There are cases with two- and three-dimensional temperature distributions where we would like to get a quick estimate of the heat transfer rate. For example, with a hot water pipe buried underground, the temperature in the soil is obviously two-dimensional. The concept of thermal resistance may be applied in such cases by relating the heat transfer rate to the geometry and thermal conductivity of the soil.
Consider the heat transfer rate from the inner surface to the outer surface of a long hollow block (Figure 1) satisfying the following conditions.
• Inner and outer surfaces are at uniform temperatures of T1 and T2 respectively • Steady state
• No internal energy generation • Constant properties
• Only one homogeneous material
T h e t e m p e r a t u r e d i s t r i b u t i o n i s t w o -dimensional. The heat flow to the outer surface of the block is given by
where n represents the outward normal to the outer surface and the integration is over the outer surface. D e f i n i n g t h e d i m e n s i o n l e s s t e m p e r a t u r e t h e i n t e g r a l c a n b e expressed as
where
θ
is 0 on the inner surface and 1 at the outer surface. The value of∂
θ/∂
n depends only on the dimensions of the block and can be evaluated for different dimensions. The magnitude of the integral, , is known as the conduction shape factor and is denoted by S (with dimension of length). Conduction shape factors for a few cases are given in Table 1:Figure 1 Heat transfer in a block with two-dimensional temperature distribution q
q″ Ad k n ∂ ∂TdA
– = = θ = (T1–T) T⁄( 1–T2) n ∂ ∂TdA
– (T1–T2) n ∂ ∂θdA
(T1–T2)S = = q = Sk T( 1–T2) ∂θ ∂n⁄ ( ) Ad
.
Table 1: Conduction Shape Factors
Configuration Shape factor S (m) Restrictions 1 Edge of two adjoining walls 0.54W W > L/5 2 Corner of three adjoining walls 0.15L L << length and width of wall 3 Isothermal rectangular block in a semi-infinite body with one face parallel to surface of the body
4 Thin rectangu-lar plate buried in a semi-infi-nite medium with an isother-mal surface d = 0 W > L 5 Cylinder of length L cen-tered inside a solid square of W and length L L >> W 2.756L 1 d W ---+ ln 0.59 – H d ---- 0.078 L»W H d, , πW 4W⁄L ( ) ln ---2πL 0.54W R⁄ ( ) ln
---6 Isothermal cyl-inder buried in a semi-infinite medium L >> R L >> R d >3R d >> R L >> d 7 Horizontal cyl-inder midway between semi-infinite paral-lel, isothermal surfaces at the same tempera-ture 8 Isothermal sphere in a semi -infinite medium 9 Isothermal sphere in an infinite medium 10 Isothermal sphere in a semi-infinite medium with an insulated surface
Table 1: Conduction Shape Factors
Configuration Shape factor S (m) Restrictions 2πL cosh–1(d R⁄ ) ---2πL 2d R⁄ ( ) ln ---2πL L R⁄ ( ) 1[ – ln(L 2d⁄ )⁄ln(L R⁄ )] ln ---2πL 4d R⁄ ( ) ln ---4πR 1–R 2d⁄ ---4πR 4πR 1+R 2d⁄
---Example 1
Consider the heat transfer rate from a hot water tube embedded in the floor, used to heat a building. We consider the floor with just one tube although in actual cases there are a number of tubes in the floor.
A 10-mm diameter (O.D.) hot water tube is embedded in 150-mm thick concrete floor whose upper and lower surfaces are at 20 oC as shown in the figure . Determine the heat transfer rate if the tube is 2-m long and the surface of the tube is at 80 oC.
Given
(a) Outer diameter of the tube (2R) = 0.01 m
(b) Thickness of concrete floor (2d) = 0.15 m
(c) Length of tube (L) = 2 m
(d) Temperature of floor surface (Ts) = 20 oC (e) Temperature of tube surface (Tw) = 80 oC
Find
The heat transfer rate from the tube surface 11 Hemisphere on the surface of semi-infinite medium 12 Conduction between two isothermal cyl-inders of length L, in an infi-nite medium L >> R1, R2 L >> d
Table 1: Conduction Shape Factors
Configuration Shape factor S (m) Restrictions 2πR 2πL cosh–1 d 2 R 12 – –R22 2R1R2 ---
---Figure 2 Hot water tube to heat floor of a building
Assumptions
(1) Steady state(2) Width of the concrete slab is much greater than the depth, and the slab may be considered to be semi-infinite.
(3) Constant properties
Solution
For air-dried concrete (2000 kg/m3), k = 0.896 W/m K. From the definition of conduction shape factor, . From Table 1:, entry 7,
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Example 2
Wa t e r a t 0 . 1 k g / s a n d a m e a n temperature of 80 oC flows in a 50-mm wide, 50-mm high, and 5-m long rectangular stainless steel (AISI 304) block, with a 10-mm diameter- passage. Air at 20 oC flows over the block with a surface heat transfer coefficient of 40 W/m2 oC. Assuming that the surface heat transfer coefficient on the inner surface is very large, determine the heat t r a n s f e r r a t e f r o m t h e b l o c k t o t h e surroundingsGiven
(a) Diameter of the passage (2R) = 0.01m
(b) Block dimensions: 0.05 m × 0.05 m × 5 m
(c) Outside air temperature (T∞) = 20 oC
(d) Outer surface heat transfer coefficient (h) = 40 W/m2 oC (e) Material of the block: Type 304 stainless steel
Find
The heat transfer rate from the block to the surrounding air
Assumptions
(1) Steady state.
(2) Outer and inner surfaces of the block are at uniform temperatures.
q = Sk T( w–Ts) S 2πL 4d R⁄ ( ) ln --- 2×π 2× 4×0.075⁄0.005 ( ) ln --- 3.069 m = = = q = 3.069×0.896×(80–20) = 165 W
Figure 3 A rectangular block with a 10-mm diameter passage and the corresponding thermal circuit
(3) As the inner surface heat transfer coefficient is very large, the inner surface is at approximately 80 oC.
(4) Negligible heat transfer from the end surfaces of the block.
Solution
The heat transfer rate can be computed from the thermal circuit. The conduction resistance is found through the appropriate conduction shape factor and added to the resistance associated with the outer surface. From the resistance circuit:
The thermal conductivity of type 304 stainless steel (k) = 14.9 W/m K. Since W = 0.05 m and L = 5 m, L >> W, the conduction shape factor given in Table 1:, entry 5, is used.
Comment
Water enters the tube at 80 oC but because of the heat transfer to the surroundings, the temperature of the water decreases in the direction of flow. This may result in the inner surface temperature not being uniform. The change in the water temperature, is found from
At 80 oC the specific heat of water is 4193 J/kg K. Substituting the values we find that is approximately 5 oC. If the water enters at 82.5 oC, it exits at 77.5 oC. The deviation from the assumed mean temperature of 80 oC is not significant and neglecting the axial variation of the surface temperature is acceptable.
q Ti–T∞ 1 Sk --- 1 hoAo ---+ ---= S 2πL 0.54W R --- ln ---= 1 Sk ---0.54×0.05 0.005 --- ln 2×π 5 14.9× × --- 3.603×10 K/W–3 = = 1 hoAo --- 1 40×(4×0.05×5) --- 0.025 K/W = = q 80–20 3.603×10–3+0.025 --- 2098 W = = q = m· cp(Tin–Tout) Tin–Tout ( )
