Respuesta :
Answer:
The answer to the question is;
The size of the commercial steel pipe that is required is 8.5 in diameter pipe
Based on standard pipe sizes the next size is 10 in. schedule 140 will be required.
Explanation:
To solve the question, we use the steady flow energy equation as follows
[tex]\frac{P_A}{\rho g} +\frac{v^2_A}{2g} +z_A=\frac{P_B}{\rho g} +\frac{v^2_B}{2g} +z_B + 0.5*\frac{v_1^{2} }{2g}+K*\frac{v_1^{2} }{2g}[/tex]
Where
[tex]P_A[/tex] = Pressure at the opened end of tank A
[tex]P_B[/tex] = Pressure at the opened end of tank B
[tex]z_A[/tex] = Elevation of tank A
[tex]z_B[/tex] = Elevation of tank B
[tex]v_A[/tex] = Velocity of flow in tank A
[tex]v_B[/tex] = Velocity of flow in tank B
v₁ = Velocity of flow in the pipe
K = Sum of losses coefficients for inlet and fittings = 0.8 for the rounded inlet and outlet and 6 for the opened globe valve = 6.8
[tex]0.5*\frac{v^2_1}{2g}[/tex] = Head losses
[tex]P_A = P_B[/tex] = Atmospheric pressure, and with large reservoirs [tex]v_A[/tex] and [tex]v_B[/tex] will be negligible then
[tex]z_A - z_B = (6.8 +0.5)*\frac{v_1^{2} }{2g} = (7.3)*\frac{v_1^{2} }{2g}[/tex]
4.2 ft = 1.28 m
g = 9.81 m/s²
v₁² =3.44 m²/s²
v₁ = 1.855 m/s
This gives the size of the pipe as
since Q = v·A
Where
A = area and
Q = Volume flow rate = 2.4 cfs = 6.796×10⁻² m³/s
v = Velocity = 1.855 m/s
A = Q/A = (6.796×10⁻² m³/s)/(1.855 m/s) = 3.66×10⁻² m² = 0.394 ft²
The size of the commercial steel pipe that is required is 0.394 ft² = 56.79 in² = 8.5 in diameter pipe
