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1)
FIND: (a) Calculate the internal energy generation rate
, by applying an overall energy balance to the wall, (b) Determine the coefficients a, b, and c, by applying the boundary conditions to the prescribed form of the temperature distribution.
KNOWN: Steady-state conduction with uniform internal energy generation in a plane wall; temperature distribution has quadratic form. Surface at x=0 is prescribed and boundary at x = L is insulated.
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction with constant properties and uniform internal generation, and (3) Boundary at x = L is adiabatic.
ANALYSIS:
(a) The internal energy generation rate can be calculated from an overall energy balance on the wall as shown in the schematic below.
(b) The coefficients of the temperature distribution, , can be evaluated by applying the boundary conditions at x = 0 and x = L.
Boundary condition at x=0, convection surface condition:
Boundary condition at x=L, adiabatic or insulated surface:
Find a from temperature at x=0:
2) Consider a 0.8m tall, 1.5m wide double-pane window consisting of 4mm thick layers of glass (k=0.78W/mK) seperated by non-convecting air (k=0.026 W/mK). Determine the steady state heat rate if the room is maintained at 20 degrees and the outer surface of the outer pane of glass is -10 degrees.
First, determine the thermal resistances of each layer:
Sum these series resistances to get Rtotal = 0.33K/W. The temperature difference is 30 degrees, so using Q = T/R, The heat rate Q = 91W.
3) A 3m high and 5m wide wall consists of long 16cmx22cm cross section horizontal bricks (k=0.72W/mK) seperated by 3cm thick plaster layers (k=0.22 W/mK). There are also 2cm thick plaster layers on each side of the brick and a 3cm rigid foam (k=0.026) on the inner side of the wall. The inner surface of the wall is 20 degrees and the outer surface of the plaster is -10 degrees. Calculate the heat rate through the wall.
Again, like in question two, one must determine the thermal resistance of each section to tackle this problem. However, the thermal resistance are not arranged as simple sheets - the plaster and brick orientation is constructed in parallel and therefore their thermal resistances should be considered as such. This leaves the thermal circuit:
Where R1 represents the foam, R2 and R6 represents the left side plaster respectively , R3 and R5 represents the plaster up to 1.5cm immediately above and below the brick respectively and R4 represents the brick:
Finding the equivalent resistance of the parallel thermal resistors gives Rp = 0.97K/W. Adding this resistance to R2, R6 and R1 gives Rtotal = 6.31K/W. For a temperature difference of 30 degrees, this gives a heat rate through the wall of 4.75W.
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