(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
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
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