HEAT TRANSFER
HEAT TRANSFER DEFINITION • Heat is a form of energy that moves from one body to another due to temperature differences. • Temperature difference is the driving force that produces heat transfer. • 2 applications of heat transfer: Ø Conductor (heat exchanger, boiler, condenser, evaporator etc) Ø Insulator (pipeline insulator)
Heat transfer mechanism Convection Conduction Radiation
FOURIER'S LAW Fourier's law - The law of heat conduction • states that the rate of heat transfer through a material is proportional to the negative gradient in the temperature and to the area, at right angles to that gradient, through which the heat flows. Where: k = thermal conductivity (W/mK) A = cross-sectional area, m2 t = temperature, K x = thickness, m
EXAMPLE 1 The inner surface of a plane brick wall is at 40C and outer surface is at 20C. Calculate the rate of heat transfer per m² of surface area of the wall, which is 250 mm thick. The thermal conductivity of the brick is 0.52W/m.K .
NEWTON'S LAW OF COOLING Q hA(t t) = w − The heat transfer from the solid surface to the fluid can be described by Newton's law of cooling. It states that the heat transfer, from a solid surface of area A, at a temperature Tw, to a fluid of temperature T, is: Where: h = convection coefficient (W/m2K) A = surface area, m2 tw = surface temperature, K t = ambient temperature, K
Heat Transfer Between Two Fluids Through The Wall Heat from fluid A to the wall = Through the wall = From fluid A to the wall Heat resistance, RT Heat loss, Q = TA -TB RT
EXAMPLE 2 The mild steel tank of wall thickness 10 mm contains water at 90C. Calculate the rate of heat loss per m² of tank surface area when the atmospheric temperature is 15C. The thermal conductivity of mild steel is 50 W/mK and the heat transfer coefficients for the inside and outside of the tank are 2800 W/m2K and 11 W/m2K respectively. Calculate also the temperature of the outside surface of the tank.
Resistance of water film, h A 1 R 1 1 = W 3.571 10 K W K 2800 1 1 R -4 1 = = Resistance of insulating firebrick k A x R 2 2 2 = Resistance of air gap (fluid film), h A 1 R A 2 = W 0.091 K W K 11 1 1 R3 = = RT Rate of heat loss per m² of surface area = 0.82 kW
The Composite Wall
EXAMPLE 3 A furnace wall consists of 125 mm wide refractory brick and 125 wide insulating firebrick separated ay an air gap. The outside wall is covered with a 12 mm thickness of plaster. The inner surface of the wall is at 1100C and the room temperature is 25C Calculate the rate at which heat is lost per m² of wall surface. The heat transfer coefficient from the outside wall surface to the air in the room is 17 W/m2K and the resistance to the heat flow of the air gap is 0.16 K/W. The thermal conductivity of refractory brick, insulating firebrick, and plaster are 1.6, 0.3, and 0.14 W/mK respectively. Calculate also each interface temperature, and the temperature of the outside surface of the wall.