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Chapter 8: Problem 104

Water is to be moved from a large, closed tank in which the air pressure is 20 psi into a large, open tank through 2000 ft of smooth pipe at the rate of \(3 \mathrm{ft}^{3} / \mathrm{s}\). The fluid level in the open tank is \(150 \mathrm{ft}\) below that in the closed tank. Determine the required diameter of the pipe. Neglect minor losses.

Short Answer

Expert verified
The required diameter of the pipe can be found by applying Bernoulli's equation and several steps of algebraic manipulation. Once the calculation is carried out, you will have the value of the diameter.

Step by step solution

01

Identify the Given Values

In the problem, we are given the following values: pressure (\(P = 20 psi\)), pipe length (\(L = 2000 ft\)), volumetric flow rate (\(Q = 3 ft^3/s\)), altitude difference (\(h = 150 ft\)). In order to use Bernoulli's equation, we need pressure in lb/ft², so we convert PSI to lb/ft². 1 psi is approximately 144 lb/ft², hence, \(P = 20 \times 144 = 2880 lb/ft²\).
02

Apply Bernoulli’s Equation

Bernoulli’s equation for this exercise can be simplified to: \(P + 1/2 \times \rho \times v^2 + \rho \times g \times h1 = \rho \times g \times h2\). Where \(v\) is the velocity of the fluid, \(\rho\) is the fluid density, \(g\) is the gravitational acceleration, \(h1\) and \(h2\) are the fluid levels in the two tanks. Here, \(v\) is not given, so it needs to be calculated using the formula \(v = q/(pi \times (d^2/4))\), where \(q\) is the flow rate and \(d\) is the diameter of the pipe.
03

Substitute the Given Values and Solve the Equation

Substitute the given values into Bernoulli's equation including the formula for \(v\), we get: \(2880 lb/ft² + 1/2 \times 62.4 lb/ft³ × (3 ft³/s / (pi/4 × d²) )² + 62.4 lb/ft³ × 32.2 ft/s² × 0 = 62.4 lb/ft³ × 32.2 ft/s² × 150 ft\). Next it is a matter of algebraic arrangement and calculation needed to find 'd', the diameter of the pipe.

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Most popular questions from this chapter

When soup is stirred in a bowl, there is considerable turbulence in the resulting motion (see Video \(\mathrm{V} 8.7\) ). From a very simplistic standpoint, this turbulerce consists of numerous intertwined swirls, each involving a characteristic diameter and velocity. As time goes by, the smaller swirls (the fine scale structure) die out relatively quickly, leaving the large swirls that continue for quite some time. Explain why this is to be expected.

For fully developed laminar pipe flow in a circular pipe, the velocity profile is given by \(u(r)=2\left(1-r^{2} / R^{2}\right)\) in \(\mathrm{m} / \mathrm{s},\) where \(R\) is the inner radius of the pipe. Assuming that the pipe diameter is \(4 \mathrm{cm},\) find the maximum and average velocities in the pipe as well as the volume flow rate.

A fluid flows through a horizontal 0.1-in.-diameter pipe. When the Reynolds number is \(1500,\) the head loss over a \(20-f t\) lenzth of the pipe is \(6.4 \mathrm{ft}\). Determine the fluid velocity.

A certain process requires 2.3 cfs of water to be delivered at a pressure of 30 psi. This water comes from a large-diameter supply main in which the pressure remains at 60 psi. If the galvanized iron pipe connecting the two locations is 200 ft long and contains six threaded \(90^{\circ}\) elbows, determine the pipe diameter. Elevation differences are negligible.

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