Question

Steam enters a turbine operating at steady state with a mass flow of 10 kg/s, a specific enthalpy of 3100 kJ/kg, and a velocity of 30 m/s. At the exit, the specific enthalpy is 2300 kJ/kg and the velocity is 45 m/s. The elevation of the inlet is 3 m higher than at the exit. Heat transfer from turbine to its surroundings occur at rate of 1.1 kJ per kg of steam flowing. Let g 9.81 m/s. Determine the power developed by the turbine, in kW.

Answer 1

Answer:7989.86KW

Explanation:

Given data

$h_1$=3100 KJ/kg

$\dot{m}$=10kg/s

$V_1$=30m/s

$Z_1$=3m

$h_2$=2300KJ/kg

$V_2$=45m/s

$Z_2$=0

using steady flow energy equation which is

$\dot{m}$$\left (h_1+gZ_1+\frac{V_1^2}{g}\right )$+$\dot{Q}$=$\dot{m}$$\left (h_2+gZ_2+\frac{V_2^2}{g}\right )\+$$\dot{W}$

$10\left (3100+\frac{9.81\times 3}{1000}+\frac{30^2}{9.81\times 1000}\right )$+$\dot{Q}$=$10\left (2300+\frac{9.81\times 0}{1000}+\frac{45^2}{9.81\times 1000}\right )$+$\dot{W}$

$10\left ( \left ( 3100-2300\right )+\frac{30^{2}-45^{2}}{2\times 9.81\times 1000}+\frac{9.82\times 3}{1000}\right )$-$1.1\times 10$=$\dot{W}$

$\dot{W}$=7989.8673 KW