Answer to Question #263461 in Molecular Physics | Thermodynamics for den

Question

Steam enters a turbine stage with an enthalpy of 3628 kJ/kg at 70 m/s and leaves at the

same stage with an enthalpy of 2846 kJ/kg at 124 m/s. Calculate the work per kg done by

the steam.

Answer 1

$H_1=3628kj/kg\\H_2=2846kj/kg\\v_1=70m/sec\\v_2=124m/sec$

$E_{in}=E_{out}$

$KE_1+H_1=KE_2+H_2+w$

$w=H_1-H_2+\frac{1}{2}(mv_1^2-mv_1^2)$

$$w=3628-2846+\frac{1}{2\times1000}(124^2-70^2)$$

$$w=782+5.238=0.787238kJ/kg$$