Question

In the figure below, this “double” nozzle discharges water (at 10°C, density= 1000 kg/m3) into the atmosphere at a rate of 0.50 m3/s. The pressure at the inlet is to be 315612 Pa. If the nozzle is lying in a horizontal plane. Jet A is 10 cm in diameter, jet B is 12 cm in diameter, and the pipe (1) is 30 cm in diameter. The x-component of force (Rx) acting through the flange bolts is required to hold the nozzle in place is:

Answer 1

Given data :

p = 315612 Pa

`$V_1=7.07 \ m/sec$`

At exit of B,

p =
`$P_(atm)$`

`$V_B = 26.1 \ m/sec$`

At exit of A,

`p=P_(atm)`

`$V_(A) = 26.1 \ m/s$`

We need to determine X component of force (
`$R_x$`) to hold in its place.

From figure,

`$\sum F_x = m_0'V_(0x) - m_iV_(ix) $`

`$=F_x+P_1A_1\sin 30=-mVA-mV_1 \sin 30$`

`$=F_x=-pA_1\sin 30-m_AV_AA-m_B \sin30$`

Substitute all the values,

`$=F_x=[-315612 * (\pi)/(4)(0.3)^2 \sin 30]-[26.1 * 1000 * 26.1 (\pi)/(4)(0.1)^2]-[7.07 * 1000* 0.5 \sin 30]$`
`$=F_x = -11154.64-5350.21-1767.28$`

`$F_x = -18.2733 \ kN$`

Therefore, the force required to hold the nozzle in its place along horizontal direction.

`$F_x = -18.2733 \ kN$`