a, What is the exit velocity of the water from the nozzle and what is the veloci

Question

a, What is the exit velocity of the water from the nozzle and what is the velocity through the connecting pipe?

b, What is the pressure in the tank, Pt, in units of psi?

A large pressurized tank is used to shoot water (gamma water= 62.4 lb/ft^3) out of a nozzle into the atmosphere. The nozzle diameter is 6 inches and the flow rate out of the nozzle is 15.03 ft^3/s. The pipe connecting the tank to the nozzle has a uniform diameter of 12 inches. Between the surface of the water and the nozzle, the head loss is 18 ft. D = 6in Q = 15.03 ft^3/s

Explanation / Answer

p1 = Pt ,v1 = 0 , neglecting, z1 = 14 ft

p2 = Patm = 14.69 Psi v2 = ? z2 = 20 ft

Head loss = 18 ft

Q = v2 *Area

15.03 = v2 *pi *d^2/4 , 6inch = 0.5ft

15.03 = v2 *pi*(0.5)^2/4

speed through nozzle , v2 = 76.547 ft/s

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writing bernoulli's equation between water level and nozzle exit

Pt/yw + v1^2/2g + z1 = Patm/yw + v2^2/2g +z2 + Hl

Pt/62.4 + 14 = 14.69/62.4 + 76.547^2/2*32.2 + 20 + 18

Pt = 7189 Psi

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