Problem 1: A tank shown in Fig. 1 is filled with two different fluids. The pressure at the top of the tank is 1000 kPa. Calculate the pressure at the bottom of the tank. Table 1 Density and depth of each fluid column Pi P2 1500 kg/m³ 200 kg/m³ hi 2500 m h₂ 500 m TROPJ Pi P2 1000 KPa h₁ h₂

Problem 1: A tank shown in Fig. 1 is filled with two different fluids. The pressure at the top of the tank is 1000 kPa. Calculate the pressure at the bottom of the tank. Table 1 Density and depth of each fluid column Pi P2 1500 kg/m³ 200 kg/m³ hi 2500 m h₂ 500 m TROPJ Pi P2 1000 KPa h₁ h₂

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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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.

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to gage pressure if it is not specified.
Problem 1: A tank shown in Fig. 1 is filled with two different fluids. The pressure at the top of the tank is 1000 kPa.
Calculate the pressure at the bottom of the tank.
Table 1 Density and depth of each fluid column
P1
P2
1500 kg/m³
200 kg/m³
hı
2500 m
h₂
500 m
Pi
P2
1000 KPa
h₁
h2
Fig. 1 Vertical tank

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