Question

Water is discharged through an elbow nozzle as shown below. PB - Patm ds The exit velocity VB = 30 ft/s, the inlet diameter da = 0.5 ft, the exit diameter dB = 0.25 ft. For water density, use p = 32.2 lb/ft = 1.94 lb/ft. Assume steady flow. Neglect the weight of the nozzle and the water in the nozzle.

Answer 1

(1)

Mass flow rate is given as;

$\dot{m} = \rho A_BV_B = (1.94)\left [ \frac{\pi}{4}(0.25)^2 \right ](30)$

$\Rightarrow \dot{m} = 2.86\;\;slug/s$

...(Answer)

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(2)

Inlet velocity is given as (by continuity equation);

$V_A = V_B \left [ \frac{d_B}{d_A} \right ]^2=(30) \left [ \frac{0.25}{0.5} \right ]^2$

→ VA = 7.5 ft/s

...(Answer)

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(3)

Internal gage pressure will be (by Bernoull's equation);

$P_A = \rho \left [\frac{V_B^2-V_A^2}{2} \right ]$

$\Rightarrow P_A = (1.94) \left [\frac{30^2-7.5^2}{2} \right ]$

$\Rightarrow P_A = 818\;\;lbf/ft^2$

...(Answer)

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(4)

Force reaction exerted by the flange will be;

$\sum F_x = \dot{m}[V_Bcos60^o-V_A]$

$\Rightarrow P_AA_A+R_x= \dot{m}[V_Bcos60^o-V_A]$

$\Rightarrow (818)\left [ \frac{\pi}{4}(0.5)^2 \right ]+R_x= (2.86)[30cos60^o-7.5]$

$\Rightarrow R_x = -139\;\;lbf$

Force acting on the flange will be equal and opposite to this reaction force. Hence;

$\Rightarrow F_x = 139\;\;lbf\;\;(\rightarrow)$

...(Answer)

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(5)

Force reaction exerted by the flange will be;

$\sum F_y = \dot{m}[V_Bsin60^o-0]$

$\Rightarrow 0+R_y= \dot{m}[V_Bsin60^o-0]$

$\Rightarrow R_y= (2.86)[30sin60^o]$

$\Rightarrow R_y =74.4\;\;lbf$

Force acting on the flange will be equal and opposite to this reaction force. Hence;

$\Rightarrow F_y = -74.4\;\;lbf\;\;(\downarrow)$

...(Answer)