/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 135 Waste heat recovery from the exh... [FREE SOLUTION] | 91Ó°ÊÓ

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Waste heat recovery from the exhaust (flue) gas of a melting furnace is accomplished by passing the gas through a vertical metallic tube and introducing saturated water (liquid) at the bottom of an annular region around the tube. The tube length and inside diameter are 7 and \(1 \mathrm{~m}\), respectively, and the tube inner surface is black. The gas in the tube is at atmospheric pressure, with \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{v})\) partial pressures of \(0.1\) and \(0.2 \mathrm{~atm}\), respectively, and its mean temperature may be approximated as \(T_{g}=1400 \mathrm{~K}\). The gas flow rate is \(\dot{m}=2 \mathrm{~kg} / \mathrm{s}\). If saturated water is introduced at a pressure of \(2.455 \mathrm{~ b a r s , ~ e s t i m a t e ~ t h e ~ w a t e r ~ f l o w ~ r a t e ~ s i t}\) which there is complete conversion from saturated liquid at the inlet to saturated vapor at the outlet. Thermophysical properties of the gas may be approximated as \(\mu=530 \times 10^{-7} \mathrm{~kg} / \mathrm{s}-\mathrm{m}, \mathrm{k}=0.091 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). and \(P_{r}=0.70\).

Short Answer

Expert verified
The water flow rate at which there is complete conversion from saturated liquid at the inlet to saturated vapor at the outlet is approximately \(0.63 \, \mathrm{kg/s}\).

Step by step solution

01

1. Calculate the heat transfer rate from the gas to the water

To calculate the heat transfer rate (Q), we need to find the convection heat transfer coefficient (h) and the surface area of the tube (A). The formula for calculating the convection heat transfer coefficient is: \[h = k \frac{(P_{r})^{2/3}}{\mu * D}\] where \(k\) is the gas thermal conductivity, \(P_{r}\) is the gas Prandtl number, \(\mu\) is the gas dynamic viscosity, and \(D\) is the tube inner diameter. Using the provided values, we can calculate the convection heat transfer coefficient h: \[h = 0.091 \frac{(0.70)^{2/3}}{530 \times 10^{-7} * 1}\] \[h \approx 71.9 \, \mathrm{W} / \mathrm{m}^2 \cdot \mathrm{K}\] Now, let's calculate the surface area of the tube (A): \[A = \pi D L\] where \(L\) is the tube length. \[A = \pi * 1 * 7\] \[A \approx 21.99 \, \mathrm{m}^2\] Now, we can calculate the heat transfer rate (Q): \[Q = h * A * (T_{g} - T_s)\] where \(T_g\) is the mean gas temperature, and \(T_s\) is the saturation temperature of water at the given pressure of 2.455 bars. From the steam tables, we find that \(T_s\) is approximately 134.15°C or 407.3 K. \[Q = 71.9 * 21.99 * (1400 - 407.3)\] \[Q \approx 1.457 \times 10^6 \, \mathrm{W}\]
02

2. Calculate the water flow rate

Now that we have the heat transfer rate (Q), we can calculate the water flow rate (m_dot) by dividing the heat transfer rate by the latent heat of vaporization of water at the given pressure: \[m_\text{dot} = \frac{Q}{h_{fg}}\] where \(h_{fg}\) is the latent heat of vaporization of water at the given pressure of 2.455 bars. From the steam tables, we find that \(h_{fg}\) is approximately 2300 kJ/kg. \[m_\text{dot} = \frac{1.457 \times 10^6}{2300 \times 10^3}\] \[m_\text{dot} \approx 0.63 \, \mathrm{kg/s}\] So, the water flow rate at which there is complete conversion from saturated liquid at the inlet to saturated vapor at the outlet is approximately \(0.63 \, \mathrm{kg/s}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convection Heat Transfer Coefficient
Understanding the convection heat transfer coefficient is essential for solving problems related to heat exchange between a fluid and a surface. It's a measure of how easily heat is transferred from the surface to the fluid or vice versa. The coefficient, represented as 'h' in equations, depends on the properties of the fluid, the velocity of the fluid, and the characteristics of the surface.

