MAE 204 Thermodynamics HW #6: Piston-Cylinder, Mixing Chamber Analysis

This document presents the complete solutions for MAE 204 Homework #6, a mechanical engineering assignment focusing on thermodynamics. The assignment includes two primary problems. The first problem analyzes air compression within a closed piston-cylinder system, requiring students to draw a schematic, state assumptions, and write the energy balance equation. It then asks for the determination of work and heat transfer under both isothermal and polytropic process conditions (n=1.3). The second problem involves a well-insulated water mixing chamber operating at steady state, with two inlets and one outlet. Students must determine the phase description, temperature, and specific volume of the outlet water, considering given mass flow rates, temperatures, pressures, and the use of a mixing paddle. The solutions demonstrate the application of thermodynamic principles, including energy balance, and the use of thermodynamic tables for property determination.

MAE 204 Fall 2019 HW #6
Due at 4:00pm on 10/8/19. Turn in during lecture or to Jarvis 326.
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Problem 1 (15 pts)
Consider air being compressed in a closed piston-cylinder from an initial specific volume of 1 m3/kg and
100 kPa to 1 MPa. Assume that specific heats vary with temperature.
a) Draw a schematic of the process, state appropriate assumptions, and write the energy balance
expression using the assumptions.
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The following air standard-assumptions are made:
I. The liquid used is air
II. Ideal gas is used in the system
III. It’s a assume that the process used is reversible
The Energy balance equation is shown bellow
∆ U =−∆ KE+ ∆ PE=Q−W
Q= heat transfer to the system
W=work done
Air
Piston
System
Boundary
Part a
Air
Piston System
boundary
Part b

∆ U =change of internal enegy
b) Determine the work into the system and heat out of the system, both in kJ/kg, assuming the process
is isothermal.
W=∫
1
2
pdv=¿∫
1
2
c
v dv= p 1 v 1 ln ( v 2
v 1 ) n=1¿
W=p1v1ln ( v 2
v 1 ¿
=100* 1ln ( 0.1
1 ¿=-230.259kj/kg
Work output is equal to heat out of the system
W¿ p 2 v 2 ln ( v 1
v 2 )=1000∗1 ln ( 1
0.1 )=230.259 kj/kg

c) Determine the work into the system and heat out of the system, both in kJ/kg, assuming the process
is polytrophic with n =1.3.
For the polytrophic process:
Wpolytonic=∫
1
2
Pdv=∫
1
2
C V−ndV = P 2 v2− p 1 v 1
1−n
Our p1=100kpa
P2=1000kpa
V1=1 M3
V2=0.1 M 3
N=1.3
W= ∫ pdv
p vn=100 kpa . m3=constant
p= 100 kpa . m3
v1.3
W= ∫
v1
v2
100 kpa . m3
v1.3 dv
W=(100kpa. m3 ¿∫
1
0.1
1
v1.3 dv
W=(100kpa.m3 ¿[lnv1.3] 0.1
1
100 ×−o .95
Ans=-95KJ
Problem 2 (15 pts)
A well-insulated water mixing chamber with two inlets and one outlet operates at steady state. One inlet
is at 5 MPa and 800◦C at a mass flow rate of 5 kg/s while the other inlet is at 60◦C and 2 kg/s. Assuming a

5-kW mixing paddle is used to mix the water and all pressures are equal, determine the phase
description, temperature, and specific volume of the outlet water.
1 Energy entering the system = Energy leaving the system
M1 +M2=M
Mh 1+ M 2 h 2=M 3 h 3
Energy in =Energy out
Mass 1 + Mass2=mass of the mixtures
∆ ke=0
∆ pe=0
Q=0
h=enthalpy
W=5KW
Mass of mixtures =mass of steam +mass of liquid
Q−W =(mass flow rate out × entalpy out)+¿
And Q=0
−¿W =¿ ¿
_W= ( Ms 1+ Ms 2 ) hout− ( Ms 1h 1+ms 2 h 2 )
800℃=4137.7kj
60=℃=1213.4 kj
-5 ¿ ( 5+2 ) h−ou t ( 5 ×4137.7 ) + ( 2 ×1213.4 )
7h-out=23110.3
h-out=3301.4kj/kg
From enthalpy table
2 Temperature is =420℃
3 Specific volume
pressure= V 2
ρ
V 2=5000 kpa × 1000 m3
kg

V=√5000000
=2236.07m3 /kg
REFFENCES
Oono, Y., & Paniconi, M. (1998). Steady state thermodynamics. Progress of Theoretical Physics
Supplement, 130, 29-44.