1. NON-FLOW PROCESS J2006/4/1
UNIT 4
NON - FLOW PROCESS
OBJECTIVES
General Objective : To understand and apply the concept of non-flow process in
thermodynamics
Specific Objectives : At the end of the unit you will be able to:
 define and describe the differences between the flow and the
non-flow processes
 identify heat and work in reversible process
 define and calculate the following non-flow processes :
• constant temperature (Isothermal)
• adiabatic

2. NON-FLOW PROCESS J2006/4/2
INPUT
4.0 INTRODUCTION
What is a non
flow
process?
O
nce a fluid has entered a system, it may be possible for it to undergo a series
of processes in which the fluid does not flow. An example of this is the
cylinder of an internal combustion engine. In the suction stroke, the working
fluid flows into the cylinder in which it is then temporarily sealed. Whilst the
cylinder is sealed, the fluid is compressed by the piston moving into the cylinder,
after which heat energy is supplied so that the fluid possesses sufficient energy to
force the piston back down the cylinder, causing the engine to do external work. The
exhaust valve is then opened and the fluid is made to flow out of the cylinder into the
surroundings. Processes which are undergone by a system when the working fluid
cannot cross the boundary are called non-flow process. This process occurs during
the compression and the working stroke as mentioned in the above example (refer to
Fig. 4.0).
4.1 Differences Between The Flow and Non-flow processes
4.1.1 Flow Process
SUCTION WORKING
STROKE COMPRESSION STROKE
EXHAUST
STROKE STROKE
Figure 4.0 The cycle of an internal combustion engine

3. NON-FLOW PROCESS J2006/4/3
In an open system, not only the energy transfers take place across the
boundary, the fluid may also cross the boundary. Any process undergone by
an open system is called a flow process. This process may be sub-divided
into an unsteady flow process and steady flow process. The general equation
is shown below,
C12 C2
gZ1 + u1 + P1v1 + + Q = gZ 2 + u 2 + P2 v 2 + 2 + W
2 2
4.1.2 Non-flow process
In a close system, although energy may be transferred across the boundary in
the form of work energy and heat energy, the working fluid itself never
crosses the boundary. Any process undergone by a close system is referred to
as the non-flow process.
If the fluid is undergoing a non-flow process from state (1) to state (2) then
the terms from the general equation for p1V1 and p2V2 (which represent the
amount of work energy required to introduce and expel the fluid from the
system) will be zero, since the fluid is already in the system, and will still be
in the system at the end of the process. For the same reason, the changes in
kinetic and potential energies of the fluid will also be zero. Thus the equation
becomes
U1 + Q = U2 + W
or, U2 – U1 = Q –W (4.1)
In words, this equation states that in a non-flow process, the change in the
internal energy of the fluid is equal to the nett amount of heat energy
supplied to the fluid minus the nett amount of work energy flowing from the
fluid.
This equation is known as the non flow energy equation, and it will now be
shown how this may apply to the various non-flow processes.
4.2 Constant temperature (Isothermal) process (pV = C)
If the change in temperature during a process is very small then that process may be
approximated as an isothermal process. For example, the slow expansion or
compression of fluid in a cylinder, which is perfectly cooled by water may be
analysed, assuming that the temperature remains constant.

4. NON-FLOW PROCESS J2006/4/4
P
1
W
2
W
v
v1
Q
v2
Figure 4.2 Constant temperature (Isothermal) process
The general relation properties between the initial and final states of a perfect gas are
applied as follows:
p1V1 p 2V2
=
T1 T2
If the temperature remains constant during the process, T1 = T2 and the above
relation becomes
p1V1 = p 2V2
From the equation we can know that an increase in the volume results in a decrease
in the pressure. In other words, in an isothermal process, the pressure is inversely
proportional to the volume.
Work transfer:
Referring to the process represented on the p – V diagram in Fig.4.2 it is noted that
the volume increases during the process. In other words the fluid is expanding. The
expansion work is given by
2
W = ∫ pdV
1
2
c
=∫ dV (since pV = C, a constant)
1
V

