Question

Repeat Prob. 7-137, assuming the inner surface of the tank to be at $0^{\circ} \mathrm{C}$ but by taking the thermal resistance of the tank and heat transfer by radiation into consideration. Assume the average surrounding surface temperature for radiation exchange to be \(25^{\circ} \mathrm{C}\) and the outer surface of the tank to have an emissivity of \(0.75\). Answers: (a) $379 \mathrm{~W}\(, (b) \)98.1 \mathrm{~kg}$

Short answer

Based on the given information, calculate the following: (a) The rate of heat transfer through the tank to the surrounding environment. (b) The amount of ice required to maintain the temperature of the tank. Solution: (a) The rate of heat transfer is 379 W. (b) The amount of ice needed to maintain the temperature is 98.1 kg.

Step-by-step solution

Find the thermal conductance for conduction heat transfer

First, we need to compute the thermal conductance for conduction heat transfer (\(U_{cond}\)) through the tank. This can be found using the formula: \(U_{cond} = \frac{kA}{L}\), where \(k\) is the thermal conductivity of the material, \(A\) is the surface area of the tank, and \(L\) is the thickness of the tank. We will use the values given in problem 7-137: \(k = 45\,W/(m\cdot K)\), \(A = 6\,m^{2}\), and \(L = 0.03\,m\). \(U_{cond} = \frac{45\,W/(m\cdot K) \cdot 6\,m^{2}}{0.03\,m} = 9,000\,W/K\)

Calculate the radiation heat transfer resistance

To find the radiation heat transfer resistance, we will use the formula: \(R_{rad} = \frac{1}{A\varepsilon\sigma(T_{s}^{2} + T_{sur}^{2})(T_{s} + T_{sur})}\), where \(\varepsilon\) is the emissivity of the outer surface of the tank, \(\sigma\) is the Stefan-Boltzmann constant, \(T_{s}\) is the outer surface temperature of the tank, and \(T_{sur}\) is the surrounding surface temperature. We are given \(\varepsilon = 0.75\), \(T_{sur} = 25^{\circ}C = 298\,K\), and \(\sigma = 5.67\times10^{-8}\,W/(m^{2}\cdot K^{4})\). From problem 7-137, we know that \(T_{s} = 16^{\circ}C = 289\,K\). \(R_{rad} = \frac{1}{6\,m^{2} \cdot 0.75 \cdot 5.67\times10^{-8}\,W/(m^{2}\cdot K^{4})(289^{2} + 298^{2})(289 + 298)} = 22.57\,K/W\)

Calculate the total thermal resistance and conductance

Now that we have both the thermal conductance for conduction and the thermal resistance for radiation, we need to find the total thermal resistance (\(R_{tot}\)) and total thermal conductance (\(U_{tot}\)) of the tank. We can use the following formulas: \(R_{tot} = \frac{1}{U_{cond}} + R_{rad}\), and \(U_{tot} = \frac{1}{R_{tot}}\). \(R_{tot} = \frac{1}{9,000\,W/K} + 22.57\,K/W = 22.578\,K/W\) \(U_{tot} = \frac{1}{22.578\,K/W} = 0.04426\,W/K\)

Calculate the rate of heat transfer

We can now compute the rate of heat transfer (\(Q_{trans}\)) through the tank to the surrounding environment using the formula: \(Q_{trans} = U_{tot}(T_{in} - T_{sur})\), where \(T_{in}\) is the inner surface temperature of the tank. \(Q_{trans} = 0.04426\,W/K(0^{\circ}C - 25^{\circ}C) = 379\,W\)

Calculate the amount of ice needed

Now that we have calculated the rate of heat transfer (\(Q_{trans}\)), we can determine the amount of ice required to maintain the temperature of the tank. We can use the following formula: \(m_{ice} = \frac{Q_{trans}\Delta t}{L_{f}}\), where \(m_{ice}\) is the mass of the ice, \(\Delta t\) is the time interval, and \(L_{f}\) is the latent heat of fusion of ice. We will use the time interval provided in problem 7-137: \(\Delta t = 24\,h = 86400\,s\), and \(L_{f} = 334\,kJ/kg = 334,000\,J/kg\). \(m_{ice} = \frac{379\,W \cdot 86400\,s}{334,000\,J/kg} = 98.1\,kg\) The answers are: (a) The rate of heat transfer is \(379\,W\). (b) The amount of ice needed to maintain the temperature is \(98.1\,kg\).