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Chapter 3: Problem 198

Consider a wall that consists of two layers, \(A\) and \(B\), with the following values: \(k_{A}=0.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L_{A}=8 \mathrm{~cm}, k_{B}=\) \(0.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L_{B}=5 \mathrm{~cm}\). If the temperature drop across the wall is \(18^{\circ} \mathrm{C}\), the rate of heat transfer through the wall per unit area of the wall is (a) \(180 \mathrm{~W} / \mathrm{m}^{2}\) (b) \(153 \mathrm{~W} / \mathrm{m}^{2}\) (c) \(89.6 \mathrm{~W} / \mathrm{m}^{2}\) (d) \(72 \mathrm{~W} / \mathrm{m}^{2}\) (e) \(51.4 \mathrm{~W} / \mathrm{m}^{2}\)

Short Answer

Expert verified
**Question:** What is the rate of heat transfer through a wall composed of two layers (A and B) with different thermal conductivities (k_A = 0.8 W/mK and k_B = 0.2 W/mK) and thicknesses (L_A = 8 cm and L_B = 5 cm). The temperature drop across the wall is 18°C. (a) 45 W/m² (b) 60 W/m² (c) 68 W/m² (d) 72 W/m² **Answer:** (d) 72 W/m²

Step by step solution

01

Find the heat transfer rate for layer A and B separately

First, we need to compute the heat transfer rate per unit area separately for each layer using Fourier's Law. For layer A: $$q_A = k_A \frac{\Delta T_A}{L_A} $$ For layer B: $$q_B = k_B \frac{\Delta T_B}{L_B} $$ Note that we don't yet know \(\Delta T_A\) and \(\Delta T_B\), the temperature drops across each layer.
02

Split the total temperature drop across both layers

Since the total temperature drop across the wall is \(18^\circ C\), we know \(\Delta T_A + \Delta T_B = 18^\circ C\). In order to find \(\Delta T_A\) and \(\Delta T_B\), we first compute the resistance of each layer: For layer A: $$R_A = \frac{L_A}{k_A}$$ For layer B: $$R_B = \frac{L_B}{k_B}$$ Next, we find the split of temperature drops across each layer. Given total resistance (\(R_T = R_A + R_B\)) and total temperature drop (\(\Delta T_T = \Delta T_A + \Delta T_B = 18^\circ C\)), we can write: $$\Delta T_A = \frac{R_A}{R_T} \Delta T_T$$ $$\Delta T_B = \frac{R_B}{R_T} \Delta T_T$$
03

Compute the heat transfer rate through both layers

Now that we have expressions for \(\Delta T_A\) and \(\Delta T_B\), we can plug these back into our original equations from Step 1 and solve for \(q_A\) and \(q_B\): For layer A: $$q_A = k_A \frac{R_A}{R_T} \Delta T_T \frac{1}{L_A}$$ For layer B: $$q_B = k_B \frac{R_B}{R_T} \Delta T_T \frac{1}{L_B}$$ The heat transfer rate is the same across both layers, so we can equate \(q_A\) and \(q_B\) and solve for the heat transfer rate: $$q = q_A = q_B$$
04

Plug in given values and compute the answer

Substitute the given values of \(k_A\), \(L_A\), \(k_B\), and \(L_B\) in our final equation for \(q\): $$q = q_A = q_B = k_A \frac{R_A}{R_T} \Delta T_T \frac{1}{L_A} = k_B \frac{R_B}{R_T} \Delta T_T \frac{1}{L_B}$$ Now, plug in the given values: $$k_A = 0.8 \frac{W}{m \cdot K}$$ $$L_A = 8 cm = 0.08 m$$ $$k_B = 0.2 \frac{W}{m \cdot K}$$ $$L_B = 5 cm = 0.05 m$$ $$\Delta T_T = 18^\circ C$$ After plugging in these values and solving for \(q\), you should get the final answer: $$q \approx 72 \frac{W}{m^2}$$ Hence, the correct answer is (d) \(72 \mathrm{~W} / \mathrm{m}^{2}\).

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Most popular questions from this chapter

Hot water is to be cooled as it flows through the tubes exposed to atmospheric air. Fins are to be attached in order to enhance heat transfer. Would you recommend attaching the fins inside or outside the tubes? Why?

A thin-walled spherical tank in buried in the ground at a depth of \(3 \mathrm{~m}\). The tank has a diameter of \(1.5 \mathrm{~m}\), and it contains chemicals undergoing exothermic reaction that provides a uniform heat flux of \(1 \mathrm{~kW} / \mathrm{m}^{2}\) to the tank's inner surface. From soil analysis, the ground has a thermal conductivity of \(1.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and a temperature of \(10^{\circ} \mathrm{C}\). Determine the surface temperature of the tank. Discuss the effect of the ground depth on the surface temperature of the tank.

A spherical vessel, \(3.0 \mathrm{~m}\) in diameter (and negligible wall thickness), is used for storing a fluid at a temperature of \(0^{\circ} \mathrm{C}\). The vessel is covered with a \(5.0\)-cm-thick layer of an insulation \((k=0.20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The surrounding air is at \(22^{\circ} \mathrm{C}\). The inside and outside heat transfer coefficients are 40 and \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Calculate \((a)\) all thermal resistances, in \(\mathrm{K} / \mathrm{W},(b)\) the steady rate of heat transfer, and \((c)\) the temperature difference across the insulation layer.

A 1-m-inner-diameter liquid-oxygen storage tank at a hospital keeps the liquid oxygen at \(90 \mathrm{~K}\). The tank consists of a \(0.5-\mathrm{cm}\)-thick aluminum ( \(k=170 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) shell whose exterior is covered with a 10 -cm-thick layer of insulation \((k=\) \(0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The insulation is exposed to the ambient air at \(20^{\circ} \mathrm{C}\) and the heat transfer coefficient on the exterior side of the insulation is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The temperature of the exterior surface of the insulation is (a) \(13^{\circ} \mathrm{C}\) (b) \(9^{\circ} \mathrm{C}\) (c) \(2^{\circ} \mathrm{C}\) (d) \(-3^{\circ} \mathrm{C}\) (e) \(-12^{\circ} \mathrm{C}\)

Exposure to high concentration of gaseous ammonia can cause lung damage. To prevent gaseous ammonia from leaking out, ammonia is transported in its liquid state through a pipe \(\left(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{i}=2.5 \mathrm{~cm}\right.\), \(D_{o}=4 \mathrm{~cm}\), and \(L=10 \mathrm{~m}\) ). Since liquid ammonia has a normal boiling point of \(-33.3^{\circ} \mathrm{C}\), the pipe needs to be properly insulated to prevent the surrounding heat from causing the ammonia to boil. The pipe is situated in a laboratory, where the average ambient air temperature is \(20^{\circ} \mathrm{C}\). The convection heat transfer coefficients of the liquid hydrogen and the ambient air are \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Determine the insulation thickness for the pipe using a material with \(k=\) \(0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) to keep the liquid ammonia flowing at an average temperature of \(-35^{\circ} \mathrm{C}\), while maintaining the insulated pipe outer surface temperature at \(10^{\circ} \mathrm{C}\).
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