5. NON-FLOW PROCESS J2006/4/5
2
dV
= c∫
1
V
2
dV
= p1V1 ∫
1
V
V2  larger volume 
= p1V1 ln  smaller volume 
V1  
V2
= mRT1 ln (since p1V1 = mRT1)
V1
p1 V2 p
= mRT1 ln (since = 1) (4.2)
p2 V1 p2
Note that during expansion, the volume increases and the pressure decreases. On the
p – V diagram, the shaded area under the process line represents the amount of work
transfer.
Since this is an expansion process (i.e. increasing volume), the work is done by the
system. In other words the system produces work output and this is shown by the
direction of the arrow representing W.
Heat transfer:
Energy balance to this case is applied:
U1 + Q = U2 + W
For a perfect gas
U1 = mcvT1 and U2 = mcvT2
As the temperature is constant
U1 = U2
Substituting in the energy balance equation,
Q=W (4.3)
Thus, for a perfect gas, all the heat added during a constant temperature process is
converted into work and the internal energy of the system remains constant.
For a constant temperature
process
W==Q
or
W==Q

6. NON-FLOW PROCESS J2006/4/6
4.3 Adiabatic process (Q = 0)
If a system is thermally well insulated then there will be negligible heat
transfer into or out of the system. Such a system is thermally isolated and a
process within that system may be idealised as an adiabatic process. For
example, the outer casing of steam engine, steam turbines and gas turbines
are well insulated to minimise heat loss. The fluid expansion process in such
machines may be assumed to be adiabatic.
P
1
W
2
W
v
v1
Thermal insulation
v2

7. NON-FLOW PROCESS J2006/4/7
Figure 4.3 Adiabatic (zero heat transfer) process
For a perfect gas the equation for an adiabatic process is
pVγ = C
Cp
where γ = ratio of specific heat =
Cv
The above equation is applied to states 1 and 2 as:
p1V1γ = p 2V2γ
γ
p 2  V1 
=  (4.4)
p1 V2 
Also, for a perfect gas, the general property relation between the two states is given
by the equation below
p1V1 p 2V2
= (4.5)
T1 T2
By manipulating equations 4.4 and 4.5 the following relationship can be determined:
γ −1
γ −1
T2  p 2  γ V 
(4.6)
=  =  1
T1  p1   V2 
By examining equations 4.4 and 4.6 the following conclusion for an adiabatic
process on a perfect gas can be drawn:
An increase in volume results in a decrease in pressure.
An increase in volume results in a decrease in temperature.
An increase in pressure results in an increase in temperature.
Work transfer:
Referring to the process represented on the p-V diagram (Fig.4.3) it is noted that the
volume increases during the process.

8. NON-FLOW PROCESS J2006/4/8
In other words, the fluid expanding and the expansion work is given by the formula:
2
W = ∫ pdV
1
2
c
=∫ γ
dV (since pVγ = C, a constant)
1V
2
dV
= c∫ γ
1 V
p1V1 − p 2V2
= [larger pV- small pV] (4.7)
γ −1
Note that after expansion, p2 is smaller than p1. In the p – V diagram, the shaded area
under the process represents the amount of work transfer.
As this is an expansion process (i.e. increase in volume) the work is done by the
system. In other words, the system produces work output and this is shown by the
direction of the arrow representing W (as shown in Fig 4.3).
Heat transfer:
In an adiabatic process, Q = 0.
Applying an energy balance to this case (Fig.4.3):
U1 - W = U2
W = U1 – U2
Thus, in an adiabatic expansion the work output is equal to the decrease in internal
energy. In other words, because of the work output the internal energy of the system
decreases by a corresponding amount.
For a perfect gas, U1 = mcvT1 and U1 = mcvT1
On substitution
W = mcv(T1-T2) [larger T- smaller T] (4.8)
We know
cp- cv = R
or
R
cv =
γ −1

