Answer to Question #140169 in Mechanics | Relativity for krithi

Question

A jet of water is discharging at a constant rate of 0.06 m 3 /s from the upper tank (Fig. 2). If the jet diameter at

section 1 is 10 cm, what forces will be measured by scales A and B? Assume the empty tank weighs 1335 N, the cross-sectional area of the tank is 0.37 m 2 , h= 30 cm, and H =2.74 m.(density of water at 60°F = 983.2 kg/m)

Answer 1

Q = 0.06 m³/s

d = 0.1 m

$A = \frac{π}{4}d^2$

$A = 7.85 \times 10^{-3} \;m^2$

$V = \frac{Q}{A}$

$V = \frac{0.06}{7.85 \times 10^{-3}} = 7.64 \;m/s\; (horizontal \;velocity)$

When water hits the lower tank, vertical velocity

$V_v = 0 + gt$

$V_v^2 = 2gβ$

β = H = 2.74 m

$V_v = \sqrt{2 \times 9.81 \times 2.74}$

$V_v = 7.33 \;m/s$

Force at point A = force due to weight of water (Fw) + Force due to impulse of vertical velocity + weight of empty tank

$F_w = 0.37 \times 0.3 \times 9.81 = 1.08 \;kN$

$F_{impulse} = ρAV_v^2$

$F_{impulse} = 0.9832 \times 7.85 \times 10^{-3} \times 7.33^2 = 0.414 \;kN$

Total force

$F_A = 1.08 + 0.414 + 1.335 = 2.829 \;kN$

Force at B = Force due to impulse of jet dur to horizontal component of velocity.

$F_B = ρAV_h^2$

$F_B = 0.9832 \times 7.85 \times 10^{-3} \times 7.64^2 = 0.45 \;kN$