Answer:
a. h₁ =0.0008m = 0.8mm is the height of water that flow out
the change in height of water in the tank = h - h₁ = 2 - 0.0008 = 1.9992m
b. t =31.93 sec
Explanation:
a) When 1.00 gal of water flows out of the tank, what is the change in the height of the water in the tank?
given:,
cylindrical water tank with diameter, D - 2.40 m
the tank is 2.00 m above the ground.
depth of the water in the tank is 2.00 m.
diameter of hole, d - 0.520 cm
firstly, we need to calculate the volume of water in the tank:
Volume = πr²h
=(pi * D * D)/4 * height of water
=(3.142 * 2.4 * 2.4)/4 * 2 = 4.52*2 = 9.05m³
1.0 gal of water is equivalent to = 0.0038m³
the volume of 1gal is 0.0038 = A *h
the area of the tank is calculated above as 4.52m²
therefore, 0.0038 = 4.52 *h
h₁ =0.0008m = 0.8mm is the height of water that flow out
the change in height of water in the tank = h - h₁ = 2 - 0.0008 = 1.9992m
b) How long does it take you to collect 1.00 gal of water in the bucket?
1.0 gal of water is equivalent to = 0.0038m³
to calculate for Volume flow, Q for a draining tank
Q = Cd * A *[tex]\sqrt{2gH}[/tex]
where Cd is a discharge coefficient, and is given by 0.9 for water
A is the area of the small hole = (pi * D * D)/4 = (pi * 0.0052 * 0.0052)/4
A = 0.0000212m²
H= height of the hole from the tank water level= 2m - 0.0052 = 1.9948m
g = 9.8m/s2
Q = 0.9 * 0.0000212 *[tex]\sqrt{2*9.8 * 1.9948}[/tex]
Q = 0.0000191 * 6.253 = 0.000119 m³/s
Q = V/t
Qt = V
t = V/Q = 0.0038m³/0.000119 m³/s
t =31.93 sec
(a) The change in the height of the water is 1.99 m.
(b) The time taken for 1 gal of water to enter the bucket is 28.47 s.
The given parameters:
The volume of water in cylindrical tank is calculated as follows;
[tex]V = \pi r^2 h\\\\ V = \pi (1.2)^2 \times 2\\\\ V = 9.05 \ m^3 [/tex]
When 1 gallon of water flows out, the change in the height of the water is calculated as follows;
1 gal = 0.003785 m³
[tex]V_1 - V_2 = \pi r^2 (h_1 -h_2)\\\\ h_1 - h_2 = \frac{V_1 - V_2}{\pi r^2} \\\\ h_1 - h_2 = \frac{9.05 -0.003785}{\pi (1.2)^2} \\\\ h_1 - h_2 = 1.99 \ m[/tex]
The time taken for 1 gal of water to enter the bucket is calculated as follows;
[tex]\frac{V}{t} = Av[/tex]
where;
The speed of the water when it hits the bottom of the bucket is calculated as follows;
[tex]v^2 = u^2 + 2gh\\\\ v^2 = 0 + 2gh\\\\ v= \sqrt{2gh} \\\\ v = \sqrt{2 \times 9.8 \times 2} \\\\ v = 6.26 \ m/s[/tex]
The area of the hole is calculated as follows;
[tex]A = \pi r^2\\\\ A = 3.142 \times (\frac{0.0052}{2} )^2\\\\ A = 2.124 \times 10^{-5} \ m^2[/tex]
The time taken for 1 gal of water to enter the bucket is calculated as;
[tex]t = \frac{V}{Av} \\\\ t = \frac{0.003785}{2.124 \times 10^{-5} \times 6.26} \\\\ t = 28.47 \ s[/tex]
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