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A tank is partitioned by a divider as shown in the figure below. The tank is filled by water (SG=1) and mercury (SG-13.55) and the cross-sectional area ratio A₁/AL=5:1. a) Find the gage pressure in the air trapped on the left side b) To bring the water and mercury free surface even, what pressure would the air on the left to be pumped to c) If the right side is sealed airtight, what pressure would the air on the left to be pumped to to bring the water and mercury free surface even (assume the air trapped on the right side behaves isothermally, i.e., the ideal gas law applies) Cross-sectional Cross-sectional Area A₁ Area AR 1m 2.9 m Air Water Mercury 3m 3m

A tank is partitioned by a divider as shown in the figure below. The tank is filled by water (SG=1) and mercury (SG-13.55) and the cross-sectional area ratio A₁/AL=5:1. a) Find the gage pressure in the air trapped on the left side b) To bring the water and mercury free surface even, what pressure would the air on the left to be pumped to c) If the right side is sealed airtight, what pressure would the air on the left to be pumped to to bring the water and mercury free surface even (assume the air trapped on the right side behaves isothermally, i.e., the ideal gas law applies) Cross-sectional Cross-sectional Area A₁ Area AR 1m 2.9 m Air Water Mercury 3m 3m

Elements Of Electromagnetics
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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please solve and explain part C deeply  so I can understand 

**Problem Statement:** A tank is partitioned by a divider as shown in the figure. The tank is filled with water (SG=1) and mercury (SG=13.55), and the cross-sectional area ratio \(A_R/A_L = 5:1\). a) Find the gage pressure in the air trapped on the left side. b) To bring the water and mercury free surfaces even, what pressure would the air on the left to be pumped to? c) If the right side is sealed airtight, what pressure would the air on the left to be pumped to bring the water and mercury free surface even (assume the air trapped on the right side behaves isothermally, i.e., the ideal gas law applies)? **Diagram Explanation:** The diagram shows a vertical tank split into two sections by a divider: - **Left Section:** - Contains air above a water column. - Water column height: \(1 \, \text{m}\). - Below the water is mercury, with a total fluid column height of \(3 \, \text{m}\). - **Right Section:** - Contains mercury with a total fluid height of \(3 \, \text{m}\). - Cross-sectional area labeled as \(A_R\). Both sides of the tank have the same height of mercury. **Cross-sectional Areas:** - Left side: \(A_L\) - Right side: \(A_R = 5 \times A_L\) **Objective:** Solve for the air pressure in different scenarios to equalize the fluid levels across the partition.
Transcribed Image Text:**Problem Statement:** A tank is partitioned by a divider as shown in the figure. The tank is filled with water (SG=1) and mercury (SG=13.55), and the cross-sectional area ratio \(A_R/A_L = 5:1\). a) Find the gage pressure in the air trapped on the left side. b) To bring the water and mercury free surfaces even, what pressure would the air on the left to be pumped to? c) If the right side is sealed airtight, what pressure would the air on the left to be pumped to bring the water and mercury free surface even (assume the air trapped on the right side behaves isothermally, i.e., the ideal gas law applies)? **Diagram Explanation:** The diagram shows a vertical tank split into two sections by a divider: - **Left Section:** - Contains air above a water column. - Water column height: \(1 \, \text{m}\). - Below the water is mercury, with a total fluid column height of \(3 \, \text{m}\). - **Right Section:** - Contains mercury with a total fluid height of \(3 \, \text{m}\). - Cross-sectional area labeled as \(A_R\). Both sides of the tank have the same height of mercury. **Cross-sectional Areas:** - Left side: \(A_L\) - Right side: \(A_R = 5 \times A_L\) **Objective:** Solve for the air pressure in different scenarios to equalize the fluid levels across the partition.
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