Question

1.)What volume of H2O will escape per minute from an open tank through an opening 3.0 m in diameter that is 5.0 m below the H2O level in the tank?

2.) A hole of area 1 mm^2 in the pipe near the lower end of a large H2O storage tank & a stream H2O shoots from it. If the topof the H2O in the tank is 20 m above the point of the tank, how much H2O escapes in one (1) second?

Answer 1

Answer on Question #72582-Physics-Other

1.) What volume of H2O will escape per minute from an open tank through an opening 3.0 cm in diameter that is 5.0 m below the H2O level in the tank?

Solution

We can use the Bernoulli equation:

$$\frac {\rho v ^ {2}}{2} = \rho g h$$ $$v = \sqrt {2 g h}$$

The flow rate is

$$\frac {d V}{d t} = v A = v \frac {\pi d ^ {2}}{4} = \sqrt {2 (9 . 8) (5)} \left(\frac {\pi}{4} (0. 0 3) ^ {2}\right) = 0. 0 0 7 \frac {m ^ {3}}{s} = 0. 4 2 \frac {m ^ {3}}{m i n}.$$

Answer: $0.42\frac{m^3}{min}$

2.) A hole of area $1 \, \text{mm}^2$ in the pipe near the lower end of a large H2O storage tank & a stream H2O shoots from it. If the top of the H2O in the tank is $20 \, \text{m}$ above the point of the tank, how much H2O escapes in one (1) second?

Solution

We can use the Bernoulli equation:

$$\frac {\rho v ^ {2}}{2} = \rho g h$$ $$v = \sqrt {2 g h}$$

The flow rate is

$$\frac {d V}{d t} = v A = \sqrt {2 (9 . 8) (2 0)} (1 0 ^ {- 6}) = 2 \cdot 1 0 ^ {- 5} \frac {m ^ {3}}{s}.$$ $$m = \rho \frac {d V}{d t} t = (1 0 0 0) (2 \cdot 1 0 ^ {- 5}) (1) = 0. 0 2 k g = 2 0 g.$$

Answer: $0.02\mathrm{kg} = 20\mathrm{g}$

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