Answer to Question #235871 in Molecular Physics | Thermodynamics for Unknown346307

Question

. A turbine, operating under steady-flow conditions, receives 4500 kg of

steam per hour. The steam enters the turbine at a velocity of 2800 m/min, an elevation of 5.5 m

and a specific enthalpy of 2800 kJ/kg. It leaves the turbine at a velocity of 5600 m/min, an elevation

of 1.5 m and a specific enthalpy of 2300 kJ/kg. Heat losses from the turbine to the surroundings

amout to 16000 kJ/h.

Determine the power output of the turbine.

Answer 1

Gives

Steam rate flow

$\dot{m}=\frac{4500}{3600}=1.25kg/sec$

Steam velocity at inlet

$v_1=\frac{2800}{60}=46.67m/sec$

Velocity of exis $v_2=\frac{5600}{60}=93.34m/sec$

Elevation at inlet $z_1=5.5m$

Elevation at exis $z_2=1.5m$

$$Q=16000kJ/hr=\frac{16000}{3600}=4.45KJ/sec$$

Enthalpy

$h_1=2800KJ/kg$

$h_2=2300KJ/kg$

Turbine operates under steady flow condition

$$Q-w_s=h_2-h_1+\frac{1}{2}(v_2^2-v_1^2)+g(z_2-z_1)J/kgsec$$

$$4.45-w_s=(2300-2800)+\frac{(93.34^2-46.67^2)}{2\times1000}+\frac{9.8(1.5-5.5)}{1000}$$

$$w_s=4.45-[(2300-2800)+\frac{(93.34^2-46.67^2)}{2\times1000}+\frac{9.8(1.5-5.5)}{1000}]$$

$$w_s=4.45+500+3.2671-0.0392=507.67KW$$