Answer:
160 Kg
Explanation:
This problem is solved by considering the mass balance. We know the density of water, which we will assume constant , the volume of water originally in the tank and the flows of water entering and leaving the tank.
Let m₀ mass of water initially in the tank
m₁ mass of water entering the tank and
m₂ mass of water exiting the tank
V tank volume
then
m₀ = D x V = 1000 Kg x 0.30 m³ = 300 Kg
m₁ = D Vwater in x time = 1000 Kg/m³ x 5 x 10⁻³ m /min x 20 min = 100 Kg
m₂ = D x Area Hose x Vel exit x time =
= 1000 Kg /m³ x 3.1416 x ( 0.02 m)² x 0.5 m/s x ( 20 min x 60 s/min)
= 240 Kg
mass of water initially + mass water in - mass water out =
(300 + 100 - 240) Kg = 160 Kg
The flow rate out of and into the tank of 0.5 m/s and 5 L/min give the
amount of water in the tank after 20–minutes as approximately 211.5 L
Solution:
The rate at which water is withdrawn from the tank, [tex]Q_{out}[/tex] in L/minute is
found as follows;
[tex]Q_{out}[/tex] = π·r² × v
Where;
[tex]r = \dfrac{2 \, cm}{2} = 1 \, cm = \mathbf{ 0.01 \, m}[/tex]
v = 0.5 m/s
Multiply by 60 × 1000 to convert m³/s to L/minute
Which gives;
[tex]Q_{out}[/tex] = π × 0.01² × 0.5 × 60 × 1000 ≈ 9.425 L/min
The volume that leaves the tank in 20 minutes is therefore;
[tex]V_{out}[/tex] ≈ 20 min × 9.425 L/min = 188.5 L
The volume of water that enters the tank, [tex]\mathbf{V_{in}}[/tex], is found as follows;
[tex]V_{in}[/tex] = 20 min × 5 L/min = 100 L
The volume of water in the tank after 20 minutes is therefore;
V = V₀ + [tex]\mathbf{V_{in}}[/tex] - [tex]\mathbf{V_{out}}[/tex]
Which gives;
V = V₀ + [tex]V_{in}[/tex] - [tex]V_{out}[/tex]
V = 300 L + 100 L - 188.5 L = 211.5 L
Therefore;
Learn more about the flow rate of a fluid here:
https://brainly.com/question/6858718
