applications of first law

8/3/2019 Applications of First Law

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Q.1. Write the steady flow energy equation arid point out the significance of various

terms involved.

Fig. shows a steady flow process

Let us consider an open system through which the working substance flows at a steady rate.

Supposing the fluid enters at (1) (1) and leaves at (2)(2)Total energy entering the system,

Then according to 1st law of thermodynamics,

This equation is known as steady flow energy equation (SFE).Steady flow energy equation for unit mass of fluid,

Where h1, h2 = enthalpy of fluid entering and leaving the system

q = heat supplied to system in J/kg

w = work delivered by system.


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Q.2. 15kg of air per minute is delivered by a centrifugal compressor. The inlet and

outlet velocities are 12 m/s and 90 m/s, the pressures are 1 bar and 8 bar and specificvolumes 0.5 m /kg and 0.14 m /kg. The increase in enthalpy of air passing through the

compressor is 150 kJ/kg and heat loss to the surroundings is 900 kJ/min.

Find (a) Motor power required to drive the compressor.

(b) Ratio of inlet and outlet pipe diameter. Assume that inlet and discharge at the same

level.

Increase in enthalpy of air passing through compressor = 150 kJ/kg

Also, heat lost to surroundings =700 kJ/min

1. Motor power required to drive the compressor:Applying SFEE

So the motor required to drive the compressor = 48.18 kW

2. Ratio of inlet and outlet pipe diameter


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Q 3 In a certain steady flow process, 12kg per minute enter at a pressure of 1 4 bar,

density 25kg/m , velocity 120m/s and internal energy 920kJ/kg The fluid properties at exit

are pressure 56 bar density 5kg/m , velocity 180 m/s and internal

energy 720kJ/kg. During the process, the fluid rejects 6.0kJ/s of heat and rises through

60m. Determine the work done during the process in kW.

Applying SFEE


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But for unit mass

from equations (1)

Q. 4. A stream of gases at 2.5bar and 150 m/sis passed through the turbine of a jet

engine. The stream comes out of turbine at 2bar and 300 m/s. The process may be assumedadiabatic. The enthalpy of gas at entry is 260 kJ/kg than at the exit. Determine the capacity

of turbine if gas flow is 4 kg/s.

Ans. P1 =7.5 bar = 7.5 x N/m

P2 = 2 bar = 2 x N/m

C1 = 150 m/s, C2 = 300 m/s

h1 = 260 + h2; m =4kg/s

assuming Z1 = Z2Considering SFEE for unit mass

But Q = 0. [as given process is adiabatic)


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Power capacity of turbine = mW

= 4 x 226.2

= 905 kJ/s= 905 kW

Q. 5. Write down a general equation for steady flow system and simplify it when

applied to

1. Gas turbine.

2. Steam nozzle.

Ans. SFEE:

Gas Turbine: A turbine is a device which transforms thermal energy of a flowing fluid into

useful work. This output of turbine may be used to run a generator to produce electricity.

In the turbine,

1. Turbine is insulated, so that no heat enters or leaves the system i.e. q = 0.

2. The changes in RE. are negligible i.e. Z1 = Z2.

3. The changes in K.E. are negligible i.e.Applying steady flow energy equations per Gas/Steam in unit mass

For turbine, if,

q = 0Z1 = Z2

h1= h2 + w

w = h1 h2

This shows that work s done by the system due to decrease in enthalpy of the system.


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Nozzle is a device with varying x-sectional area through which the pressure energy of

flowing fluid decreases and the K.E. increases.The main aim is to produce a jet of very high velocity

In the nozzle,

1 Nozzle is insulated so that no heat enters or leaves the system i e q = 0

2 No work is being done in the nozzle, w = 03 Changes in P.E. are negligible i.e. Z1 = Z2

Applying steady flow energy equation per unit mass

Q. 6. At the inlet to a nozzle, the enthalpy of fluid passing is 2800 kJ/kg and the velocity

is 50 m/s. At the discharge of enthalpy is 2600 kJ/kg. The nozzle is horizontal and there is

no heat loss. Find

(a) Velocity at the exit of nozzle.(b) Mass flow rate if inlet area is 900 cm and specific volume is 0.187 m /kg.

(c) Exit area of nozzle if specific volume at exit is 0.498 m /kg.

Ans. Enthalpy of fluid at inlet, h1 = 2800 kJ/kg

Enthalpy of fluid at discharge, h2 = 2600 kJ/kgVelocity of fluid at inlet, C1 = 50 m/s

SFEE


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as nozzle is horizontal and with no heat loss

i.e. Z1 = Z2, q = 0, w = 0

C2 = 634.43 m/s

Inlet area of nozzle, A1 = 900 cm

Q.7. The following data pertaining to a steam power plant are given for each state

corresponding to Fig. shown below. Determine heat transfer in each process and turbine

work.


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Process 1

2 for boiler:

for boiler system, w = 0, , Z1Z0

Process 23 for Turbine Work:

w = (h2h3) = 30002300 = 700 kJ/kg

Process 34 for heat exchanged system

Process 4

1 for compressor:

Q = (h1 h4) + w

= (3020200) + 4


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= 2820 + 4 = 2824 kJ/kg

=20 + 02100 + 2824

=724kJ/kg

W.D. for turbine = 700 kJ/kg.

Q.8. A perfect gas flows steadily through a horizontal cooler. The mass flow rate :S 1

kg/s. The pressure and temperature are 2 bar and 400 K at entry and 1.5 bar and 280 K at

exit, respectively. The cross-sectional areas at entry and exit are each 0.01 m .

Using the data given below; determine

(a) velocities at entry and exit

(b) the heat transfer rate from gas

Cv = 0.161kJ/kgK, R = 0.2 kJ/kg K.

Ans. m = 1kg/s

P1 = 2 x N/m , T1 = 400 K

P2 = 1 5 x N/m , T2 = 280 KA1 = A2 = 0.01 m

1 Velocities at entry and exit:

R = Cp Cv

0.2 = Cp0.161

Cp = 0.361 kJ/kg K

P1V1 = mRT1


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Heat transfer rate,

Q = 43320W

Q = 43.32kW

Q. 9. Show that the work done in a steady flow process is given by

Ans. SFEE

In differential form, the above equation (1) can be changed to

From 1st law of thermodynamics for closed system

= du + pdV

(2) becomes

If changes in P.E. and K.E. are negligible,


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Q.10. A steady operating pneumatic motor develops a shaft power of 0.1 kW when

supplied with dry air at a pressure of 10 bar and at a temperature of 300 K and exhausting

at 1 bar. The motor is adiabatic and the isentropic efficiency (i.e. the ratio of the actual

enthalpy drop to the isentropic drop from the same initial state to the final pressure) is 0.7.

The change in specific KE between supply and exhaust negligible. Determine the

temperature of the air leaving the motor (C) and the mass flow

rate.


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(T2300) = 0.7 (155300)

T2 = 198.9 K

applications of first law

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