Assignment On Energy Balance Equation

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.