50 h = Pbot =Vw dt 8₂-50¹ ftTonly unknownonly50 psigwaterglycerin5.g=1.26HW:'50'Sg= LeJg= 1.268₁=1.26xX62.4 16Local atmospleric pressure = 780 mm Hgat the bottom reads 50 psiggageTemperature of water and glycerinis 68°F X = 624 11,89= PaDetermine the depth of water (din the tank.

Answered: Image /qna-images/answer/1a95c7d3-eb0d-40df-8ab1-6595c17c4f85.jpg

50 h = Poot = wd+ √₂-50 ft T only unknown 50 рзід 17 water glycerin 5.g=1.26 Jg=1.26 d Local gage = 780 mm Hg atmospheric Die pressure = at the bottom reads 50 psig Temperature of water and glycerin is 687 16244,8₂ =F depth of water I Determine the depth of in the tank.

50 h = Poot = wd+ √₂-50 ft T only unknown 50 рзід 17 water glycerin 5.g=1.26 Jg=1.26 d Local gage = 780 mm Hg atmospheric Die pressure = at the bottom reads 50 psig Temperature of water and glycerin is 687 16244,8₂ =F depth of water I Determine the depth of in the tank.

Refrigeration and Air Conditioning Technology (MindTap Course List)

8th Edition

ISBN:9781305578296

Author:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson

Publisher:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson

Chapter1: Heat, Temperature, And Pressure

Section: Chapter Questions

Problem 17RQ: Convert 22C to Fahrenheit.

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Concept explainers

Properties of Fluids

The term fluid represents a type of substance that can easily change its shape under any external pressure/force and can flow. Fluids can acquire any shape in which it is stored and can resist normal stresses but cannot resist shear stress in rest condition. Whenever external shear stress is applied to fluids, it starts to flow. The fluids may be liquid or gas. Example- water, air, etc.

Question

50 h = Pbot =Vw dt 8₂-50¹ ft
T
only unknown
only
50 psig
water
glycerin
5.g=1.26
HW:
'50'
Sg= Le
Jg= 1.268₁
=1.26x
X62.4 16
Local atmospleric pressure = 780 mm Hg
at the bottom reads 50 psig
gage
Temperature of water and glycerin
is 68°F X = 624 11,89
= Pa
Determine the depth of water (d
in the tank.

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