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

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.

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

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.  

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

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

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

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