Fluid Mechanics Test 2

Concept:

Young’s Modulus of elasticity (Y):

It is the ratio of Longitudinal (tensile or compressive) stress (σ) to the longitudinal strain (ϵ) is defined as Young’s modulus and is denoted by the symbol Y. (it is valid within the elastic limit)

\(i.e.,\;Y = \frac{{Langitudinal\;stress\;\left( \sigma \right)}}{{Langitudinal\;strain\;\left( \epsilon \right)}} = \frac{{\frac{F}{A}}}{{\frac{{{\rm{\Delta }}L}}{L}}}\)

The bulk modulus of elasticity (Β): It is the ratio of Hydraulic (compressive) stress (p) to the volumetric strain (ΔV/V) is defined as Bulk modulus and is denoted by the symbol K. (it is valid with in the elastic limit)

\(i.e.,\;B = - \frac{p}{{\frac{{{\rm{\Delta }}V}}{V}}}\)

And a unit of B is N/m2 or Pa (Pascal) since 1 N/m2 = 1 Pa

Modulus of rigidity or shear modulus of elasticity (η): It is the ratio of tangential stress to the shearing strain θ is defined as modulus of rigidity or shear modulus of elasticity and it is denoted by the symbol of η.

\(i.e.,\;\eta = \frac{{tangential\;stress}}{{shearing\;strain}} = \frac{{\frac{F}{A}}}{\theta }or\frac{{FL}}{{A{\rm{\Delta }}x}}\)

Explanation:

Given,

P₁ = 0.40 MPa, P₂ = 12.3 MPa, dV/V = 1.2%

dP = 12.3 – 0.40 = 11.9 MPa

\(\frac{{dV}}{V} = 1.2\% = - 0.012\)

Bulk modulus of elasticity,

\(k =- \frac{{dP}}{{\left( {\frac{{dV}}{V}} \right)}} = \frac{{11.9}}{{0.012}} = 991.67\ MPa\)

Hence the modulus of elasticity of the liquid is 991.7 MPa

Note most students get confused about the application of different modulus of elasticity, thus each modulus of elasticity can be summarized as 

Young's modulus ⇒ Used in linear expansion

Bulk Modulus/ Compressibility ⇒ Used for Volumetric expansion 

Rigidity modulus ⇒ Used to measure the change in the shape of an object due to deforming force 

Concept:

Force exerted by the jet on a moving plate is given by

F = ρ A (v - u)²

where ρ is the density of flowing fluid, A is the cross-sectional area of the jet, v is the velocity of the jet and u is the velocity of the moving plate.

Calculation:

Given:

ρ = 1000 kg/m³, v = 20 m/s, u = 10 m/s, A = 0.02 m²

F = ρ A (v - u)²

F = 1000 × 0.02 × (20 - 10)² = 2000 N

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

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

Concept:

Given:

h_L = constant in both the pipes

\({h_l} = \frac{{fL{Q^2}}}{{12{D^5}}}\)      ---(1)

Calculation:

Given:

Pipe-1:

L₁ = 100 m, D₁ = 200 mm, f₁ = 0.015

Pipe – 2

D₂ = 400 mm, f₂ = 0.012, L₂ =?

Q = Constant in both the pipes

Now,

From   (1),

\({h_l} = \frac{{{f_1}{L_1}{Q^2}}}{{12D_1^5}}\ =\frac{{{f_2}{L_2}{Q^2}}}{{12D_2^5}}\)

\(\frac{{0.015\: \times \:100 \:\times\: {Q^2}}}{{12\: \times \:{{0.2}^5}}} = \frac{{0.012\: \times \:{L_2} \:\times\: {Q^2}}}{{12 \:\times \:{{0.4}^5}}}\)

∴ L₂ = 4000 m

Now,

We know

\(f = \frac{{64}}{{Re}}\)

For pipe-1 (D = 200 mm, f = .015)

\(0.015 = \frac{{64}}{{Re}}\)

Re = 4266 (>2300), so it is turbulent flow

For pipe-2 (D = 400 mm, f = .012)

\(0.012 = \frac{{64}}{{Re}}\)

Concept:

Q = nAV

Q = Discharge, nA = Effective area

V = Velocity of the jet = \({C_v}\sqrt {2gH} \)

Calculation:

Given:

n = 5 × 2 × 4 = 40, H = Head developed = 250 m

C_v = 0.985, Q = 40 m³/s

\(V = {C_v}\sqrt {2gH} = 0.985 \times \sqrt {2 \times 9.81 \times 250} = 68.985\;m/s\)

Now,

Q = nAV

\(Q = n\frac{{\pi {D^2}}}{4}V\)

\(40 = 40 \times \frac{{\pi {D^2}}}{4} \times 68.985\)

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:

Hydraulic efficiency of the Pelton wheel is given by:

\({\eta _{h}} = \frac{{Runner\;power}}{{Kinetic\;energy\;per\;second}} = \frac{{2 \times \left( {{V_1} - u} \right) \times (1 + k\cos \phi )u}}{{V_1^2}}\)

For maximum efficiency,

When u = V₁/2

\((\eta _h)_{max} = \left( {\frac{{1 + k\cos \phi }}{2}} \right)\)

k = friction factor for blades

Calculation:

Given:

ϕ = 60°

As friction is neglected.

