/*! 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 38 An air heater for an industrial ... [FREE SOLUTION] | 91Ó°ÊÓ

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An air heater for an industrial application consists of an insulated, concentric tube annulus, for which air flows through a thin-walled inner tube. Saturated steam flows through the outer annulus, and condensation of the steam maintains a uniform temperature \(T_{s}\) on the tube surface. Consider conditions for which air enters a 50 -mmdiameter tube at a pressure of \(5 \mathrm{~atm}\), a temperature of \(T_{m, i}=17^{\circ} \mathrm{C}\), and a flow rate of \(\dot{m}=0.03 \mathrm{~kg} / \mathrm{s}\), while saturated steam at \(2.455\) bars condenses on the outer surface of the tube. If the length of the annulus is \(L=5 \mathrm{~m}\), what are the outlet temperature \(T_{m, o}\) and pressure \(p_{o}\) of the air? What is the mass rate at which condensate leaves the annulus?

Short Answer

Expert verified
The outlet temperature of the air (\(T_{m, o}\)) can be found by applying energy balance equations on the air and steam: \(Q = \dot{m}c_p(T_{m, o} - T_{m, i})\). The mass rate at which condensate leaves the annulus (\(m_{s}\)) can be obtained from the energy balance on the steam and annulus: \(Q = m_{s}h_{fg}\). The outlet pressure of the air (\(P_o\)) can be determined using the pressure drop along the annulus: \(P_o = P_i - \Delta P\).

Step by step solution

01

Determine air properties at the inlet

First, we need to determine air properties at the inlet. From the ideal gas law, we can find the specific volume (\(v_{i}\)) of the air at the inlet: \(v_{i} = \dfrac{RT_{m, i}}{P_{i}}\) Here, R is the specific gas constant for air (\(287 \mathrm{~J/kg\cdot K}\)), \(T_{m, i}\) is the temperature of air at the entrance (\(17^{\circ} \mathrm{C}\) or \(290 \mathrm{~K}\)), and \(P_{i}\) is the pressure at the entrance (\(5 \mathrm{~atm}\) or \(506625 \mathrm{~Pa}\)). Using these values, we can now calculate the specific volume of air at the entrance.
02

Determine steam properties

Next, we need to determine the properties of the saturated steam, which condenses on the outer surface of the tube. From the steam table, we can find the saturation temperature, latent heat of vaporization, and specific volume of steam for the given pressure of \(2.455 \mathrm{~bar}\) (or \(245500 \mathrm{~Pa}\)). Here, we assume that the steam condensation is maintaining a uniform temperature \(T_{s}\) at the tube surface.
03

Apply energy balance to the air

Now, we will apply the energy balance equation to the air in the inner tube, considering the heat transfer between air and steam: \(Q = \dot{m}c_p(T_{m, o} - T_{m, i})\) Here, \(Q\) is the heat transfer between air and steam, \(\dot{m}\) is the mass flow rate of the air (\(0.03 \mathrm{~kg/s}\)), and \(c_p\) is the specific heat capacity of air at constant pressure (\(1005 \mathrm{~J/kg\cdot K}\)). We need to find \(Q\) using the given data and apply this equation to get the outlet temperature of the air, \(T_{m, o}\).
04

Apply energy balance to the steam and annulus

From the energy balance on the steam and annulus, we can write: \(Q = m_{s}h_{fg}\) Here, \(m_{s}\) is the mass rate at which condensate leaves the annulus and \(h_{fg}\) is the latent heat of vaporization for the saturated steam at the given pressure (determined in Step 2). Using the heat transfer \(Q\) from the air (determined in Step 3), we can solve this equation for \(m_{s}\), the mass rate at which condensate leaves the annulus.
05

Determine the outlet pressure of the air

Finally, to find the outlet pressure of the air, we can use the following equation: \(P_o = P_i - \Delta P\) Here, \(\Delta P\) is the pressure drop in the air along the length of the annulus. The pressure drop in this case can be determined using the friction factor, Reynolds number, and the Darcy-Weisbach equation. Once we find the pressure drop, we can then use this equation to find the outlet pressure of the air, \(P_o\).

