Superheated steam at an average temperature \(200^{\circ} \mathrm{C}\) is transported through a steel pipe \(\left(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{o}=8.0 \mathrm{~cm}\right.\), \(D_{i}=6.0 \mathrm{~cm}\), and \(L=20.0 \mathrm{~m}\) ). The pipe is insulated with a 4-cm thick layer of gypsum plaster \((k=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The insulated pipe is placed horizontally inside a warehouse where the average air temperature is \(10^{\circ} \mathrm{C}\). The steam and the air heat transfer coefficients are estimated to be 800 and \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Calculate \((a)\) the daily rate of heat transfer from the superheated steam, and \((b)\) the temperature on the outside surface of the gypsum plaster insulation.

Step by step solution

01

Calculate the Thermal Resistance of Steel Pipe

Find the thermal resistance of the steel pipe using the formula and given values: \(R_{\text{steel}} = \frac{\ln\left(\frac{8\,\text{cm}}{6\,\text{cm}}\right)}{2 \pi (50\, \mathrm{W/m \cdot K})(20\,\text{m})}\) \(R_{\text{steel}} = \frac{\ln\left(\frac{4}{3}\right)}{2 \pi (50\, \mathrm{W/m \cdot K})(20\,\text{m})} = 9.74 \times 10^{-5}\, \mathrm{K} / \mathrm{W}\)

02

Calculate the Thermal Resistance of Gypsum Plaster Insulation

Similarly, find the thermal resistance of the gypsum plaster insulation using the formula and given values: \(R_{\text{gypsum}} = \frac{\ln\left(\frac{12\,\text{cm}}{8\,\text{cm}}\right)}{2 \pi (0.5\, \mathrm{W/m \cdot K})(20\,\text{m})}\) \(R_{\text{gypsum}} = \frac{\ln\left(\frac{3}{2}\right)}{2 \pi (0.5\, \mathrm{W/m \cdot K})(20\,\text{m})} = 0.00379\, \mathrm{K} / \mathrm{W}\)

03

Calculate Temperature Difference

Calculate the temperature difference between steam and air required for heat transfer: \(\Delta T = (200^{\circ}\mathrm{C}) - (10^{\circ}\mathrm{C}) = 190\,\text{K}\)

04

Calculate Overall Thermal Resistance

Now, we will find the combined thermal resistance of both layers: \(R_{\text{total}} = R_{\text{steel}} + R_{\text{gypsum}} = 9.74 \times 10^{-5}+ 0.00379 \,\mathrm{K}/\mathrm{W} = 0.00389\, \mathrm{K} / \mathrm{W}\)

05

Calculate Heat Transfer Rate

To determine the rate of heat transfer, we will use the formula: \(q = \frac{\Delta T}{R_{\text{total}}}\) Plug in the values: \(q = \frac{190\, \mathrm{K}}{0.00389\, \mathrm{K} / \mathrm{W}} = 48817\, \mathrm{W}\)

06

(a) Daily Rate of Heat Transfer

Now, to find the daily rate of heat transfer, convert the rate from watts to kilowatt-hours and then multiply by the number of hours in a day (24 hours): \(\text{Heat Transfer Rate} = \frac{48817\, \mathrm{W}}{1000\, \mathrm{W/kW}} \times 24\, \mathrm{hr/day} = 1171.6\, \mathrm{kWh/day}\)

07

Calculate Temperature of Outer Surface

Next, we will find the temperature on the outer surface of the gypsum plaster insulation. We know that \(q = U A \Delta T\). We can find \(U\) by using the formula: \(U = \frac{1}{R_{\text{total}}}\) Now, we can find the temperature difference between outside and inside of the insulation: \(\Delta T_{\text{gypsum}} = \frac{q}{U A} = \frac{48817\, \mathrm{W}}{\frac{1}{0.00389\, \mathrm{K/W}} \times 2\pi \left(\frac{12\,\text{cm}}{100\,\text{cm/m}}\right) 20\,\text{m}} = 38\,\mathrm{K}\)

08

(b) Temperature on the Outside Surface of Gypsum Plaster Insulation

Now, using the temperature difference, we can find the temperature of the outer surface of the gypsum plaster insulation: \(T_{\text{outer}} = T_{\text{steam}} - \Delta T_{\text{gypsum}} = 200^{\circ}\mathrm{C} - 38\,\mathrm{K} = 162^{\circ}\mathrm{C}\) So, \((a)\) the daily rate of heat transfer from the superheated steam is 1171.6 kWh/day, and \((b)\) the temperature on the outside surface of the gypsum plaster insulation is 162°C.

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