A firehose must be able to shoot water to the top of a building 35 m tall when aimed straight up. Water enters the hose at a steady rate of 0.500 m3 / s and shoots out of a round nozzle. (a) What is the maximum diameter this nozzle can have? (b) If the diameter of the hose is 10 cm, what is the pressure in the hose 1.00 m below the nozzle?(adsbygoogle = window.adsbygoogle || []).push({});

(Vyf)^2 = (Vyi)^2 + 2ay(yf-yi)

J = AV

A = (pie)r^2

P1 + 0.5pv1^2 + pgh1 = P2 + 0.5pv2^2 + pgh2

Found that maximum diameter in part (a) was 15.6 cm. This means the nozzle diameter is more than 50% larger than the hose diameter. When I applied Bernoulli's equation, I came up with a negative number for the pressure in part (b) like this... P1 = (1.01*10^5) + 500(26.19^2 - 63.66^2) + 9800 = -1572539.75 pa. I'm not sure how to explain it, but I think it has something to do with the fact that the water ejected from the narrow hose would remain at the same velocity, because the is no force causing it to conform to the larger dimensions of the nozzle. Therefore, the pressure would be simply Patm + pgh = 110800 pa?

**Physics Forums - The Fusion of Science and Community**

# Bernoulli's equation limitations?

Have something to add?

- Similar discussions for: Bernoulli's equation limitations?

Loading...

**Physics Forums - The Fusion of Science and Community**