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Charra [1.4K]
3 years ago
10

Steam at a pressure of 100 bar and temperature of 600 °C enters an adiabatic nozzle with a velocity of 35 m/s. It leaves the noz

zle at a pressure of 80 bar with a quality 85 %. From the information given, calculate the exit velocity.
Engineering
1 answer:
butalik [34]3 years ago
4 0

Answer:

Exit velocity V_2=1472.2 m/s.

Explanation:

Given:

At inlet:

P_1=100 bar,T_=600°C,V_1=35m/s

Properties of steam at 100 bar and 600°C

        h_1=3624.7\frac{KJ}{Kg}

At exit:Lets take exit velocity V_2

We know that if we know only one property inside the dome  then we will find the other property by using steam property table.

Given that dryness or quality of steam at the exit of nozzle  is 0.85 and pressure P=80 bar.So from steam table we can find the other properties.

Properties of saturated steam at 80 bar

   h_f= 1316.61\frac{KJ}{Kg} ,h_g= 2757.8\frac{KJ}{Kg}

So the enthalpy of steam at the exit of turbine  

h_2=h_f+x(h_g-h_f)\frac{KJ}{Kg}

h_2=1316.61+0.85(2757.8-1316.61)\frac{KJ}{Kg}

 h_2=2541.62\frac{KJ}{Kg}

Now from first law for open system

h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w

In the case of adiabatic nozzle Q=0,W=0

3624.7+\dfrac{35^2}{2000}+0=2541.62+\dfrac{(V_2)^2}{2000}+0

V_2=1472.2 m/s

So Exit velocity V_2=1472.2 m/s.

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Answer:

Material K has a modulus of elasticity E=3.389× 10¹¹ Pa

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Length=L = 40 ft =12.192 m

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