GATE Past Year Questions: Viscous Flow | Fluid Mechanics for Mechanical Engineering PDF Download

From Blausius equation,


& we know x2 > x1
∴ t_w / x₁ = t_w / x₂

Couette flow is steady incompressible, laminar flow between one fixed plate and other moving with constant velocity.

 

Above equation shows that


% change in pressure drop

= 80.24% ≌ 80%

For fully developed in compressible viscous flow,

= 2/3 x 6 = 4m/s
 

Above equation shows that DP is independent of velocity.

Reynolds number, R_{e = 
Discharge, Q = AV = 
From above,
∴ Reynolds No.= 
Reynolds No. = 4Q/πd         ........(1)
From (1), Re ∝ 1/2
Reynolds number is large in the Narrow pipe.}

For original configuration
l₁ = l
d₁ = d
Q₁ = Q
For new configuration
l_{2 =} 21
d₂ = 2d
Q₂ = 2Q
Head loss due to friction,

where f = friction factor
l = Length of pipe
v = Average velocity
d = Diameter pipe

Discharge: 
or 

Applying Bernoulli’s equation at sections A and B, we get

or 

Now, the ratio


Substituting the Value of 

we get

Given: n = 7.4 × 10^-7 m²/s r = 0.88 × 1000 = 880
v = 0.5 m/s

du = V – O – V
dy = h

Also 
where F = mg
A = L<sup>2

By (1) (2), we get</sup>

Given, v = velocity of water = 10 × 10^–3 kg/sm
Now Re= 

                                     .....since = v₂ = 1
R_e = 1000 Now,
Darcy’s friction factor,

Head loss in needle = h_t = 

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


Since z₁ = z₂ and P₂ = 0


= 499.95 + 3199.7 = 3699.65
Now force required on plunger = P₁ × A<sub>1

= 0.3 N</sub>

ρ_water =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

Since z₁ = z₂ and P₂ = 0

Applying continuity equation at points (i) and (ii), we have
A₁v₁ = A₂v<sub>2

= 100 v</sub>1
<sub>= 100 × 0.01 = 1 m/s Now from equation (i),

= 499.95 N/m2 Force required on plunger = P</sub>1 _{× v}1

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


P₁ = P₂ = Patm V₁ = 0
V₂ = ?
Z₂ = H – h

u_max = 2 u_avg
u_max = 2u_avg
1.5 = 2 u_avg