Question

Sixty identical drippers, each with a hole of diameter 1.00 mm, are used to water a yard. If the water in the main pipe of diameter 2.54 cm is flowing at speed 3.00 cm/s, (a) how much water is used in one hour and (b) how fast the water is coming out the drippers

Answer 1

Answer:

a

   $V = 5.30 *10^{-2} \ m^3$

b

   $v_1 = 0.3127 \ m/s$

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  $d = 1.0 \ mm = 0.001 \ m$  

   The diameter of the main pipe is  $d_m = 2.5 \ cm = 0.025 \ m$

    The speed at which the water is flowing is  $v = 3.00 \ cm/s = 0.03 \ m/s$

Generally the amount of water used in  one hour = 3600 seconds is mathematically represented as

          $V = A * v * 3600$

Here A is the area of the main pipe with value

         $A = \pi * \frac{d^2}{4}$

=>       $A = 3.142 * \frac{0.025^2}{4}$

=>        $A = 0.0004909 \ m^2$

So  

=>   $V = 0.0004909 * 0.03 * 3600$

=>  $V = 5.30 *10^{-2} \ m^3$

Generally the area of the drippers is mathematically represented as

       $A_1= n * \pi \frac{d^2}{4}$

=>    $A_1 = 60 * 3.142 * \frac{0.001 ^2}{4}$

=>    $A_1 = 4.713 *10^{-5} \ m^2$

Generally from continuity equation we have that  

         $Av = A_1 v_1$

=>      $0.0004909 * 0.03 = 4.713 *10^{-5} * v_1$

=>   $v_1 = 0.3127 \ m/s$