Answer to Question #166005 in Molecular Physics | Thermodynamics for Ajay Yadav

Question

A steam turbine operates under steady flow conditions. It

receives 7200 kJ/h of steam from the boiler. The steam enters

the turbine at enthalpy of 2800 kJ/kg, a velocity of 4000 m/min.

and an elevation of 4 m. The steam leaves the turbine at enthalpy

of 2000 kJ/kg, a velocity of 8000 m/min. and an elevation of

1 m. Due to radiation heat losses from the turbine to the

surrounding amount to 1580 kJ/h. Calculate the output of the turbine

Answer 1

Formula

$Q - W = m [(ΔH_{o} - ΔH_{i} ) + (\frac{V_{o}^{2} - V_{i}^{2}}{2}) + g(h_{o}-h_{i})]$

Q = heat loss = 0.43 KJ/S

W = Work output

m = mass flow rate = 2Kg/S

$ΔH_{o} =$ output enthalphy = 2000 kJ/kg

$ΔH_{i} =$ inlet enthalphy = 2800 kJ/kg

$V_{o}^{2}=$ 133.33 M/S

$V_{i}^{2}=$ 66.66 M/S

$h_{o}=$height at outlet = 1 M

$h_{i}=$ height at inlet = 4 M

g = accel. due to gravity = 10 M/S²

$0.43 - W = 2 [(2000 - 2800 ) + (\frac{133.33^{2} - 66.66^{2}}{2 \times1000}) + \frac{10(1-4)}{1000} ]$

$0.43 - W = -1600 + 26.66 - 0.03$

$0.43 - W = -1573.37$

$-W = -1573.37-0.43$

$W= 1573.8$

Output = 1573.8 KJ/S

Note: the division by 1000 is to express the terms in KJ/S.