Answer:
a
[tex]V = 5.30 *10^{-2} \ m^3[/tex]
b
[tex]v_1 = 0.3127 \ m/s[/tex]
Explanation:
From the question we are told that
The number of identical drippers is n = 60
The diameter of each hole in each dripper is [tex]d = 1.0 \ mm = 0.001 \ m[/tex]
The diameter of the main pipe is [tex]d_m = 2.5 \ cm = 0.025 \ m[/tex]
The speed at which the water is flowing is [tex]v = 3.00 \ cm/s = 0.03 \ m/s[/tex]
Generally the amount of water used in one hour = 3600 seconds is mathematically represented as
[tex]V = A * v * 3600[/tex]
Here A is the area of the main pipe with value
[tex]A = \pi * \frac{d^2}{4}[/tex]
=> [tex]A = 3.142 * \frac{0.025^2}{4}[/tex]
=> [tex]A = 0.0004909 \ m^2[/tex]
So
=> [tex]V = 0.0004909 * 0.03 * 3600[/tex]
=> [tex]V = 5.30 *10^{-2} \ m^3[/tex]
Generally the area of the drippers is mathematically represented as
[tex]A_1= n * \pi \frac{d^2}{4}[/tex]
=> [tex]A_1 = 60 * 3.142 * \frac{0.001 ^2}{4}[/tex]
=> [tex]A_1 = 4.713 *10^{-5} \ m^2[/tex]
Generally from continuity equation we have that
[tex]Av = A_1 v_1[/tex]
=> [tex]0.0004909 * 0.03 = 4.713 *10^{-5} * v_1[/tex]
=> [tex]v_1 = 0.3127 \ m/s[/tex]
