Recent content by Lee Harrington

I'm not sure what you mean, so your saying let Pout equal to Patm as there is no gauge pressure on exit? My understanding is that if there is a gauge pressure on exit you would need to have been told that in order to calculate Pin

sorry, being lazy pin/ρ + Vin2/2 + gz = pout/ρ + Vout2/2 + gz pout=0 pin/ρ + Vin2/2 = + Vout2/2 isolating pin pin = ρ((Vout2 - Vin2)/2)

converting mass flow rate to volume flow rate and using Q=VA: Vin=1.98m/s ; Vout=6.97m/s pin=999((6.972 - 1.982)/2)=22308 Pa

Also for the next part, do you continue to assume that pout=0 ?

ok, so using bernoulli's equation: pin = ρ((Vout2 - Vin2)/2) ? And that's the gauge pressure?

Homework Statement Water flows through a circular pipe with a 180° horizontal elbow and exits to the atmosphere through a nozzle as shown in Fig. Q3. The diameter of the pipe is 300 mm and the diameter of the nozzle exit is 160 mm. The density of water is 999 kg/m3 . The mass flow rate of...