SOLVED PROBLEMS
Combined Convection and Radiation
Problem 1: A surface is at 200°C and is exposed to surroundings at 60°C
and convects and
radiates heat to the surroundings. The convection coefficient is 80W/m2K.
The radiation factor is one. If the heat is conducted to the surface through
a solid of conductivity 12 W/mK, determine the temperature gradient at
the surface in the solid.
Solution:
Heat convected + heat radiated = heat conducted considering 1m2,
4
h(T1 – T2) + σ(T – T 4) = – kdT/dx
Therefore, 80(200 – 60) + 5.67 {[(200 + 273)/100]4 – [(60 + 273)/100]4}
= – 12 dT/dx
Therefore dT/dx = – (11200 + 2140.9)/12 = – 1111.7°C/m.
Problem 2: Heat is conducted through a material with a temperature
gradient of – 9000 °C/m. The conductivity of the material is 25W/mK. If
this heat is convected to surroundings at 30°C with a convection
coefficient of 345W/m2K, determine the surface temperature.
If the heat is radiated to the surroundings at 30°C determine the surface
temperature.
Solution: In this case only convection and conduction are involved.
– kAdT/dx = hA(T1 – T2). Considering unit area,
– 25 × 1 × (– 9000) = 345 × 1 (T1 – 30)
Therefore, T1 = 682.17°C
In this case conduction and radiation are involved.
Heat conducted = Heat radiated
– 25 × 1 × (– 9000) = 5.67 [(T1/100)4 – (303/100)4]
Therefore, T1 = 1412.14K = 1139°C.
Problem 3: There is a heat flux through a wall of 2250W/m2. The same is
dissipated to the
surroundings by convection and radiation. The surroundings is at 30°C.
The convection
coefficient has a value of 75W/m2K. For radiation F = 1. Determine the
wall surface temperature.
Solution: For the specified condition, Consider unit area.
The heat conducted = heat convected + heat radiated
Using the rate equations, with absolute temperature
2250 = [(T2 – 303)/(1/75 × 1] + 5.67 × 1[(T2/100)4 – (303/100)4]
= 75T2 – 22725 + 5.67(T2/100)4 – 477.92
or, (T2/100)4 + 13.2275T2 – 4489.05 = 0.
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, This equation can be solved only by trial. It may be noted that the
contribution of (T2/100)4 is small and so the first choice of T2 can be a little
less than 4489/13.227 = 340K. The values of the reminder for T2 = 300,
310, 320, 330 are given below:
Assumed value 300 310 320 330 330.4 330.3
of T2
Remainder – 439.80 – 296.2 – 15.1 – 5.38 0.484 – 0.98
So, the temperature T2 is near 330K. By one more trial T2 is obtained as
330.4K or
57.4°C.
Check: Q = 75(330.4 – 303) + 5.69(3.3044 – 3.034)
= 2047.5 + 206 = 2253.5 W.
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Combined Convection and Radiation
Problem 1: A surface is at 200°C and is exposed to surroundings at 60°C
and convects and
radiates heat to the surroundings. The convection coefficient is 80W/m2K.
The radiation factor is one. If the heat is conducted to the surface through
a solid of conductivity 12 W/mK, determine the temperature gradient at
the surface in the solid.
Solution:
Heat convected + heat radiated = heat conducted considering 1m2,
4
h(T1 – T2) + σ(T – T 4) = – kdT/dx
Therefore, 80(200 – 60) + 5.67 {[(200 + 273)/100]4 – [(60 + 273)/100]4}
= – 12 dT/dx
Therefore dT/dx = – (11200 + 2140.9)/12 = – 1111.7°C/m.
Problem 2: Heat is conducted through a material with a temperature
gradient of – 9000 °C/m. The conductivity of the material is 25W/mK. If
this heat is convected to surroundings at 30°C with a convection
coefficient of 345W/m2K, determine the surface temperature.
If the heat is radiated to the surroundings at 30°C determine the surface
temperature.
Solution: In this case only convection and conduction are involved.
– kAdT/dx = hA(T1 – T2). Considering unit area,
– 25 × 1 × (– 9000) = 345 × 1 (T1 – 30)
Therefore, T1 = 682.17°C
In this case conduction and radiation are involved.
Heat conducted = Heat radiated
– 25 × 1 × (– 9000) = 5.67 [(T1/100)4 – (303/100)4]
Therefore, T1 = 1412.14K = 1139°C.
Problem 3: There is a heat flux through a wall of 2250W/m2. The same is
dissipated to the
surroundings by convection and radiation. The surroundings is at 30°C.
The convection
coefficient has a value of 75W/m2K. For radiation F = 1. Determine the
wall surface temperature.
Solution: For the specified condition, Consider unit area.
The heat conducted = heat convected + heat radiated
Using the rate equations, with absolute temperature
2250 = [(T2 – 303)/(1/75 × 1] + 5.67 × 1[(T2/100)4 – (303/100)4]
= 75T2 – 22725 + 5.67(T2/100)4 – 477.92
or, (T2/100)4 + 13.2275T2 – 4489.05 = 0.
Downloaded by Areeba Fatima ()
, This equation can be solved only by trial. It may be noted that the
contribution of (T2/100)4 is small and so the first choice of T2 can be a little
less than 4489/13.227 = 340K. The values of the reminder for T2 = 300,
310, 320, 330 are given below:
Assumed value 300 310 320 330 330.4 330.3
of T2
Remainder – 439.80 – 296.2 – 15.1 – 5.38 0.484 – 0.98
So, the temperature T2 is near 330K. By one more trial T2 is obtained as
330.4K or
57.4°C.
Check: Q = 75(330.4 – 303) + 5.69(3.3044 – 3.034)
= 2047.5 + 206 = 2253.5 W.
Downloaded by Areeba Fatima ()
