Fluid Mechanics Test 1

Bulk modulus is defined as the ratio of compressive stress to Volumetric strain. It is a measure of the compressibility of a liquid.

\(k = - \frac{{dP}}{{\left( {\frac{{dV}}{V}} \right)}}\)

\({a_x} = u\frac{{du}}{{dx}} + v\frac{{du}}{{dy}} + w\frac{{du}}{{dz}} + \frac{{du}}{{dt}}\)

= (-2x) × (2) = 4x

\({a_y} = u\frac{{dv}}{{dx}} + v\frac{{dv}}{{dy}} + w\frac{{dv}}{{dz}} + \frac{{dv}}{{dt}}\)

= 0 + 2y × 2 + 0 + 0 = 4y

Bulk modulus of elasticity

k = Δp/(Δv/v) = 18/1/100

= 1800 MN/m²

Height of capillary rise is given by,

Concept:

Pascal Law states that at any point in a fluid the intensity of pressure is equal in all the directions if and only if shear stresses acting on the  fluid is zero.

Newton’s law of viscosity:

\(\tau = \frac{{\mu du}}{{dy}}\)   

Where

τ is the shear stress on plate

du/dy is velocity gradient

μ is dynamic viscosity of fluid.

Explanation:

Statement- I -­ It is valid for ideal fluid in motion.

 Since fluid is in motion, u ≠ 0 but

 For ideal fluid, μ = 0 ⇒ τ = 0

Therefore, Pascal’s law is valid for ideal fluids even in state of motion.

Statement II- It is valid when real fluids with rigid body motion.

Rigid body motion means every layer of fluid has uniform velocity.

For real fluid in motion mean that μ ≠ 0 and u ≠ 0

But, du = 0 ⇒ τ = 0

∴ For Real fluids Pascal’s law is also valid with rigid body motion.

The soap bubble has two interfaces
∴ Work done
= 0.04 x 4π x (6 x 10^-2)² x 2
= 36.2 x 10^-4 Nm

Pressure inside the liquid droplet is given by
 Δp = 4σ/d  

σ is the surface tension = 0.075 N/m = (0.075/100) N/cm
d is the diameter = 0.075 mm= (0.075/10) cm

Therefore, pressure = [4 x (0.075/100)] / [0.075/10]
= 0.4 N/cm²

Specific volume(v) is defined as the volume(V ) per unit mass(m).

v = v⁄m = 1 / m⁄v = 1⁄p

where p is the mass density. Thus, if p1 > p2, the relation between the specific volumes v1 and v2

will be represented by v1 < v2.

Concept:

\(V = \sqrt {2g{h_a}} \)

Calculation:

Given:

ρ_a = 1.2 kg/m³, deflection of the manometer = 24 mm

Now,

ρ_ah_a = ρ_wh_w

1.2 × h_a = 1000 × 24 × 10^-3

h_a = 20

\(= \sqrt {2 \times 10 \times 20} \)

Manometric head/ gross lift is defined as the head against which a centrifugal pump has to work.

It can be expressed as:

H_m = Suction Head (h_s) + Delivery Head (h_d) + Friction head loss in the suction pipe (h_fs) + Friction head loss in the Delivery pipe (h_fd) + Velocity head of water in delivery pipe (V²/2g)

Here Gross lift = \(h + {h_f} + \frac{{v_d^2}}{{2g}}\)

The separation point S is determined from the condition

\(\rm{{\left( {\frac{{{\partial u}}}{{{\partial y}}}} \right)_{y = o}} = 0}\)

So, If

\(\rm{{\left( {\frac{{{\partial u}}}{{{\partial y}}}} \right)_{y = 0}} < 0 \Rightarrow flow\;has\;separated}\) 

\(\rm{\begin{array}{l} {\left( {\frac{{{\partial u}}}{{{\partial y}}}} \right)_{y = 0}} \Rightarrow \rm{The\;flow\;is\;on\;verge\;of\;separation}\\ {\left( {\frac{{{\partial u}}}{{{\partial y}}}} \right)_{y = 0}} > 0 \Rightarrow \rm{Flow\;will\;not\;separate }\end{array}}\)

