HEAT TRANSFER: solutions to radiation tutorial problems

1. For both hot plate (1) and cold plate (2), X = 2 m, Y = 2 m, H = 1 m
® F12 = F21 = 0.415253284 (from formula)

Let surroundings be imaginary surface 3.
Both F11 and F22 are zero (surfaces are flat).
® F13 = F23 = 1 - 0.415253284 = 0.584746716 (by rule of addition)
Defining variables T1 = 1273.15 K
  T2 = 773.15 K
  T3 = 293.15 K
  e1 = 1 (black body)
  e2 = 1 (black body)
  s = 5.67 ´ 10-8 W/m2K4
  A1 = 4 m2
  A2 = 4 m2

Heat transferred from hot plate (1) to cold plate (2)
Q’12 = A1F12e1s[(T1)4 - (T2)4]
= 4 ´ 0.415253284 ´ 1 ´ 5.67 ´ 10-8 ´ [1273.154 - 773.154]
= 213 790.56 Watts

Heat transferred from hot plate (1) to surroundings (3)
Q’13 = A1F13e1s[(T1)4 - (T3)4]
= 4 ´ 0.584746716 ´ 1 ´ 5.67 ´ 10-8 ´ [1273.154 - 293.154]
= 347 461.48 Watts

Heat transferred from cold plate (2) to surroundings (3)
Q’23 = A2F23e1s[(T2)4 - (T3)4]
= 4 ´ 0.584746716 ´ 1 ´ 5.67 ´ 10-8 ´ [773.154 - 293.154]
= 46 408.30 Watts

Net loss from hot plate = 213 790.56 + 347 461.48 = 561 252.04 Watts
Net gain by cold plate = 213 790.56 - 46 408.30 = 167 382.26 Watts
Net loss to surroundings = 46 408.30 + 347 461.48 = 393 869.78 Watts


2. First let the view factors be calculated:

A fourth (imaginary) surface needs to be added: a circular disc at the base of the hemisphere, 0.5 metres radius and parallel to the other disc 0.5 metres below.

Surface 4 is flat, therefore all radiation leaving its upper side goes to 2: ® F42 = 1.0

By reciprocity, A4F42 = A2F24

Surfaces 4 and 1 are parallel coaxial discs

The fraction of radiation from the hemisphere that hits the solid disc is the product of the two view factors just calculated.
® F21 = 0.5 ´ 0.381966 = 0.190983

By addition, F41 + F43 = 1
® F43 = 1 - 0.381966 = 0.618034

As surfaces 1 and 4 are geometrically identical with respect to each other,
® F13 = 0.618034

Defining variables T1 = 673.15 K
T2 = 1123.15 K
T3 = 288.15 K
e1 = 1 (black body)
e2 = 1 (black body)
s = 5.67 ´ 10-8 W/m2K4
A1 = p (0.5)2 m2
A2 = 2p (0.5)2 m2

Heat transferred from hot hemisphere (2) to warm plate (1)
Q’21 = A2F21e2s[(T2)4 - (T1)4]
= 2p (0.25) ´ 0.190983 ´ 1 ´ 5.67 ´ 10-8 ´ [1123.154 - 673.154]
= 23 574.965 Watts

Heat transferred from warm plate (1) to cold surroundings (3)
Q’13 = A1F13e1s[(T1)4 - (T3)4]
= p (0.25) ´ 0.618034 ´ 1 ´ 5.67 ´ 10-8 ´ [673.154 - 288.154]
= 5461.26 Watts

Net gain by cold plate = 23 574.965 - 5461.26 = 18113.609 Watts


3. Name hot plate (1), cold plate (2) and walls (3).
As surfaces 1 and 2 are geometrically identical with respect to each other,
® F23 = F13

By reciprocity, A2F23 = A3F32

For walls to reach an equilibrium temperature, their net heat loss should be zero.
® Q’13 = Q’32
A1F13e1s[(T1)4 - (T3)4] = A3F32e3s[(T3)4 - (T2)4]

Substituting for view factors, inserting areas, accounting for black surfaces and simplifying:
(4) [(T1)4 - (T3)4] = 0.5 (8) [(T3)4 - (T2)4]
(1273.15)4 - (T3)4 = (T3)4 - (773.15)4
(1273.15)4 + (773.15)4 = 2(T3)4
2(T3)4 = 2.985 ´ 1012 K4
(T3)4 = 1.492 ´ 1012 K4
T3 = 1105.27 K º 832.12°C

