An engineer wishes to heat air for a shopping mall with 100,000 kJ of energy per
ID: 2992596 • Letter: A
Question
An engineer wishes to heat air for a shopping mall with 100,000 kJ of energy per hour. One option is to use steam (available at 100 deg c and 1 atm pressure with a specific enthalpy of 48.168 kJ/mole. When the steam gives up its energy to heat the air, it condenses from a vapour to a liquid. The liquid consate now has a temperature of 100 deg c, a pressure of 1 atm and a much lower specific enthalpy of 7.533 kJ/mol
1. Assuming there is a minimal cooling of the condensate after condesing, how many kilograms of steam per hour must be used to heat the air
2. What is the heating power required in kW
[molecular weight of water is 18 g/mol]
Please include full working (equations, methods, and substituion so I can see where you are coming from
Explanation / Answer
(1) specific enthalpy of steam = 48.168 KJ/mol specific enthalpy of water = 7.533 KJ/mol net heat lost by steam when it condenses from vapor to liquid = (48.168-7.533) = 40.635 KJ/mol mass of steam needed= 100,000/40.635 = 2460.933 mol = 44296.8 g = 44.3 kg (2) heating power needed = (energy/time) =(100,000/3600) = 27.78 kW