HEAT TRANSFERSteady State Conduction

ProblemThe top and bottom surfaces of a solid cylindrical rod are maintained at constant temperatures of 20°C and 95°C while the side surface is perfectly insulated. The rate of heat transfer through the rod is to be determined for the cases of copper, steel, and granite rod.Assumptions.1 )Heat conduction is steady and one-dimensional. 2) Thermal conductivity is constant. 3) There is no heat generation.Properties. The thermal conductivities are given to be k = 380 W/m⋅°C for copper, k = 18 W/m⋅°C for steel, and k = 1.2 W/m⋅°C for granite.

Analysis.Noting that the heat transfer area (the area normal to the direction of heat transfer) is constant, the rate of heat transfer along the rod is determined from

where L = 0.15 m and the heat transfer area A is:

Discussion: The steady rate of heat conduction can differ by orders of magnitude, depending on the thermal conductivity of the material.

ProblemA wall is constructed of two layers of sheetrock with fiberglass insulation in between. The thermal resistance of the wall is to be determined.Assumptions. 1) Heat transfer through the wall is one-

dimensional. 2) Thermal conductivities are constant.Properties. The thermal conductivities are given to be: ksheetrock = 0.10 Btu/h⋅ft⋅°F and kinsulation = 0.020 Btu/h⋅ft⋅°F.Analysis (a) The surface area of the wall is not

given and thus we consider a unit surface area (A = 1 ft2). Then the R-value of insulation of the wall becomes equivalent to its thermal resistance, which is determined from.

The outer surface of a 0.2m-thick concrete wall is kept at a temperature of -5°C, while the inner surface is kept at 20°C. The thermal conductivity of the concrete is 1.2W/(mK). Determine the heat loss through a wall 10m long and 3m high.

PROBLEM

GIVEN- 10m long, 3m high, and 0.2m thick concrete wall- Thermal conductivity of the concrete (k) = 1.2 W/(m k)- Temperature of the inner surface (Ti) = 20°C- Temperature of the outer surface (To) = -5°C

FIND- The heat loss through the wall (qk)

SOLUTION:

The rate of heat loss through the wall is given by the equation:

Heat is transferred at a rate of 0.1 kW through glass wool insulation (density = 100 kg/m3) of 5 cm thickness and 2 m2 area. If the hot surface is at 70°C, determine the temperature of the cooler surface.

PROBLEM

GIVEN- Glass wool insulation with a density (p) = 100 kg/m3

- Thickness (L) = 5 cm = 0.05 m- Area (A) = 2 m2

- Temperature of the hot surface (Th) = 70°C- Rate of heat transfer (qk) = 0.1 Kw = 100 W

FIND- The temperature of the cooler surface (TC)

SOLUTION:

The rate of heat transfer is given by the equation:

Solving this for TC