Answer:
[tex]K_v=12.34[/tex]
Explanation:
Given;
For orifice, loss coefficient, K₀ = 10
Diameter, D₀ = 45 mm = 0.045 m
loss coefficient of the orifice, Ko = 10
Diameter of the gate valve, Dy = 1.5D₀ = 1.5 × 0.045 m = 0.0675 m
Total head drop, Δhtotal=25 m
Discharge, Q = 10 l/s = 0.01 m³/s
Now,
the velocity of flow through orifice, Vo = Discharge / area of the orifice
or
Vo = [tex]\frac{0.01}{\frac{\pi}{4}0.045^2}[/tex]
or
Vo = 6.28 m/s
also,
the velocity of flow through gate valve, [tex]V_v[/tex] = Discharge / area of the orifice
or
[tex]V_v[/tex] = [tex]\frac{0.01}{\frac{\pi}{4}0.0675^2}[/tex]
or
[tex]V_v[/tex] = 2.79 m/s
Now,
the total head drop = head drop at orifice + head drop at gate valve
or
25 m = [tex]K_o\frac{V_o^2}{2g}+K_v\frac{V_v^2}{2g}[/tex]
where,
[tex]K_v[/tex] is the loss coefficient for the gate valve
on substituting the values, we get
25 m = [tex]10\frac{6.28^2}{2\times 9.81}+K_v\frac{2.79^2}{2\times9.81}[/tex]
or
[tex]K_v\frac{2.79^2}{2\times9.81}[/tex] = 4.898
or
[tex]K_v=12.34[/tex]
