fa khara ny
Fall 2012/2013.
Level: 3 Semester: 5 Code: 230
Title: Thermodynamics 1
PHAROS UNIVERSITY
FACULTY OF ENGINEERING
MECHANICAL ENGINEERING Dept.
Date: 24/11/2012
9th week Exam
Duration: 1
Hour
Steam Tables and Charts are allowed to be used.
Illustrate your answer with neat sketches, whenever it is possible.
1.
A rigid tank contains 10 kg of air at 150 kPa and
20oC. More air is added to the tank until the pressure
and temperature rise to 250 kPa and 30oC,
respectively. Show this filling process on p-V
diagram and determine the amount of air added to
the tank.
(Take for air cp = 1.005 kJ/kg-K, cv = 0.718 kJ/kg-K)
m2-m1
m1 = 10 kg
p1 = 150 kPa
T1 = 20 oC
p2 = 250 kPa
T2 = 30 oC
2.
A piston-cylinder device contains 0.8 kg of steam at
300oC and 1 MPa. Steam is cooled at constant
pressure until one-half of the mass condenses. Show
the process on p-v and T-s phase diagrams, and then
determine:
(a)
(b)
(c)
The final temperature of the steam.
The volume change through the cooling
process.
The amount of heat rejected through this
process
Q
3.
A piston-cylinder device, initially, contains 0.5 m3 of
nitrogen gas at 400 kPa and 27°C. An electric heater
within the device is turned on and is allowed to pass a
current of 2A for 5 min from a 120-V source.
Nitrogen expands at constant pressure, and a heat loss
of 2800 J occurs during the process. Represent this
process on p-v and T-s diagrams and determine:
2800 J
2A
120 V
a)
Mass of gas inside the cylinder.
b)
The added heat by electric heater.
c)
The final temperature.
( Take for nitrogen gas: Cp = 1.039 kJ/kg.K, Cv = 0.743 kJ/kg.K)
Exam Committee : 1. Prof Dr M. G. Wasel
2.
Page 1 of 4
fa khara ny
Problem 1
Given:
V1=V2, m1=10 kg, p1=150 kPa, T1=293 K, p2=250
kPa, T2=303 K, Cp=1.005 kJ/kg.K, Cv=0.718 kJ/kg.K
p
Required:
∆m = m2 – m1
p2
Solution:
𝑅 = 𝐶𝑝 − 𝐶𝑣 = 1.005 − 0.718 = 0.287
Since
𝑘𝐽
𝑘𝑔 𝐾
𝑝1 𝑉1 = 𝑚1 𝑅 𝑇1
Then
p1
𝑚1 𝑅 𝑇1 10 × 0.287 × 293
=
= 5.61 𝑚3
𝑝1
150
Since V2=V1
𝑉1 =
v 1= v 2
Then, applying equation of state at the end of filling
process yields to:
𝑚2 =
𝑝2 𝑉2
250 × 5.61
=
= 16.12 𝑘𝑔
𝑅 𝑇2 0.287 × 303
Mass of added air ∆m = m2 - m1 = 16.12 - 10 = 6.12 kg
Page 2 of 4
v
fa khara ny
Problem 2
p
Given:
m=0.8 kg, T1=300°C, p1=10 bar, p2=p1=10 bar,
mf,2=0.5 kg, mtotal, x2=0.5
T1 =300oC
Required:
p-V and T-s diagrams, T2, ∆V=V2-V1
from
tables
(B):
0.
and
1
2
=
bar,
T2 =180oC
x
Solution:
At
p1=10
10 bar
5
Ts@p=10bar=179.9°C
v
Since T1 > Ts , then the steam at condition 1
is superheated steam. From table (C) at p1=10
bar and T1=300°C:
3
condition
dryness fraction =
1
300oC
v1=0.258 m /kg.
Considering
T
2,
𝑚𝑔
by
definition
of
180oC
10 bar 2
, the steam at condition
2 is wet steam and its dryness fraction x2=0.5.
0.5
From Tables (B) at p2=10 bar:
x=
𝑚𝑔 +𝑚𝑓
and
T2=Ts=179.9°C and vf=0.0011 m3/kg
s
3
vg=0.1943 m /kg
Accordingly;
𝑣2 = (1 − 𝑥2 )𝑣𝑓 + 𝑥2 𝑣𝑔 = 0.5(0.0011) + 0.5(0.1943) = 0.0977𝑚3 /𝑘𝑔
The change of volume through this cooling process:
∆𝑉 = 𝑉2 − 𝑉1 = 𝑚(𝑣2 − 𝑣1 ) = 0.8(0.0977 − 0.258) = −0.128𝑚3
Problem 3
Page 3 of 4
fa khara ny
Given:
V1 = 0.5 m3, p1 = p2 = 400 kPa, T1 = 27°C, I = 2 A, Volt = 120 V,
t = 5 min., Qlost = 2800 J =2.8 kJ,
Required:
a) mgas
b) Qelectric
𝐶𝑝 = 1.039
𝑘𝐽
𝑘𝑔.𝐾
, 𝐶𝑣 = 0.743
𝑘𝐽
𝑘𝑔.𝐾
c) T2
Solution:
(a)
𝑅 = 𝐶𝑝 − 𝐶𝑣 = 1.039 − 0.743 = 0.296
2800 J
2A
𝑘𝐽
𝑘𝑔. 𝐾
120 V
𝑝1 𝑉1 = 𝑚 𝑅 𝑇1
𝑚=
𝑝1 𝑉1
𝑅 𝑇1
=
400×0.5
0.296×(273+27)
= 2.252 𝑘𝑔
(b)
𝑄𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 = 𝑉 × 𝐼 × 𝑡 = 2 × 120 × (5 × 60)
= 72000 𝐽 = 72 𝑘𝐽
p
400
kPa
1
2
(c)
v
𝑄12 − 𝑊12 = 𝑈2 − 𝑈1
𝑄12 = 𝑄𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 − 𝑄𝑙𝑜𝑠𝑡 = 72 − 2.8 = 69.2 𝑘𝐽
2
𝑊12 = ∫ 𝑝 𝑑𝑉 = 𝑝(𝑉2 − 𝑉1 )
T
2
1
𝑄12 = 𝑊12 + (𝑈2 − 𝑈1 ) = 𝐻2 − 𝐻1
= 𝑚 𝐶𝑝 (𝑇2 − 𝑇1 )
1
69.2 = 2.252 × 1.039 × (𝑇2 − 300)
s
𝑇2 = 329.6 𝐾 = 56.6 °𝐶
Page 4 of 4