P3.43 Water at 20 C flows through a 5-cm-diameter pipe that has a 180 vertical bend, as in Fig. P3.43. The total length of pipe between flanges 1 and 2 is 75 cm. When the weight flow rate is 230 N/s, p1

F_x = 750.7 N

Explanation:

Given:

- Length of the pipe between flanges L = 75 cm

- Weight Flow rate is flow (W) = 230 N/c

- P_1 = 165 KPa

- P_2 = 134 KPa

- P_atm = 101 KPa

- Diameter of pipe D = 0.05 m

Find:

The total force that the flanges must withstand F_x.

Solution:

- Use equation of conservation of momentum.

(P_1 - P_a) * A + (P_2 - P_a) * A - F_x = flow (m) * (V_2 - V_1)

- From conservation of mass:

A*V_1 = A*V_2

V_1 = V_2 (but opposite in directions)

- Hence,

(P_1 - P_a) * A + (P_2 - P_a) * A - F_x = - 2*flow (m) * V_1

flow (m) = flow (W) / g

p*A*V_1 = flow (W) / g

V_1 = flow (W) / g*p*A

Hence,

(P_1 - P_a) * A + (P_2 - P_a) * A - F_x = - 2*flow (W) ^2 / g^2*p*A

Hence, compute:

64*10^3 * pi*0.05^2 / 4 + 33*10^3 * pi*0.05^2 / 4 - F_x = - 2 * (230/9.81) ^2 / 997*pi*0.05^2 / 4

125.6 + 64.7625 - F_x = - 560.33

F_x = 750.7 N