A fountain sends a stream of water straight up into the air to a maximum height of 4.21 m. The effective area of the pipe feeding the fountain is 4.07 x 104 m². Neglecting air resistance and any viscous effects, determine how many gallons per minute are being used by the fountain. (1 gal = 3.79 x 10³ m³) Number Units

Principles of Physics: A Calculus-Based Text
5th Edition
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Chapter15: Fluid Mechanics
Section: Chapter Questions
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The second image shows how I worked it out. I'm suspecting I miscalculated on very last step. Please help me see where I'm missing this one. Thanks 

 

 

 

 

A fountain sends a stream of water straight up into the air to a maximum height of 4.21 m. The effective area of the pipe feeding the
fountain is 4.07 x 104 m². Neglecting air resistance and any viscous effects, determine how many gallons per minute are being used by
the fountain. (1 gal = 3.79 x 10³ m³)
Number
Units
Transcribed Image Text:A fountain sends a stream of water straight up into the air to a maximum height of 4.21 m. The effective area of the pipe feeding the fountain is 4.07 x 104 m². Neglecting air resistance and any viscous effects, determine how many gallons per minute are being used by the fountain. (1 gal = 3.79 x 10³ m³) Number Units
Given Data:
The maximum height of the water is: H = 4.21m
The cross-sectional area of the pipe is: A = 4.07 x 104 m²
The expression for the velocity to attain the maximum height is given:
V=√(2gh)
Substitute the value in the above equation:
√√2 x 9.81 m/s² x 4.21 mV = 9.088 m/s
Step 2:
The expression for the flow rate of the water is: Q=A*V
Substitute value in the above equation:
Q=4.07 x 10-4 m² x 9.088 m/s
60s
1min
Q= 36.99 x 10-4 m³/s x X
1 gal
3.79 x 10-3 m³
Q = 585.59 x 10¹gal/min
Transcribed Image Text:Given Data: The maximum height of the water is: H = 4.21m The cross-sectional area of the pipe is: A = 4.07 x 104 m² The expression for the velocity to attain the maximum height is given: V=√(2gh) Substitute the value in the above equation: √√2 x 9.81 m/s² x 4.21 mV = 9.088 m/s Step 2: The expression for the flow rate of the water is: Q=A*V Substitute value in the above equation: Q=4.07 x 10-4 m² x 9.088 m/s 60s 1min Q= 36.99 x 10-4 m³/s x X 1 gal 3.79 x 10-3 m³ Q = 585.59 x 10¹gal/min
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