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
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=820-5.379=814.62J\/kg"
"w=0.815kJ\/kg"
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