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Chapter 8: Problem 36

Water flows downward through a vertical 10 -mm-diameter galvanized iron pipe with an average velocity of \(5.0 \mathrm{m} / \mathrm{s}\) and exits as a free jet. There is a small hole in the pipe \(4 \mathrm{m}\) above the outlet. Will water leak out of the pipe through this hole, or will air enter into the pipe through the hole? Repeat the problem if the average velocity is \(0.5 \mathrm{m} / \mathrm{s}\)

Short Answer

Expert verified
With the average velocities of both \(5.0 \mathrm{m} / \mathrm{s}\) and \(0.5 \mathrm{m} / \mathrm{s}\), air will enter the pipe through the hole due to the lower pressure inside the pipe compared to the atmospheric pressure.

Step by step solution

01

Calculating the Dynamic Pressure

The average velocity \( v \) in the pipe is \(5.0 \mathrm{m} / \mathrm{s}\) and the dynamic pressure (\(P_d\)) can be calculated as \(P_d = 0.5\rho v^2\), where \(\rho\) is the fluid density. Using \(\rho = 1000 \mathrm{kg/m^3}\) for water, we find \(P_d = 0.5*1000*5.0^2 = 12500 \mathrm{Pa}\) (Pascal).
02

Calculating the Hydrostatic Pressure

The hole is located \(4 \mathrm{m}\) above the outlet. The hydrostatic pressure (\(P_h\)) at this height can be calculated as \(P_{h} = \rho g h\), where g is the gravitational acceleration constant (\(9.81 \mathrm{m/s^2}\)). Therefore, \(P_{h} = 1000*9.81*4 = 39240 \mathrm{Pa}\).
03

Comparing with Atmospheric Pressure

Now we add dynamic and hydrostatic pressures to get the total pressure at the hole which is \(P = P_d + P_h = 12500 + 39240 = 51740 \mathrm{Pa}\). The atmospheric pressure is approximately \(101325 \mathrm{Pa}\). As our calculated pressure at the hole is lesser than the atmospheric pressure, air will enter into the pipe through the hole.
04

Repeat for \(v = 0.5 m/s\)

If we repeat the calculations for an average velocity of \(0.5 m/s\), we now find \(P_d = 0.5*1000*0.5^2 = 125 \mathrm{Pa}\) and the total pressure at the hole is \(P = 125 + 39240 = 39365 \mathrm{Pa}\), which is again far lesser than the atmospheric pressure. Air will still enter into the pipe through the hole. Thus, the reduction in velocity does not cause water to flow outward from the hole.

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Most popular questions from this chapter

A person with no experience in fluid mechanics wants to estimate the friction factor for 1 -in.-diameter galvanized iron pipe at a Reynolds number of 8,000 . The person stumbles across the simple equation of \(f=64 / \mathrm{Re}\) and uses this to calculate the friction factor. Explain the problem with this approach and estimate the error.

During a heavy rainstorm, water from a parking lot completely fills an 18 -in.-diameter, smooth, concrete storm sewer. If the flowrate is \(10 \mathrm{ft}^{3} / \mathrm{s}\), determine the pressure drop in a \(100-\mathrm{ft}\) horizontal section of the pipe. Repeat the problem if there is a \(2-f t\) change in elevation of the pipe per 100 ft of its length.

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For oil \(\left(S G=0.86, \mu=0.025 \mathrm{Ns} / \mathrm{m}^{2}\right)\) flow of \(0.2 \mathrm{m}^{3} / \mathrm{s}\) through a round pipe with diameter of \(500 \mathrm{mm}\), determine the Reynolds number. Is the flow laminar or turbulent?

For laminar flow in a round pipe of diameter \(D,\) at what distance from the centerline is the actual velocity equal to the average velocity?
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