An evaporative cooling tower (see the figure below) is used to cool water from 107 to 75 °F. Water enters the tower at a rate of m1 = 275000 Ibm/hr. Dry air (no water vapor) flows into the tower at a rate of ṁ3 = rate of wet air flow out of the tower is m2 = 150200 lbm/hr, determine 145000 Ibm/hr. If the %3D (a) the rate of water evaporation in Ibm/hr, (b) the rate of cooled water flow in lbm/hr. Wet air m = m, Warm water m = m, Dry air Cooled water n = m,

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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Please answer these 2 questions. Only final answer needed. Please fast.
An evaporative cooling tower (see the figure
below) is used to cool water from 107 to 75 °F.
Water enters the tower at a rate of m1
Ibm/hr. Dry air (no water vapor) flows into the
tower at a rate of mz = 145000 Ibm/hr. If the
rate of wet air flow out of the tower is m2 =
= 275000
150200 Ibm/hr, determine
(a) the rate of water evaporation in Ibm/hr,
(b) the rate of cooled water flow in Ibm/hr.
Wet air
m = m,
Warm water
m = m,
Dry air →
m = m,
Cooled
water
(a) mw
i
Ibm/hr
(b) m =
i
Ibm/hr
Transcribed Image Text:An evaporative cooling tower (see the figure below) is used to cool water from 107 to 75 °F. Water enters the tower at a rate of m1 Ibm/hr. Dry air (no water vapor) flows into the tower at a rate of mz = 145000 Ibm/hr. If the rate of wet air flow out of the tower is m2 = = 275000 150200 Ibm/hr, determine (a) the rate of water evaporation in Ibm/hr, (b) the rate of cooled water flow in Ibm/hr. Wet air m = m, Warm water m = m, Dry air → m = m, Cooled water (a) mw i Ibm/hr (b) m = i Ibm/hr
Current Attempt in Progress
As shown in the figure below, at the entrance to a
3-ft-wide channel the velocity distribution is
uniform with a velocity V. Further downstream the
velocity profile is given by u = 4y - 2y², where u is in
ft/s and y is in ft. Determine the value of V.
Assume h = 0.86 ft.
y
h
u = 4y – 2y2
0.75 ft
V =
i
ft/s
Transcribed Image Text:Current Attempt in Progress As shown in the figure below, at the entrance to a 3-ft-wide channel the velocity distribution is uniform with a velocity V. Further downstream the velocity profile is given by u = 4y - 2y², where u is in ft/s and y is in ft. Determine the value of V. Assume h = 0.86 ft. y h u = 4y – 2y2 0.75 ft V = i ft/s
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