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Key Concepts

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

Energy Balance Equations
In the realm of heat and mass transfer, energy balance equations are crucial for determining the amount of heat exchange between systems. An energy balance is a mathematical representation of the conservation of energy principle, which states that energy cannot be created or destroyed, only transferred or converted from one form to another.

In the context of the given problem, we apply an energy balance to both the air and the steam in the annulus. For the air, the heat gained or lost is given by the equation \(Q = \dot{m}c_p(T_{m, o} - T_{m, i})\), where \(\dot{m}\) represents the mass flow rate of air, \(c_p\) is the specific heat capacity, \(T_{m, o}\) is the outlet temperature, and \(T_{m, i}\) is the inlet temperature. This equation allows us to solve for the outlet temperature once the amount of heat transfer \(Q\) is known.

Similarly, the heat lost by the condensing steam can also be described by an energy balance equation: \(Q = m_{s}h_{fg}\), where \(m_{s}\) is the mass rate at which the condensate leaves the annulus and \(h_{fg}\) is the latent heat of vaporization. This relationship helps us calculate the mass rate of the condensate using the heat transfer determined from the air's energy balance.

These energy balance equations are fundamental tools in calculating temperature change and mass loss in heating and cooling processes, which are significant in various engineering applications.
Specific Volume Calculation
The specific volume \(v\) of a substance is a thermo-physical property representing the volume occupied by a unit mass of the substance. It is essentially the inverse of density and is a critical component in analyzing and understanding heat transfer and thermodynamic processes.

In the exercise, the specific volume calculation for the air at the inlet involves the ideal gas law, given by \(v_{i} = \frac{RT_{m, i}}{P_{i}}\), where \(R\) is the specific gas constant, \(T_{m, i}\) is the inlet temperature, and \(P_{i}\) is the inlet pressure. This calculation is important because it helps determine other properties of air that are necessary for further calculations, including the energy balance and the subsequent steps required to solve for outlet temperature and pressure drop.

Calculating the specific volume accurately is crucial in engineering applications, as it impacts the design and efficiency of heating, ventilation, and air conditioning (HVAC) systems, power plants, and various processes within chemical and mechanical engineering fields.
Saturated Steam Properties
Saturated steam is steam that is in equilibrium with liquid water at the same temperature and pressure, meaning it cannot absorb further heat without changing into a superheated state. In the exercise, the saturated steam provides a constant heat source for the air flowing through the tube and is defined by the properties at the given pressure of \(2.455 \text{bars}\).

The properties of saturated steam—such as temperature, specific volume, and latent heat of vaporization—are obtained from steam tables or charts. These properties are essential for calculating the amount of heat transfer taking place during the condensation process.

Understanding saturated steam properties is vital for the correct design and analysis of boilers, heat exchangers, and any system where steam is used for heat transfer. Students should be familiar with navigating steam tables and interpreting the values for accurate measurements and calculations in thermal systems processes.

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Most popular questions from this chapter

Consider a thin-walled, metallic tube of length \(L=1 \mathrm{~m}\) and inside diameter \(D_{i}=3 \mathrm{~mm}\). Water enters the tube at \(\dot{m}=0.015 \mathrm{~kg} / \mathrm{s}\) and \(T_{m, i}=97^{\circ} \mathrm{C}\). (a) What is the outlet temperature of the water if the tube surface temperature is maintained at \(27^{\circ} \mathrm{C}\) ? (b) If a \(0.5-\mathrm{mm}\)-thick layer of insulation of \(k=0.05\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) is applied to the tube and its outer surface is maintained at \(27^{\circ} \mathrm{C}\), what is the outlet temperature of the water? (c) If the outer surface of the insulation is no longer maintained at \(27^{\circ} \mathrm{C}\) but is allowed to exchange heat by free convection with ambient air at \(27^{\circ} \mathrm{C}\), what is the outlet temperature of the water? The free convection heat transfer coefficient is \(5 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\).