Maximum efficiency is:

\((\eta _h)_{max} = \left( {\frac{{1 + k\cos \phi }}{2}} \right)\)

\((\eta _h)_{max} = \left( {\frac{{1 + \cos 60^\circ }}{2}} \right) = \frac{{1 + \frac{1}{2}}}{2} = \frac{3}{4} \times 100 = 75\% \)

Concept:

For a Newtonian fluid,

\(Shear~stress\left( \tau \right) = dynamic~visocity\left( \mu \right) \times rate~of~shear~strain\left( {\frac{{du}}{{dy}}} \right)\)

For the above situation, 

\(\tau = \mu \times \frac{V}{y} \)

\(\frac{{mg\sin \theta }}{A} = \mu \times \frac{V}{y}\)

where, A = Area in contact of a Newtonian fluid, V = Terminal velocity of the block

Calculation:

Given:

Edge of the block, a = 100 mm = 0.1 m , weight, W = 1 kN, θ = 30°, y = 0.02 mm, \(\mu = 0.2~\frac{{Ns}}{{{m^2}}}\)

Shear stress acting at the face of the cube in contact with fluid,

\(\tau = \mu \times \frac{V}{y} \)

\(\frac{{mg\sin \theta }}{A} = \mu \times \frac{V}{y}\)

\(\frac{{1000 \times 0.5}}{{0.1 \times 0.1}} = 0.2 \times \frac{V}{{0.02 \times {{10}^{ - 3}}}}\)

V = 5 m/s

Concept:

Hydraulic press works on the principle of Pascal’s law,

The pressure intensity on plunger will be transmitted equally as pressure intensity at ram.

Since no frictional losses, W_in = W_output ⇒ F_p × h_P = F_R × h_R

Where h is the movement of plunger/ram.

Mechanical advantage = F_output/F_input = F_R/F_P

Calculation:

Given:

D_P = 5 cm = 0.05 m; D_R = 40 cm = 0.4 m; P_base ≤ 260 kPa; h = 1 m (from base);

A_P = π × 0.05²/4 = 1.96 × 10^-3 m²; A_R = π × 0.4²/4 = 0.1256 m²

The maximum weight to be lifted is limited by the base of press as the pressure intensity applied at plunger will equally transmit to the base also.

⇒ P_plunger + ρgh = P_base ⇒ P_plunger + ρgh ≤ 260 kPa

⇒ P_plunger + 10³ × 10 × 1 ≤ 260 × 10³ ⇒ P_plunger ≤ 250 kPa

∴ Maximum pressure that can be applied at plunger = 250 kPa

⇒ P_ram (max) = 250 kPa

⇒ F_ram (max) = W_max = P_ram (max) × A_R = 250 × 0.1256 = 31.4 kN (Option 1)

Now,

Always P_plunger = P_ram

⇒ \(\frac{{{F_P}}}{{{A_P}}} = \frac{{{F_R}}}{{{A_R}}} \Rightarrow \frac{{{F_R}}}{{{F_P}}} = \frac{{{A_R}}}{{{A_P}}}\)

⇒ Mechanical advantage = 0.1256/00196 = 64 (Option 2 is wrong)

Now,

If F_P = 400 N, F_R = M.A × F_P = 400 × 64 = 25.6 kN (Option 3)

Since the fluid is enclosed, the volume displaced in plunger side will be equal to volume displaced in ram side.

⇒ A_P h_P = A_R h_R ⇒ h_R = h_P / M.A = 20/64 = 3.12 mm (Option 4)

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:

According to Newton’s law of viscosity, shear stress is given by:

\(\tau = \mu \times \frac{{du}}{{dy}} = \;\mu \times \frac{{d\theta }}{{dy}}\;\)

τ = shear stress, μ = coefficient of viscosity or absolute viscosity (or dynamic viscosity)

\(\frac{{du}}{{dy}} = Velocity\;gradient\)

\(\frac{{d\theta }}{{dt}}\; = Rate\;of\;angular\;deformation\;or\;Rate\;of\;shear\;strain\)

Calculation:

Given:

Velocity profile 

\(u\left( y \right) = \;{U_\infty } \times \sin \left( {\frac{{\pi y}}{{2\delta }}} \right),\;0 \le y \le \delta \)

\(\tau = \mu \times \frac{{du}}{{dy}}\)

\(\tau = \mu \times \frac{{du}}{{dy}} = \mu \times \;{U_\infty } \times\frac{\partial }{{\partial y}}\left( {\sin \left( {\frac{{\pi y}}{{2\delta }}} \right)} \right)\)

\( = \mu \times {U_\infty } \times \cos \left( {\frac{{\pi y}}{{2\delta }}} \right) \times \frac{\pi }{{2\delta }}\)

As we have to find the shear stress at the wall,

y = 0

\({\tau _{wall}} = \mu \times {U_\infty } \times \frac{\pi }{{2\delta }} = \frac{{\pi \mu {U_\infty }}}{{2\delta }}\)

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

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

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

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