Fluid Mechanics and Hydraulic Machinery Miscellaneous
 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
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
 Neglect losses in the cylinder and assume fully developed laminar viscous flow throughout the needle: the Darcy friction factor is 64/Re. where Re is the Reynolds number. Given that the viscosity of water is 1.0 × 10^{3} kg/ms, the force F in newtons required on the plunger is

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Given, v = velocity of water = 10 × 10^{–3} kg/sm
Now , Re = ρv_{2}d v
... since = v_{2} = 1
Re = 1000
Now, Darcy’s friction factor,4f = 64 = 64 = 0.064 Re 1000 Head loss in needle = h_{t} = flv^{2} 2gD = 0.064 × 0.1 × (1)^{2} = 0.3265 2 × 9.8 × 0.001
Applying Bernoulli’s equation at points 1 and 2, we haveP_{1} + v_{1}² + z_{1} = P_{2} + v_{2}² + z_{2} + h_{1} ρg 2g ρg 2g
Since z_{1} = z_{2} and P_{2} = 0∴ P_{1} = v_{2}²  v_{1}² + h_{l} ρg 2g P_{1} = ρ (v_{2}²  v_{1}²) + ρgh_{l} 2 P_{1} = 1000 [ (1)²  (0.01)²] + 1000 × 9.8 × 0.3265 2
= 499.95 + 3199.7 = 3699.65
Now force required on plunger = P_{1} × A_{1}= 3699.65 × π × (0.01)^{2} = 0.3 N 4
Correct Option: C
Given, v = velocity of water = 10 × 10^{–3} kg/sm
Now , Re = ρv_{2}d v
... since = v_{2} = 1
Re = 1000
Now, Darcy’s friction factor,4f = 64 = 64 = 0.064 Re 1000 Head loss in needle = h_{t} = flv^{2} 2gD = 0.064 × 0.1 × (1)^{2} = 0.3265 2 × 9.8 × 0.001
Applying Bernoulli’s equation at points 1 and 2, we haveP_{1} + v_{1}² + z_{1} = P_{2} + v_{2}² + z_{2} + h_{1} ρg 2g ρg 2g
Since z_{1} = z_{2} and P_{2} = 0∴ P_{1} = v_{2}²  v_{1}² + h_{l} ρg 2g P_{1} = ρ (v_{2}²  v_{1}²) + ρgh_{l} 2 P_{1} = 1000 [ (1)²  (0.01)²] + 1000 × 9.8 × 0.3265 2
= 499.95 + 3199.7 = 3699.65
Now force required on plunger = P_{1} × A_{1}= 3699.65 × π × (0.01)^{2} = 0.3 N 4
 A cubic block of side L and mass M is dragged over an oil film across table by a string connects to a hanging block of mass m as shown in figure. The Newtonian oil film of thickness h has dynamic viscosity μ and the flow condition is laminar. The acceleration due to gravity is g. The steady state velocity V of block is

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τ = μ du dy
du = V – O – V
dy = hτ = u V ...(1) h Also , τ = F A
where F = mg , A = L^{2}∴ τ = mg ......(2) L^{2}
By (1) (2), we getV = mgh uL^{2}
Correct Option: C
τ = μ du dy
du = V – O – V
dy = hτ = u V ...(1) h Also , τ = F A
where F = mg , A = L^{2}∴ τ = mg ......(2) L^{2}
By (1) (2), we getV = mgh uL^{2}
 A smooth pipe of diameter 200 mm carries water. The pressure in the pipe at section S1 (elevation: 10 m) ie 50 kPa, At section S2 (elevation: 12 m) the pressure is 20 kPa and velocity is 2 m/s. Density of water is 1000 kg/m^{3} and acceleration due to gravity is 9.8 ms^{–2}, Which of the following is TRUE

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At section S1
P_{1} = 50 kPa
z_{1} = 10 m
D = 200 mm,
l = 1000 kg/m^{3}
V = 2 m/s
g = 9.8 m/s^{2}
At section S2
P_{2} = 20 kPa
z_{2} = 12 m
Applying Bernoulli’s equation at two sections, we haveP_{1} + z_{1} = P_{2} + z_{2} + h_{2} ρg ρg
where h_{2} = Loss of head from s_{1} to s_{2}∴ 50 × 10^{3} + 10 = 20 × 10^{3} + 12 + h_{2} 10^{3} × 9.81 10^{3} × 9.81
⇒ h_{2} = 1.06
Correct Option: C
At section S1
P_{1} = 50 kPa
z_{1} = 10 m
D = 200 mm,
l = 1000 kg/m^{3}
V = 2 m/s
g = 9.8 m/s^{2}
At section S2
P_{2} = 20 kPa
z_{2} = 12 m
Applying Bernoulli’s equation at two sections, we haveP_{1} + z_{1} = P_{2} + z_{2} + h_{2} ρg ρg
where h_{2} = Loss of head from s_{1} to s_{2}∴ 50 × 10^{3} + 10 = 20 × 10^{3} + 12 + h_{2} 10^{3} × 9.81 10^{3} × 9.81
⇒ h_{2} = 1.06