In the case of our exercise, we assume a steady state of heat transfer and that the properties of the gas remain constant. This assumption allows us to use the simplified formula:
\[\begin{equation}h = k \frac{(P_{r})^{2/3}}{ewline \mu * D}ewline newline \end{equation}\]Here, 'k' stands for the thermal conductivity of the gas, 'Pr' is the Prandtl number, '\mu' is the dynamic viscosity, and 'D' is the tube diameter. Since all those properties are given, we can easily calculate the heat transfer coefficient, which is the first step in determining the heat transfer rate from the hot gas to the water. The coefficient is higher when the fluid, in this case the gas, has a high thermal conductivity and lower viscosity, and when the Prandtl number indicates efficient heat transfer due to the fluid's momentum and thermal diffusivities.
Water Flow Rate
The water flow rate in heat transfer applications is crucial because it determines how effectively the heat is absorbed or released by the water. In our exercise, the goal is to find the rate at which the water can absorb all the heat from the gas and vaporize fully without leaving any residual liquid behind.

The water flow rate ('m_dot') is calculated by dividing the calculated heat transfer rate ('Q') by the latent heat of vaporization ('h_fg'). This relationship is vital because the latent heat of vaporization is the amount of energy required to change a unit mass of a substance from liquid to vapor without changing its temperature. So by knowing this energy requirement and the heat being transferred, the mass of water turning into vapor over time can be computed.
\[\begin{equation}m_\text{dot} = \frac{Q}{h_{fg}}ewline newline \end{equation}\]High flow rates are needed when there's a lot of heat to absorb, which is exactly what we find in industrial processes like waste heat recovery described in our problem. When dealing with such sizable amounts of heat, understanding and accurately calculating the correct flow rate is essential for designing an efficient system.
Latent Heat of Vaporization
Latent heat of vaporization, represented as 'h_fg' in calculations, is the amount of heat that is required to convert a unit mass of a liquid into vapor at constant temperature and pressure. It's a crucial factor in the calculations for heat transfer operations involving phase changes, such as boiling or condensation.

In this exercise, knowing the latent heat of vaporization at the specific pressure given (2.455 bars) allows us to calculate how much energy from the waste heat will be needed to vaporize the entering water. This energy, obtained from steam tables, is a unique property of every fluid and greatly influences the design and analysis of heating and cooling systems.
\[\begin{equation}h_{fg} (2.455 \text{ bars}) \approx 2300 \text{ kJ/kg}ewline newline \end{equation}\]Understanding this concept is important not only for the exercise but also for real-world applications, such as in the operation of boilers, refrigerators, and air conditioning systems, where the efficient use of energy relies on accurate calculations involving the latent heat of vaporization.

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Most popular questions from this chapter

Most architects know that the ceiling of an ice-skating rink must have a high reflectivity. Otherwise, condensation may occur on the ceiling, and water may drip onto the ice, causing bumps on the skating surface. Condensation will occur on the ceiling when its surface temperature drops below the dew point of the rink air. Your assignment is to perform an analysis to determine the effect of the ceiling emissivity on the ceiling temperature, and hence the propensity for condensation. The rink has a diameter of \(D=50 \mathrm{~m}\) and a height of \(L=10 \mathrm{~m}\), and the temperatures of the ice and walls are \(-5^{\circ} \mathrm{C}\) and \(15^{\circ} \mathrm{C}\), respectively. The rink air temperature is \(15^{\circ} \mathrm{C}\), and a convection coefficient of \(5 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\) characterizes conditions on the ceiling surface. The thickness and thermal conductivity of the ceiling insulation are \(0.3 \mathrm{~m}\) and \(0.035 \mathrm{~W} / \mathrm{m}-\mathrm{K}\), respectively, and the temperature of the outdoor air is \(-5^{\circ} \mathrm{C}\). Assume that the ceiling is a diffuse-gray surface and that the Walls and ice may be approximated as blackbodies. (a) Consider a flat ceiling having an emissivity of \(0.05\) (highly reflective panels) or \(0.94\) (painted panels). Perform an energy balance on the ceiling to calculate the corresponding values of the ceiling temperature. If the relative humidity of the rink air is \(70 \%\), will condensation occur for either or both of the emissivities? (b) For each of the emissivities, calculate and plot the ceiling temperature as a function of the insulation thickness for \(0.1 \leq r \leq 1 \mathrm{~m}\). Identify conditions for which condensation will occur on the ceiling.

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