Consider a 300-L storage tank of a solar water heating system initially filled with warm water at 45°C. Warm water is withdrawn from the tank through a 2-cm diameter hose at an average velocity of 0.5 m/s while cold water enters the tank at 20°C at a rate of 5 L/min. Determine the amount of water in the tank after a 20-minute period. Assume the pressure in the tank remains constant at 1 atm.

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

oeerivona oeerivona · Answer 2 · 2022-03-18T21:58:14+01:00

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

How can the amount of water in the tank be calculated?

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The rate at which water is withdrawn from the tank, [tex]Q_{out}[/tex] in L/minute is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;

The volume of water in the tank after 20 minutes is approximately 211.5 L

Learn more about the flow rate of a fluid here:

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