A cylindrical hot water heater of cross sectional areaA heater =0.2 m 2 is fille

Question

A cylindrical hot water heater of cross sectional areaAheater=0.2 m2 is filled with water at aheight h=1.25 m. I need to empty the water heater from a valveplaced at the bottom of the heater which has an opening ofAvalve =3 cm2 when it is fully oppened.

a) Find an expression of the speed V of thewater coming out of the fully opened valve at the time the waterlevel in the tank is h .

Hint: We have discussed this problem in class: UseBernoulli’s principle to calculate the speedV in terms of the acceleration of gravityg and the water level h. g=10 m/s2

b) Calculate the numerical value of the speed Vof the water when the water level in the tank is at h=1.25 m.

c) Calculate the flow of the water out of the valve, when thewater level is at h=1.25 m.

d) Now assume that the average flow in the valve duringthe time it took me to empty the tank is half of what youcalculated in part (c) above.

How long did it take me to empty the water heater?

Explanation / Answer

p + 1/2 v2 + g h =constant      Bernoulli's equation p0 + g h = p0 + 1/2 v2       since v is nearlyzero at the top and h = 0 at the bottom of the tank (p0= atmospheric pressure) v = (2 g h)      Torricelli'sequation v = ( 2 * 10 * 1.25) = 5 m / sec V = A x     the volume ofwater when x flows thru an area A R = V / t = A x / t = Av     where R is the rate of water flow out ofthe tank R = 3 * 10E-4 * 5 = 1.5 * 10E-3 m3 /sec        1 cm2 = 1* 10E-4 m2 Rav = 7.5 * 10E-4 m3 /sec        average rate offlow V = h A = 1.25 * .02 = .025m3        volume ofwater in tank t = .025 / 7.5 * 10E-4 = 250 / 7.5 = 33.3 sec

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