Department of Mechanical Engineering
ME 322 – Mechanical Engineering
Thermodynamics
Lecture 17
Unsteady State (Transient) Analysis
What is a Transient Process?
A transient process occurs whenever the time derivative
related to the system gain (or loss) is not zero ...
Q W 

i

Vi 2
g 
m i  hi 
 zi  
2 gc gc 

 mi   m e 
i
e

e

Ve2
g  dE
m e  he 
 ze   G
2 gc gc  dt

dmsys
dt
Emptying
a tank
Filling a
tank
2
Department of Mechanical Engineering
ME 322 – Mechanical Engineering
Thermodynamics
Example – Problem 6.61
3
Example – Filling a Tank
Given: Filling a rigid tank with oxygen
T  20C
m1  0 kg
m 2  15 kg
Use the ideal gas
model for oxygen
properties
Q12
Find: The heat transfer required to fill the tank isothermally
4
T  20C
Example
m1  0 kg
The First Law for the system is,
m 2  15 kg
Q12
Q W 

i

Vi 2
g 
m i  hi 
 zi  
2 gc gc 


Q  m i hi 
e

Ve2
g  dEG
m e  he 
 ze  
2 gc gc  dt

dEG
dt
dU sys
dEG d
 U  KE  PE G 
dt
dt
dt
 Q  m i hi 
5
dU sys
dt
T  20C
Example
So far, the following expression for
the system has been developed,
Q  m i hi 
m1  0 kg
m 2  15 kg
Q12
dU sys
dt
Some observations ...
• The problem asks for the heat transferred (total energy)
but the First Law has a heat transfer rate
• How do I know what the mass flow rate into the tank is?
Is this a constant value?
• How do I represent the derivative on the RHS?
6
T  20C
Example
The mass flow rate in is contained in
the conservation of mass,
mi  me 
m1  0 kg
m 2  15 kg
dmsys
dt
Making this substitution into the First Law,
Q
dmsys
dt
hi 
dU sys
dt
Multiply both sides by dt,
Qdt  hi dmsys  dU sys
7
Q12
T  20C
Example
m1  0 kg
This equation can be integrated over
the time of the filling process,
 Qdt  
t
0
m2
m1
hi dmsys 
m 2  15 kg
Q12

U2
dU sys
U1
 Qdt  Q
t
12
0


m2
m1
U2
U1
8
hi dmsys  hi  m 2  m1 
dU sys  U 2  U1  m 2u2  m1u1
T  20C
Example
m1  0 kg
Substituting and rearranging,
m 2  15 kg
Q12  hi  m 2  m1   m 2u2  m1u1
Q12
Q12  m 2  u2  hi 
The oxygen is being modeled as an ideal gas. Therefore,
hi  ui  Pv
i i  ui  RTi
Substitution into the First Law results in,
Q12  m 2 u2   ui  RTi   
9
 Q12  m 2 RTi
T  20C
Example
Finally, doing the calculations ...
m1  0 kg
m 2  15 kg
Q12
Q12   m 2 RTi
What is Ti? There is a valve between the tank and the line.
However, we know that a valve is isenthalpic. Since enthalpy
is constant and the substance is being modeled as an ideal
gas, the throttling process is also isothermal. Therefore,

kJ 
Q12   15 kg   0.260
  20  273.15  K
kg-K 

Table C.13b
Q12  1143 kJ 
10
The heat transfer is from the oxygen
(the assumed direction was wrong)