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Please double check before rejecting this question. If it needs to be rejected, please explain why as I cannot see how this is a breach of the honor code. This is a questions from the previous year's exam at my university in Engineering Science. I have submitted a solution given to us for study purposes as proof that I will not be graded on this assessment. In this solution, the formula (2gh)1/2 is used to find the solution, but I am on familliar with Bernoulli's Principle (P=1/2pv^2+pgh), and I was not able to find a solution using this. My solution incurred an error when I found that I had two unkown variables left that I could not break down in any meaningful way. (Velocity being the desired variable, Pressure being the problematic variable). Pressure = Force x Area, but I don't know enough about the dimensions of the tank or tap to be able to understand this.

Please double check before rejecting this question. If it needs to be rejected, please explain why as I cannot see how this is a breach of the honor code. This is a questions from the previous year's exam at my university in Engineering Science. I have submitted a solution given to us for study purposes as proof that I will not be graded on this assessment. In this solution, the formula (2gh)1/2 is used to find the solution, but I am on familliar with Bernoulli's Principle (P=1/2pv^2+pgh), and I was not able to find a solution using this. My solution incurred an error when I found that I had two unkown variables left that I could not break down in any meaningful way. (Velocity being the desired variable, Pressure being the problematic variable). Pressure = Force x Area, but I don't know enough about the dimensions of the tank or tap to be able to understand this.

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 double check before rejecting this question. If it needs to be rejected, please explain why as I cannot see how this is a breach of the honor code.

This is a questions from the previous year's exam at my university in Engineering Science. I have submitted a solution given to us for study purposes as proof that I will not be graded on this assessment.

In this solution, the formula (2gh)1/2 is used to find the solution, but I am on familliar with Bernoulli's Principle (P=1/2pv^2+pgh), and I was not able to find a solution using this. My solution incurred an error when I found that I had two unkown variables left that I could not break down in any meaningful way. (Velocity being the desired variable, Pressure being the problematic variable). Pressure = Force x Area, but I don't know enough about the dimensions of the tank or tap to be able to understand this.

Thank you for your help.

5c) 7.3m above the tap 1 Bernoulli's Principle: P = Pv². + pgh p≈ 998kg/m³ 1 P≈ (998.v)² +998-9.8 - 7.3 2 1 P≈ - (998v)² +71396.92 2
Transcribed Image Text:5c) 7.3m above the tap 1 Bernoulli's Principle: P = Pv². + pgh p≈ 998kg/m³ 1 P≈ (998.v)² +998-9.8 - 7.3 2 1 P≈ - (998v)² +71396.92 2
c) A round rainwater tank is open at the top and a tap is positioned a distance 7.3 m below the water level on the side of the tank. Calculate the velocity of the water exiting the tap. Assume the water level in the tank reduces with negligible velocity (since the diameter of the tap is much smaller than the diameter of the tank). (2 marks) Diagram - (1 mark) Bernoulli's or Toricelli's theorem - flow velocity = (2gh)¹/2 = (2 x 9.81 m/s² x 7.3 m)¹/2 = 12 m/s to 2 s.f. (1 mark)
Transcribed Image Text:c) A round rainwater tank is open at the top and a tap is positioned a distance 7.3 m below the water level on the side of the tank. Calculate the velocity of the water exiting the tap. Assume the water level in the tank reduces with negligible velocity (since the diameter of the tap is much smaller than the diameter of the tank). (2 marks) Diagram - (1 mark) Bernoulli's or Toricelli's theorem - flow velocity = (2gh)¹/2 = (2 x 9.81 m/s² x 7.3 m)¹/2 = 12 m/s to 2 s.f. (1 mark)
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