A shower head has 25 circular openings each with radius 1mm. The shower head is connected to a pipe with a radius of 0.8cm. If the speed of water in the pipe is 3.0 m/s, what is its speed as it exits the shower head opening?

Solution:

Let the radii of the shower head opening and the pipe be r_s and r_p respectively.

Let the speed of water in the pipe and as it exits the shower head be v_p and v_s respectively.

Given:

r_s=1 mm =10⁻³m,  

r_p=0.8 cm =8×10⁻³m and

v_p=3.0 m/s.

Cross-sectional area of the pipe, A_p=πr_p ²=6.4×10⁻⁵π m².

⇒ Volumetric flow rate of water in the pipe is A_pv_p=6.4×10⁻⁵π×3 m³/s =19.2×10⁻⁵π m³/s.

Cross-sectional area of each shower head opening, A_s=πr_s ²=10⁻⁶πm².

⇒ Volumetric flow rate of water through each shower head opening is A_sv_s=10⁻⁶πv_s m³/s.

There are 25 shower head openings.

⇒ Volumetric flow rate of water through the shower head is A_sv_s×25=2.5×10⁻⁵πv_s m³/s.

Since water is not compressible, the volumetric flow rate through the shower head must be the same as the volumetric flow rate through the pipe.

⇒2.5×10⁻⁵πv_s=19.2×10⁻⁵π.

⇒v_s=7.68 m/s.

⇒ The speed of water as it exits the shower head is 7.68 m/s.