Question

the tank shown below 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 volumetric flowrate of 595 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. formula 5 points

Explanation:

Step1: Convert time to seconds

15 minutes = 15×60 s = 900 s

Step2: Calculate volume of water inflow

The volumetric - flow rate of inflow $Q_{in}=595\ cm^{3}/s$. The volume of water inflow $V_{in}=Q_{in}\times t$, where $t = 900 s$. So $V_{in}=595\times900=535500\ cm^{3}$.

Step3: Calculate volume of water outflow

The cross - sectional area of the drain pipe $A=\pi r^{2}$, with $r=\frac{d}{2}=\frac{4}{2}=2\ cm$. So $A=\pi\times(2)^{2}=4\pi\ cm^{2}$. The velocity of water exiting the drain $v = 37.4\ cm/s$. The volumetric - flow rate of outflow $Q_{out}=A\times v=4\pi\times37.4\ cm^{3}/s$. The volume of water outflow $V_{out}=Q_{out}\times t=(4\pi\times37.4)\times900\ cm^{3}$. $V_{out}=4\times3.14\times37.4\times900 = 421968\ cm^{3}$.

Step4: Calculate net volume of water in the tank

The net volume of water in the tank $V = V_{in}-V_{out}$. $V=535500 - 421968=113532\ cm^{3}$.

Step5: Convert volume to mass

The density of water $
ho=997.1\ kg/m^{3}=0.9971\ g/cm^{3}$. Using the formula $m=
ho V$, we have $m = 0.9971\times113532\ g$. $m = 113202.7572\ g\approx113.2\ kg$.

Answer:

113.2