Question

A fountain sends a stream of water straight up into the air to a maximum height of 4.87 m. The effective area of the pipe feeding the fountain is 5.71 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#37461, Physics, Other

Question:

A fountain sends a stream of water straight up into the air to a maximum height of $4.87\mathrm{m}$. The effective area of the pipe feeding the fountain is $5.71 \times 10^{-4}\mathrm{m}^2$. 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:

The law of conservation of energy:

$$T + U = \text{const}$$

where $T = \frac{mv^2}{2}$ - kinetic energy of water, m - mass, v - speed

$U = mgh$ - potential energy of water, $h$ is maximum height

$$0 + mgh = \frac{m v_0^2}{2} + 0$$

Therefore, initial velocity of water equals:

$$v_0 = \sqrt{2gh} = 9.77 \frac{m}{s}$$

Volume of the water equals:

$$V = A v_0 t$$

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

Volume per 1 minute equals:

$$V = 9.77 \frac{m}{s} \cdot 5.71 \cdot 10^{-4} \, m^2 \cdot 60 \, s = 0.335 \, m^3$$

In gallons:

$$V = \frac{0.237}{3.79 \cdot 10^{-3}} = 88.3 \, \text{gallons}$$

Answer: 88.3 gallons