problem 3 - iowa state university...problem 3.137 carbon dioxide (co 2) is compressed in a...
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PROBLEM 3.136
PROBLEM 3.137
Carbon dioxide (CO2) is compressed in a piston-cylinder assembly from p1 = 0.7 bar, T1 = 280 K
to p2 = 11 bar. The initial volume is 0.262 m3. The process is described by pV1.25
= constant.
Assuming ideal gas behavior and neglecting kinetic and potential energy effects, determine the
work and heat transfer for the process, each in kJ, for (a) constant specific heats evaluated at 300
K, and (b) data from Table A-23. Compare the results and discuss.
KNOWN: Data are provided for the polytropic compression of carbon dioxide in a piston-
cylinder assembly.
FIND: Determine the work and heat transfer for the process using (a) constant specific heats, (b)
data from Table A-23.
SCHEMATIC AND GIVEN DATA:
ANALYSIS: Work is evaluated from W12 =
=
(See Example 2.1 for
details). For the polytropic process
V2 =
V1 =
(0.262 m3) = 0.02892 m
3
The work is
W =
= 53.89 kJ (in)
carbon
dioxide
(CO2)
p1 = 0.7 bar
T1 = 280 K
V1 = 0.262 m3
pV1.25
= constant
p2 = 11 bar
p
v
.
. 0.7 bar
11 bar
280 K
T2
ENGINEERING MODEL: (1) The carbon
dioxide within the piston-cylinder assembly is
a closed system. (2) The process is polytropic,
with pV1.25 = constant. (3) Volume change is
the only work mode. (4) Kinetic and potential
energy effects can be neglected. (5) The
carbon dioxide is modeled as an ideal gas with
(a) constant specific heats, and (b) with
variable specific heats.
PROBLEM 3.137 (CONTINUED)
Reducing the energy balance gives: Q12 = m(u2 – u1) + W12
The mass is
m = p1V1/RT1 =
= 0.3467 kg
and the final temperature is
T2 = p2V2/mR =
= 485.7 K
(a) From Table A-20 for CO2 at 300 K; cv = 0.657 kJ/kg∙K. Using constant specific heats, the
energy balance becomes
Q = mcv(T2 – T1) + W12 = (0.3467 kg)(0.657 kJ/kg∙K)(485.7 – 280)K + ( 53.89 kJ)
= 7.035 kJ (out)
(b) Using data from Table A-23: 1 = 6369 kJ/kmol and interpolating; 2 = 13004 kJ/kmol
Q = m
+ W12
= (0.3467)
+ ( kJ) = kJ (out)
In this case, the assumption of constant specific heat results in a value for Q that differs from the
variable specific heat value obtained using data from Table A-23 by approximately 334 %. The
assumption of constant specific heat evaluated at 300 K is questionable in this case.
PROBLEM 3.132
Revised 10-14
Revised 10-14
Problem 3.139 Air contained in a piston-cylinder assembly undergoes two processes in series ,
as shown in Fig. P3.139. Assuming ideal gas behavior for the air, the work and heat transfer for
the overall process, each in kJ/kg.
KNOWN: Air contained in a piston-cylinder assembly undergoes two processes in series.
FIND: For the overall process of the air, find the work and the heat transfer, in kJ/kg.
Using the ideal gas equation of state, the mass is
m = p1V1/RT1 =
= 0.2323 kg
So, W12/m = (40 kJ)/(0.2323 kg) = 172.2 kJ/kg
For 2-3, T is constant and p = mRT/V, so
W23 =
= mRT
= mRT2 ln (V3/V2)
0
1
2
0 0.1 0.2 0.3 0.4 0.5 0.6
ANALYSIS: For 1-2, the work determined using Eq. 2.17 is
W12 =
= p(V2 – V1) = (2 bar)(0.3 – 0.1) m
3/kg
= 40 kJ
pressure is constant
V (m3)
p
(bar)
1
3
2
ENGINEERING MODEL: (1) The air
in the piston-cylinder is the closed
system. (2) The air is modeled as an
ideal gas. (3) Kinetic and potential
energy effects are ignored.
Air
. .
.
Isothermal
process
T1 = 300 K
Revised 10-14
Revised 10-14
Problem 3.139 (Continued)
Process 1-2 is at constant pressure, so
= p1 = p2 =
→ T2 = (V2/V1)T1 = (0.3/0.1)(300 K) = 900 K
Thus
W23/m = RT2 ln (V3/V2) = (8.314/28.97 kJ/kg∙K)(900 K) ln (0.6/0.3) = 179.0 kJ/kg
The total work is W/m = W12/m + W23/m = 172.2 + 179.0 = 351.2 kJ/kg
An energy balance for the overall process is obtained as follows: ΔKE + ΔPE + ΔU = Q – W
Q = ΔU + W = m(u3 – u1) + W → Q/m = (u3 – u1) + W/m
With data from Table A-22
Q/m = (674.58 – 214.07)kJ/kg + 351.2 kJ/kg = 811.7 kJ/kg
1 Note that the heat transfers Q12 and Q23 could be determined from the respective energy
balances, as follows
Q12 = (u2 – u1) + W12/m = (674.58 – 214.07) + (172.2) = 632.7 kJ/kg
and
Q23 = (u3 – u2) + W23/m = (0) + 179.0 = 179.0 kJ/kg
So, the total for the overall process is Q = Q12 + Q23 = 632.7 + 179.0 = 811.7 kJ/kg
which is the same result as above, as expected.
1
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