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An open cylindrical tank of diameter 2 m contains water, as shown in the following 5 figure. (a) Derive an analytical expression for dh/dt in terms of Q_{1} Q_{2} and Q3. (b) If h is constant and given V₁ = 3 m/s and Q3=10 litres per second, determine V₂.?
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An open cylindrical tank of diameter 2 m contains water, as shown in t...
Problem Statement

An open cylindrical tank of diameter 2 m contains water, as shown in the following 5 figure. (a) Derive an analytical expression for dh/dt in terms of Q1, Q2 and Q3. (b) If h is constant and given V₁ = 3 m/s and Q3=10 litres per second, determine V₂. Explain in details.


Solution


Part (a)

We can find dh/dt by applying the continuity equation which states that the mass flow rate into a system must equal the mass flow rate out of the system, assuming steady state conditions. In this case, the system is the cylindrical tank and the water flowing in and out are represented by Q1, Q2 and Q3.


Let's assume that the height of the water in the tank is h. The volume of water in the tank can be found using the formula for the volume of a cylinder:


V = πr²h


where r is the radius of the tank, which is half of the diameter. So, r = 1 m.


Now, let's differentiate both sides of the equation with respect to time t:


dV/dt = πr²(dh/dt)


The left-hand side represents the rate of change of volume of water in the tank with respect to time. This is equal to the sum of the rates at which water is flowing in and out of the tank:


dV/dt = Q1 - Q2 - Q3


where Q1 is the rate of inflow of water through the pipe, Q2 is the rate of outflow of water through the hole in the bottom of the tank, and Q3 is the rate of outflow of water through the pipe.


Substituting the above equation into the equation for the rate of change of volume, we get:


πr²(dh/dt) = Q1 - Q2 - Q3


Dividing both sides by πr², we get the final expression for dh/dt:


dh/dt = (Q1 - Q2 - Q3)/(πr²)


Part (b)

Given that h is constant, we can assume that the rate of change of height with respect to time is zero (dh/dt = 0).


Using the formula for the volume of a cylinder, we can relate the height of the water in the tank to the volume of water:


V = πr²h


Substituting the values of r and h, we get:


V = π(1)²h = πh


Now, we can differentiate both sides of the equation with respect to time t:


dV/dt = π(dh/dt)


Since dh/dt = 0, we get:


dV/dt = 0


This means that the volume of water in the tank is constant and is equal to the sum of the rates
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An open cylindrical tank of diameter 2 m contains water, as shown in the following 5 figure. (a) Derive an analytical expression for dh/dt in terms of Q_{1} Q_{2} and Q3. (b) If h is constant and given V₁ = 3 m/s and Q3=10 litres per second, determine V₂.?
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An open cylindrical tank of diameter 2 m contains water, as shown in the following 5 figure. (a) Derive an analytical expression for dh/dt in terms of Q_{1} Q_{2} and Q3. (b) If h is constant and given V₁ = 3 m/s and Q3=10 litres per second, determine V₂.? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An open cylindrical tank of diameter 2 m contains water, as shown in the following 5 figure. (a) Derive an analytical expression for dh/dt in terms of Q_{1} Q_{2} and Q3. (b) If h is constant and given V₁ = 3 m/s and Q3=10 litres per second, determine V₂.? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An open cylindrical tank of diameter 2 m contains water, as shown in the following 5 figure. (a) Derive an analytical expression for dh/dt in terms of Q_{1} Q_{2} and Q3. (b) If h is constant and given V₁ = 3 m/s and Q3=10 litres per second, determine V₂.?.
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