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

The value is
`F_t =  76024 \  N`

Step-by-step 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 (d^2)/(4)`

`A = 3.142 (0.7^2)/(4)`

`A =0.385 \ m^2`

Generally the exit cross-sectional area is mathematically represented as

`A_e =  \pi (d_e^2)/(4)`

`A_e = 3.142 (0.42^2)/(4)`

`A_e =0.139\ m^2`

Generally from the continuity equation

`v_1 * A  =  v_2 * A_e`

=>
`v_2  =  (v_1 * A)/(A_e)`

=>
`v_2  =  (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]`