The discharge rate of water (Q) that flows across the exit of the pipe in 1.00 seconds is 0.200 [tex]m^3/s[/tex].
Given the following data:
- Elevation of point 1 = 10.0 m
- Elevation of point 2 = 2.00 m
- Elevation of point 3 = 2.00 m
- Cross-sectional area at point 2 = [tex]4.80\times 10^{-2} \;m^2[/tex]
- Cross-sectional area at point 3 = [tex]1.60\times 10^{-2} \;m^2[/tex]
To find the discharge rate of water (Q) that flows across the exit of the pipe in 1.00 seconds assuming that Bernoulli's equation applies:
In this exercise, we would apply Torricelli's Law, which states that the speed of the efflux in an apparatus, under the force of gravity is directly proportional to the square root of the height of the fluid above it.
Mathematically, Torricelli's Law is given by the formula:
[tex]V =\sqrt{2gh}[/tex]
Where:
- g is the acceleration due to gravity.
- h is the height of the apparatus.
- V is the speed of liquid.
Substituting the given parameters into the formula, we have;
[tex]V =\sqrt{2 \times 9.8 (10-2)}\\\\V =\sqrt{19.6 (8)}[/tex]
Speed, V = 12.52 m/s
Now, we can find the discharge rate of the pipe:
[tex]Q = V \times A\\\\Q = 12.52 \times 1.60\times 10^{-2}[/tex]
Discharge rate, Q = 0.200 [tex]m^3/s[/tex]
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