A hot fluid passes through a thin-walled tube of \(10-\mathrm{mm}\) diameter and 1-m length, and a coolant at \(T_{\infty}=25^{\circ} \mathrm{C}\) is in cross flow over the tube. When the flow rate is \(\dot{m}=18 \mathrm{~kg} / \mathrm{h}\) and the inlet temperature is \(T_{m, i}=85^{\circ} \mathrm{C}\), the outlet temperature is \(T_{m \rho}=78^{\circ} \mathrm{C}\). Assuming fully developed flow and thermal conditions in the tube, determine the outlet temperature, \(T_{m, a}\) if the flow rate is increased by a factor of 2 . That is, \(\dot{m}=36 \mathrm{~kg} / \mathrm{h}\), with all other conditions the same. The thermophysical properties of the hot fluid are \(\rho=\) \(1079 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=2637 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \mu=0.0034 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\), and \(k=0.261 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

One way to cool chips mounted on the circuit boards of a computer is to encapsulate the boards in metal frames that provide efficient pathways for conduction to supporting cold plates. Heat generated by the chips is then dissipated by transfer to water flowing through passages drilled in the plates. Because the plates are made from a metal of large thermal conductivity (typically aluminium or copper), they may be assumed to be at a temperature, \(T_{s, c p^{-}}\) (a) Consider circuit boards attached to cold plates of height \(H=750 \mathrm{~mm}\) and width \(L=600 \mathrm{~mm}\), each with \(N=10\) holes of diameter \(D=10 \mathrm{~mm}\). If operating conditions maintain plate temperatures of \(T_{\text {s.tp }}=32^{\circ} \mathrm{C}\) with water flow at \(\dot{m}_{1}=0.2 \mathrm{~kg} / \mathrm{s}\) per passage and \(T_{m, i}=7^{\circ} \mathrm{C}\), how much heat may be dissipated by the circuit boards? (b) To enhance cooling, thereby allowing increased power generation without an attendant increase in system temperatures, a hybrid cooling scheme may be used. The scheme involves forced airflow over the encapsulated circuit boards, as well as water flow through the cold plates. Consider conditions for which \(N_{\mathrm{cb}}=10\) circuit boards of width \(W=350 \mathrm{~mm}\) are attached to the cold plates and their average surface temperature is \(T_{s, \text { do }}=47^{\circ} \mathrm{C}\) when \(T_{s, \text { ep }}=32^{\circ} \mathrm{C}\). If air is in parallel flow over the plates with \(u_{\infty}=10 \mathrm{~m} / \mathrm{s}\) and \(T_{\infty}=7^{\circ} \mathrm{C}\), how much of the heat generated by the circuit boards is transferred to the air?

Fluid enters a thin-walled tube of \(5-\mathrm{mm}\) diameter and \(2-\mathrm{m}\) length with a flow rate of \(0.04 \mathrm{~kg} / \mathrm{s}\) and a temperature of \(T_{m, i}=85^{\circ} \mathrm{C}\). The tube surface is maintained at a temperature of \(T_{s}=25^{\circ} \mathrm{C}\), and for this operating condition, the outlet temperature is \(T_{m, o}=31.1^{\circ} \mathrm{C}\). What is the outlet temperature if the flow rate is doubled? Fully developed, turbulent flow may be assumed to exist in both cases, and the fluid properties may be assumed to be independent of temperature.

The final step of a manufacturing process in which a protective coating is applied to the inner surface of a circular tube involves passage of dry, atmosphere air through the tube to remove a residual liquid associated with the process. Consider a coated 5-m-long tube with an inner diameter of \(50 \mathrm{~mm}\). The tube is maintained at a temperature of \(300 \mathrm{~K}\), and the residual liquid exists as a thin film whose corresponding vapor pressure is \(15 \mathrm{~mm}\) Hg. The molecular weight and diffusion coefficient of the vapor are \(\Lambda_{\alpha}=70 \mathrm{~kg} / \mathrm{kmol}\) and \(D_{A B}=10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), respectively. Air enters the tube at a mean velocity of \(0.5 \mathrm{~m} / \mathrm{s}\) and a temperature of \(300 \mathrm{~K}\). (a) Estimate the partial pressure and mass density of vapor in the air exiting the tube. (b) What is the rate of liquid removal from the tube in \(\mathrm{kg} / \mathrm{s}\) ?

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