Two circular water ( = 62.4 lbm/ft3) jets of 1 in. diameter (d = 1 in.) issue from this unusual nozzle. If the efflux speed is 80.2 ft/s, what force is required at the flange to hold the nozzle in place? The pressure in the 4 in. pipe (D = 4 in.) is 43 psig.

the force required are

Explanation:

Data given

V2 = V3 = 80.2 ft/s

P1 = 43 psig = 6192 Ib/ft^2

note:  P2 = P3 = Patm = 0 Ib/ ft^2

D2 = D3 = 1 inch

D1 = 4 inches

p = 62.4 Ibm/ft3 = 1.94 slug/ft3

calculate the cross sectional Areas of the water jets

A 1 =  = 0.0873 ft2

A2 = = 0.005454 ft2

calculate the mass flow rate

M2 = M3 = P *A2*V2 = 1.94 * 0.005454 * 80.2 = 0.8486 slug/s

calculate the mass balance

M1 = M2 + M2 = 0.8486 + 0.8486 = 1.6972 Ibm/ s

next calculate the velocity at each section

velocity at section 1  of pipe  

=  

V1 =   = 10.02 ft/s

calculate the linear momentum

F = m ( Vout - Vin )

along the x - direction

Fx + (P1 A1) - (P2 A2) + (P3 A3) sin 30 = (M2 V2) + M3 ( -V3 sin 30) - (M1 V1)

Fx + 540.56 = 68.06 -34.03 - 17

Fx = -523.5 Ibf

Along the y - direction

Fy + P3 A3 COS 30 = M3 ( -V3 COS 30 ) - 0

Fy + 0 = 0.8486 ( -80.2 cos 30 )

Fy = - 58.94 Ibf

answer: make you irritable

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