Fluid Mechanics Test 4

1. Pipe flow:

Re < 2000 → Laminar

Re > 4000 → turbulent

2000 < Re > 4000 → transition

2. Parallel plates:

Re < 1000 → Laminar

Re > 2000 → turbulent

1000 < Re < 2000 → transition

3. Open channel Flow: 

Re < 500 → Laminar

Re > 2000 → turbulent

500 < Re < 2000 → transition

4. For flow through soil:

Re < 1 → Laminar

1< Re < 2 → transition

Average velocity for laminar flow through circular pipe,

\(\bar U = \frac{1}{{8\mu }}\left( { - \frac{{\partial P}}{{\partial x}}} \right).{R^2}\)

Maximum velocity for laminar flow through circular pipe,

\({U_{max}} = \frac{1}{{4\mu }}\left( { - \frac{{\partial P}}{{\partial x}}} \right).{R^2}\)

∴ \(\frac{{{U_{max}}}}{{\bar U}} = 2\)

Concept:

For turbulent flow,

\(\frac{{{v_{max}} - v}}{{{v_*}}} = 5.75\;{\log _{10}}\left( {\frac{R}{y}} \right)\)        ---(1)

\({{\rm{V}}_{\rm{*}}} = {\rm{\;shear\;velocity\;}} = \sqrt {\frac{{{\tau _w}}}{\rho }} \)

Calculation:

Given:

D = 130 mm = 0.13 m, v_max = 2.5 m/s, v = 1.5 m/s, r = 35 mm = 0.035 m

Now,

\(y = R - r = \left( {\frac{{0.13}}{2}} \right) - \left( {0.035} \right)\)

y = 0.03

Using the relation from equation (1)

\(\frac{{2.5 - 1.5}}{{{v_*}}} = \left( {5.75} \right){\log _{10}}\left( {\frac{{0.065}}{{0.03}}} \right)\)

∴ V_* = 0.5179 m/s

Now,

\({v_*} = \sqrt {\frac{{{\tau _w}}}{\rho }}\)

\({v_*} = 0.5179 = \sqrt {\frac{{{\tau _w}}}{{{{10}^3}}}} \)

Concept: Using Hagen-Poiseuille equation

\(\frac{{{{\rm{P}}_1} - {{\rm{P}}_2}}}{{{\rm{\rho g}}}} = {\rm{\;}}\frac{{32{\rm{\mu \bar uL}}}}{{{\rm{\rho g}}{{\rm{D}}^2}}}\)

where, \(\frac{{{{\rm{P}}_1} - {{\rm{P}}_2}}}{{{\rm{\rho g}}}}\) = difference in pressure head

μ → dynamic viscosity of fluid flowing

L → Length of pipe

ρ → density of fluid

D → diameter of pipe

ū → Mean velocity of flow

Calculation:

μ = 0.2 \(\frac{{{\rm{NS}}}}{{{{\rm{m}}^2}}}\)

ρ = 0.85 × 1000 = 850 kg/m³

L = 500 m

\(\frac{{{{\rm{P}}_1} - {{\rm{P}}_2}}}{{{\rm{\rho g}}}} = 70{\rm{m}}\)

\(70 = \frac{{32{\rm{\mu \bar uL}}}}{{{\rm{\rho g}}{{\rm{D}}^2}}}\)

\(70 = \frac{{32 \times 0.2 \times {\rm{\bar u\;}} \times 500}}{{850 \times 9.81 \times {{\rm{D}}^2}}}\)

\({\rm{\bar u}} = \frac{{\rm{Q}}}{{\rm{A}}}\)

\({\rm{\bar u}} = \frac{{4 \times 4}}{{1000 \times {\rm{\pi }} \times {{\rm{D}}^2}}}\)

\({\rm{\bar u}} = \frac{{16}}{{1000 \times {\rm{\pi }} \times {{\rm{D}}^2}}} = \frac{{5.09 \times {{10}^{ - 3}}}}{{{{\rm{D}}^2}}}\)

\(70 = \frac{{32 \times 0.2 \times 5.09 \times {{10}^{ - 3}} \times 500}}{{{{\rm{D}}^4} \times 9.81 \times 850}}\)

\(70 = \frac{{1.953 \times {{10}^{ - 3}}}}{{{D^4}}}\)

D = 72.68 mm ≅ 73 mm

Concept:

When the water flowing in a pipe is suddenly brought to rest by closing the valve or by any similar case, there will be a sudden rise in pressure due to the momentum of water being destroyed. This rise in pressure called water hammer pressure.

This hammer pressure depends on whether the valve is closed rapidly or gradually and it also depends on elastic characteristics of pipe material and fluid flow.

Let T_c is the time required by pressure wave for round trip i.e. forward and back again and it is given as

\({T_c} = \frac{{2L}}{C}\)

C is the velocity of the pressure wave and L is the length of pipe

Let T_a is the time required for closure of the valve.

Based on the relation between T_a and T_c, following cases arises:

Case 1:  Rapid Closure of Valve and pipe is rigid:

If T_a < T_c, then closure is called Rapid closure.

Water hammer pressure, P_h = ρCV

Where,

\(C = \;\sqrt {\frac{K}{\rho }} \)

K is the bulk modulus of elasticity of the fluid flowing.

ρ is the density of the fluid

V is the velocity of fluid flowing

Case 2:  Rapid Closure of Valve and pipe is flexible:

If Ta < Tc, then closure is called Rapid.

Water hammer pressure, P_h = ρCV

Where

\(C = \;\sqrt {\frac{{{K_c}}}{\rho }} \)

And K_c is given as:

\(\frac{1}{{{K_c}}} = \frac{1}{K} + \frac{D}{{tE}}\;\)

K is the bulk modulus of elasticity of fluid flowing

ρ is the density of the fluid

V is the velocity of fluid flowing

E is the modulus of elasticity

t is the thickness of the pipe

D is the diameter of the pipe

Case 3:  Gradual Closure of Valve

If T_a > T_c, then closure is called gradual.

Water hammer pressure, P_h = ρVL/T_a

L is the length of pipe

ρ is the density of fluid

V is the velocity of fluid flowing

Calculation:

Given:

L = 3km, V = 1.2 m/s, T_a = 4 sec, K = 2 × 10⁹ N/m², ρ = 1000 kg/m³ , P_S = 100 KPa

\(\therefore C = \sqrt {\frac{K}{\rho }} = \sqrt {\frac{{2 \times {{10}^9}}}{{1000}}} = 1414.2\;m/s\) 

 \({T_C} = \frac{{2L}}{C} = \frac{{2 \times 3 \times {{10}^3}}}{{1414.2}} = 4.24\;sec\)

∵ T_a < T_c ⇒ Closure is rapid.

It is also given that the pipe is rigid

∴ “Case I” will be used.

The rise in pressure due to water hammer, P­_H is:

P_H = ρCV

P_H = 10³ × 1414.2 × 1.2 = 1697 kPa