Fluid Mechanics and Hydraulic Machinery Miscellaneous

Fluid Mechanics and Hydraulic Machinery

1. The discharge velocity at the pipe exit in figure is

1. By applying Bernoulli’s equation at (1) and (2), we get

 P1 + V1² + Z1 = P2 + V2² + Z2 ρg 2g ρg 2g

P1 = P2 = Patm
V1 = 0
V2 = ?
Z2 = H – h
 ∴ H = V2² + H - h 2g

V2 = √2gh

Correct Option: B

By applying Bernoulli’s equation at (1) and (2), we get

 P1 + V1² + Z1 = P2 + V2² + Z2 ρg 2g ρg 2g

P1 = P2 = Patm
V1 = 0
V2 = ?
Z2 = H – h
 ∴ H = V2² + H - h 2g

V2 = √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/m3. The plunger is pushed in at 10 mm/s and the water comes out as a jet.

1. Assuming ideal flow, the force F in newtons required on the plunger to push out the water is

1. ρwater =1000 kg/m3
Velocity at points 1 = velocity of plunger = 10 mm/ s = 0.014 m/s

Applying Bernoulli’s equation at points 1 and 2, we have

 P1 + v1² + z1 = P2 + v2² + z2 ρg 2g ρg 2g

Since z1 = z2 and P2 = 0
 P1 = v2² - v1² ρq 2g 2g

 P1 = ρ (v2² - v1²) ...(i) 2

Applying continuity equation at points (i) and (ii), we have
A1 v1 = A2 v2
 ⇒ v2 = A1 v1 A2

 v2 = [ (π / 4) × (0.01)2 ] v1 [ (π / 4) × (0.001)2 ]

= 100 v1 = 100 × 0.01 = 1 m/s
Now from equation (i),
 P1 = 1000 [ (1)² - (0.01)² ] = 499.95 N / m² 2

Force required on plunger = P1 × v1
 = 499.95 × 11 × (0.01)2 = 0.04 N 4

Correct Option: B

ρwater =1000 kg/m3
Velocity at points 1 = velocity of plunger = 10 mm/ s = 0.014 m/s

Applying Bernoulli’s equation at points 1 and 2, we have

 P1 + v1² + z1 = P2 + v2² + z2 ρg 2g ρg 2g

Since z1 = z2 and P2 = 0
 P1 = v2² - v1² ρq 2g 2g

 P1 = ρ (v2² - v1²) ...(i) 2

Applying continuity equation at points (i) and (ii), we have
A1 v1 = A2 v2
 ⇒ v2 = A1 v1 A2

 v2 = [ (π / 4) × (0.01)2 ] v1 [ (π / 4) × (0.001)2 ]

= 100 v1 = 100 × 0.01 = 1 m/s
Now from equation (i),
 P1 = 1000 [ (1)² - (0.01)² ] = 499.95 N / m² 2

Force required on plunger = P1 × v1
 = 499.95 × 11 × (0.01)2 = 0.04 N 4

1. 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

1. Given, v = velocity of water = 10 × 10–3 kg/sm

 Now , Re = ρv2d v

... since = v2 = 1
Re = 1000
Now, Darcy’s friction factor,
 4f = 64 = 64 = 0.064 Re 1000

 Head loss in needle = ht = flv2 2gD

 = 0.064 × 0.1 × (1)2 = 0.3265 2 × 9.8 × 0.001

Applying Bernoulli’s equation at points 1 and 2, we have
 P1 + v1² + z1 = P2 + v2² + z2 + h1 ρg 2g ρg 2g

Since z1 = z2 and P2 = 0
 ∴ P1 = v2² - v1² + hl ρg 2g

 P1 = ρ (v2² - v1²) + ρghl 2

 P1 = 1000 [ (1)² - (0.01)²] + 1000 × 9.8 × 0.3265 2

= 499.95 + 3199.7 = 3699.65
Now force required on plunger = P1 × A1
 = 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 = ρv2d v

... since = v2 = 1
Re = 1000
Now, Darcy’s friction factor,
 4f = 64 = 64 = 0.064 Re 1000

 Head loss in needle = ht = flv2 2gD

 = 0.064 × 0.1 × (1)2 = 0.3265 2 × 9.8 × 0.001

Applying Bernoulli’s equation at points 1 and 2, we have
 P1 + v1² + z1 = P2 + v2² + z2 + h1 ρg 2g ρg 2g

Since z1 = z2 and P2 = 0
 ∴ P1 = v2² - v1² + hl ρg 2g

 P1 = ρ (v2² - v1²) + ρghl 2

 P1 = 1000 [ (1)² - (0.01)²] + 1000 × 9.8 × 0.3265 2

= 499.95 + 3199.7 = 3699.65
Now force required on plunger = P1 × A1
 = 3699.65 × π × (0.01)2 = 0.3 N 4

1. 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

1.  τ = μ du dy

du = V – O – V
dy = h
 τ = u V ...(1) h

 Also , τ = F A

where F = mg , A = L2
 ∴ τ = mg ......(2) L2

By (1) (2), we get
 V = mgh uL2

Correct Option: C

 τ = μ du dy

du = V – O – V
dy = h
 τ = u V ...(1) h

 Also , τ = F A

where F = mg , A = L2
 ∴ τ = mg ......(2) L2

By (1) (2), we get
 V = mgh uL2

1. 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/m3 and acceleration due to gravity is 9.8 ms–2, Which of the following is TRUE

1. At section S1
P1 = 50 kPa
z1 = 10 m
D = 200 mm,
l = 1000 kg/m3
V = 2 m/s
g = 9.8 m/s2
At section S2
P2 = 20 kPa
z2 = 12 m
Applying Bernoulli’s equation at two sections, we have

 P1 + z1 = P2 + z2 + h2 ρg ρg

where h2 = Loss of head from s1 to s2
 ∴ 50 × 103 + 10 = 20 × 103 + 12 + h2 103 × 9.81 103 × 9.81

⇒ h2 = 1.06

Correct Option: C

At section S1
P1 = 50 kPa
z1 = 10 m
D = 200 mm,
l = 1000 kg/m3
V = 2 m/s
g = 9.8 m/s2
At section S2
P2 = 20 kPa
z2 = 12 m
Applying Bernoulli’s equation at two sections, we have

 P1 + z1 = P2 + z2 + h2 ρg ρg

where h2 = Loss of head from s1 to s2
 ∴ 50 × 103 + 10 = 20 × 103 + 12 + h2 103 × 9.81 103 × 9.81

⇒ h2 = 1.06