So, \(\frac{u}{v} = \frac{5}{2}\left( {\frac{y}{\delta }} \right) - \frac{1}{2}{\left( {\frac{y}{\delta }} \right)^2} + \frac{2}{3}{\left( {\frac{y}{\delta }} \right)^3}\)

\(\rm{\begin{array}{l} \frac{{\partial u}}{{\partial y}} = \frac{{5v}}{{2\delta }} - \frac{{yv}}{{{\delta ^2}}} + \frac{{2{y^2}}}{{{\delta ^3}}}\\ {\left( {\frac{{{\partial _u}}}{{{\partial _y}}}} \right)_{y = 0}} = \frac{5}{{2\delta }}v > 0 \end{array}}\)

⇒ The flow will not separate.

Concept:

A Stream line is an imaginary curve drawn through a flowing fluid in such a way that tangent to it any point gives the direction of instantaneous velocity at that point.  The equation of stream line for 2 D flow is given as:

\(\frac{{{\rm{dx}}}}{{\rm{u}}} = \frac{{{\rm{dy}}}}{{\rm{v}}}\) 

Where, u and v are the x and y components of the velocity.

Calculation:

Given:

V = (5x) î - (5y) ĵ

u = 5x and v = -5y

The equation of streamline is given as

\(\frac{{dx}}{{5x}} = \frac{{dy}}{{ - 5y}} \Rightarrow \smallint \frac{{dy}}{y} = - \smallint \frac{{dx}}{x}\)

⇒ In y = -In x + C

⇒ In (xy) = C’ ⇒ xy = λ                                   [C, C’, λ are constants]

If the stream line pases through point (1,1)  i.e. x = 1 and y = 1 ⇒ λ = 1

∴ The equation of sream line is xy = 1

In general, surface tension decreases when temperature increases because cohesive forces decrease with an increase of molecular thermal activity. The influence of the surrounding environment is due to the adhesive action of liquid molecules that they have at the interface.

Concept:

For pipe flow, the dynamic similarity will be obtained if the Reynolds number in the model and prototype are equal.

(Re)_Model = (Re)_Prototype

\(\frac{{{\rho _m}{V_m}{D_m}}}{{{\mu _m}}} = \frac{{{\rho _p}{V_p}{D_p}}}{{{\mu _p}}}\)

\(\frac{{{V_m}{D_m}}}{{{\nu _m}}} = \frac{{{V_p}{D_p}}}{{{\nu _p}}}\)

Calculation:

Given: ν_oil = 1.0 × 10^-5 m²/s, v_w = 0.89 × 10^-6 m²/s, D_oil = 0.5 m, D_w = 20 mm = 0.02 m, V_oil = 10 m/s, V_w = ?

\({\left( {\frac{{VD}}{v}} \right)_P} = {\left( {\frac{{VD}}{v}} \right)_m}\)

\(\frac{{10 \times 0.5}}{{1.0 \times {{10}^{ - 5}}}} = \frac{{{V_w} \times 0.02}}{{0.89 \times {{10}^{ - 6}}}}\)

Concept

If the velocity field represents the flow, it must satisfy the continuity equation, given as:

\(\frac{{\partial u}}{{\partial x}} + \;\frac{{\partial v}}{{\partial y}} = 0\)

Where u and v are the x and y components of the velocity.

If it satisfies, the rate of shear deformation is given as:

\(\theta = \frac{1}{2}\left( {\;\frac{{\partial v}}{{\partial x}} + \;\frac{{\partial u}}{{\partial y}}} \right)\)

And the angular velocity in z-direction is given as,

\(\omega = \frac{1}{2}\left( {\;\frac{{\partial v}}{{\partial x}} - \;\frac{{\partial u}}{{\partial y}}} \right)\)

Calculation:

u = 3y – y² and v = 3x – x²

\(\frac{{\partial u}}{{\partial x}} = 0\;\;and\;\frac{{\partial v}}{{\partial y}} = 0\)

\( \Rightarrow \frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} = 0 + 0 = 0\)

i.e. equation of continuity is satisfied, so, it represents a flow field.

