Fluid Mechanics and Hydraulic Machinery Miscellaneous
Direction: A syringe with a frictionless plunger contains water and has at its end a 100 mm long needle of 1 mm diameter. The internal diameter of the syringe is 10 mm. Water density is 1000 kg/m^{3}. The plunger is pushed in at 10 mm/s and the water comes out as a jet.
 Assuming ideal flow, the force F in newtons required on the plunger to push out the water is

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ρ_{water} =1000 kg/m^{3}
Velocity at points 1 = velocity of plunger = 10 mm/ s = 0.014 m/s
Applying Bernoulli’s equation at points 1 and 2, we haveP_{1} + v_{1}² + z_{1} = P_{2} + v_{2}² + z_{2} ρg 2g ρg 2g
Since z_{1} = z_{2} and P_{2} = 0P_{1} = v_{2}²  v_{1}² ρq 2g 2g P_{1} = ρ (v_{2}²  v_{1}²) ...(i) 2
Applying continuity equation at points (i) and (ii), we have
A_{1} v_{1} = A_{2} v_{2}⇒ v_{2} = A_{1} v_{1} A_{2} v_{2} = [ (π / 4) × (0.01)^{2} ] v_{1} [ (π / 4) × (0.001)^{2} ]
= 100 v_{1} = 100 × 0.01 = 1 m/s
Now from equation (i),P_{1} = 1000 [ (1)²  (0.01)² ] = 499.95 N / m² 2
Force required on plunger = P_{1} × v_{1}= 499.95 × 11 × (0.01)^{2} = 0.04 N 4
Correct Option: B
ρ_{water} =1000 kg/m^{3}
Velocity at points 1 = velocity of plunger = 10 mm/ s = 0.014 m/s
Applying Bernoulli’s equation at points 1 and 2, we haveP_{1} + v_{1}² + z_{1} = P_{2} + v_{2}² + z_{2} ρg 2g ρg 2g
Since z_{1} = z_{2} and P_{2} = 0P_{1} = v_{2}²  v_{1}² ρq 2g 2g P_{1} = ρ (v_{2}²  v_{1}²) ...(i) 2
Applying continuity equation at points (i) and (ii), we have
A_{1} v_{1} = A_{2} v_{2}⇒ v_{2} = A_{1} v_{1} A_{2} v_{2} = [ (π / 4) × (0.01)^{2} ] v_{1} [ (π / 4) × (0.001)^{2} ]
= 100 v_{1} = 100 × 0.01 = 1 m/s
Now from equation (i),P_{1} = 1000 [ (1)²  (0.01)² ] = 499.95 N / m² 2
Force required on plunger = P_{1} × v_{1}= 499.95 × 11 × (0.01)^{2} = 0.04 N 4
 The discharge velocity at the pipe exit in figure is

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By applying Bernoulli’s equation at (1) and (2), we getP_{1} + V_{1}² + Z_{1} = P_{2} + V_{2}² + Z_{2} ρg 2g ρg 2g
P_{1} = P_{2} = P_{atm}
V_{1} = 0
V_{2} = ?
Z_{2} = H – h∴ H = V_{2}² + H  h 2g
V_{2} = √2gh
Correct Option: B
By applying Bernoulli’s equation at (1) and (2), we getP_{1} + V_{1}² + Z_{1} = P_{2} + V_{2}² + Z_{2} ρg 2g ρg 2g
P_{1} = P_{2} = P_{atm}
V_{1} = 0
V_{2} = ?
Z_{2} = H – h∴ H = V_{2}² + H  h 2g
V_{2} = √2gh
 For laminar flow through along pipe, the pressure drop per unit length increases

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P_{1}  P_{2} = 32μUL D² Correct Option: C
P_{1}  P_{2} = 32μUL D²
 In fully developed laminar flow in the circular pipe, the head loss due to friction is directly proportional to _____ (mean velocity/square of the mean velocity)

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f ∝ Mean velocity
Correct Option: A
f ∝ Mean velocity
 Fluid is flowing with an average velocity of V through a pipe of diameter d. Over a length of L,
the head loss is given by fLV² .The friction factor, f for laminar flow in terms of Reynolds 2gD
number (Re) is _______ (fill in the blanks)

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For Laminar flow,
f = 64 Re
Correct Option: B
For Laminar flow,
f = 64 Re