Answer to Question #71018 in Molecular Physics | Thermodynamics for maged

Question

A steady flow of steam enters a condenser with a specific enthalpy of 2300 kJ/kg and a velocity of 350 m/s. The condensate leaves the condenser with a specific enthalpy of 160 kJ/kg and a velocity of 70 m/s. Calculate the heat transfer to the cooling fluid per kilogram of steam condensed.

Answer 1

The steady state energy equation.

$Q-W = Δ(h+ \frac{V^2}{2} +gz) \\ Q -0 = Δ(h+ \frac{V^2}{2} +g(0)) \\ Q = Δ(h+ \frac{V^2}{2}) \\ = (2300 \times 10^3-160 \times 10^3) + \frac{350-70}{2} \\ = 2198800 \;J/kg \\ = 2198.8 \;kJ/kg$

But the heat transferred to the cooling fluid will be negative because the heat is lost from the system. Therefore,

Q= -2198.8 kJ/kg