Homework 8_solutions - Homework 10 Prof Hodges 58 homework...

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Homework 10 Prof. Hodges 58 homework points You are designing a water supply piping system for a new housing development. The fire department requires a flow velocity out of the nozzle of 20 m/s when the hose is 25 m above the hydrant and a hose length of 40 m is attached. The nozzle cross-sectional area is 1/5th of the hose cross-sectional area . The hose diameter is 10 cm. The head loss in the hose is estimated from h L(hose) = 0.3L V hose 2 2g , where L is the length of the hose. The head loss in the nozzle is estimated from h L(nozzle) = 0.1 V nozzle 2 2g A. SIMPLE PROBLEMS 1. (4 points) Determine the minimum gage pressure required at a fire hydrant to meet the fire department’s needs. Make sure you sketch the system and label the components you use in your equations. Sketch the system Because the hose and nozzle have head losses, we must use the energy equation (not the Bernoulli equation) to solve this problem. The upstream position (1) is the hydrant, the downstream position (3) is the free jet of water outside of the nozzle. The position (2) is just before the nozzle (which we will need for other parts of this homework). We can write the energy equation as P 1 ρ g + z 1 + V 1 2 2g = P 3 ρ g + z 3 + V 3 2 2g + h L(hose) + h L(nozzle) hydrant hose nozzle 25 m x z 1 2 3 https://www.coursehero.com/file/10923745/Homework-08-CE-319F-solutions/ This study resource was shared via CourseHero.com
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Noting that P 3 is a free jet, we have P 3 =0 (gage pressure atmospheric). Then substituting our relationships for the velocity, we get P 1 ρ g = z 3 z 1 V 1 2 2g + V 3 2 2g + 0.3L V 1 2 2g + 0.1 V 3 2 2g Note: be very careful to put the correct subscripts on the head loss terms. As given in the problem statement, the head loss in the hose is a function of the hose velocity (V 1 ) whereas the head loss in the nozzle is a function of the nozzle velocity (V 3 ) We note that conservation of mass for incompressible flow implies that Q 1 = Q 3 , so that V 1 A 1 = V 3 A 3 or V 1 = V 3 A 3 A 1 we can apply this in the energy equation as P 1 ρ g = z 3 z 1 A 3 A 1 2 V 3 2 2g + V 3 2 2g + 0.3L A 3 A 1 2 V 3 2 2g + 0.1 V 3 2 2g then regroup terms to obtain P 1 ρ g = z 3 z 1 + V 3 2 2g 1.1 + A 3 A 1 2 0.3L 1 Solving for pressure we have P 1 = ρ g z 3 z 1 + V 3 2 2g 1.1 + A 3 A 1 2 0.3L 1 Leaving V 2 as the only unknown we have P 1 = 9810 N m 3 25m + V 3 2 2 9.81 m s 2 1.1 + 1 5 2 0.3 40 ( ) 1
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