Answer to Question #128840 in Mechanics | Relativity for Ariyo Emmanuel

Question #128840
A steady flow of steam enters a condenser with a specific enthalpy of 2300 kJ/kg and a velocity of 350m/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.
Expert's answer

Steady flow energy equation equation states that for a given volume in steady state the incoming energy is equal to outgoing energy.

"Q = W + \\Delta H + \\Delta K"

where Q is heat, W is work, H is enthalpy and K is kinetic energy. This equation per unit of mass becomes "q = w + \\Delta h + \\frac{ (v_2^2-v_1^2)}{2}" . By the condition of the task, "w = 0" (no external work done).

"q = (160-2300)\\cdot10^3 + \\frac{70^2-350^2}{2}= -2 \\;198 \\;800 \\approx -2.2 \\cdot 10^{6}" J = - 2.2 MJ.

The heat is negative because the fluid is cooling.

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