Respuesta :
Answer:
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