Now,

\(\frac{{\partial v}}{{\partial x}} = 3 - 2x\;\;and\;\frac{{\partial u}}{{\partial y}} = 3 - 2y\)

\(\theta = \frac{1}{2}\left( {3 - 2x + 3 - 2y} \right) = \frac{1}{2}\left[ {6 - 2\left( {x + y} \right)} \right]\)

\(\theta \left( {x,\;y} \right) = \frac{1}{2}\left( {6 - 2\left( {x + y} \right)} \right)\)

θ (2, 3) \(= \frac{1}{2}\left[ {6 - 2\left( {2 + 3} \right)} \right]\)

θ(2, 3) = -2 units

Again,

\(\omega = \frac{1}{2}\left( {\left( {3 - 2x} \right) - \left( {3 - 2y} \right)} \right)\)

⇒ \(\omega = \frac{1}{2}\left( {2y - 2x} \right) = y - x\)

∴  ω(2, 3) = 3 – 2 = 1 units

∴ θ = -2 units and ω = 1 units.  

Surface tension

• It is the property of a liquid by virtue of which it tries to minimize its free surface area.
• It happens due to the cohesive force between liquid molecules. 

Capillary action is caused by the combination of cohesive forces of the liquid and the adhesive forces between the liquid and tube material. Cohesion and adhesion are two types of intermolecular forces.

Given data; L = 5 m, B = 2 m, V = 4 m/s, ρ = 1.25 kg/m³

ν = 1.5 × 10^-5 m²/s

\(R{{e}_{L}}=\frac{\rho VL}{\mu }=\frac{VL}{\nu }=\frac{\left( 4 \right)\left( 5 \right)}{1.5\times {{10}^{-5}}}=13.33\times {{10}^{5}}\)

Re_L = 13.33 × 10⁵ > 5 × 10⁵

⇒ Flow is turbulent.

Average drag coefficient (C_D) \(=\frac{1}{L}\mathop{\int }_{0}^{L}{{C}_{D,x}}dx\)

\(\Rightarrow {{C}_{D}}=\frac{1}{L}\mathop{\int }_{0}^{L}\frac{0.059}{Re_{x}^{1/5}}dx=\frac{0.07375}{Re_{L}^{1/5}}\)

\({{C}_{D}}=\frac{0.07375}{{{\left( 13.33\times {{10}^{5}} \right)}^{1/5}}}=0.004393\)

\({{C}_{D}}=\frac{{{F}_{D}}}{\frac{1}{2}\rho A{{V}^{2}}}\Rightarrow {{F}_{D}}=\left( 0.004393 \right).\frac{1}{2}.\left( 1.25 \right)\left( 10 \right)\left( 16 \right)\)

Shear Stress (N/m²) = Dynamic Viscosity x Velocity Gradient (1/s) Dimensions of Dynamic Viscocity = M L^-1 T^-1

Velocity Gradient = Velocity(v)/ Distance(y) = m/s-m = 1/s and 1N = kg-m /s²

If for any liquid there, is less attraction for solid molecules or in other words the cohesion predominates, than the liquid will not wet the solid surface and the liquid surface will be depressed at the point of contact with the result the liquid surface is concave downward and the angle of contact ‘θ’ is greater than 90°.

When a droplet is separated initially from the surface of the main body of liquid, then due to surface tension, there is net inward force exerted over the entire surface of the droplet which causes the surface of the droplet to contract from all sides and results in increasing the internal pressure within the droplet the contraction of the droplet continues till the inward force due to the surface tension is in balance with the internal pressure and the droplet forms into sphere which is the shape for minimum surface area.

Adhesion- Two different particles binded together-Mixing of milk with water

Cohesion- Two same particles binded together-Mixing of water with water.

Let us suppose that the level of liquid has risen (or fallen) by 'h' above (or below) the general liquid surface when a tube of diameter 'd' is inserted in the liquid. For equilibrium of vertical forces acting on the mass of liquid lying above (or below) the general liquid level, the weight of liquid column ‘h’ (or the total internal pressure in the case of capillary depression) must be balanced by the force, at surface of the liquid, due to surface tension σ. Thus, equating these two forces we have