9. NON-FLOW PROCESS J2006/4/9
Substituting in equation 4.8
mR (T1 − T2)
W = (4.9)
γ −1
But, mRT2 = p2V2 and mRT1 = p1V1
Then the expression for the expansion becomes
p1V1 − p 2V2
W = (4.10)
γ −1
Referring to the process represented on the p-V diagram it is noted that during this
process the volume increases and the pressure decreases. For a perfect gas, equation
4.6 tells that a decrease in pressure will result in a temperature drop.
Example 4.1
In an industrial process, 0.4 kg of oxygen is compressed isothermally from
1.01 bar and 22o C to 5.5 bar. Determine the work done and the heat transfer
during the process. Assume that oxygen is a perfect gas and take the molecular
weight of oxygen to be M = 32 kg/kmole.
Solution to Example 4.1
Data: m = 0.4 kg; p1= 1.01 bar; t1= 22oC
p2 = 5.5 bar; W=? Q=?
From the equation
R
R= 0
M
8314
=
32
= 260 J/kgK
= 0.260 kJ/kgK

10. NON-FLOW PROCESS J2006/4/10
For an isothermal process
Work input,
p2
W = mRTln
p1
5.5
= 0.4 x 0.260 x ( 22 + 273) ln
1.01
= 52 kJ
In an isothermal process all the work input is rejected as heat.
Therefore, heat rejected, Q = W = 52 kJ
Example 4.2
In a thermally insulated reciprocating air compressor, air at 0.98 bar and 20oC
is compressed into one sixth of its original volume. Determine the pressure and
temperature of the air after compression. If the compressor cylinder contains
0.05 kg of air, calculate the required work input. For air, take γ = 1.4 and
cv = 0.718 kJ/kgK.
Solution to Example 4.2
Data : p1 = 0.98 bar; T1= 20 + 273 = 293 K
V2 1
= ; m = 0.05 kg; W = ?
V1 6
As the cylinder is well insulated the heat transfer is negligible and the process may
be treated as adiabatic.
Considering air as a perfect gas
γ
p 2  V1 
From equation 4.4, = 
p1 V2 
p2 = 0.98 x 61.4

11. NON-FLOW PROCESS J2006/4/11
= 12 bar
γ −1
T2 V 1 
From equation 4.6, = 
T1  V2 
T2 = 293 x 60.4
= 600 K
= 327oC
Re-writing equation 4.8 for an adiabatic compression process
W = mcv(T2-T1) [larger T- smaller T]
= 0.05 x 0.718 (600-293)
= 11 kJ

12. NON-FLOW PROCESS J2006/4/12
Activity 4
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE
NEXT INPUT…!
4.1 In the cylinder of a large engine, 1.0 kg of carbon dioxide at 527o C and
20 bar expands isothermally to a pressure of 1.4 bar. What is the final volume
of the gas?
Take R = 189 Nm/kgK for carbon dioxide.
4.2 1 kg of nitrogen (molecular weight 28) is compressed reversibly and
isothermally from 1.01 bar, 20oC to 4.2 bar. Calculate the work done and the
heat flow during the process. Assume nitrogen to be a perfect gas.
4.3 Air at 1.02 bar, 22oC, initially occupying a cylinder volume of 0.015 m3, is
compressed reversibly and adiabatically by a piston to a pressure of 6.8 bar.
Calculate the final temperature, the final volume, and the work done on the
mass of air in the cylinder.

