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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 AR/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)

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 AR/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)

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 explain part b and c

1. A tank is partitioned by a divider as shown in the figure below. 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 surface even, what pressure would the air on the left need to be pumped to? c) If the right side is sealed airtight, what pressure would the air on the left need to be pumped to in order 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 cross-sectional view of a tank that is divided into two sections by a vertical partition. - The left section contains a column of air above a column of water, with a total height of 3 meters. The water column is 1 meter high. - The right section contains a column of mercury with a height of 3 meters. - Both sections have different cross-sectional areas, denoted by \(A_L\) for the left and \(A_R\) for the right, with an area ratio of \(A_R/A_L = 5:1\).
Transcribed Image Text:1. A tank is partitioned by a divider as shown in the figure below. 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 surface even, what pressure would the air on the left need to be pumped to? c) If the right side is sealed airtight, what pressure would the air on the left need to be pumped to in order 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 cross-sectional view of a tank that is divided into two sections by a vertical partition. - The left section contains a column of air above a column of water, with a total height of 3 meters. The water column is 1 meter high. - The right section contains a column of mercury with a height of 3 meters. - Both sections have different cross-sectional areas, denoted by \(A_L\) for the left and \(A_R\) for the right, with an area ratio of \(A_R/A_L = 5:1\).
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