Don't let this problem hose you

Classical Mechanics Level 4

A pumper fire truck is pumping water through a hose. The hose connects to the truck at 1 m above sea level and the hose is held on top of a ladder at 10 m above sea level. The base of the hose, where it connects to the truck, has a diameter of 10 cm, and the nozzle, where the water comes out, has a diameter of 2 cm. The pressure at the base of the hose is 700,000 Pa. How much mechanical energy in Joules does one gram of water gain as it goes from the bottom to the top of the hose?

Details and assumptions

  • The acceleration due to gravity is − 9.8 m / s 2 -9.8~m/s^2 .
  • The density of water is 1 g / c m 3 1~g/cm^3 .
  • You may neglect any viscosity for water.
  • 1 atmosphere is 101,325 Pa.


The answer is 0.6.

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David Mattingly by David Mattingly

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2 solutions

Pratik Satish
Dec 23, 2013

Using bernoulli's equation, the change in mech energy is equal to change in(PV) . Since the pressure at the top where water comes out is atmospheric pressure the answer is (700000-101325)*(10pow-6)=0.598

3 Helpful 0 Interesting 0 Brilliant 0 Confused

...isn't it the drop 9 m, or is it?

Luciano Riosa - 7 years, 4 months ago
Milly Choochoo
Jan 1, 2014

Apply Bernoulli's Principle:

P o + 1 2 ρ v o 2 + ρ g y o = P f + 1 2 ρ v f 2 + ρ g y f P_o + \frac{1}{2}\rho v_o^2 + \rho g y_o = P_f + \frac{1}{2}\rho v_f^2 + \rho g y_f

( 700 , 000 P a ) + 1 2 ( 1000 k g m 3 ) ( 0 m 2 s 2 ) + ( 1000 k g m 3 ) ( g ) ( 1 m ) = ( 101 , 325 P a ) + 1 2 ( 1000 k g m 3 ) ( v m 2 s 2 ) + ( 1000 k g m 3 ) ( g ) ( 10 m ) (700,000 Pa) + \frac{1}{2}(1000 \frac{kg}{m^3})(0 \frac{m^2}{s^2}) + (1000 \frac{kg}{m^3})(g)(1 m) = (101,325 Pa) + \frac{1}{2}(1000 \frac{kg}{m^3})(v \frac{m^2}{s^2}) + (1000 \frac{kg}{m^3})(g)(10 m)

Solve for v 2 v^2 :

v 2 = 1020.77 m 2 s 2 v^2 = 1020.77 \frac{m^2}{s^2}

Calculate total mechanical energy of one gram of water at the top and the bottom, and then subtract them:

E = P E + K E = m g y + 1 2 m v 2 E = PE + KE = mgy + \frac{1}{2}mv^2

E b = ( . 001 k g ) ( g ) ( 1 m ) E_b = (.001 kg)(g)(1 m)

E t = ( . 001 k g ) ( g ) ( 10 m ) + 1 2 ( . 001 k g ) ( 1020.77 m 2 s 2 ) E_t = (.001 kg)(g)(10 m) + \frac{1}{2}(.001 kg)(1020.77 \frac{m^2}{s^2})

E t − E b ≈ . 59868 J ≈ . 6 J E_t - E_b \approx .59868 J \approx \boxed{.6 J}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

This question made me want to apply the conservation of volume flow rate, but you must realize that the velocity at the bottom is zero, so you can't do that as simply as you (I) might think.

This is, however, just one way to solve it, and a pretty simple one at that. I'm pretty sure there are many ways to solve this.

Milly Choochoo - 7 years, 5 months ago

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why would the velocity at the bottom be zero?

Calvin Chau - 7 years, 4 months ago

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I believe it's the same concept as pushing a block up an incline. At the bottom, its velocity is zero, and the force is what pushes it up.

Milly Choochoo - 7 years, 4 months ago

...isn't it the drop 9 m, or is it?

Luciano Riosa - 7 years, 4 months ago

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