Thermodynamics

1. 1[Date]
NAME:
COURSE & YR LEVEL:
SCORE TIME: DAYS:
DATE:
I. Solve the following problems. Write your final answers in four (4) decimal places. In your complete solutions,
enclose your final answers with correct units. One page per problem.
1. Steam is supplied to a fully loaded 100 hp turbine at 200 psia with 𝑈1 = 1,163 𝐵𝑇𝑈
𝑙𝑏⁄ , 𝑣1 = 265
𝑓𝑡3
𝑙𝑏⁄ ,
and velocity = 400
𝑓𝑡3
𝑠⁄ . Exhaust is at 1 psia with 𝑈2 = 925 𝐵𝑇𝑈
𝑙𝑏⁄ ,𝑣2 = 295
𝑓𝑡3
𝑙𝑏⁄ , and velocity =
1,100
𝑓𝑡3
𝑠⁄ . The heat loss from the steam in the turbine is 10 𝐵𝑇𝑈
𝑙𝑏⁄ . Neglect potential energychange.
Determine the following: (a) Work steady flow per lb; (b) steam flow rate.
∆𝐾𝐸 = (
1
2
) (
1
32.2
) [(1,100)2 − (400)2](
1 𝐵𝑇𝑈
778 𝑓𝑡 − 𝑙𝑏
)
∆𝐾𝐸 = 20.9567
𝐵𝑇𝑈
𝑙𝑏
∆𝑈 = 925
𝐵𝑇𝑈
𝑙𝑏
− 1,163
𝐵𝑇𝑈
𝑙𝑏
∆𝑈 = −238
𝐵𝑇𝑈
𝑙𝑏
∆𝑊𝑓 = {[(1
𝑙𝑏
𝑓𝑡2
)(
12 𝑖𝑛
1 𝑓𝑡
)
2
(295
𝑓𝑡3
𝑙𝑏
)] − [(200
𝑙𝑏
𝑓𝑡2
) (
12 𝑖𝑛
1 𝑓𝑡
)
2
(265
𝑓𝑡3
𝑙𝑏
)]}(
1 𝐵𝑇𝑈
778 𝑓𝑡 − 𝑙𝑏
)
∆𝑊𝑓 = −43.4961
𝐵𝑇𝑈
𝑙𝑏
−10
𝐵𝑇𝑈
𝑙𝑏
= 0 + 20.9567
𝐵𝑇𝑈
𝑙𝑏
− 238
𝐵𝑇𝑈
𝑙𝑏
− 43.4961
𝐵𝑇𝑈
𝑙𝑏
+ 𝑊𝑠
𝑊𝑠 = 250.5394
𝐵𝑇𝑈
𝑙𝑏
100 ℎ𝑝 = 250.5394
𝐵𝑇𝑈
𝑙𝑏
∙ 𝑚̇ (
𝑙𝑏
ℎ𝑟
) ∙
ℎ𝑝
2,545
𝐵𝑇𝑈
ℎ𝑟
𝑚̇ = 1,015.8083
𝑙𝑏
ℎ𝑟

2. 2[Date]
NAME:
COURSE & YR LEVEL:
SCORE TIME: DAYS:
DATE:
II. A fluid system undergoes a non-flow frictionless process from 4.5 𝑓𝑡3 to a final volume of 1.5 𝑓𝑡3 in
accordance with the defining relation 𝑃 =
60
𝑣
+ 30 𝑝𝑠𝑖𝑎, where volume is in 𝑓𝑡3. Duringthe process, the
system rejects 20 BTUof heat. Determine the (a) Change of enthalpy and (b) internal energy.
𝑊𝑛 = ∫ 𝑃𝑑𝑣
2
1
𝑊𝑛 = 60 ∫
1
𝑣
𝑑𝑣 + ∫30𝑑𝑣
𝑊𝑛 = 60 ln (
𝑣2
𝑣1
) + 30( 𝑣2 − 𝑣1)
𝑊𝑛 = 60 ln(
1.5 𝑓𝑡3
4.5 𝑓𝑡3
) + 30(1.5 𝑓𝑡3 − 4.5 𝑓𝑡3)
𝑊𝑛 = (−155.9167 𝑝𝑠𝑖𝑎) ∙ 𝑓𝑡 ∙
144 𝑖𝑛2
𝑓𝑡2 ∙
1 𝐵𝑇𝑈
778 𝑓𝑡 − 𝑙𝑏
𝑊𝑛 = 28.8586 𝐵𝑇𝑈
𝑄 = ∆𝑈 + 𝑊𝑛
∆𝑈 = 𝑄 − 𝑊𝑛
∆𝑈 = −20 𝐵𝑇𝑈 − (−28.8586 𝐵𝑇𝑈)
∆𝑈 = 8.8586 𝐵𝑇𝑈
∆𝐻 = ∆𝑈 + ∆𝑊
∆𝐻 = 8.8586 𝐵𝑇𝑈 + (−28.8586 𝐵𝑇𝑈)
∆𝐻 = −20 𝐵𝑇𝑈

3. 3[Date]
NAME:
COURSE & YR LEVEL:
SCORE TIME: DAYS:
DATE:
III. Air flow steadily at the rate of 0.5
𝑘𝑔
𝑠⁄ through an air compressor. Entering at 7 𝑚
𝑠⁄ speed,100 kPa
pressure and 0.95 𝑚3
𝑘𝑔⁄ specificvolume and leaving at 5 𝑚
𝑠⁄ , 700 kPa and 0.19 𝑚3
𝑘𝑔⁄ . The internal
energyof the air leaving is 90
𝑘𝐽
𝑘𝑔⁄ greater than that of the air entering.Cooling water in the compressor
jacket absorbs heat from the air at the rate of 58
𝑘𝐽
𝑠⁄ . Compute for (a) 𝑊𝑠 , (b) ∆𝐻
𝑄 = ∆𝐾𝐸 + ∆𝑃𝐸 + ∆𝑈 + ∆𝑊𝑓 + 𝑊𝑠
∆𝐾𝐸 = (
1
2
)(0.5
𝑘𝑔
𝑠
)[(5
𝑚2
𝑠
) − (7
𝑚2
𝑠
)](
1
1,000
)
∆𝐾𝐸 = −6 𝑥 10−3
𝑘𝐽
𝑠
∆𝑊𝑓 = (700 𝑘𝑃𝑎) (0.19
𝑚3
𝑘𝑔
)(0.5
𝑘𝑔
𝑠
) − (1,000 𝑘𝑃𝑎) (0.95
𝑚3
𝑘𝑔
)(0.5
𝑘𝑔
𝑠
)
∆𝑊𝑓 = 19
𝑘𝐽
𝑠
∆𝑈 = 90
𝑘𝐽
𝑘𝑔
(0.5
𝑘𝑔
𝑠
)
∆𝑈 = 45
𝑘𝐽
𝑠
∆𝐻 = 45
𝑘𝐽
𝑠
+ 19
𝑘𝐽
𝑠
∆𝐻 = 64
𝑘𝐽
𝑠
= 64 𝑘𝑊
−58
𝑘𝐽
𝑠
= −6 𝑥 10−3
𝑘𝐽
𝑠
+ 0 + 64
𝑘𝐽
𝑠
+ 𝑊𝑠
𝑊𝑠 = −121.9940
𝑘𝐽
𝑠
= −121.9940 𝑘𝑊