Question

A fountain sends a stream of water straight up into the air to a maximum height of 4.33 m. The effective area of the pipe feeding the fountain is 5.82 x 10-4 m2. Neglecting air resistance and any viscous effects, determine how many gallons per minute are being used by the fountain. (1 gal = 3.79 x 10-3 m3)

Answer 1

Answer on Question#37649, Physics, Other

Question:

A fountain sends a stream of water straight up into the air to a maximum height of 4.33 m. The effective area of the pipe feeding the fountain is 5.82 x 10-4 m². Neglecting air resistance and any viscous effects, determine how many gallons per minute are being used by the fountain. (1 gal = 3.79 x 10-3 m³)

Answer:

$T + U = \text{const}$ - the law of conservation of energy;

$T$ - kinetic energy of water,

$U$ - potential energy of water,

$$T = \frac{m v^2}{2}$$ $$U = m g h$$

$h$ - maximum height,

$m$ - mass, $v$ - speed.

In our case: $0 + mgh = \frac{m v_0^2}{2} + 0$

Therefore, initial velocity of water equals: $v_0 = \sqrt{2gh} = 9.21\frac{m}{s}$

Volume of the water equals: $V = A v_0 t$

$A$ - effective area of the pipe, $t$ - time.

Full volume per 1 minute equals: $V = 9.21\frac{m}{s} 5.82 \times 10^{-4} \, m^2 \times 60 \, s = 0.322 \, m^3$

Full volume in gallons: $V = \frac{0.322}{3.79 \times 10^{-3}} = 85 \, \text{gallons}$

Answer: 85 gallons