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Energy Balances
S.GunabalanAssociate ProfessorMechanical Engineering DepartmentBharathiyar College of Engineering & TechnologyKaraikal - 609 609.e-Mail : [email protected]
Part - 2
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Steady Flow Energy equation
1 +1
2
+ 1 +
= 2 +
2
2
+ 2 +
Energy Balances on Open Systems
is Velocity
111
= 22
2
This is Equation of Continuity
Velocity of flow
Mass Balance
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Study Flow process in NOZZLE
Device which increases the velocity
1 +1
2 + 1 +
= 2 +2
2 + 2 +
1 +1
2 = 2 +
2
2
No heat transfer & work transfer
No potential energy change
The inlet velocity of the nozzle is very small compared
to the outlet velocity ie V2 >>>> V1
So V1 is negligible and thus V1 = 0
1 +1
2 = 2 +
22
1 = 2 +2
2
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Study Flow process in Diffusor
Device which increases the Pressure
1 +1
2 + 1 +
= 2 +2
2 + 2 +
1 +1
2 = 2 +
2
2
No heat transfer & work transfer
No potential energy change
The inlet velocity of the diffuser is very high compared
to the outlet velocity ie V1 >>>> V2
So V2 is negligible and thus V2 = 0
1 +1
2 = 2 +
22
1 +1
2 = 2
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Study Flow process in Throttle valve
Device which increases the Pressure
1 +1
2 + 1 +
= 2 +2
2 + 2 +
1 +1
2 = 2 +
2
2
No heat transfer & work transfer
No potential energy change
The inlet velocity and outlet velocity are same
V1 - V2 = 0 (or negligible)
1 +1
2 = 2 +
22
=
Write down the steady flow
Equation and reduce it for
throttling process (4 marks)
(Apr 2013)
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Study Flow process in Turbine and
CompressorDevice which increases the Pressure
1 +1
2 + 1 +
= 2 +2
2 + 2 +
1 +1
2 = 2 +
2
2 +
No heat transfer & there is a Definite Work transfer
No potential energy change
If The inlet velocity and outlet velocity are negligible
1
+ 1
2 =
2
+ 2
2 +
= +
= +
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Prob-1
A steam power plant, steam flow steadily through a
0.2m diameter pipeline from a boiler to a turbine.
At the boiler end, the steam conditions are p =
4MPa, T = 400C, h= 3213.6 KJ/kg, v = 0.073 m3/kgAt the turbine end, the steam conditions are p =
3.5MPa, t = 392C, h= 3202.6 KJ/kg, v = 0.084 m3/kg
There is a heat loss 8.5 KJ/kg from the pipeline.
Calculate the steam flow rate(Apr 2013)
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Prob-1
A steam power plant,
steam flow steadily through a 0.2m diameter pipeline
boiler --->> turbine.
At the boiler end
p = 4MPa, T = 400C, h= 3213.6 KJ/kg, v = 0.073 m3/kgAt the turbine end
p = 3.5MPa, t = 392C, h= 3202.6 KJ/kg, v = 0.084 m3/kg
heat loss 8.5 KJ/kg (dQ/dm) so take as -ve.
Calculate the steam flow rate (w kg/s)( Apr 2013)
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Heat loss from the pipeSo
= - 8.5 KJ/kg
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=
1 =
2 ------ (1)
1 + 1
2 + 1 +
= 2 + 2
2 + 2 +
------ (2)
(pipe Constant cross section)
1 = 0.073 m3/kg2 = 0.084 m3/kg
h1= 3213.6 KJ/kg
h2= 3202.6 KJ/kg
= - 8.5 KJ/kg
Substitute eqn.(1) in (2)
Get, v1 and even V2
=
1 = 0.073 m3
/kgSubstitute in below equation
get w
Mass flow rate w =
Pipe diameter = 0.2m
In steady flow equation mass flow rate is constant
Mass flow rate w =
=
This is Equation of Continuity
Velocity of flow m/s2
Specific Volume m3
/kg
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Prob-2
Air at a temperature of 15 passes through aheat exchanger at a velocity of 30 m/s, where
its temperature is raised to 800 .It is then
enters a turbine with the same velocity of30m/s and expands until the temperature falls
to 650 . On leaving the turbine, the air istaken at a velocity of 60m/s to a nozzle where
it expands until the temperature has fallen to500 .If the air flow rate is 2 Kg/S
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Prob-1
Find
a) The rate of heat transfer to the air at the
heat exchanger
b) The power output from the turbine
assuming no heat loss
c) The velocity at the exit of the nozzle,
assume no heat lossTake the enthalpy of air as Cpt, Cp = 1.0005
KJ/KgK (Nov-2010)
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Ans: 1 +1
2 + 1 +
= 2 +
2
2 + 2 +
w1 1 +
+ 1 + 12 = w2 2 +
+ 2 + 12
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Ans: a)
w1 1 +
+ 1 + 12 = w2 2 +
+ 2 + 12
w1 1 + 12 = w2 2 air flow rate is w 1= w2 = 2 Kg/SSpecific enthalpy
= 2 1 = 2 1 Find
12 = w2 2 w1 1 W2 = w112 = w(h2h1) = w 2 1
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Ans:b) The power output (W) from the
turbine assuming no heat loss (Q =0)w2 2 +
+ 2 + 23 = w3 3 +
+ 3 + 23
w2 2 +
= w3 3 +
+ 23
w 2 3 + w = 23wcp 2 3 + w
= 23
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Ans:c) The velocity at the exit of the
nozzle, assume no heat loss (Q =0)w2 2 +
+ 2 + 23 = w3 3 +
+ 3 + 23
w2 3 +
= w3 4 +
(3 4) +3
2 = 4
2
(3 4) +3
2 =
2
= ?
