Question

we start with a tank that has a single - drain pipe at the bottom, with a diameter equal to 4 centimeters. water is fed to the tank from the faucet with a certain flow rate of 395 cm³/sec. the velocity of water exiting the tank drain at the bottom is equal to 37.4 cm/sec. if the tank is dry at the start, how much water in kilograms will be held in the tank after 15 minutes? assume the density of water to be 997.1 kg/m³. report your answer to one decimal place.

Explanation:

Step1: Calculate the volume - flow rate of water leaving the tank

The cross - sectional area of the drain pipe $A=\pi r^{2}$, where $r = \frac{d}{2}=\frac{4}{2}=2$ cm $=0.02$ m. The velocity of water leaving the tank $v = 37.4$ cm/s $=0.374$ m/s. The volume - flow rate $Q_{out}=A\times v=\pi\times(0.02)^{2}\times0.374\ m^{3}/s$.
$Q_{out}=\pi\times4\times10^{-4}\times0.374\ m^{3}/s\approx4.7\times10^{-4}\ m^{3}/s$.

Step2: Calculate the time in seconds

The time $t = 15$ minutes. Since $1$ minute $= 60$ seconds, $t=15\times60 = 900$ s.

Step3: Calculate the volume of water that has drained out

The volume of water that has drained out $V_{out}=Q_{out}\times t=4.7\times10^{-4}\times900\ m^{3}\approx0.423\ m^{3}$.

Step4: Calculate the mass of water that has drained out

The density of water $
ho = 997.1\ kg/m^{3}$. Using the formula $m=
ho V$, the mass of water that has drained out $m=
ho V_{out}=997.1\times0.423\ kg\approx422\ kg$.

Answer:

$422$ kg