Answer:
a) 1.542 m/s
b) t = 38.91 mins
Explanation:
Given data:
Diameter of tank ( D ) = 3 m
height of tank ( initial ) ( h = z1 ) = 2m
Diameter of sharp edged orifice = 0.1 m
pipe length ( L ) = 100 m
coefficient of friction of pipe ( f ) = 0.015
∝1 = ∝2 = 1
A) calculate initial velocity from the tank
since the fluid is open to the atmosphere ; p1 = p2 = Patm ( atm pressure )
the initial velocity of the tank can be determined with the equation below
V = [tex]\sqrt{\frac{2gZ1}{1 + f \frac{L}{d} + K _{L} } }[/tex] ---- ( 1 )
where : Z1 = 2m , f = 0.015, L = 100m , Kl = 0.5m , d = 0.1 m , g = 9.81m/s^2
input given values into equation 1 above
V = 1.542 m/s
b) Determine time required to empty tank
Given velocity = [tex]\sqrt{\frac{2gZ1}{1 + f \frac{L}{d} + K _{L} } }[/tex] and Flow rate = [tex]\sqrt{\frac{2gZ1}{1 + f \frac{L}{d} + K _{L} } }* \frac{\pi }{4} *d^2[/tex]
differentiate the flow rate ( dt ) and then integrate the equation to get the required expression
t = [tex]\frac{2D^2}{d^2} \sqrt{\frac{(1+f\frac{L}{d}+K_{L})Z_{1} }{2g} }[/tex] -------- ( 2 )
where : : Z1 = 2m , f = 0.015, L = 100m , Kl = 0.5m , d = 0.1 m , g = 9.81m/s^2, D = 3m
input values into equation 2
t = 38.91 mins
