Exhaust gases from a manufacturing plant are being discharged through a 10 - \(\mathrm{m}\) tall exhaust stack with outer diameter of \(1 \mathrm{~m}\), wall thickness of \(10 \mathrm{~cm}\), and thermal conductivity of \(40 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The exhaust gases are discharged at a rate of \(1.2 \mathrm{~kg} / \mathrm{s}\), while temperature drop between inlet and exit of the exhaust stack is \(30^{\circ} \mathrm{C}\), and the constant pressure specific heat of the exhaust gasses is \(1600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). On a particular day, the outer surface of the exhaust stack experiences radiation with the surrounding at \(27^{\circ} \mathrm{C}\), and convection with the ambient air at \(27^{\circ} \mathrm{C}\) also, with an average convection heat transfer coefficient of \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Solar radiation is incident on the exhaust stack outer surface at a rate of \(150 \mathrm{~W} / \mathrm{m}^{2}\), and both the emissivity and solar absorptivity of the outer surface are 0.9. Assuming steady one-dimensional heat transfer, (a) obtain the variation of temperature in the exhaust stack wall and (b) determine the inner surface temperature of the exhaust stack.

Step by step solution

01

Determine the heat transfer rate for the exhaust gas

The heat transfer rate for the exhaust gas can be determined using the given mass flow rate, specific heat, and temperature drop. The equation for heat transfer rate is: $$q_{gas} = \dot{m} c_p \Delta T$$ Where: - \(q_{gas}\) is the heat transfer rate in the exhaust gas (W), - \(\dot{m}\) is the mass flow rate of the gas (\(1.2 \, \mathrm{kg/s}\)), - \(c_p\) is the specific heat at constant pressure (\(1600 \, \mathrm{J/kg \cdot K}\)), - and \(\Delta T\) is the temperature drop between the inlet and exit of the exhaust stack (\(30^\circ \mathrm{C}\)). Calculating \(q_{gas}\) yields: $$q_{gas} = (1.2 \, \mathrm{kg/s}) (1600 \, \mathrm{J/kg \cdot K}) (30^\circ \mathrm{C}) = 57600 \, \mathrm{W}$$

02

Determine the heat transfer rate due to solar radiation

The heat transfer rate due to solar radiation can be calculated using the given solar radiation incident on the surface and the solar absorptivity. The equation for heat transfer due to solar radiation is: $$q_{solar} = A_s I_s \alpha_s$$ Where: - \(q_{solar}\) is the heat transfer rate due to solar radiation (W), - \(A_s\) is the surface area of the stack exposed to solar radiation (\(\pi d_h L\) , where \(d_h\) is the outer diameter and \(L\) is the height of the stack), - \(I_s\) is the solar radiation intensity (\(150 \, \mathrm{W/m^2}\)), - and \(\alpha_s\) is the solar absorptivity (0.9). Calculating \(q_{solar}\) yields: $$q_{solar} = (\pi (1 \, \mathrm{m})(10 \, \mathrm{m})) (150 \, \mathrm{W/m^2}) (0.9) = 4244 \, \mathrm{W}$$

03

Determine the heat transfer rate due to convection

The heat transfer rate due to convection can be calculated using the given convection heat transfer coefficient, the outer surface area, and the temperature difference between the outer surface and the ambient air. The equation for heat transfer due to convection is: $$q_{conv} = h A_s \Delta T_{conv}$$ Where: - \(q_{conv}\) is the heat transfer rate due to convection (W), - \(h\) is the convection heat transfer coefficient (\(8 \, \mathrm{W/m^2 \cdot K}\)), - and \(\Delta T_{conv}\) is the temperature difference between the outer surface and the ambient air ((\(T_s - 27^\circ \mathrm{C}\))).

04

Determine the energy balance at the surface

The sum of the heat transfer rates of the exhaust gas, solar radiation, and convection should equal the heat transfer rate of radiation from the outer surface (Stephens-Boltzmann law). The heat transfer rate of radiation is: $$q_{rad} = \epsilon_s \sigma A_s (T_s^4-T_\infty^4)$$ Where: - \(q_{rad}\) is the heat transfer rate due to radiation (W), - \(\epsilon_s\) is the emissivity of the surface (0.9), - \(\sigma\) is the Stephens-Boltzmann constant (\(5.67 \times 10^{-8}\,\mathrm{W/m^{2} \cdot K^{4}}\)), - \(T_s\) is the outer surface temperature (K), - and \(T_\infty\) is the surrounding temperature (\(27^\circ \mathrm{C}\) or \(300^\circ \mathrm{K}\)). Now, write the energy balance at the surface: $$q_{gas} + q_{solar} = q_{conv} + q_{rad}$$ $$57600 \, \mathrm{W} + 4244 \, \mathrm{W} = h A_s (T_s - 27^\circ \mathrm{C}) + \epsilon_s \sigma A_s (T_s^4-T_\infty^4)$$ Solve for \(T_s\): $$T_s = 134.9^\circ \mathrm{C} = 408^\circ \mathrm{K}$$

05

Determine the heat conduction through the wall

The heat conduction can be calculated using the thermal conductivity and the temperature difference between the inner and outer surface: $$q_{cond} = k A \frac{\Delta T}{L_w}$$ Where: - \(q_{cond}\) is the heat transfer rate due to conduction (W), - \(k\) is the thermal conductivity of the wall material (\(40 \, \mathrm{W/m \cdot K}\)), - \(A\) is the annular area of the wall that has the heat flow (\(\pi L (d_o - d_i)\)), - \(\Delta T\) is the temperature difference between the inner surface temperature \(T_i\) and the outer surface temperature \(T_s\), - \(L_w\) is the wall thickness (\(0.1\,\mathrm{m}\)), - \(d_o\) is the outer diameter (\(1\,\mathrm{m}\)), - and \(d_i\) is the inner diameter (\(0.8\,\mathrm{m}\)).

06

Determine the inner surface temperature of the exhaust stack

Finally, we can solve for the inner surface temperature using the heat conduction equation and the given temperature at the outer surface: $$q_{gas} + q_{solar} = q_{cond}$$ Plug in the values for \(q_{gas}\), \(q_{solar}\), \(k\), \(A\) and \(L_w\) and solve for \(T_i\): $$57600 \, \mathrm{W} + 4244 \, \mathrm{W} = 40 \, \mathrm{W/m \cdot K} (\pi (10\,\mathrm{m})(1\,\mathrm{m} - 0.8\,\mathrm{m})) \frac{T_i - 134.9^\circ \mathrm{C}}{0.1\,\mathrm{m}}$$ $$T_i = 601.9^\circ \mathrm{C}$$ In conclusion, the temperature distribution in the exhaust stack wall is from \(408\,K (134.9^\circ \mathrm{C})\) at the outer surface to \(875\,K (601.9^\circ \mathrm{C})\) at the inner surface. The inner surface temperature of the exhaust stack is \(601.9^\circ \mathrm{C}\).

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