Question

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.

Short answer

Answer: The inner surface temperature of the exhaust stack is ${601.9}^{\circ }\mathrm{C}$.

Step-by-step solution

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

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 ($150\mathrm{W}/{\mathrm{m}}^2$), - and ${\alpha }_s$ is the solar absorptivity (0.9). Calculating $q_{solar}$ yields: $$q_{solar}=(\pi (1\mathrm{m})(10\mathrm{m}))(150\mathrm{W}/{\mathrm{m}}^2)(0.9)=4244\mathrm{W}$$

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: $$q_{conv}=h{A}_s\mathrm{\Delta }{T}_{conv}$$ Where: - $q_{conv}$ is the heat transfer rate due to convection (W), - $h$ is the convection heat transfer coefficient ($8\mathrm{W}/{\mathrm{m}}^2\cdot \mathrm{K}$), - and $\mathrm{\Delta }{T}_{conv}$ is the temperature difference between the outer surface and the ambient air (($T_s-{27}^{\circ }\mathrm{C}$)).

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

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: $$q_{cond}=kA\frac{\mathrm{\Delta }T}{L_w}$$ Where: - $q_{cond}$ is the heat transfer rate due to conduction (W), - $k$ is the thermal conductivity of the wall material ($40\mathrm{W}/\mathrm{m}\cdot \mathrm{K}$), - $A$ is the annular area of the wall that has the heat flow ($\pi L(d_o-d_i)$), - $\mathrm{\Delta }T$ 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 ($0.8\mathrm{m}$).

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 $T_i$: $$57600\mathrm{W}+4244\mathrm{W}=40\mathrm{W}/\mathrm{m}\cdot \mathrm{K}(\pi (10\mathrm{m})(1\mathrm{m}-0.8\mathrm{m}))\frac{T_i-{134.9}^{\circ }\mathrm{C}}{0.1\mathrm{m}}$$ $$T_i={601.9}^{\circ }\mathrm{C}$$ In conclusion, the temperature distribution in the exhaust stack wall is from $408K({134.9}^{\circ }\mathrm{C})$ at the outer surface to $875K({601.9}^{\circ }\mathrm{C})$ at the inner surface. The inner surface temperature of the exhaust stack is ${601.9}^{\circ }\mathrm{C}$.

Key concepts

Convection

Convection is one of the key mechanisms of heat transfer, and it involves the transfer of heat by the physical movement of a fluid, such as air or a liquid. This process plays a major role in cooling or heating systems, where fluid movement is essential. In our exercise, the exhaust stack experiences convection with ambient air.

The rate of heat transfer by convection depends on the convection heat transfer coefficient, the surface area, and the temperature difference between the surface and fluid. The formula used is:$$q_{conv}=h{A}_s\mathrm{\Delta }{T}_{conv}$$

• Convection Heat Transfer Coefficient (

  h): This is a measure of the fluid’s ability to transfer heat to or from a surface.

• Surface Area (

  A_s): The area over which heat transfer takes place.

• Temperature Difference ($\mathrm{\Delta }{T}_{conv}$): The difference in temperature between the surface and the surrounding fluid.

The exercise indicates the ambient air around the stack is at 27°C, with a convection heat transfer coefficient of 8 W/m²·K, facilitating the heat transfer away from the stack into the air.

Conduction

Conduction is the process by which heat is transferred through a solid material, from a region of higher temperature to a lower temperature region. It occurs at the molecular level, as excited particles transfer energy to neighboring ones. This fundamental concept can be observed in our exhaust stack’s metal wall.

The heat conduction through the stack's wall is governed by the equation:$$q_{cond}=kA\frac{\mathrm{\Delta }T}{L_w}$$

• Thermal Conductivity (

  k): A property of the material that indicates its ability to conduct heat. Higher values mean more efficient heat conduction.

• Surface Area (

  A): The area through which heat is conducted.

• Temperature Gradient ($\mathrm{\Delta }T$): The difference in temperature across the material’s thickness.
• Wall Thickness (

  L_w): The distance through which heat must travel.

In our exercise, the exhaust stack's walls are 10 cm thick, with a thermal conductivity of 40 W/m·K. This affects how efficiently heat travels from the hot exhaust gases inside to the cooler outer surface.

Radiation

Radiation is a heat transfer mode that occurs through electromagnetic waves, without the need for a physical medium. It can transfer energy across distances, even through the vacuum of space. In our exhaust stack scenario, radiation interactions occur with the surrounding environment around the stack surface.

The formula for calculating radiation heat transfer is expressed as:$$q_{rad}={ϵ}_s\sigma A_s(T_s^4-T_{\mathrm{\infty }}^4)$$

• Emissivity (

  ${ϵ}_s$): A measure of how effectively a surface emits thermal radiation. It ranges between 0 and 1.

• Stefan-Boltzmann Constant (

  $\sigma$): A physical constant used in black body radiation calculations ($5.67\times {10}^{-8}{\text{W/m}}^2\cdot {\text{K}}^4$).

• Surface Temperature (

  $T_s$) and Ambient Temperature (

  $T_{\mathrm{\infty }}$): These determine the temperature differential driving radiation transfer.

The stack's surface exchanges heat with its environment at 27°C, and with an emissivity of 0.9, it can radiatively transfer heat effectively.

Thermal Conductivity

Thermal conductivity is a material property indicating the ability to conduct heat. It plays a vital role in determining how efficiently heat is transferred through a material when exposed to temperature differences. That property is crucial for heat transfer calculations in solid materials, such as the exhaust stack's wall.

In our exercise, the thermal conductivity of the exhaust stack wall is given as 40 W/m·K. This high value means the wall is quite efficient in conducting heat from the inside to the outside, contributing to the overall heat loss from the exhaust gases.

When dealing with problems involving thermal conduction, ensure you account for:

• Material type and structure – some materials like metals conduct heat better than non-metals.
• Temperature gradient – greater temperature differences increase heat transfer rates.
• Thickness – thinner materials conduct heat more quickly.

Thermal conductivity is essential in analyses of temperature variations, making calculations accurate in the scenario of exhaust gas discharge.

Exhaust Gases

Exhaust gases are by-products of combustion processes, often consisting of carbon dioxide, water vapor, and other substances. In industrial scenarios, these gases need to be expelled safely. The challenge lies in managing their heat content, as in our exhaust stack case.

Heat transfer involving exhaust gases often involves all forms of heat transfer, being:

• Convection: Moves heat through the gases inside the stack.
• Conduction: Transfers heat through the stack walls.
• Radiation: Transfers heat from the outer stack surface to the environment.

The rate of heat transfer from exhaust gases can be calculated using:$$q_{gas}=\dot{m}{c}_p\mathrm{\Delta }T$$where $\dot{m}$ is the mass flow rate, $c_p$ is the specific heat, and $\mathrm{\Delta }T$ is the temperature drop.

Our exercise specifies a mass flow rate of 1.2 kg/s, specific heat of 1600 J/kg·K, and a temperature drop of 30°C. Understanding these concepts helps ensure safe and efficient exhaust system design.