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IrinaK [193]
3 years ago
5

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,
Physics
1 answer:
iragen [17]3 years ago
6 0

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]

         

     

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