Answer to Question #228859 in Molecular Physics | Thermodynamics for Thorfinn

Question

Steam enters a turbine stage with an enthalpy of 3620 kJ/kg at 75 m/s and leaves the same stage with an enthalpy of 2800 kJ/kg at 128 m/s. Calculate the work done by steam. 

W = ____ kJ/kg

Answer in 2 decimal places

Answer 1

Gives

$H_1=3620kJ/kg\\H_2=2800kJ/kg\\v_1=75m/sec\\v_2=128m/sec$

$m=\frac{1}{1000}kg$

Now

$E_{in}=E_{out}$

$KE_1+H_1=KE_2+H_2+w$

$w=(H_1-H_2)+\frac{1}{2}m(v_1^2-v_2^2)$

$w=(3620-2800)+\frac{1}{2\times1000}(75^2-128^2)$

$w=820-5.379=814.62J/kg$

$w=0.815kJ/kg$