Answer to Question #105394 in Molecular Physics | Thermodynamics for Kule

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

The steady flow energy equation per unit mass is:

Q – W = H₁ – H₂ + ((V₁)² - (V₂²)) / 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