Explanation Given information: Temperature at which shower head delivers water, T = 50°C Pressure in the head, p = 400 k P a For water at 50°C, take ρ = 988 k g / m 3 And μ = 2.09 E − 3 k g / m . s Further we need to assume that the shower head is a poor diffuser. So, the pressure in the head is also about 400 kPa. Now, assume the outside pressure is sea − level standard, 101 kPa. For water at 50°C, take ρ = 988 k g / m 3 And μ = 0.548 E − 3 k g / m . s . Further assume that the shower head is a poor diffuser, so that the pressure in the head is also about 400 kPa. Then, assume the outside pressure is sea − level standard, 101 kPa. From the figure given below, estimate C d = 0.61. then with β = 0 for the small holes, each hole delivers a flow rate of: Q 1 h o l e ≈ C d A h o l e 2 Δ p ρ ( 1 − β 4 ) ≈ 0.61 ( π 4 ) ( 0.0015 ) 2 2 ( 400000 − 101000 ) 988 ( 1 − ( 0 ) 4 ) Or, Q 1 h o l e ≈ 2.65 E − 5 m 3 / s And Q t o t a l = 45 Q 1 h o l e ≈ 0.00119 m 3 s ( ≈ 19 g a l m i n ) A huge amount of expensive hot water, as this is a large flow rate. Checking again, the inlet pipe for this flow rate has R e D ≈ 183000 , so C d ≈ 0

8th Edition

ISBN: 9780073398273

Author: Frank M. White

Publisher: McGraw-Hill Education

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Chapter 6, Problem 6.142P

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Flow rate without reducer and the reducer orifice diameter that decrease the flow by 40 percent.

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Chapter 6, Problem 6.141P

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