Let\'s again consider draining my swimming pool by siphoning the water out using

Question

Let's again consider draining my swimming pool by siphoning the water out using a garden hose. My pool is 40 ft long. 20 ft wide, and has a uniform depth of 6 ft. We'll siphon the water out using a garden hose with an inside diameter of 1.0 in. The inlet end of the hose is at the bottom of the pool, and the outlet end is in the backyard. 3 ft below the bottom of the pool. What is the volume flow rate of water when we start to drain the pool? If we could hold this flow rate constant (for example, by moving the hose outlet to maintain a constant water head), how long would it take to drain the pool? I'm too lazy to move the hose, so let's assume the outlet end stays in the backyard at an elevation 3 ft below the bottom of the pool. Develop and solve an unsteady state mass balance to determine the water level in the pool as a function of time. How long will it take to completely drain the pool?

Explanation / Answer

we have point 1 as the bottom of the pool where pipe is present and point 2 as the place where water is coming out of the pipe Using bernauli's equation at these two points we get at point 1 3g + (rho*g*6+P_0)/rho where P_0 is the atmospheric pressure and rho is the density of water at point 2 v^2/2 + P_0/rho v = sqrt*(3g + 6g) = sqrt (9g) This is the velocity of water coming out of siphon Volume flow rate is given as R = rho*A/v = rho*pi*(r^2)/sqrt(9g) where r is the radius of pipe I havent put the values since the units are in old units and i dont know the conversions. Time taken to empty the tank = volume/rate T= l*b*h / R In second question I cannot figure out what will be the expression of height of water at any time t in the pool.

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