HEAT TRANSFER - PRINCIPLES & EQUIPMENT - BASIC HEAT TRANSFER CALCULATIONS PDF Print E-mail
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Wednesday, 17 February 2010 20:47
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HEAT TRANSFER - PRINCIPLES & EQUIPMENT
HEAT TRANSFER BY CONVECTION
BASIC HEAT TRANSFER CALCULATIONS
LINEAR AND CUBICAL EXPANSION
HEAT EXCHANGERS, BOILERS & FURNACES
TYPES OF HEAT EXCHANGER
EXCHANGER SHELL SIDE FLOW PATTERNS
EXCHANGER TUBE SIDE FLOW PATTERNS
FACTORS AFFECTING HEAT TRANSFER
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BASIC HEAT TRANSFER CALCULATIONS

The heat gained or lost (Sensible Heat), by a substance can be calculated using the following equation :

Q =m x c x t

Where:
Q = Quantity of heat gained or lost in Joules
m = Mass of material in kg
c = Specific Heat of material in J/kg/°C
t = Temperature change in °C


Examples 1
20 kg of water at 5 °C is mixed with 10 kg of water at 65 °C. The final temperature of the mixture is 25 °C.

What is : a). The heat gained by the 20 kg of water, and,
b). The heat lost by the 10 kg of water ?


( S.H. of water = 4180 J/kg/°C )

a). Temp. change = ( 25 - 5 ) = 20 °C
Heat gained = 20 kg x 4180 J/kg/°C x 20 °C (Note the cancelling)
= 1,6720,00 J = 1672 kJ

b). Temp. change = ( 65 - 25 ) = 40 °C
Heat lost = 10 kg x 4180 J/kg/°C x 40 °C
= 1,672,000 J = 1672 kJ

The above calculations demonstrate the Law of Conservation of Energy in that ,
Heat Lost = Heat Gained.

Examples 2
Calculate the sensible heat needed to raise the temperature of 1 kg of Ice from -10 °C to 0 °C.

(The S.H. of Ice is 2.1 kJ/kg/°C)

Q = m x c x t
= 1 x 2.1 x 10 = 21 kJ

Examples 3
Calculate the latent heat needed to change 10kg of Ice at 0 °C to water at 0 °C.

(The Latent Heat (L) of melting ice is 335 kJ/kg)

Q =m x L
= 10 x 335 = 3350 kJ



Last Updated on Wednesday, 24 February 2010 19:40