Respuesta :
Answer:
160 Kg
Explanation:
This problem is solved by considering the mass balance. We know the density of water, which we will assume constant Ā , the volume of water originally in the tank and the flows of water entering and leaving the tank.
Let mā mass of water initially in the tank
mā mass of water entering the tank Ā and
mā mass of water exiting the tank
V Ā tank volume
then
mā = D x V = 1000 Kg x 0.30 mĀ³ = 300 Kg
mā = D Vwater in x Ā time = 1000 Kg/mĀ³ x 5 x 10ā»Ā³ m /min Ā x 20 min = 100 Kg
mā = D x Area Hose x Vel exit x time =
Ā Ā Ā = 1000 Kg /mĀ³ x 3.1416 x ( 0.02 m)Ā² x 0.5 m/s x ( 20 min x 60 s/min)
Ā Ā Ā = 240 Kg
mass of water initially + mass water in - mass water out =
(300 + 100 - 240) Ā 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 20āminutes 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The rate at which water is withdrawn from the tank, [tex]Q_{out}[/tex] in L/minute is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] = ĻĀ·rĀ² Ć v
Where;
[tex]r = \dfrac{2 \, cm}{2} = 1 \, cm = \mathbf{ 0.01 \, m}[/tex]
v = 0.5 m/s
Multiply by 60 Ć 1000 to convert mĀ³/s to L/minute
Which gives;
[tex]Q_{out}[/tex] = Ļ Ć 0.01Ā² Ć 0.5 Ā Ć 60 Ć 1000 ā 9.425 L/min
The volume that leaves the tank in 20 minutes is therefore;
[tex]V_{out}[/tex] ā 20 min Ć 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 Ć 5 L/min = 100 L
The volume of water in the tank after 20 minutes is therefore;
V = Vā + [tex]\mathbf{V_{in}}[/tex] - [tex]\mathbf{V_{out}}[/tex]
Which gives;
V = Vā + [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:
https://brainly.com/question/6858718
