Answer to Question #123867 in Molecular Physics | Thermodynamics for Clara ogu

Question

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

Answer 1

Solution.

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

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

$Z_1=5m;$

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

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

$Z_2=1m;$

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

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

$m(h_1+\dfrac{C_1^2}{2}+Z_1g)+Q=m(h_2+\dfrac{C_2^2}{2}+Z_2g)+W;$

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

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

$-4.649\sdot10^3-W=-1.014\sdot10^6;$

$W=101kJ;$

Answer: $W=101kJ.$