Question

A drinking fountain shoots water about 14 cm up in the air from a nozzle of diameter 0.60 cm. The pump at the base of the unit (1.1 m below the nozzle) pushes water into 1.2 cm diameter supply pipe that goes up to the nozzle. What gauge pressure does the pump have to provide? Ignore the viscosity; your answer will therefore be an underestimate

Answer 1

Bernoullis equation has to be used to solbe this question

Pressure at the top of flow= Pressure at the nozzle

Here density rho has been denoted by p

Pressure at the top of the flow = Pa + (1/2)*p*(va^2) + pg(0.14 m + 1.1m)---------------(1)

Here velocity is 0 ,va=0 as water flows only up to that level

Pressure at the nozzle =  Pa + (1/2)*p*(vb^2) + pg(1.1m)-----------------(2)

Equating these two we get,

(1/2)*p*(vb^2) = pg(0.14)

So, vb = sqrt(2*9.8*0.14)

=1.656 m/s

Now applying Bern equation at the nozzle and pump,

Pa + (1/2)*p*(vb^2) + pg(1.1m)= Pa + (1/2)*p*(vc^2) + pg(0 m) (Taking reference height as 0 for pump)

P gauge =(1/2)*p*(vb^2 - vc^2 ) + pg(1.1m)

From continuity equation ,

Ab * vb = Ac * vc

Vc = Ab * vb / Ac = ((0.6^2)/(1.2^2) * Vb

Now,

P gauge =(1/2)*p*(vb^2)*(1- (0.6^2/1.2^2) ) + pg(1.1m)

= pg(0.14m)*3/4 + pg(1.1m) (this was obtained from initial equations already proved)

=(1.205)*(10^3) *(9.8)

=11809 Pa

=1.18 * 10 ^4 Pa