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Solution– First Law for Ideal Gases
A mass of air which acts as an ideal gas with R = 0.287 kJ/kg·K fills a piston-cylinder apparatus with an initial pressure and temperature of P 1 = 0.8 MPa and T 1 = 400K, respectively. The air then undergoes a three step path where the initial and final states are the same:
a. heating to a temperature of T 2 = 1400 K following the path equation P = a + b/V where a = 2 MPa and b = – 0.03 MPa·m^3.
b. a constant pressure decrease in volume to V 3 = V 1 , the initial volume.
c. a constant volume decrease in pressure to P 4 = P 1 , the initial pressure.
1. Find the heat transfer for path a if air
has a constant heat capacity of cv = 0.718 kJ/kg·K and cp = 1.005 kJ/kg·K.
We can use the path equation to find the work for part of the path a as pathPdV.
1
2 2 1 ln
2
1
V
V
dV aV V b V
b W PdV a
V
patha V
a
We can use the path equation to find the initial volume:
3
3
1
1 0.^025
m MPa MPa
MPa m
P a
b V
We can find the mass from the ideal gas equation of state and the given initial state data.
kg
kJ
MPa m K kg K
kJ
MPa m
RT
PV
m 0. 17422
3
3
1
1 1
At the final state the pressure and volume must satisfy the path equation and the equation of state. We can solve these two equations simultaneously for the final volume.
a
mRT b aV b mRT V V
mRT
V
b P a
2 2 2 2 2
2
2
2
3
3
3
2 0.^05
m MPa
MPa m kJ
MPa m K kg K
kJ kg
V
We can now find the work over path a.
0.
0.
0.
0.
1.
1.
1.
1.
2.
2.
0.00 0.01 0.02 0.03 0.04 0.
Pressure, P (MPa)
Volume, V (m^3 )
Quiz Four Paths
Path a Path b Path c
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MPa m MJ kJ
m
m MPa m m MPa m V
V
Wa aV V b
ln 2 0. 05 0. 025 0. 03 ln
3
3
3 3 3 3
1
2 2 1
The heat transfer along this path can be found from the first law: Qa = Ua + Wa = m(u 2 – u 1 ) + Wa. We can find the internal energy change for the ideal gas with constant heat capacity as (u 2 – u 1 ) = cv(T 2 – T 1 ). This gives the following result for the heat transfer for path a.
K K kJ
kg K
kJ Q (^) a mu u Wa mcvT T Wa kg 1400 400 29. 2067
2 1 2 1 0.^1
Qa = 154.3 kJ
2. Find the overall heat capacity for the complete path using the same constant heat capacities as in problem 1.
For the entire path the initial and final states are the same so the internal energy change is zero. (U depends only on the state; it the initial and final states are the same, the value of u at both states is the same so U = 0) This gives Q = U + W = 0 + Wa + Wb + Wc for the overall path. We have found Wa = 29.2067 kJ and we know that Wc = 0 for the constant volume step in part c of the path. The second part of the path, at constant pressure P 3 = P 2 = Pconst is found as follows: Wc = Pconst (V 3 – V 2 ). We have to find P 2 , which can be found either from the path equation or from the eqution of state. Here we will to both as a check on our calculation of V 2.
MPa m
kJ
MPa m K kg K
kJ kg
V
mRT P
MPa m
MPa m MPa V
b P a
3
3
2
2 2
3
3
2
2
So the work over path b is Wb = Pconst (V 3 – V 2 ) = (1.4 MPa)(0.025 m^3 – 0.05 m^3 ) = – 0. MPa·m 3 = – 0.035 MJ = – 35 kJ. The heat transfer for the overall path is then found to be Qtotal =
Wtotal = Wa + Wb + Wc = 29.2067 kJ + (–35 kJ) + 0 or Qtotal = – 5.794 kJ
3. Repeat problems 1 and 2, with all the data given above, but use the attached air tables instead of assuming constant heat capacity.
For all steps along the path, use of the air tables to compute the internal energy does not affect the results for pressure, volume and work found from the path equation and the equation of state. Thus, the work will be the same for a given path, regardless of how we compute the internal energy change.
For path a, we have the following equation for the first law
kJ
kg
kJ
kg
kJ Q (^) a mu u Wa muT uT Wa kg 29. 2067
2 1 2 1 0.^17422
Qa = 173.0 kJ
For the total path, where the initial and final states are the same, U, is always zero; as noted above, the work does not change. With no change in work or internal energy difference, the total
heat transfer does not change: Qtotal = – 5.794 kJ
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