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Chapter 8: Problem 129

A 0.064 -m-diameter nozzle meter is installed in a 0.097 -m-diameter pipe that carries water at \(60^{\circ} \mathrm{C}\). If the inverted air-water U-tube manometer used to measure the pressure difference across the meter indicates a reading of \(1 \mathrm{m}\), determine the flowrate.

Short Answer

Expert verified
The flow rate through the pipe is approximately \(0.01 \: m^3/s\).

Step by step solution

01

Calculating the Velocity

Start by calculating the velocity at the throat (the narrowest point in the nozzle meter) using Bernoulli's equation: \(P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2\). However, given that the pressure difference has been provided and that the area at the outlet (pipe) is greater than the area at the throat (nozzle), it's safe to assume that \(v_1\) is negligible in comparison to \(v_2\). Thus: \(v_2 = \sqrt{\frac{2 \Delta P}{\rho}}\), where \(\Delta P = \rho g h\), with \(h\) being the height difference in the manometer.
02

Calculating the Density of Water

The density of water at a given temperature can be determined from standard tables. At \(60^{\circ} \mathrm{C}\), the density of water (\(\rho\)) is approximately \(983 \: kg/m ^3\).
03

Substitution into the Velocity Equation

Substitute the known quantities into the velocity equation: \( v_2 = \sqrt{\frac{2 \cdot 983 \cdot 9.81 \cdot 1}{983}} = \sqrt{19.62} \approx 4.43 \: m/s\)
04

Calculating the Flow rate

The flow rate (\(Q\)) is the product of the velocity of the fluid \(v_2\) and the cross-sectional area at the throat (\(A = \pi d^2 / 4\)). Thus, it can be obtained by: \(Q = A \cdot v_2 = \pi (0.064)^2/4 \cdot 4.43 \approx 0.01 \: m^3/s\).

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Most popular questions from this chapter

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.

At a ski resort, water at \(40^{\circ} \mathrm{F}\) is pumped through a \(3-\mathrm{in}\). diameter, 2000 -ft-long steel pipe from a pond at an elevation of \(4286 \mathrm{ft}\) to a snow-making machine at an elevation of \(4623 \mathrm{ft}\) at a rate of \(0.26 \mathrm{ft}^{3} / \mathrm{s}\). If it is necessary to maintain a pressure of \(180 \mathrm{psi}\) at the snow-making machine, determine the horsepower added to the water by the pump. Neglect minor losses.

A 3 -in. schedule 40 commercial steel pipe (with an actual inside diameter of 3.068 in.) carries \(210^{\circ} \mathrm{F}\) SAE 40 crankcase oil at the rate of 6.0 gal/min. The oil specific gravity is \(0.89,\) and the absolute viscosity is \(6.6 \times 10^{-7} 1 \mathrm{b} \cdot \mathrm{sec} / \mathrm{ft}^{2} .\) Calculate the pipe size required to carry the same flow rate at approximately one-half the pressure drop of the 3 -in. pipe. Both pipes are horizontal.

For a given head loss per unit length, what effect on the flowrate does doubling the pipe diameter have if the flow is (a) laminar, or (b) completely turbulent?

A 3 -in. schedule 40 commercial steel pipe (with an actual inside diameter of 3.068 in.) carries 210 'F SAE 40 crankcase oil at the rate of 6.0 gal/min. The oil specific gravity is \(0.89,\) and the absolute viscosity is \(6,6 \times 10^{-7} 1 \mathrm{b} \cdot\) sec/ft \(^{2}\). For the same pressure drop, what must the acw pipe size be to carry the oil at 10.7 gal/min?
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