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Group members: Muhammad Bilal Hanif ( ME151004) Hasnain Ahmad (ME151009) Muhammad Bilal Ashraf ( ME151026) Nadeem Iqbal ( ME151049) Sufian Arshad ( ME151050)

GROUP # 04

Steady Flow Energy Equation

Presentation of

Thermodynamics-I

Topic on

Assigned by:

Dr. Ijaz Khan

What is SFEE?A steady flow process is one in

which matter and energy flow steadily.

ṁin = ṁout

Unsteady flow is:

ṁin ≠ ṁout

Derivation of Steady Flow Equation By the conservation of energy states:

Energy(in) = Energy(out)

)2

()2

( 1

21

112

22

22 gzhmgzhmWQ VV

0)2

()2

( 2

22

221

21

11 gzhmgzhmWQ VV

mmm 12since

)2

()2

( 1

21

12

22

2 gzhmgzhmWQ VV

Then;

)](2

[ 12

21

22

12 zzghhmWQ

VV

)(2 12

21

22

12 zzghhWQ

VV

Dividing the equation by m yields

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

Applications of SFEE:Nozzles and Diffusers: Nozzles and diffusers are properly shaped

ducts used to increase or decrease the speed of the

fluidq = Δh + ΔV2 / 2 + gΔz + w

q = 0 Δz = 0 w = 0

212

21then hh V

)(2 21 hh V

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

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.

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

SFEE for More Than One Outlet:

Energy(in) = Energy(out)

SeeBlow Pic.....

)(2 213

22

123

213 zzzgVhhhWQ

VV

1

2

3

1

2

3

)(2

)(2 3

23

321

22

21

21 zghWzzgVVhhQ

V

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

Equation of non-Flow Equation:- U1 + Q = U2 + W

Q = (U2 –U1) + W

Q = ∆U + WHence,Heat Transferred the boundary of system = change of internal energy + work done

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/hrP = w/m x ṁ By equation of SFFEE , we will find w/m and in above equation we get power

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.

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 = - So flow is decelerating as we negative

value.

Problem # 03 Steam is expanded isentropically in a

turbine from 30 bar 400Calculate 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.

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=} then -W=} T1 and P1 and P2 is given by this we get T2: T2=T1 So put this values in above Eq. and Required

Answer.

Problem #05 Air is expanded isentropically in a nozzle from 13.8 bar and

150To 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)+ ] By the calculating T2 from

T2=T1 Then putting values in above Eq. Get Answer

THANKS forBearing Us.

From our side it is enough.