Question

The flow rate of water through a tapered straight horizontal pipe fitting is 5 m/s. The diameter at the entrance is 0.7 m and is reduced by a factor of 0.6 at the exit. If the gauge pressure at the entrance is 350 kPa and drops by 50 Pa, the horizontal thrust on the fitting is,

Answer 1

Answer:

The  value is   $F_t = 76024 \ N$  

Explanation:

From the question we are told that

  The  flow rate is  $v_1 = 5 \ m/s$

   The  entrance diameter is  $d = 0.7 \ m$

  The  exit diameter is evaluated as $d_e = 0.6 * 0.7 = 0.42 \ m$

   The entrance gauge pressure is  $P_g = 350 \ kPa = 350*10^{3} \ Pa$

    The  droped gauge  pressure is  $P_d = 50 \ kPa = 50*10^{3} \ Pa$

 The presure at the exist is evaluated as  $P_e =(350 - 50 ) \ kPa = 300*10^{3} \ Pa$

Generally the entrance cross-sectional area is mathematically represented as

     $A = \pi \frac{d^2}{4}$

     $A = 3.142 \frac{0.7^2}{4}$

      $A =0.385 \ m^2$

Generally the exit  cross-sectional area is mathematically represented as

        $A_e = \pi \frac{d_e^2}{4}$

     $A_e = 3.142 \frac{0.42^2}{4}$

      $A_e =0.139\ m^2$

Generally from the continuity equation

    $v_1 * A = v_2 * A_e$

=>  $v_2 = \frac{v_1 * A}{A_e}$

=>  $v_2 = \frac{5 * 0.385}{0.139}$

=>  $v_2 = 13.8 m/s$

Generally the net force in horizontal axis is equivalent to the net momentum change in the horizontal direction

So

     $P_g * A + F_t - P_e * A_e = P_d [ A_e * v_2^2 - A* v_1^2]$

Here  $F_t$ is the horizontal thrust on the fitting

So  

        $350*10^{3} * 0.385 + F_t -0.385 * 0.139 = 50*10^{3} [ 0.385* 13.8^2 - 0.385* 5^2]$