Question

(a) What is the pressure drop due to the Bernoulli effect as water goes into a 2.70-cm-diameter nozzle from a 8.90-cm-diameter fire hose while carrying a flow of 41.0 L/s?

N/m²

(b) To what maximum height above the nozzle can this water rise? (The actual height will be significantly smaller due to air resistance.)

m

Answer 1

a) let

d1 = 2.70 cm - 0.027 m
d2 = 8.90 cm = 0.089 m
dV/dt = 41 L/s = 41*10^-3 m^3/s

let v1 and v2 are the speed of water through nozzle and hose.

volume flow rate,

dV/dt = A1*v1

dV/dt = (pi*d1^2/4)*v1

v1 = 4*(dV/dt)/(pi*d1^2)

= 4*41*10^-3/(pi*0.027^2)

= 71.6 m/s

now use continuty equation, A1*v1 = A2*v2

==> v2 = A1*v1/A2

= (pi*d1^2/4)*v1/(pi*d2^2/4)

= (d1/d2)^2*v1

= (0.027/0.089)^2*71.6

= 6.59 m/s

now use Bernouilli's equation

P1 + (1/2)*rho*v1^2 = P2 + (1/2)*rho*v2^2

P2 - P1 = (1/2)*rho*(v1^2 - v2^2)

= (1/2)*1000*(71.6^2 - 6.59^2)

= 2.54*10^6 pa <<<<<<<<<<---------------------Answer

b) Hmax = v1^2/(2*g)

= 71.6^2/(2*9.8)

= 261 m <<<<<<<<<<---------------------Answer