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Chapter 1: Problem 32

Air enters the duct of an air-conditioning system at \(15 \mathrm{psia}\) and \(50^{\circ} \mathrm{F}\) at a volume flow rate of \(450 \mathrm{ft}^{3} / \mathrm{min}\). The diameter of the duct is 10 inches and heat is transferred to the air in the duct from the surroundings at a rate of \(2 \mathrm{Btu} / \mathrm{s}\). Determine \((a)\) the velocity of the air at the duct inlet and (b) the temperature of the air at the exit.

Short Answer

Expert verified
Answer: (a) The velocity of the air at the duct inlet is 13.79 ft/s. (b) The temperature of the air at the exit is 52.26°F.

Step by step solution

01

Calculate the cross-sectional area of the duct

First, we need to find the cross-sectional area of the duct. Given that the diameter of the duct is 10 inches, we can first convert that to feet and then calculate the area of the duct using the formula for the area of a circle, which is \(A = \pi (d/2)^{2}\), where \(A\) is the area and \(d\) is the diameter. \(d=10 \ \mathrm{in} = \frac{10}{12} \ \mathrm{ft} = 0.833 \ \mathrm{ft}\) \(A=\pi (\frac{0.833}{2})^2 = \pi (0.4165)^2 = 0.545 \ \mathrm{ft}^{2}\)
02

Calculate the velocity of the air at the duct inlet

Now that we have the cross-sectional area of the duct, we can calculate the air velocity at the duct inlet by using the volume flow rate and the area. The formula for this relation is \(v = \frac{Q}{A}\), where \(v\) is the velocity, \(Q\) is the volume flow rate, and \(A\) is the cross-sectional area. \(v = \frac{450 \ \mathrm{ft}^{3} \ / \mathrm{min}}{0.545 \ \mathrm{ft}^{2}} = \frac{450 \ \mathrm{ft}^{3} \ / \mathrm{min}}{0.545 \ \mathrm{ft}^{2}} \cdot \frac{1 \ \mathrm{min}}{60 \ \mathrm{s}} = 13.79 \ \mathrm{ft/s}\) So the velocity of the air at the duct inlet is \(13.79 \ \mathrm{ft/s}\).
03

Calculate the temperature change due to heat transfer

To find the temperature change, we need to consider the heat transfer from the surroundings to the air. We are given that the heat transfer rate is \(2 \ \mathrm{Btu/s}\). We can use the formula \(q = mc \Delta T\) to relate heat transfer (\(q\)), mass flow rate (\(m\)), specific heat capacity (\(c\)), and temperature change (\(\Delta T\)). Then, we need to find the mass flow rate using the density and volume flow rate of the air entering the duct. From the air properties table, the density of air at \(50^{\circ} \mathrm{F}\) and \(15 \ \mathrm{psia}\) is approximately \(0.0729 \ \mathrm{lb/ft^3}\) and the specific heat capacity is approximately \(0.240 \ \mathrm{Btu/(lb ^{\circ} F)}\).
04

Calculate the mass flow rate of the air

To calculate the mass flow rate, we need to multiply the density of air and the volume flow rate: \(m = \rho Q = 0.0729 \ \mathrm{lb/ft^3} \cdot 450 \ \mathrm{ft^3/min} = 32.8 \ \mathrm{lb/min}\)
05

Calculate the temperature change due to heat transfer

Now we can use the formula \(q = mc \Delta T\) to find the temperature change: \(q = mc \Delta T \Rightarrow \Delta T = \frac{q}{mc} = \frac{2 \ \mathrm{Btu/s}}{32.8 \ \mathrm{lb/min} \cdot 0.240 \ \mathrm{Btu/(lb ^{\circ} F)}} \cdot \frac{60 \ \mathrm{s}}{1 \ \mathrm{min}} = 2.26^{\circ} \mathrm{F}\) The temperature of the air increases by \(2.26^{\circ} \mathrm{F}\) due to heat transfer.
06

Calculate the temperature of the air at the exit

Now, we can find the temperature of the air at the exit by adding the temperature change to the initial temperature: \(T_{exit} = T_{inlet} + \Delta T = 50^{\circ} \mathrm{F} + 2.26^{\circ} \mathrm{F} = 52.26^{\circ} \mathrm{F}\) So the temperature of the air at the exit is \(52.26^{\circ} \mathrm{F}\). In conclusion, (a) the velocity of the air at the duct inlet is \(13.79 \ \mathrm{ft/s}\) and (b) the temperature of the air at the exit is \(52.26^{\circ} \mathrm{F}\).

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Most popular questions from this chapter

A 3-m-internal-diameter spherical tank made of 1 -cm-thick stainless steel is used to store iced water at \(0^{\circ} \mathrm{C}\). The tank is located outdoors at \(25^{\circ} \mathrm{C}\). Assuming the entire steel tank to be at \(0^{\circ} \mathrm{C}\) and thus the thermal resistance of the tank to be negligible, determine \((a)\) the rate of heat transfer to the iced water in the tank and \((b)\) the amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24 -hour period. The heat of fusion of water at atmospheric pressure is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\). The emissivity of the outer surface of the tank is \(0.75\), and the convection heat transfer coefficient on the outer surface can be taken to be \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assume the average surrounding surface temperature for radiation exchange to be \(15^{\circ} \mathrm{C}\).

Consider a flat-plate solar collector placed on the roof of a house. The temperatures at the inner and outer surfaces of the glass cover are measured to be \(33^{\circ} \mathrm{C}\) and \(31^{\circ} \mathrm{C}\), respectively. The glass cover has a surface area of \(2.5 \mathrm{~m}^{2}\), a thickness of \(0.6 \mathrm{~cm}\), and a thermal conductivity of \(0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Heat is lost from the outer surface of the cover by convection and radiation with a convection heat transfer coefficient of \(10 \mathrm{~W} /\) \(\mathrm{m}^{2} \cdot \mathrm{K}\) and an ambient temperature of \(15^{\circ} \mathrm{C}\). Determine the fraction of heat lost from the glass cover by radiation.

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In a power plant, pipes transporting superheated vapor are very common. Superheated vapor is flowing at a rate of \(0.3 \mathrm{~kg} / \mathrm{s}\) inside a pipe with \(5 \mathrm{~cm}\) in diameter and \(10 \mathrm{~m}\) in length. The pipe is located in a power plant at \(20^{\circ} \mathrm{C}\), and has a uniform pipe surface temperature of \(100^{\circ} \mathrm{C}\). If the temperature drop between the inlet and exit of the pipe is \(30^{\circ} \mathrm{C}\), and the specific heat of the vapor is \(2190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), determine the heat transfer coefficient as a result of convection between the pipe surface and the surrounding.

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