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Chapter 8: Problem 36

Consider fully developed laminar flow in a circular pipe. If the diameter of the pipe is reduced by half while the flow rate and the pipe length are held constant, the pressure drop will \((a)\) double, \((b)\) triple, \((c)\) quadruple, \((d)\) increase by a factor of 8 , or \((e)\) increase by a factor of \(16 .\)

Short Answer

Expert verified
Short Answer: The pressure drop will increase by a factor of 16.

Step by step solution

01

Write down the Hagen-Poiseuille equation

The Hagen-Poiseuille equation is expressed as: \[\Delta P = \frac{8 \mu L Q}{\pi R^4}\] Where \(\Delta P\) is the pressure drop, \(\mu\) is the dynamic viscosity, \(L\) is the length of the pipe, \(Q\) is the flow rate, and \(R\) is the radius of the pipe.
02

Adjust equation for initial conditions

Initially, let the radius of the pipe be \(R\). The initial pressure drop can be expressed as: \[\Delta P_1 = \frac{8 \mu L Q}{\pi R^4}\]
03

Adjust equation for final conditions

When the diameter of the pipe is halved, the radius becomes \(R/2\). The final pressure drop can be expressed as: \[\Delta P_2 = \frac{8 \mu L Q}{\pi (R/2)^4}\]
04

Find the relation between initial and final pressure drop

Divide \(\Delta P_2\) by \(\Delta P_1\) to find the factor by which the pressure drop has changed: \[\frac{\Delta P_2}{\Delta P_1} = \frac{\frac{8 \mu L Q}{\pi (R/2)^4}}{\frac{8 \mu L Q}{\pi R^4}}\] Simplify the expression to find the factor: \[\frac{\Delta P_2}{\Delta P_1} = \frac{8 \mu L Q \cdot \pi R^4}{8 \mu L Q \cdot \pi (R/2)^4}\] Cancel out terms that appear in both the numerator and denominator: \[\frac{\Delta P_2}{\Delta P_1} = \frac{R^4}{(R/2)^4}\] Further simplify the expression: \[\frac{\Delta P_2}{\Delta P_1} = \frac{R^4}{(R^4/16)} = 16\]
05

Conclusion

So, the pressure drop will increase by a factor of 16 (option \((e)\)) when the diameter of the pipe is reduced by half while keeping the flow rate and the pipe length constant.

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Most popular questions from this chapter

A concentric annulus tube has inner and outer diameters of \(25 \mathrm{~mm}\) and \(100 \mathrm{~mm}\), respectively. Liquid water flows at a mass flow rate of \(0.05 \mathrm{~kg} / \mathrm{s}\) through the annulus with the inlet and outlet mean temperatures of \(20^{\circ} \mathrm{C}\) and \(80^{\circ} \mathrm{C}\), respectively. The inner tube wall is maintained with a constant surface temperature of \(120^{\circ} \mathrm{C}\), while the outer tube surface is insulated. Determine the length of the concentric annulus tube. Assume flow is fully developed.

A metal pipe \(\left(k_{\text {pipe }}=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{i, \text { pipe }}=\right.\) \(5 \mathrm{~cm}, D_{o \text {,pipe }}=6 \mathrm{~cm}\), and \(\left.L=10 \mathrm{~m}\right)\) situated in an engine room is used for transporting hot saturated water vapor at a flow rate of \(0.03 \mathrm{~kg} / \mathrm{s}\). The water vapor enters and exits the pipe at \(325^{\circ} \mathrm{C}\) and \(290^{\circ} \mathrm{C}\), respectively. Oil leakage can occur in the engine room, and when leaked oil comes in contact with hot spots above the oil's autoignition temperature, it can ignite spontaneously. To prevent any fire hazard caused by oil leakage on the hot surface of the pipe, determine the needed insulation \(\left(k_{\text {ins }}=0.95 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) layer thickness over the pipe for keeping the outer surface temperature below \(180^{\circ} \mathrm{C}\).

Water enters a 5-mm-diameter and 13-m-long tube at \(45^{\circ} \mathrm{C}\) with a velocity of \(0.3 \mathrm{~m} / \mathrm{s}\). The tube is maintained at a constant temperature of \(8^{\circ} \mathrm{C}\). The exit temperature of water is (a) \(4.4^{\circ} \mathrm{C}\) (b) \(8.9^{\circ} \mathrm{C}\) (c) \(10.6^{\circ} \mathrm{C}\) (d) \(12.0^{\circ} \mathrm{C}\) (e) \(14.1^{\circ} \mathrm{C}\) (For water, use \(k=0.607 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=6.14, v=0.894 \times\) \(10^{-6} \mathrm{~m}^{2} / \mathrm{s}, c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \rho=997 \mathrm{~kg} / \mathrm{m}^{3}\) )

Someone claims that the volume flow rate in a circular pipe with laminar flow can be determined by measuring the velocity at the centerline in the fully developed region, multiplying it by the cross-sectional area, and dividing the result by 2 . Do you agree? Explain.

Water enters a \(5-\mathrm{mm}\)-diameter and \(13-\mathrm{m}\)-long tube at \(15^{\circ} \mathrm{C}\) with a velocity of \(0.3 \mathrm{~m} / \mathrm{s}\), and leaves at \(45^{\circ} \mathrm{C}\). The tube is subjected to a uniform heat flux of \(2000 \mathrm{~W} / \mathrm{m}^{2}\) on its surface. The temperature of the tube surface at the exit is (a) \(48.7^{\circ} \mathrm{C}\) (b) \(49.4^{\circ} \mathrm{C}\) (c) \(51.1^{\circ} \mathrm{C}\) (d) \(53.7^{\circ} \mathrm{C}\) (e) \(55.2^{\circ} \mathrm{C}\) (For water, nse \(k=0.615 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=5.42, v=0.801 \times\) \(\left.10^{-6} \mathrm{~m}^{2} / \mathrm{s} .\right)\)
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