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nyoo
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"shower head" for feeding water onto a solar thermal panel
The question is homework to me, but I haven't actually seen formal physics classwork since college, 41 years ago. Please advise if this question is misplaced.
I'm building a "shower head" to trickle or spray water onto a home-made solar thermal panel. My pump will deliver 180 gallons per hour, unpressurised, to a tee fitting. Pressure at the tee is 2.6 psi. The tee is in the center of an 8' horizontal copper pipe, capped at both ends. All pipes are 3/4" L-type copper. (The wall thickness for L copper is 0.045".)
I will drill evenly-spaced holes into the 8' horizontal pipe, to spray water onto a panel. Think of it as a long skinny shower head, with a single row of holes. I'm looking for an even flow out of the sum of the holes, at greater than 90 GPH. There must be a sufficient number of holes drilled into the 8-foot length to cover the entire panel.
I've been staring at the Bernouilli's Principle, which I believe holds the answer to my question. But Bernouilli's Principle talks about gravity, and I'm not sure how to convert that for forced GPH pressure. And Bernouilli's Principle seems to assume a single hole, not a row of holes. I'm not sure how to proceed.
I'm thinking 32 holes of .125" diameter. My naive layman's logic is based on area of the pipe: (3/4)^^2 = .5625, 32 * (1/8)^^2 = .5. But I know this logic does not account for turbulence at the pin-hole openings.
As copper pipe is currently expensive, I'm trying to reduce the number of experimental iterations, by applying science.
The questions are: will the water come out of all holes evenly? Will the holes farthest from the central tee spray hardest? Will the ends not spray at all? Should I drill bigger holes?
Thanks very much for your help.
Homework Statement
The question is homework to me, but I haven't actually seen formal physics classwork since college, 41 years ago. Please advise if this question is misplaced.
I'm building a "shower head" to trickle or spray water onto a home-made solar thermal panel. My pump will deliver 180 gallons per hour, unpressurised, to a tee fitting. Pressure at the tee is 2.6 psi. The tee is in the center of an 8' horizontal copper pipe, capped at both ends. All pipes are 3/4" L-type copper. (The wall thickness for L copper is 0.045".)
I will drill evenly-spaced holes into the 8' horizontal pipe, to spray water onto a panel. Think of it as a long skinny shower head, with a single row of holes. I'm looking for an even flow out of the sum of the holes, at greater than 90 GPH. There must be a sufficient number of holes drilled into the 8-foot length to cover the entire panel.
Homework Equations
I've been staring at the Bernouilli's Principle, which I believe holds the answer to my question. But Bernouilli's Principle talks about gravity, and I'm not sure how to convert that for forced GPH pressure. And Bernouilli's Principle seems to assume a single hole, not a row of holes. I'm not sure how to proceed.
The Attempt at a Solution
I'm thinking 32 holes of .125" diameter. My naive layman's logic is based on area of the pipe: (3/4)^^2 = .5625, 32 * (1/8)^^2 = .5. But I know this logic does not account for turbulence at the pin-hole openings.
As copper pipe is currently expensive, I'm trying to reduce the number of experimental iterations, by applying science.
The questions are: will the water come out of all holes evenly? Will the holes farthest from the central tee spray hardest? Will the ends not spray at all? Should I drill bigger holes?
Thanks very much for your help.