Specific enthalpy
= 2 1 = 2 1
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Prob-3
A blower handles 1 kg/s of air at 20oC and
consumes 15KW power. The inlet and outlet
velocities of air are 100m/s and 150 m/s. find
exit air temperature, assuming adiabaticcondition. Take Cp of air 1.005 Kj/Kgk.
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Blower
Mass flow rate w = 1 kg/s
T1 = 20oC
Power or work done = 15KW
The inlet velocities V1 = 100m/s
and outlet velocities V2 = 150 m/s.
find exit air temperature
assuming adiabatic condition.
Take Cp of air 1.005 Kj/Kgk.
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Blower
General Steady Flow Energy equation for Blower
w1 1 +
+ 1 + 12 = w2 2 +
+ 2 + 12
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Steady Flow Energy equation for
Blower
No potential energy term involved in the process
w1 1 +
+ 1 + 12 = w2 2 +
+ 2 + 12
assuming adiabatic condition.
w1 1 +
+ 1 + 12 = w2 2 +
+ 2 + 12
w1 1 +
+ 12 = w2 2 +
For steady flow , w1 = w2 = ww 1 +
+ 12 = w 2 +
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Blower
For steady flow , w1 = w2 = w = 1 kg/s (given)
w 1 +
+ 12 = w 2 +
w 1
2
+ 12
= w
w (1 2) + 12 = w
1/ 1.0051000/(20 2) + 151000 = 1 kg/s
T2 = ----- oC
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Prob-4A turbine operates under steady flow conditions, receives steam at the
following state
Pressure = 1.2 Mpa
Temperature = 188oC
Enthalpy = 2785 KJ/kg
Velocity = 33.3 m/s
Elevation = 3 mSteam leaves the turbine in the following state
Pressure = 20 Mpa
Enthalpy = 2512 KJ/kg
Velocity = 100 m/s
Elevation = 0 m
Heat is lost to the surrounding at the rate of 0.29 KJ/s
Rate of steam flow through the turbine is 0.42 kg/s
What is the power output of the turbine in Kw
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w1 1 +
+ 1 12 = w2 2 +
+ 2 + 12
Heat is lost to the surrounding at the rate of 0.29 KJ/s (Q)
Rate of steam flow through the turbine is 0.42 kg/s (w1= w2)
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w1 1 +
+ 1 12 = w2 2 +
+ 2 + 12
Heat is lost to the surrounding at the rate of 0.29 KJ/s (Q)
Rate of steam flow through the turbine is 0.42 kg/s (w1= w2)
0.42kg/s 2785
+ .
+ 9.81
3 0.29/
= 0.42kg/s 2512
+ 100
2 + 0
+ 12
W = ----- KW
W = 112.6264729 KW
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Prob-5- Exercise
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a) Find the velocity at the exit of the nozzle
w1 1 +
= w2 2 +
(1 2) + 1
2 = 2
2
= ?
w1 = 2 (steady flow)
b) If the inlet area is A1 = 0.1 m2,and specific volume at inlet is = 0.187 m3/kgFind the mass flow rate
W =
c) Find the specific volume at the exit is of the nozzle is 0.498 m3/kg,Find the Exit area of the nozzle (A2)
W =
W =
=
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Reference
Rajput, R. K. 2010. Engineering thermodynamics. Jones and Bartlett
Publishers, Sudbury, Mass.
Nag, P. K. 2002. Basic and applied thermodynamics. Tata McGraw-Hill, New
Delhi.
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