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Water1 (3

This document contains solutions to two problems involving fluid statics: 1) Given the heights of water and an unknown liquid in a container, the density of the unknown liquid is calculated to be 563 kg/m3. 2) A piston separates two chambers filled with the same fluid. The gage pressures in chambers A and B are calculated, with chamber A at a pressure of 2806.5 Pa and chamber B at a negative pressure of -2098.5 Pa.

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248 views4 pages

Water1 (3

This document contains solutions to two problems involving fluid statics: 1) Given the heights of water and an unknown liquid in a container, the density of the unknown liquid is calculated to be 563 kg/m3. 2) A piston separates two chambers filled with the same fluid. The gage pressures in chambers A and B are calculated, with chamber A at a pressure of 2806.5 Pa and chamber B at a negative pressure of -2098.5 Pa.

Uploaded by

f2c95225d3
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Pressure and Fluid Statics

(3-41)page 8:
Based on height shown in the figure , find the density of unknown liquid
(𝝆water=1000 kg/m3 )

Solution :

PA =PB (pressure is the same at all points on a horizontal plane in a


given fluid )

PA=Patm+𝝆Liquid. g.H

PB=Patm+𝝆water.g.hB

Patm+𝝆Liquid.g.H=Patm+𝝆water.g.hB

-1-
𝜌𝑤𝑎𝑡𝑒𝑟 . 𝑔. ℎ𝐵 1000 × 0.45
𝜌𝐿𝑖𝑞𝑢𝑖𝑑 = = = 563𝑘𝑔/𝑚3
𝑔. 𝐻 0.8

(3-45)Page 8:

Two chambers with the same fluid at their base separated by a piston
whose weight is 25 N .Calculate the gage pressure in chambers A and B.

Solution :

𝜌𝑎𝑖𝑟 <<<< 𝜌𝐿𝑖𝑞𝑢𝑖𝑑 ⇒ ∆𝑃𝑎𝑖𝑟 ≈ 0

The gage pressure in chamber A=PE

in chamber A=PE
The gage pressure in chamber B=PD
𝐹 25
𝑃𝐶 = = = 354 𝑃𝑎
𝐴 0.32
𝜋.
4
𝑃𝐸 = 𝑃𝐶 + 𝜌. 𝑔. 0.25

𝑃𝐸 = 354 + 1000 × 9.81 × 0.25 = 2806.5 𝑃𝑎

𝑃𝐷 = 𝑃𝐶 − 𝜌. 𝑔. (0.5 − 0.25)

𝑃𝐷 = 354 − 1000 × 9.81 × 0.25 = −2098.5 𝑃𝑎

-2-
Note 1 :
the positive pressure at A pushed the water down to point E ,while
negative pressure at B pulled the water(vacuum)up to point D .

Note 2:

The pressure calculated here at points A and B are gage pressures(not


absolute pressures ):𝑃𝑎𝑏𝑠 = 𝑃𝑔𝑎𝑔𝑒 + 𝑃𝑎𝑡𝑚

𝑃𝐴𝑎𝑏𝑠 = 2806.5 + 101.3 × 103 = 104.1065 × 103 𝑃𝑎

and thus the negative pressure at B is actually equal to:

𝑃𝐵𝑎𝑏𝑠 = −2098.5 + 101.3 × 103 = 99.201 × 103 𝑃𝑎

-3-

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