Asked by Henry
A 6.0-cm diameter horizontal pipe gradually narrows to 4.0 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 32 kPa and 24 kPa, respectively. What is the volume rate of flow?
I know to use Bernoulli's Equation and the Equation of Continuity, but I can't figure out how to solve for either velocity variable.
I know to use Bernoulli's Equation and the Equation of Continuity, but I can't figure out how to solve for either velocity variable.
Answers
Answered by
drwls
P1 - P2 = (rho/2)(V2^2 - V1^2)= 8000 N/m^2 (Bernoulli equation)
(rho is the fluid density)
6^2*V1 = 4^2*V2, so V2 = (4/9)V1
(from the continuity equation)
Substitute (4/9)V1 for V2 in the first equation and solve for V1
The volume flow rate is (pi/4)D^2*V1
where D = 0.06 m
(rho is the fluid density)
6^2*V1 = 4^2*V2, so V2 = (4/9)V1
(from the continuity equation)
Substitute (4/9)V1 for V2 in the first equation and solve for V1
The volume flow rate is (pi/4)D^2*V1
where D = 0.06 m
Answered by
Anonymous
horrible
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