Question #123867
A turbine operating under steady flow condition receives 5000kg of steam per hour. The steam enter the turbine at a velocity of 3000m/min, an elevation of 5m and a specific enthalpy of 2787kj/kg, it leaves from the turbine at a velocity of 6000m/min an elevation of 1m and a specific enthalpy of 2259kj/kg. Heat losses from the turbine to the surrounding amount to 16736kj/h
Determine the power output of the turbine
1
Expert's answer
2020-06-25T09:35:01-0400

Solution.

m=5000kg/h=1.4kg/s;m=5000kg/h=1.4kg/s;

C1=3000m/min=50m/s;C_1=3000m/min=50m/s;

Z1=5m;Z_1=5m;

h1=2787kJ/kg=2.787106J/kg;h_1=2787kJ/kg=2.787\sdot10^6J/kg;

C2=6000m/min=100m/s;C_2=6000m/min=100m/s;

Z2=1m;Z_2=1m;

h2=2259kJ/kg=2.259106J/kg;h_2= 2259kJ/kg=2.259\sdot10^6J/kg;  

Q=16736kJ/h=4.649103J/s;Q=-16736kJ/h=-4.649\sdot10^3J/s;

m(h1+C122+Z1g)+Q=m(h2+C222+Z2g)+W;m(h_1+\dfrac{C_1^2}{2}+Z_1g)+Q=m(h_2+\dfrac{C_2^2}{2}+Z_2g)+W;

QW=m((h2h1)+C22C122+(Z2Z1)g);Q-W=m((h_2-h_1)+\dfrac{C_2^2-C_1^2}{2}+(Z_2-Z_1)g);

4.649103W=1.4((2.2591062.787106+10025022+(15)9.81);-4.649\sdot10^3-W=1.4((2.259\sdot10^6-2.787\sdot10^6+\dfrac{100^2-50^2}{2}+(1-5)9.81);

4.649103W=1.014106;-4.649\sdot10^3-W=-1.014\sdot10^6;

W=101kJ;W=101kJ;

Answer: W=101kJ.W=101kJ.






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Comments

shubham
14.12.21, 04:34

keep doing great work

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