In the instance of heat lost through the walls, we need to evaluate the view factors:
® F12 = F21 = 0.415253284 (from formula)

Both F11 and F22 are zero (surfaces are flat).
® F13 = F23 = 1 - 0.415253284 - 0 = 0.584746716 (by rule of addition)

The walls lose 10 kW/m2 for their 8 m2 of area:
® Q’13 = Q’32 + 80000
A1F13e1s[(T1)4 - (T3)4] = A3F32e3s[(T3)4 - (T2)4] + 80000

Simplifying view factors, and inserting areas and emissivities (black body):
4F13s[(T1)4 - (T3)4] = 8(0.5F13)s[(T3)4 - (T2)4] + 80000
4F13s[(T1)4 + (T2)4] - 80000 = 2 ´ 4F13s(T3)4

Insert Stefan-Boltzmann constants, temperatures, view factors and do arithmetic:

= 0.5 [(2.6273522 ´ 1012) + (3.5731812 ´ 1011)] - 3.0161237 ´ 1011
= 1.1907228 ´ 1012 K4
T3 = 1044.61 K º 771.46°C

To know net amount of heat transferred to cold plate, we need both Q’12 and Q’32:
Q’32 = A3F32e3s[(T3)4 - (T2)4]
= 8 ´ 0.5 ´ 0.584746716 ´ 1 ´ 5.67 ´ 10-8[(1044.61)4 - (773.15)4]
= 110 526.59 Watts

Q’12 = A1F12e1s[(T1)4 - (T2)4]
= 4 ´ 0.584746716 ´ 1 ´ 5.67 ´ 10-8[(1273.15)4 - (773.15)4]
= 301 053.18 Watts

Net heat radiated to cold plate: 110 526.59 + 301 053.18 = 411 579.77 Watts


4. Surfaces 1 and 2 are parallel coaxial discs

Surface 3 is the surrounding wall. As surface 1 is perfectly flat, F11 = 0, so by addition:
1 = F11 + F12 + F13 = 0 + 0.2864217 + F13
® F13 = 1 - 0.2864217 = 0.7135783

Defining variables T1 = 1773.15 K
T2 = 373.15 K
T3 = unknown
e1 = 1 (black body)
e2 = 1 (black body)
s = 5.67 ´ 10-8 W/m2K4
A1 = p (0.75)2 m2

Some heat will be absorbed by the wall and then emitted to the vessel. For walls to reach an equilibrium temperature, their net heat loss should be zero.
® Q’13 = Q’32
A1F13e1s[(T1)4 - (T3)4] = A3F32e3s[(T3)4 - (T2)4]

By reciprocity, A3F32 = A2F23
By geometric identity, A2F23 = A1F13
A1F13e1s[(T1)4 - (T3)4] = A1F13e3s[(T3)4 - (T2)4]

Accounting for black surfaces and simplifying:
® [(T1)4 - (T3)4] = [(T3)4 - (T2)4]
(T1)4 + (T2)4 = 2(T3)4
2(T3)4 = (1773.15)4 + (373.15)4
2(T3)4 = 9.905 ´ 1012 K4
(T3)4 = 4.952 ´ 1012 K4
T3 = 1491.766 K

Heat directly transferred from wall (3) to vessel (2)
Q’32 = A3F32e3s[(T3)4 - (T2)4] = A1F13e3s[(T3)4 - (T2)4]
= p (0.75)2 ´ 0.7135783 ´ 1 ´ 5.67 ´ 10-8 ´ [1491.7664 - 373.154]
= 635 825.67 Watts

Heat directly transferred from heater (1) to vessel (2)
Q’12 = A1F12e1s[(T1)4 - (T2)4]
= p (0.75)2 ´ 0.2864217 ´ 1 ´ 5.67 ´ 10-8 ´ [1773.154 - 373.154]
= 283 133.10 Watts

Net heat radiated to vessel: 635 825.67 + 283 133.10 = 411 579.77 Watts

If actual T3 = 10°C º 283.15K, Q’12 is unchanged, but:
Q’32 = A1F13e3s[(T3)4 - (T2)4]
= p (0.75)2 ´ 0.7135783 ´ 1 ´ 5.67 ´ 10-8 ´ [283.154 - 373.154]
= -926.63 Watts

Net heat radiated to vessel: 283 133.10 - 926.63 = 282 206.47 Watts
Percentage reduction heat to vessel:

= 55.615748 %