/*! 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 23 \( \mathrm{A}\) wet cooling towe... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 23

\( \mathrm{A}\) wet cooling tower cools warm water by spraying it into a forced dry-air flow. Some of the water evaporates in this air and is carried out of the tower into the atmosphere; the evaporation cools the remaining water droplets, which are collected at the exit pipe \((6 \text { in. } \mathrm{ID})\) of the tower. Measurements indicate the warm water mass flow rate is \(250,000 \mathrm{lb} / \mathrm{hr},\) and the cool water \(\left(70^{\circ} \mathrm{F}\right)\) flows at an average speed of 5 ft/s in the exit pipe. The moist air density is 0.065 Ib/ft \(^{3}\). Find (a) the volume flow rate \(\left(\mathrm{ft}^{3} / \mathrm{s}\right)\) and mass flow rate (lb/hr) of the cool water, (b) the mass flow rate \((\mathrm{lb} / \mathrm{hr})\) of the moist air, and (c) the mass flow rate (lb/hr) of the dry air. Hint: Google "density of moist air" for information on relating moist and dry air densities!

Short Answer

Expert verified
The volume flow rate of the cool water, mass flow rate of the moist air can be calculated using the given fluid velocity, pipe diameter and the densities. The mass flow rate of the dry air can then be obtained by the conservation of mass principle.

Step by step solution

01

Calculating the Volume Flow Rate and Mass Flow Rate of the Cool Water

Given that the cool water flows at an average speed of 5 ft/s in the exit pipe with a diameter of 6 inches (0.5 ft), we can calculate the volume flow rate \(Q\) as follows: \[Q = Area * Velocity = \pi*(D/2)^2*V = \pi*(0.5/2)^2*5 \ ft^3/s.\] Note \(蟺D^2/4\) is the formula for the cross-sectional area of a pipe, with \(D\) as the diameter and \(V\) as the fluid velocity. The mass flow rate remains 250,000 lb/hr as given, because in the absence of any chemical reactions or addition/removal of mass, the mass flow rate of the water remains consistent.
02

Figuring Out the Mass Flow Rate of the Moist Air

The volume flow rate of the moist air is equal to the volume flow rate of the cool water coming out from the pipe (since the exit pipe accommodates only these two flows). Therefore we have the moist air mass flow rate \(\dot{m}_{a_m} = \rho_{a_m}Q\) where \(\rho_{a_m}\) represents the density of the moist air. Substitute the values \(Q\) from Step 1 and \(\rho_{a_m} = 0.065 lb/ft^3\) as given in the problem. The product will indicate the mass flow rate of the moist air.
03

Calculating the Mass Flow Rate of the Dry Air

Mass conservation principle can be applied to the cooling tower operation. The sum of the mass flow rates of the cool water and the dry air equals the sum of the mass flow rates of the warm water and the moist air. Based on this, the mass flow rate of the dry air \(\dot{m}_{a_d}\) could be calculated as follows: \[\dot{m}_{a_d} = \dot{m}_{w_w} + \dot{m}_{a_m} - \dot{m}_{w_c}.\] Here, \(\dot{m}_{w_w}\) is the mass flow rate of the warm water, \(\dot{m}_{a_m}\) is the mass flow rate of the moist air we found in Step 2, and \(\dot{m}_{w_c}\) is the mass flow rate of the cool water we affirmed in Step 1.

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