A rectangular shower head has 25
circular openings of radius 1 mm
each. The pipe connecting the
shower head has a radius 6 mm. If
the water flows in the pipe at the
rate of 5m/s, then calculate the
speed of water leaving the
openings of the shower head.
Answers
Answer:
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Step-by-step explanation:
A shower head has 20 circular openings each with radius 1.0 mm. The shower head is connected to a pipe with a radius of 0.8 (I presume, cm). If the speed of water in the pipe is 3.0 m/s, what is its speed as it exits the shower head opening?
Let the radii of the shower head opening and the pipe be rs and rp respectively.
Let the speed of water in the pipe and as it exits the shower head be vp and vs respectively.
It is given that rs=1 mm =10−3 m, rp=0.8 cm =8×10−3 m and vp=3.0 m/s.
Cross-sectional area of the pipe, Ap=πr2p=6.4×10−5π m2.
⇒ Volumetric flow rate of water in the pipe is Apvp=3×6.4×10−5π m3/s =19.2×10−5π m3/s.
Cross-sectional area of each shower head opening, As=πr2s=10−6π m2.
⇒ Volumetric flow rate of water through each shower head opening is Asvs=10−6πvs m3/s.
There are 20 shower head openings.
⇒ Volumetric flow rate of water through the shower head is Asvs×20=2×10−5πvs m3/s.
Since water is not compressible, the volumetric flow rate through the shower head has to be the same as the volumetric flow rate through the pipe.
⇒2×10−5πvs=19.2×10−5π.
⇒vs=19.22=9.6 m/s.
⇒ The speed of water as it exits the shower head is 9.6 m/s.
the speed of water is 9.6 metre