During a visit to a plastic sheeting plant, it was observed that a 60 -m-long section of a 2 -in nominal \((6.03\)-cm-outerdiameter) steam pipe extended from one end of the plant to the other with no insulation on it. The temperature measurements at several locations revealed that the average temperature of the exposed surfaces of the steam pipe was \(170^{\circ} \mathrm{C}\), while the temperature of the surrounding air was \(20^{\circ} \mathrm{C}\). The outer surface of the pipe appeared to be oxidized, and its emissivity can be taken to be \(0.7\). Taking the temperature of the surrounding surfaces to be \(20^{\circ} \mathrm{C}\) also, determine the rate of heat loss from the steam pipe. Steam is generated in a gas furnace that has an efficiency of 78 percent, and the plant pays \(\$ 1.10\) per therm ( 1 therm \(=\) \(105,500 \mathrm{~kJ}\) ) of natural gas. The plant operates \(24 \mathrm{~h}\) a day 365 days a year, and thus \(8760 \mathrm{~h}\) a year. Determine the annual cost of the heat losses from the steam pipe for this facility.

Step by step solution

01

Calculate convective heat loss

To calculate the convective heat loss, we can use the formula: \(Q_{conv} = h\cdot A\cdot (T_s - T_\infty)\) where: \(Q_{conv}\) = convective heat loss (W) \(h\) = convection heat transfer coefficient (W/m²°C) \(A\) = surface area (m²) \(T_s\) = steam pipe surface temperature (°C) \(T_\infty\) = surrounding air temperature (°C) For this exercise, we are not given the value of \(h\), so we will assume it to be an average value of 10 W/m²°C for free convection from a horizontal cylinder to air, as found in standard heat transfer texts. The steam pipe has a length of 60 m and an outer diameter of 6.03 cm. To calculate the surface area, we can use the formula: \(A = L \cdot D_\text{outer} \cdot \pi\) where: \(D_\text{outer}\) = outer diameter (m) \(L\) = length of the pipe (m) Calculating the surface area: \(A = 60 \cdot 0.0603 \cdot \pi = 11.36 \mathrm{~m^2}\) Now we can determine \(Q_{conv}\): \(Q_{conv} = 10 \cdot 11.36 \cdot (170 - 20) = 10 \cdot 11.36 \cdot 150 = 17,040 \mathrm{~W}\)

02

Calculate radiative heat loss

To calculate the radiative heat loss, we can use the formula: \(Q_{rad} = \epsilon \cdot \sigma \cdot A \cdot (T_s^4 - T_\infty^4)\) where: \(Q_{rad}\) = radiative heat loss (W) \(\epsilon\) = emissivity of the pipe surface (dimensionless) \(\sigma\) = Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{~W/m^2K^4}\)) \(T_s\) and \(T_\infty\) = temperature in Kelvin Now we can convert temperatures to Kelvin and determine \(Q_{rad}\): \(T_s = 170 + 273.15 = 443.15 \mathrm{~K}\) \(T_\infty = 20 + 273.15 = 293.15 \mathrm{~K}\) \(Q_{rad} = 0.7 \cdot 5.67 \times 10^{-8} \cdot 11.36 \cdot (443.15^4 - 293.15^4) = 6,118 \mathrm{~W}\)

03

Calculate total heat loss

The total heat loss is the sum of the convective and radiative heat losses: \(Q_\text{total} = Q_{conv} + Q_{rad} = 17,040 + 6,118 = 23,158 \mathrm{~W}\)

04

Calculate the annual cost of heat losses

We can now use the total heat loss rate to determine the annual cost of this heat loss. First, we need to find the energy loss per year: \(E_\text{loss} = Q_\text{total} \cdot t_\text{operating}\) where: \(t_\text{operating}\) = operating hours per year (8760 h) \(E_\text{loss} = 23,158 \mathrm{~W} \cdot 8,760 \mathrm{~h} = 202,944,880 \mathrm{~Wh} = 202,944.88 \mathrm{~kWh}\) Next, we can calculate the energy content of the fuel used per year: \(E_\text{fuel} = \frac{E_\text{loss}}{\eta}\) where: \(\eta\) = gas furnace efficiency (0.78) \(E_\text{fuel} = \frac{202,944.880 \mathrm{~kWh}}{0.78} = 260,187.282 \mathrm{~kWh}\) Finally, we can calculate the annual cost of heat losses: \(C_\text{annual} = \frac{E_\text{fuel}}{105.5 \mathrm{~kWh/therm}} \cdot \$1.10 \mathrm{/therm}\) \(C_\text{annual} = \frac{260,187.282 \mathrm{~kWh}}{105.5 \mathrm{~kWh/therm}} \cdot \$1.10 \mathrm{/therm} = \$2,710.12\) Therefore, the annual cost of the heat losses from the steam pipe for this facility is $2,710.12.

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