Example Problem with Complete Solution

7E-3 : Work and ΔS for IGs Undergoing Isothermal, Polytropic and Adiabatic Processes 8 pts
Argon gas is compressed in a piston and cylinder device from 20 psia and 55oF to 120 psia. The compression is internally reversible and the Argon behaves as an ideal gas with a constant heat capacity of CP = (5/2) R.
Determine the work in Btu/lbm and ΔS in Btu/lbm-oR and sketch the process path on both PV and TS Diagrams assuming the compression is…
a.) Polytropic with d = 1.5
b.) Adiabatic
c.) Isothermal
 
Read : For part (a) start with the equation for PV work for internally reversible, polytropic processes for ideal gases.      When determining DS assume the heat capacity is constant. Since argon is monatomic use Cp = (5/2) R.
For part (b) determine work by applying an energy balance (where Q = 0). Assume constant heat capacity   DU = CV DT where CV = CP - R and Cp = (5/2) R. Determine T2 from an entropy balance. Recall that internally reversible, adiabatic processes are also isentropic.
For part (c) start with the definition of PV work, substitute in the ideal gas EOS for pressure and integrate.  Remember the process is isothermal (this simplifies the analysis of both the work and change in entropy).
Given: P1 20 lbf / in2 P2 120 lbf / in2
T1 55 oF m 8.4 lbm
514.67 oR Part (a) d 1.5
Find: Part (a) - (c) : W ? Btu DS ? Btu/oR
Diagram:
Assumptions: 1 - As shown in the diagram, the system is the gas.
2 - The gas is modeled as an ideal gas.
3 - The compression is internally reversible.
4 - Boundary work is the only form of work that crosses the system boundary.
5 - Changes in kinetic and potential energies are negligible.
6 - Argon has a constant heat capacity of CP = (5/2) R.
R 1.986 Btu/lbmol-oR
MW 39.948 lbm/lbmol
Equations / Data / Solve:
Part a.) In part (a) we must determine the work and the change in entropy for a polytropic process.
For polytopic, internally reversible processes with ideal gases:
Eqn 1
The problem at this point is that we do not know T2.  But, we do know that the process is polytropic !
In Lesson 7E we learned that :
Eqn 2
We can solve Eqn 2 for T2 :
Eqn 3
Substitute Eqn 3 for T2 into Eqn 1 and rearrange the result to get :
Eqn 4
Now, we can plug values into Eqns 12 & 10 or Eqn 13 : n 0.2103 lbm
T2 935.22 oR
T2 475.55 oF
W12 -351.2 Btu
DS can be determined by applying the 2nd Gibbs Equation for
ideal gases:
Eqn 5
If we assume the heat capacity is constant:
Eqn 6
Since argon is a montomic gas we can assume :
Eqn 7
CP 4.965 Btu/lbmol-oR
Subtituting CP and other values into Eqn 6 yields : DS -0.12471 Btu/oR
Part b.) Here we must determine the work and the change in entropy for an adiabatic process.
The 1st Law for an adiabatic process with negligible changes in kinetic and potential energies is :
Eqn 8
We can evaluate the change in the internal energy using CV and :
Eqn 9
Assuming the heat capacity is constant, Eqn 9 simplifies to :
Eqn 10
The following relationship applies to ideal gases :
Eqn 11
We can combine this with Eqn 7 to get :
Eqn 12
CV 2.979 Btu/lbmol-oR
At this point, the only obstacle to using Eqns 18 & 16 to evaluate W12 is that we do not know T2.
We need to make use of the fact that the process is adiabatic and internally reversible to determine T2.
An adiabatic process that is also internally reversible is isentropic : DS 0 Btu/oR
We can use this fact with the 2nd Gibbs Equation for ideal gases with constant heat capacities to determine T2 as follows :
Eqn 13
Solve Eqn 13 for T2
using ΔS = 0 :
Eqn 14
Eqn 15
Eqn 16
Hey, we already KNEW this !
Eqn 17
Where :
Eqn 18
g 1.667
T2 1053.88 oR
Now, we can plug T2 into Eqn 10 and DU12 into Eqn 8 : DU12 40.21 Btu/lbm
W12 -337.76 Btu
Part c.) Determine the work from the definition
of boundary work :
Eqn 19
For an ideal gas substitute P = nRT/V into Eqn 19 :
Eqn 20
Integrate Eqn 2 (the process is isothermal, T1 = T2 = T) :
Eqn 21
We don't know V1 or V2 but we can determine the values from the Ideal Gas EOS (T1 = T2 = T):
Eqn 22
Eqn 23
Dividing Eqn 22 by Eqn 23 we obtain :
Eqn 24
Now, substitute Eqn 24 back into Eqn 21 to get :
Eqn 25
Plug values into Eqn 25 : W12 -385.10 Btu
DS can be determined by applying the 2nd Gibbs Equation for ideal gases:
Eqn 26
For an isothermal process Eqn 26 reduces to:
Eqn 27
Now, we can plug values into Eqn 27 : DS -0.74824 Btu/oR
Verify: The ideal gas assumption needs to be verified.
We need to determine the specific volume at each state and check if
(Argon is a noble gas).
Solving the Ideal Gas EOS for molar volume yields :
Use : R 10.7316 psia-ft3 / lbmol-oR V1 = V2C 276.16 ft3/lbmol
V2A 83.64 ft3/lbmol
V2B 94.25 ft3/lbmol
The specific volume at each state is greater than 80 ft3/lbmol, therefore the ideal gas assumption is reasonable.
Answers : a.) W12 -351 Btu
DS -0.125 Btu/oR
b.) W12 -338 Btu The isentropic process requires the least work !
DS 0 Btu/oR
c.) W12 -385 Btu The isothermal process requires the most work !
DS -0.748 Btu/oR
How can ΔS be negative in parts (a) and (c) ?
Heat transfer from the system to the surroundings occurs. So, although ΔSsystem < 0, ΔSsurr > 0 by an even larger amount so that ΔSuniverse > 0 and the 2nd Law is not violated.