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

Question

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.

Answer 1

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.