Question

A spherical tank with an inner diameter of \(3 \mathrm{~m}\) is filled with a solution undergoing an exothermic reaction that generates $233 \mathrm{~W} / \mathrm{m}^{3}\( of heat and causes the surface temperature to be \)120^{\circ} \mathrm{C}$. To prevent thermal burn hazards, the tank is enclosed by a concentric outer cover that provides an air gap of \(5 \mathrm{~cm}\) in the enclosure. Determine whether the air gap is sufficient to keep the outer cover temperature below \(45^{\circ} \mathrm{C}\) to prevent thermal burns on human skin. Evaluate the properties of air in the enclosure at $80^{\circ} \mathrm{C}$ and 1 atm pressure. Is this a good assumption?

Short answer

Answer: Yes, the air gap is sufficient, as the calculated outer cover temperature is approximately \(37.46^{\circ}\mathrm{C}\), which is below the threshold of \(45^{\circ}\mathrm{C}\).

Step-by-step solution

Compute the heat generated in the tank

We are given the heat generation rate in the tank (\(233\mathrm{~W/m^3}\)) and the tank's inner diameter (\(3\mathrm{~m}\)). First, we need to find the volume of the tank. The radius of the tank (\(r_t\)) can be found using the diameter: $$r_t = \frac{d}{2} = \frac{3\mathrm{~m}}{2}=1.5\mathrm{~m}$$ Now we can find the volume of the tank (\(V_t\)) using the formula for the volume of a sphere: $$V_t = \frac{4}{3}\pi r_t^3 = \frac{4}{3}\pi (1.5\mathrm{~m})^3 \approx 14.137\mathrm{~m^3}$$ Next, we can calculate the total heat generated in the tank (\(q_t\)) using the heat generation rate: $$q_t = \left( 233\frac{\mathrm{W}}{\mathrm{m}^3} \right) \times \left( 14.137\mathrm{~m^3} \right) \approx 3293.99\mathrm{~W}$$

Calculate the heat transfer through the air gap

To do this, we first find the thermal resistance (\(R_{airgap}\)) of the air gap using the formula for the thermal resistance of a concentric sphere: $$R_{airgap} = \frac{1}{4\pi k_{air} }\left( \frac{1}{r_t}-\frac{1}{r_{cover}} \right)$$ Here, \(k_{air}\) is the thermal conductivity of the air in the gap (evaluated at \(80^{\circ}\mathrm{C}\)), \(r_t\) is the radius of the tank (\(1.5\mathrm{~m}\)), and \(r_{cover}\) is the radius of the outer cover. The air gap is \(5\mathrm{~cm}\) or \(0.05\mathrm{~m}\), so the radius of the outer cover is: $$r_{cover}=r_t + 0.05\mathrm{~m} = 1.55\mathrm{~m}$$ Given that we're evaluating the properties of air in the enclosure at \(80^{\circ}\mathrm{C}\), the thermal conductivity of air (\(k_{air}\)) can be found in the literature or using an online calculator. We'll assume \(k_{air} \approx 0.0342\mathrm{~W/(m\cdot K)}\). Now we can find the thermal resistance of the air gap: $$R_{airgap} = \frac{1}{4\pi(0.0342\mathrm{~W/(m\cdot K)}) }\left( \frac{1}{1.5\mathrm{~m}}-\frac{1}{1.55\mathrm{~m}} \right) \approx 4.052\mathrm{~m^2 \cdot K/W}$$ Since the heat generated in the tank must transfer through the air gap, we can use the thermal resistance to find the heat transfer through the air gap: $$q_{airgap} = \frac{T_t-T_{cover}}{R_{airgap}}$$ Here, \(T_t\) is the surface temperature of the tank, which is given as \(120^{\circ}\mathrm{C}\), and \(T_{cover}\) is the temperature of the outer cover, which we want to find.

Determine the temperature of the outer cover

Now we can solve for \(T_{cover}\) using the equation above: $$T_{cover} = T_t - q_{airgap}\cdot R_{airgap} = 120^{\circ}\mathrm{C} - (3293.99\mathrm{~W}) \times (4.052\mathrm{~m^2 \cdot K/W}) \approx 37.46^{\circ}\mathrm{C}$$

Check the assumption for air properties

Since the temperature of the outer cover is calculated as \(37.46^{\circ}\mathrm{C}\), which is close to the given temperature of air at \(80^{\circ}\mathrm{C}\), our assumption of using the properties of air at this temperature is appropriate. Since the outer cover temperature is \(37.46^{\circ}\mathrm{C}\), which is below the threshold of \(45^{\circ}\mathrm{C}\), we can conclude that the air gap is sufficient to keep the outer cover temperature below the threshold and prevent thermal burns.