The aim of this lab is to determine the average convective heat transfer coefficient for forced convection of a fluid (air) past a copper tube, which is used as a heat transfer model. Introduction The general definition for convection may be summarized to this definition “energy transfer between the surface and fluid due to temperature difference” and this energy transfer by either forced (external, internal flow) or natural convection. Heat transfer by forced convection generally makes use of a fan, blower, or pump to provide high velocity fluid (gas or liquid).
The high-velocity fluid results in a decreased thermal resistance across the boundary layer from the fluid to the heated surface. This, in turn, increases the amount of heat that is carried away by the fluid. [1] Theory Background [2] Considering the heat lost by forced convection form the test rod. The amount of heat transferred is given by (1) Where = rate of heat transfer, unknown value. ?= film heat transfer coefficient, this is what we need to found out. A= area for heat transfer, this is the area of the cross section area of test section. T= temperature of the copper rod, the temperature after heating.
Ta= temperature of air, surrounding temperature. So, in any period of time, dt, then the fall in temperature, dT, will be given as: (2) Where m= mass of copper rod, cp= specific heat of the copper rod, J/kgK Eliminating Q from (1) and (2) then Since Ta is constant, dT=d(T-Ta) Integrating gives: At t = 0, T=To, hence C1 = ln(T-To), hence: Or Therefore a plot of ln((T-Ta)/(Tmax-Ta))) against t should give a straight line of gradient from which the heat transfer coefficient, ? , can be found. To find the velocity of air passing the rod, first the velocity upstream must be found.
The Essay on Heat Transfer
Practice Problems Set – 1 MEC301: Heat Transfer Q.1 The slab shown in the figure is embedded on five sides in insulation materials. The sixth side is exposed to an ambient temperature through a heat transfer coefficient. Heat is generated in the slab at the rate of 1.0 kW/m3. The thermal conductivity of the slab is 0.2 W/m-K. (a) Solve for the temperature distribution in the slab, noting any ...
