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) \(-12^{\circ} \mathrm{C}\)

Step by step solution

01

Calculate Aluminum Resistance

Using the formula for the resistance of a cylinder, calculate the resistance through the aluminum shell: \(R_{al} = \frac{ln(r_2/r_1)}{2 \pi \; l \; k_{al}}\) Where: \(r_1\) is the inner radius of the aluminum shell (0.5 m) \(r_2\) is the outer radius of the aluminum shell (0.5 m + 0.005 m) \(l\) is the length of the aluminum shell (1 m) \(k_{al}\) is the thermal conductivity of the aluminum (170 W/m⋅K)

02

Calculate Insulation Resistance

Calculate the resistance through the insulation: \(R_{ins} = \frac{ln(r_3/r_2)}{2 \pi \; l \; k_{ins}}\) Where: \(r_3\) is the outer radius of the insulation (0.5 m + 0.005 m + 0.1 m) \(k_{ins}\) is the thermal conductivity of the insulation (0.02 W/m⋅K)

03

Compute Overall Resistance

Calculate the overall conduction resistance: \(R_{cond} = R_{al} + R_{ins}\) Step 2: Perform energy balance with convection resistance

04

Compute Convection Resistance

Calculate the convection resistance on the exterior side of the insulation: \(R_{conv} = \frac{1}{h \; A}\) Where: \(h\) is the heat transfer coefficient (5 W/m²⋅K) \(A\) is the surface area of the exterior side of the insulation (\(2 \pi \; r_3 \; l\))

05

Heat Energy Balance

Perform a heat energy balance by equating the heat transfer rate through conduction and convection: \(Q = \frac{T_{ins} - T_{liq}}{R_{cond}} = \frac{T_{amb} - T_{ins}}{R_{conv}}\) Where: \(T_{liq}\) is the temperature of the liquid oxygen (90 K or -183.15°C) \(T_{ins}\) is the temperature of the exterior surface of the insulation (unknown) \(T_{amb}\) is the ambient temperature (20°C) Step 3: Solve for the exterior surface temperature

06

Rearrange Energy Balance Equation

Rearrange the heat energy balance equation to solve for the exterior surface temperature: \(T_{ins} = \frac{T_{liq} \; R_{conv} + T_{amb} \; R_{cond}}{R_{conv} + R_{cond}}\)

07

Calculate Exterior Surface Temperature

Calculate the exterior surface temperature of the insulation: \(T_{ins} = \frac{(-183.15 \mathrm{~°C}) \; R_{conv} + (20 \mathrm{~°C}) \; R_{cond}}{R_{conv} + R_{cond}}\) Once we calculated \(T_{ins}\), we can verify which one of the given options is closest to the obtained value: (a) \(13^{\circ} \mathrm{C}\) (b) \(9^{\circ} \mathrm{C}\) (c) \(2^{\circ} \mathrm{C}\) (d) \(-3^{\circ} \mathrm{C}\) (e) \(-12^{\circ} \mathrm{C}\)

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