Question

In a cogeneration plant, combustion gases at 1 atm and \(800 \mathrm{~K}\) are used to preheat water by passing them through 6-m-long, 10-cm-diameter tubes. The inner surface of the tube is black, and the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) in combustion gases are \(0.12\) atm and \(0.18\) atm, respectively. If the tube temperature is \(500 \mathrm{~K}\), determine the rate of radiation heat transfer from the gases to the tube.

Short answer

Answer: The rate of radiation heat transfer from the combustion gases to the tube, denoted by \(q\), can be calculated by following these steps: 1. Calculate the emissivity of the combustion gases (\(ε_{gas}\)) using Hottel's correlation. 2. Calculate the heat transfer coefficient for radiation (\(h_R\)) using the formula: \(h_R = ε_{gas} \cdot σ \cdot (T_{gas}^4 - T_{tube}^4) / (T_{gas} - T_{tube})\). 3. Determine the heat transfer rate (\(q\)) using the formula: \(q = h_R \cdot A \cdot (T_{gas} - T_{tube})\), where \(A\) represents the area of the tubes, calculated as \(A = πDL\). Once all the necessary values are calculated and substituted into the equation, the value of \(q\) will represent the rate of radiation heat transfer from the combustion gases to the tube.

Step-by-step solution

Calculate the emissivity of the combustion gases.

The emissivity of the combustion gases can be estimated using the Hottel's correlation: \(ε_{gas} = ε_{\mathrm{CO}_{2}} + ε_{\mathrm{H}_{2}\mathrm{O}} - \epsilon_{\mathrm{CO}_{2}}ε_{\mathrm{H}_{2}\mathrm{O}}\) where: \(ε_{\mathrm{CO}_{2}} = 1 - exp\left[-1.22 \cdot 10^{3} \cdot P_{\mathrm{CO}_{2}}x_{\mathrm{CO}_{2}}L\right]\) \(ε_{\mathrm{H}_{2}\mathrm{O}} = 1 - exp\left[-1.73 \cdot 10^{3} \cdot P_{\mathrm{H}_{2}\mathrm{O}}x_{\mathrm{H}_{2}\mathrm{O}}L\right]\) Here, \(P_{\mathrm{CO}_{2}}\) and \(P_{\mathrm{H}_{2}\mathrm{O}}\) denote the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\) in the combustion gases, respectively. The value of \(L\) represents the length of the tubes. The values are given by: \(P_{\mathrm{CO}_{2}} = 0.12 \,\text{atm}\) \(P_{\mathrm{H}_{2}\mathrm{O}} = 0.18\, \text{atm}\) \(L = 6\,\text{m}\) We can calculate the emissivity of the combustion gases as follows: \(ε_{\mathrm{CO}_{2}} = 1 - exp[-1.22 \cdot 10^{3} \times 0.12 \times 6]\) \(ε_{\mathrm{H}_{2}\mathrm{O}} = 1 - exp[-1.73 \cdot 10^{3} \times 0.18 \times 6]\) \(ε_{gas} = ε_{\mathrm{CO}_{2}} + ε_{\mathrm{H}_{2}\mathrm{O}} - \epsilon_{\mathrm{CO}_{2}}ε_{\mathrm{H}_{2}\mathrm{O}}\)

Calculate the heat transfer coefficient for radiation.

The heat transfer coefficient for radiation is given by: \(h_R = ε_{gas} \cdot σ \cdot (T_{gas}^4 - T_{tube}^4) / (T_{gas} - T_{tube})\) Here, \(σ\) denotes the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} W/m^2K^4\)), and \(T_{gas}\) and \(T_{tube}\) are the temperatures of the combustion gases and tube, respectively. We have: \(T_{gas} = 800\,\text{K}\) \(T_{tube} = 500\,\text{K}\) Substituting the values, we get: \(h_R = ε_{gas} \cdot 5.67 \times 10^{-8} \cdot (800^4 - 500^4) / (800 - 500)\)

Determine the heat transfer rate.

The heat transfer rate can be calculated using the formula: \(q = h_R \cdot A \cdot (T_{gas} - T_{tube})\) First, calculate the area of the tubes, where \(A = πDL\), wherein \(D\) represents the diameter of the tubes. \(D = 0.10\,\text{m}\) \(L = 6\,\text{m}\) \(A = π \times 0.10 \times 6\,\text{m}^2\) Finally, calculate the heat transfer rate: \(q = h_R \cdot A \cdot (800 - 500)\) The calculated \(q\) value will represent the rate of radiation heat transfer from the combustion gases to the tube.

Key concepts

Radiation Heat Transfer

Radiation heat transfer is a key process where energy is emitted or absorbed in the form of electromagnetic waves or particles. Unlike conduction or convection, radiation does not require a medium to propagate, making it an essential concept in understanding how energy moves from one place to another through space. In our cogeneration plant example, radiation allows the heat to transfer from hot combustion gases, which are at 800 K, to the cooler tube at 500 K without any direct contact. This process is efficient for heating applications, as it solely relies on the heat difference between objects.

Emissivity

Emissivity is a measure of a material's ability to emit energy as radiation. It is a crucial property in determining how well a surface can transfer heat by radiation when given the chance. The emissivity of a perfect black body is 1, which means it emits all the radiation it receives. In real-life applications, materials have an emissivity less than 1 and their heat-emitting efficiency depends on that value.
In our exercise, we calculate the emissivity of combustion gases as a mix of components like CO₂ and H₂O using Hottel's correlation, which considers each component's contribution and interaction. The combined effect determines how effectively the gases can radiate heat to the surrounding surfaces.

Hottel's Correlation

Hottel's correlation provides an empirical way to estimate the emissivity of gas mixtures, particularly in combustion processes. It is based on the idea that the total emissivity of a gas can be derived from the emissivities of its individual components and their overlap. The formula used in the solution for our cogeneration plant scenario is:
\[ε_{gas} = ε_{ ext{CO}_2} + ε_{ ext{H}_2 ext{O}} - ε_{ ext{CO}_2} ε_{ ext{H}_2 ext{O}}\]
This accounts for the combined effect of carbon dioxide and water vapor, two prevalent gases in combustion, to estimate how effectively they emit infrared radiation. Understanding such correlations makes it simpler to design and analyze systems where gas radiation heat transfer is significant, improving system efficiency and performance.

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is a fundamental principle of thermal radiation that quantifies the power radiated from a black body in terms of its temperature. It states that the total radiant heat power emitted is proportional to the fourth power of the body's absolute temperature. The law is mathematically expressed as:
\[E = σT^4\]
where \(σ\) is the Stefan-Boltzmann constant, approximately \(5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4\), and \(T\) is the absolute temperature in Kelvin. In our exercise, this law helps calculate the heat transfer rate by radiation from the combustion gases to the tube. By applying this law, we derive the heat transfer coefficient, which is then used to determine how much heat is transferred through radiation based on the temperature difference between the gases and the tube.