13. NON-FLOW PROCESS J2006/4/13
Feedback to Activity 4
4.1 Data: m = 1.0 kg; T1= 527 + 273 = 800 K
p1 = 20 bar; p2= 1.4 bar; V2 = ?
Carbon dioxide is a perfect gas and we can apply the following characteristic
gas equation at state 1.
p1V1 = mRT1
mRT1
V1 =
p1
1x189 x800
=
20 x10 5
= 0.0756 m3
Applying the general property relation between state 1 and 2
p1V1 p 2V2
=
T1 T2
For an isothermal process T1 = T2
Hence,
p1V1 = p 2V2
20
V2 = x 0.0756
1 .4
V2 = 1.08 m3

14. NON-FLOW PROCESS J2006/4/14
4.2 Data: m=1kg; M= 28 kg/kmole p1 = 1.01 bar;
T1 = 20 + 272 = 293 K; p2 = 4.2 bar
From equation
R
R= 0
M
8.314
=
28
= 0.297 kJ/kgK
The process is shown on a p-v diagram below. When a process takes place
from right to left on a p-v diagram the work done by the fluid is negative.
That is, work is done on the fluid.
p
4.2
pV=C
1.01
v
From equation 4.2
p1
W = mRT1 ln
p2
1.01
= 1 x 0.297x293x ln
4.2
= -124 kJ/kg
For an isothermal process for a perfect gas,
Q = W = -124 kJ/kg

15. NON-FLOW PROCESS J2006/4/15
4.3 Data: p1=1.02 bar; T1=22 + 273 = 295 K;
v1= 0.015 m3; p2= 6.8 bar
From equation 4.6
γ −1
T2  p 2  γ
= 
T1  p1 
(1.4 −1) / 1.4
 6.8 
T2 = 295 x  
 1.02 
= 507.5 K
(where γ for air = 1.4)
i.e. Final temperature = 507.5 – 273 = 234.5oC
From equation 4.4
γ 1/ γ
p 2  V1  v1  p 2 
=  or = 
p1 V2  v 2  p1 
1 / 1.4
0.015  6.8 
∴ = 
v2  1.02 
i.e. Final volume
v2 = 0.0038 m3
For an adiabatic process,
W = u1 – u2
and for a perfect gas,
W = cv(T1- T2)
= 0.718(295-507.5)
= - 152.8 kJ/kg
i.e. work input per kg = 152.8 kJ
The mass of air can be found using equation pV = mRT
p1v1 1.02 x10 5 x0.015
∴ m= = = 0.018kg
RT1 0.287 x10 3195
i.e. Total work done = 0.0181 x 152.8 = 2.76 kJ

16. NON-FLOW PROCESS J2006/4/16
SELF-ASSESSMENT
You are approaching success. Try all the questions in this self-assessment section
and check your answers with those given in the Feedback to Self-Assessment on the
next page. If you face any problem, discuss it with your lecturer. Good luck.
1. 0.05 m3 of a perfect gas at 6.3 bar undergoes a reversible isothermal process
to a pressure of 1.05 bar. Calculate the heat flow to or from the gas.
2. Nitrogen (molecular weight 28) expands reversibly in a perfectly thermally
insulated cylinder from 2.5 bar, 200oC to a volume of 0.09 m3. If the initial
volume occupied was 0.03 m3, calculate the work done during the expansion.
Assume nitrogen to be a perfect gas and take cv = 0.741 kJ/kg K.
3. A mass of 0.05 kg of air at a temperature of 40oC and a pressure of 1 bar is
compressed adiabatically to 5 bar. Determine the following:
a) final temperature
b) final volume
c) work transfer
d) heat transfer
e) change in internal energy
4. A quantity of gas occupies a volume of 0.3 m3 at a pressure of 100 kN/m2 and
a temperature of 20oC. The gas is compressed isothermally to a pressure of
500 kN/m2 and then expanded adiabatically to its initial volume.
For this quantity of gas determine the following:
a) the heat received or rejected (state which) during the compression,
b) the change of internal energy during the expansion,
c) the mass of gas.

17. NON-FLOW PROCESS J2006/4/17
Feedback to Self-Assessment
Have you tried the questions????? If “YES”, check your answers now.
1. 56.4 kJ
2. 9.31 kJ
3. 222.7oC, 14230 cm3, 6.56 kJ input, 0 kJ, 6.56 kJ increase.
4. – 48.3 kJ (heat rejected), -35.5 kJ, 0.358 kg
CONGRATULATIONS!!!!
You can now move on to Unit
5…