5过程为绝热作功过程，已知

p1=0.1MPa、t1=27℃，求其热效率，再计算同

2.压力为1MPa、温度为200℃的水蒸气，以20m/s

1MPa、200℃时，h1=2827.3kJ/kg，

v1=0.2059m3/kg；0.6MPa、170℃时，h2=2782.6kJ/kg，v2=0.3257m3/kg。试求：①出口处气流速度；②当进口速度近似取作零时，出口速度为多少？百分误差多少？③进出口截面面积比A1/A2。

1 圆筒形捏瘪后，体积缩小；水是不可压缩流体，所以气体的体积变小，瓶盖内气

2 克劳修斯说法：热不可能自发地、不付代价地、从低温物体传到高温物体。

3 w=?(pv)+w t , h=u+pv, ?h+w t =?u+w

4 干球温度>湿球温度>露点温度

5 设压缩前后压力为p 1、p 2，则p A =221p p ，p B =221p p

6 范德瓦尔方程为2v

a

b v RT p --=

，a 、b 称为范德瓦尔常数，与R 类似，决定于工质的性质。数学上，拐点有0=??? ????c T v p 和022=???? ????c

T v p ，所以 ()023

2=+--=???

????c

c c T v a b v RT v p c ()0624322=--=???? ????c c

c T v a b v RT v p c 解得：a =2

2

236427c c C

C v p p T R =

b =

C C p RT 8=3

c

v 对于水：范德瓦尔常数a =2

3c c v p =3?22.12?106?0.003172= 666.845(Pa ?m 6/kg 2)

3

00317

.03=

c v =0.001057(m 3/kg) 7 w t =kw, ?h=k ?u

8 假定进口压力为p 1，背压为p b ，当p b 从p 1向下降时，

9 所谓理想气体，是一种假想的实际上不存在的气体，其分子是一些弹性的、不占体

10 热源：工质可以从中吸取热量或向其放出热量的理想物体，其唯一特征是温度，分

1 2

1v v

=20, v 3=v 2, q 23=250kJ/kg, p 4=p 3, q 34=250kJ/kg, p 5=p 1=0.1MPa, T 1=300.15℃

η=1-34

2351

q q q +, q 51=c p (T 1-T 5)

R=287J/(kg ?K), c p =1004J/(kg ?K), c v =717J/(kg ?K)

v 1=11p RT =6

101.015.300287??=0.86143m 3/kg, v 2=v 1/20=0.04307m 3/kg T 2=11

2

1

T v

v k -???

? ??=201.4-1?300.15=994.83K

T 3=q 23/c v +T 2=250000/717+994.83=1343.51K

p 3=33v RT =04307.051.1343287?=8.9526MPa=p 4

T 4=q 34/c p +T 3=250000/1004+1343.51=1592.51K

T 5=T 44

.114.11

45

9526.81

.051.1592--?

?

? ???=?

??

? ??k

k p p =440.94K

q 51= c p (T 1-T 5)=1.004?(300.15-440.94)=141.36kJ/kg

η=1-342351

q q q +=1-141.36/(250+250)=0.7173

ηc =1-51.159215.300141-=T T =0.8115

2 β2=

1

6

.012=

p p =0.6>βcr , 为渐缩喷管。 c 2=()()232

1

2120106.27823.282722+?-=+-c h h =299.667m/s c 2’=()()321106.27823.282722?-=-h h =298.998m/s, 相差0.669m/s, 百分误差为0.669/299.667=0.22%