Question

The steady flow rate of steam through an adiabatic turbine is 2.5 kg/s . The steam at 600 ^oc and 10 bar enters the turbine through pipeline 10 cm in diameter. The steam exits the turbine through at pipeline 25 cm in diameter at a temperature 400 ^oc and pressure of 1 bar.

1. State the steady flow energy equation
2. At the inlet conditions of the turbine, obtain from steam tables

                The saturation temperature of steam at 10 bar

                 Specific volume

                 Specific internal energy

                 Specific enthalpy

Calculate

1. The degree of superheat
2. The cross sectional area of inlet pipeline
3. Inlet velocity

3. At the outlet conditions of the turbine, from the steam tables, read

The saturation temperature of steam at 1 bar

Specific volume

Specific internal energy

Specific enthalpy

Calculate

1. The degree of superheat
2. The cross sectional area of inlet pipeline
3. Inlet velocity

4. Calculate the change in enthalpy across turbine
5. Calculate the change in kinetic energy across the turbine
6. Calculate the work obtained from the turbine

Answer 1

SOLUTION.

To find =Adiabatic Turbine is operating at steady state ,we have to Calculate outlet conditions and work from turbine etc.

From steam table data we obtained following values .

Property	inlet,1	outlet,2
Temp,K	873	673
Pressure ,barr	10	1
Tsat, °C	453	372
Sp volume,m³/kg,v	0.4011	3.1027
Sp Internal energy, kJ/kg,u	3297	2968
Sp enthalpy kJ/kg ,h	3698	3278
Degree of superheat=(Tsup-Tsat)	420 K=147°C	301 K =28°C
Cross section area=π/4 *D²,m²	0.007853	0.04908
Velocity, V=m*v/(A)	127.7 m/sec	158 m/sec

Mass flow rate =m1=m2=2.5 kg/sec

A) energy balance over the steady state Turbine ,

Ein-Eout =∆E ......kJ

In terms of rate ,

e in - eout =∆E/dt. (KW).....E=e/time

At Steady state b

ein-eout =0 (kW)

ein =eout.

m(h1+V1²/2) = W + m(h2+V2²/2)

A ) change in enthalpy across turbine =

m *(h1-h2) =2.5(3698-3278)=1050 (kg/sec * kJ/kg)

Change in enthalpy=1050 kW.

B)

Change in kinetic energy

Energy at inlet=m*V1²/2

Energy at outlet= m*V2²/2

∆Ke = m(V2²-V1²)/2 =2.5 kg/sec *(158²-127²)/2=11043 kW.

C) Work done in the Turbine.

W = m(h1+V1²/2 -h2 +V2²/2)

W=2.5 (kg/sec)*{3698-3278+158²/2 -127²/2} kJ/kg

W = 4397.5 kW.

W=work obtained from turbine =4397.5 kW

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