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Chapter 8: Problem 107

A water tank (open to the atmosphere) contains water to a depth of \(5 \mathrm{m} .\) A 25 -mm-diameter hole is punched in the bottom. Modeling the hole as square-edged, estimate the flow rate (L/s) exiting the tank. If you were to stick a short section of pipe into the hole, by how much would the flow rate change? If instead you were to machine the inside of the hole to give it a rounded edge \((r=5 \mathrm{mm})\), by how much would the flow rate change?

Short Answer

Expert verified
1. The flow rate with a square-edged hole is calculable with Torricelli's theorem and the area of the hole cross-section. 2. A section of pipe does not alter the flow rate. 3. Machining a rounded edge increases the flow rate, represented by the contraction coefficient in the flow rate formula.

Step by step solution

01

Calculate the initial flow rate from the tank (square-edged hole)

First, calculate the flow rate with the square-edged hole using Torricelli鈥檚 theorem, which states that the speed \(v\) of efflux of a fluid under the force of gravity from a hole in a large container is given by \(v = \sqrt{2gh}\), where \(g\) is the acceleration due to gravity and \(h\) is the height of the water above the centre of the hole. The cross-sectional area \(A\) of the hole is given by \(\pi (D/2)^2\), where \(D\) is the diameter of the hole. Then the flow rate \(Q\) is calculated by \(Q = A \cdot v\).
02

Calculate the flow rate with a section of pipe

When a section of pipe is added, the flow rate does not change, since the water pressure, gravitational pull, and size of the opening remain the same. Thus, the flow rate remains at the value calculated in step 1.
03

Calculate the flow rate with a rounded edge

When the hole is machined to give it a rounded edge, the flow rate increases due to reduced fluid friction. This scenario is represented by a contraction coefficient (\(C_c\)) which is greater than 1. For a hole with a rounded edge (where the radius of the rounded edge \(r\) is half the hole's diameter, \(D\)), \(C_c\) is approximately 0.98. The flow rate is then calculated by \(Q = C_c \cdot A \cdot v\)

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