02 Part6 Energy Balance

7/28/2019 02 Part6 Energy Balance

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Energy Balances

S.GunabalanAssociate ProfessorMechanical Engineering DepartmentBharathiyar College of Engineering & TechnologyKaraikal - 609 609.e-Mail : [email protected]

Part - 2


2/28

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


3/28

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


4/28

Study Flow process in Diffusor

Device which increases the Pressure

1 +1

2 + 1 +

= 2 +2

2 + 2 +

1 +1

2 = 2 +

2

2



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


5/28

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



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)


6/28

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


If The inlet velocity and outlet velocity are negligible

1

+ 1

2 =

2

+ 2

2 +

= +

= +


7/28

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)


8/28

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)


9/28

Heat loss from the pipeSo

= - 8.5 KJ/kg


10/28

=

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


11/28

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


12/28

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)


13/28

Ans: 1 +1

2 + 1 +

= 2 +

2

2 + 2 +

w1 1 +

+ 1 + 12 = w2 2 +

+ 2 + 12


14/28

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


15/28

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


16/28

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


17/28

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.


18/28

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.


19/28

Blower

General Steady Flow Energy equation for Blower

w1 1 +

+ 1 + 12 = w2 2 +

+ 2 + 12


20/28

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 +


21/28

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


22/28

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


23/28

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)


24/28

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


25/28

Prob-5- Exercise


26/28


27/28

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.

Documents

02 Part6 Energy Balance