A large, cylindrical water tank with diameter 2.40 m is on a platform 2.00 m above the ground. The vertical tank is open to the
1 answer:
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 *
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 *
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
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