This document provides information about a group presentation on the steady flow energy equation (SFEE). It defines steady and unsteady flow, derives the SFEE, and discusses its applications to nozzles, diffusers, steam turbines, throttling valves, heat exchangers, and systems with multiple outlets. It also covers the non-flow energy equation and provides examples of using the SFEE to solve problems involving turbines, duct flow, isentropic expansion in a turbine, air compression in a nozzle, and isentropic expansion in a nozzle.

2. 
Group members:
 Muhammad Bilal Hanif ( ME151004)
 Hasnain Ahmad (ME151009)
 Muhammad Bilal Ashraf ( ME151026)
 Nadeem Iqbal ( ME151049)
 Sufian Arshad ( ME151050)
2
GROUP # 04

3. Steady Flow Energy
Equation
Presentation of
Thermodynamics-I
Topic on
Assigned by:
Dr. Ijaz Khan

4. What is SFEE?
A steady flow process is one in
which matter and energy flow
steadily.
ṁin = ṁout
Unsteady flow is:
ṁin ≠ ṁout

5. Derivation of Steady Flow Equation
By the conservation of energy states:
Energy(in) = Energy(out)
  )
2
()
2
( 1
2
1
112
2
2
22 gzhmgzhmWQ
VV

0)
2
()
2
( 2
2
2
221
2
1
11   gzhmgzhmWQ
VV


6. mmm   12since
)
2
()
2
( 1
2
1
12
2
2
2 gzhmgzhmWQ 
VV

Then;
)](
2
[ 12
2
1
2
2
12 zzghhmWQ 


VV

)(
2
12
2
1
2
2
12 zzghhWQ 


VV
Dividing the equation by m yields

7. Uniform-Flow Processes
At any instant during the process,
the state of control volume is
uniform
The fluid properties may differ
from one inlet or exit to another,
 But the fluid flow at an inlet or
exit is uniform and steady

8. Applications of SFEE:
Nozzles and Diffusers:
 Nozzles and diffusers are properly shaped
ducts
 used to increase or decrease the speed of the
fluid
q = Δh + ΔV2 / 2 + gΔz + w
q = 0 Δz = 0 w = 0
21
2
2
1
then hh V
)(2 21 hh V

9. Steam Turbine:
from q = Δh + ΔV2 / 2 + gΔz + ws
if q = 0; Δc = 0; Δz = 0
then ws = -Δh
= h1 - h2
 A turbine is a device
 With rows of blades mounted on a shaft
 Could be rotated about its axis

10. Throttling Valves:
From q = Δh + ΔV2 / 2 + gΔz + ws
as q = 0; ws = 0; z = 0 ; ΔV2 / 2 =0
then Δh =0
That is h1 = h2
A throttling valve is a device
Used to cause a pressure drop in a flowing
fluid.
It does not involve any work.

11. Heat exchangers:
From q = Δh + ΔV2 / 2 + gΔz + ws
as q = 0; ws = 0; Δ z = 0;ΔV2 / 2 =0
Δh =0
That is ∑hin = ∑ hexit
It is a device
Hot fluid stream exchanges heat
with a cold fluid stream
Without mixing with each other

12. SFEE for More Than One Outlet:
Energy(in) = Energy(out)
See
Blow
Pic
.
.
.
.
.
)(
2
213
2
2
1
2
3
213 zzzg
V
hhhWQ 


VV
1
2
3
1
2
3
  )(
2
)(
2
3
2
3
321
2
2
2
1
21 zghWzzg
VV
hhQ 




 

V

13. Non-Flow Energy Equation:-
When the fluid in closed system
It is not passing through the
system boundary
The flow terms in SFEE will not
apply
In derivation of it u neglect PV
and
 velocity

14. Equation of non-Flow Equation:-
U1 + Q = U2 + W
Q = (U2 –U1) + W
Q = ∆U + W
Hence,
Heat Transferred the boundary of system
= change of internal energy + work done

15. Problem # 01 :-
Steam enter a turbaine with velocity of 16m/sec and sp. enthalpy 2990kj/kg.
The steam leaves the turbine with velocity of 37m/s and sp. enthalpy
2530kJ/kg . The heat lost to the surrounding as the steam passes through the
turbine is 25kJ/kg. The steam flow rate is 324000kg/hour. Determine the work
out in kw (power).
Given;
ṁ = 32400kg/hr
P = w/m x ṁ
By equation of SFFEE , we will find w/m
and in above equation we get power

16. Bernoullie’s Equation:
 By the help SFEE we can derive Bernoullie’s
equation
 If there is no change in internal energy
 Q = 0 ,
 W = 0
Then Equation will remain of Bernoullie’s Eq.

17. Problem# 02
 Steam flows along a horizontal duct . At one point in the duct
the pressure of steam is 1 bar and the temperature is 400℃. At
a second point some distance from first , the pressure is 1.5 bar
and temp. is 500℃. Assuming flow is be friction less and
adiabatic . Determine whether the flow is accelerating or
decelerating.
 Solve:
 In SFEE we put the values like w = 0, Q
= 0, PE = 0
 Then we will get ;
 h2-h1 = -∆𝑉2
2
 So flow is decelerating as we negative value.

18. Problem # 03
 Steam is expanded isentropically in a turbine
from 30 bar 400℃ 𝑡𝑜 4 𝑏𝑎𝑟. Calculate the
work done per unit mass flow of steam.
Neglect changes as K.E and P.E.
 Solve:
 Q = 0, K.E = P.E = 0,
 Hence ;
 W = m [h2-h1]
 By table , we get h1 = 3230.9kJ/kg, h2 =
2750kJ/kg & m = 1
 Hence we get;
 Answer by putting values in above Eq.

19. Problem# 04
 A compressor takes in air at 1 bar and 20℃.and discharge into line. The average
air velocity in the line at appoint close the discharge is 0.7m/s and discharge
pressure is 3.5 bar. Assuming isentropically, calculate the work input to
compressor .Assume that air inlet velocity is very small.
 Slove:
 By putting the value Q=0 & P.E=0 in SFEE Eq. we get this relation:
-W=𝑚{∆ℎ +
∆𝑉2
2
} then
 -W=𝑚{𝐶𝑝. ∆𝑇 +
∆𝑉2
2
}
 T1 and P1 and P2 is given by this we get T2:
 T2=T1(
𝑃2
𝑃1
)
𝑛−1
𝑛
 So put this values in above Eq. and Required
Answer.

20. Problem #05
 Air is expanded isentropically in a nozzle from 13.8 bar and 150℃. To
a presume of 6.9 bar. The inlet velocity to the nozzle is very small
and process occur under steady flow. Calculate the exit velocity from
the nozzle knowing that the nozzle is laid in a horizontal plane and
that the inlet is 10m/s.
Solve:
 By putting these Q = 0 , P.E = 0, w = 0 .So SFEE is:
 0= m[Cp(T2-T1)+
∆𝑉2
2
]
 By the calculating T2 from
T2=T1(
𝑃2
𝑃1
)
𝑛−1
𝑛
 Then putting values in above Eq.
 Get Answer

21. THANKS for
Bearing Us.
From our side it is
enough.