Question

Exhaust gases from a manufacturing plant are being discharged through a 10 -m-tall exhaust stack with outer diameter of \(1 \mathrm{~m}\). The exhaust gases are discharged at a rate of \(0.125 \mathrm{~kg} / \mathrm{s}\), 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 gases is \(1600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). On a particular calm day, the surrounding quiescent air temperature is \(33^{\circ} \mathrm{C}\). Solar radiation is incident on the exhaust stack outer surface at a rate of $500 \mathrm{~W} / \mathrm{m}^{2}$, and both the emissivity and solar absorptivity of the outer surface are \(0.9\). Determine the exhaust stack outer surface temperature. Assume the film temperature is \(60^{\circ} \mathrm{C}\).

Short answer

The outer surface temperature of the exhaust stack is approximately 210.73°C.

Step-by-step solution

Calculate the exhaust stack surface area

The exhaust stack is a cylinder with height of 10 m and diameter of 1 m. The surface area \(A\) of a cylinder is given by \(A=2\pi rh\), where \(r\) is the radius and \(h\) is the height. So, we have: \(A = 2\pi \times 0.5 \times 10 = 31.42 \mathrm{~m^2}\)

Determine the heat transfer rate due to convection

The temperature drop \(\Delta T\) between the inlet and exit of the exhaust stack is given: \(\Delta T = 30^{\circ} \mathrm{C}\) The heat transfer rate \(Q_\mathrm{conv}\) due to convection can be calculated by using the mass flow rate (\(\dot{m}\)), the specific heat of the exhaust gases (\(c_P\)), and temperature drop \(\Delta T\): \(Q_\mathrm{conv} = \dot{m} \times c_P \times \Delta T = 0.125\mathrm{~kg/s} \times 1600\mathrm{~J/(kg\cdot K)} \times 30^{\circ}\mathrm{C} = 6000\mathrm{~W}\)

Calculate the heat transfer rate due to radiation

We will use the energy balance equation to find the heat transfer rate due to radiation. The energy balance equation is: \(Q_\mathrm{conv} - Q_\mathrm{rad} = Q_\mathrm{abs}\) But first, we need to find the value for the emissivity \(\epsilon\) and the Stefan-Boltzmann constant \(\sigma\). We are given the emissivity of the outer surface as 0.9. The Stefan-Boltzmann constant value is: \(\sigma = 5.67\times10^{-8} \mathrm{~W/(m^{2}\cdot K^{4})}\) Next, we need to convert the film temperature to Kelvin, which is given as: \(T_\mathrm{film} = 60^{\circ}\mathrm{C} + 273.15 = 333.15\mathrm{~K}\) Now we can calculate the heat transfer rate due to radiation \(Q_\mathrm{rad}\) using the following formula: \(Q_\mathrm{rad} = \epsilon \sigma A(T_\mathrm{s}^{4} - T_\mathrm{film}^{4})\)

Compute the heat absorbed due to solar radiation

To calculate the heat absorbed by the exhaust stack due to solar radiation, we need the solar absorptivity \(\alpha\), which is given as 0.9. The heat absorbed per unit area can be calculated as: \(Q_\mathrm{abs,per\,unit\,area} = \alpha \cdot I = 0.9 \times 500\mathrm{~W/m^{2}} = 450\mathrm{~W/m^{2}}\) The total heat absorbed \(Q_\mathrm{abs}\) can be calculated by multiplying the heat absorbed per unit area with the total surface area of the exhaust stack: \(Q_\mathrm{abs} = Q_\mathrm{abs,per\,unit\,area} \times A = 450\mathrm{~W/m^{2}} \times 31.42\mathrm{~m^{2}} = 14,139\mathrm{~W}\)

Apply the energy balance equation to find the outer surface temperature

Now we can use the energy balance equation to find the outer surface temperature \(T_s\): \(Q_\mathrm{conv} - Q_\mathrm{rad} = Q_\mathrm{abs}\) Substituting values: \(6000\mathrm{~W} - 0.9 \times 5.67\times10^{-8}\mathrm{~W/(m^{2}\cdot K^{4})} \times 31.42\mathrm{~m^{2}} (T_\mathrm{s}^{4} - 333.15\mathrm{~K}^{4}) = 14,139\mathrm{~W}\) Solve for \(T_\mathrm{s}\): \((T_\mathrm{s}^{4} - 333.15\mathrm{~K}^{4}) = \frac{14,139\mathrm{~W} - 6000\mathrm{~W}}{0.9 \times 5.67\times10^{-8}\mathrm{~W/(m^{2}\cdot K^{4})} \times 31.42\mathrm{~m^{2}}} = 5.79\times10^{8}\mathrm{~K^{4}}\) \(T_\mathrm{s}^{4} = 5.79\times10^{8}\mathrm{~K^{4}} + 333.15\mathrm{~K}^{4} = 2.15\times10^{9}\mathrm{~K^{4}}\) \(T_\mathrm{s} = \sqrt[4]{2.15\times10^{9}\mathrm{~K^{4}}} = 483.88\mathrm{~K}\) Convert back to Celsius: \(T_\mathrm{s} = 483.88\mathrm{~K} - 273.15 = 210.73^{\circ}\mathrm{C}\) So, the exhaust stack outer surface temperature is approximately \(210.73^{\circ} \mathrm{C}\).