Answer to Question #82455 in Molecular Physics | Thermodynamics for Abraham

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 flow energy equation per unit mass is:

Q – W = H1 – H2 + (V_1^2-V_2^2)/2 + h2 – h1; where

Q = heat transfer across boundary per unit mass;

W = external work done by system per unit mass;

H = fluid enthalpy;

V = fluid velocity;

h = fluid height.

As it is no data for external work and fluid height: W = 0, h2 – h1= 0.

Hence,

Q = (2300 – 160) × 10³ + (〖350〗^2-〖70〗^2)/2 = 2198800 J/kg = 2.2 MJ/kg