Fluid Mechanics Test 2

Concept:

P_abs = P_gauge + P_atm

Calculation:

Given:

h = 5 m

P_atm = 0.75 m of mercury = ρ_Hg × g × 0.75 = 13600 × 9.81 × 0.75 = 100062 Pa

P_gauge = ρ_water × g × h

P_gauge = 1000 × 9.81 × 5 = 49050 Pa

P_abs = P_gauge + P_atm

P_abs = 49050 + 100062

P_abs = 149112 Pa = 149112 N/m² = 0.149 N/mm² ≃ 0.15 N/mm²

Concept:

Expression for capillary rise or fall

\(h = \frac{{4\sigma \cos \theta }}{{\rho gd}}\)

Calculation:

Given: θ = 130°, σ = 0.52 N/m, d = 2.5 × 10^-3m

\(h = \frac{{4\; \times \;0.52\; \times\; \cos 130^\circ }}{{13600\; \times \;9.81\; \times \;2.5\; \times \;{{10}^{ - 3}}}}\)

∴ h = - 0.4 cm

Note: - ve sign indicates capillary depression (or full)

Concept:

\({\left( {\frac{{\partial u}}{{\partial y}}} \right)_{y = 0}} \Rightarrow + {\rm{ve\;}} \Rightarrow {\rm{\;stick\;to\;surface}}\)

\({\left( {\frac{{\partial u}}{{\partial y}}} \right)_{y = 0}} \Rightarrow 0{\rm{\;}} \Rightarrow {\rm{\;on\;verge\;of\;separation}}\)

\({\left( {\frac{{\partial u}}{{\partial u}}} \right)_{y = 0}} \Rightarrow - {\rm{ve\;}} \Rightarrow flow~is~seperated\)

Calculation:

\(\frac{u}{U} = 2{\left( {\frac{y}{\delta }} \right)^2} - {\left( {\frac{y}{\delta }} \right)^3}\)

\(\frac{{\partial u}}{{\partial y}} = U\left[ {4\left( {\frac{4}{\delta }} \right) - 3{{\left( {\frac{y}{\delta }} \right)}^2}} \right]\)

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

∴ Flow is on the verge of separation.

Concept:

\({h_L} = \frac{{fL{Q^2}}}{{12{D^5}}}\)

\(Q\; \propto {\rm{\;}}\frac{1}{{\sqrt f }}\;\;\; - - - \left( 1 \right)\)

Calculation:

Here f is misjudged by 25 %, it means f can be 1.25f or 0.75f

Case: 1

Let it be 1.25f

\({Q_1} = \frac{k}{{\sqrt f }}\)

\({Q_2} = \frac{k}{{\sqrt {1.25f} }}\)

\(Error = \frac{{{Q_2} - {Q_1}}}{{{Q_1}}} \times 100 = - 11\;\% \)

Case: 2

Let it be 0.75f

\({Q_1} = \frac{k}{{\sqrt f }}\)

\({Q_2} = \frac{k}{{\sqrt {0.75f} }}\)

\(Error = \frac{{{Q_2} - {Q_1}}}{{{Q_1}}} \times 100\)

∴ Error = 15.47%

Explanation:

Pump is defined as a device which transfer the input mechanical energy of a motor or a of an engine into pressure energy or kinetic energy or both of a liquid.

Centrifugal pump is type of dynamic pressure pump.it also known as velocity pump.

\(Q = 0.0125\frac{{{m^3}}}{s}\)

Shaft power or rating of motor (P_S) = 5 kW

\(density\;\left( \rho \right) = 1000\frac{{kg}}{{{m^3}}}\)

Head (H) = 30 m

\(\eta = \;\frac{{Power\;output\;}}{{shaft\;power}}\)

\(\eta = \;\frac{{\rho \times g \times Q \times H}}{{{P_S}}}\)

\(\eta = \;\frac{{1000 \times 9.81 \times 0.0125 \times 30}}{{5000}}\)

η = 0.73575 ≈ 0.7360

Concept:

Pascal’s 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 fluid are zero.

Newton’s law of viscosity:

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

Where,

τ is the shear stress on the 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

Concept:

For any type of flow, the shear stress at wall,

\(\tau = - \left( {\frac{{dp}}{{dx}}} \right)\frac{R}{2}\;\)

\(\therefore \tau = \frac{{\rho g{h_L}}}{L} \times \frac{R}{2}\)

\(\tau = \frac{{1000 \times 9.81 \times 10}}{{100}} \times \frac{{0.1}}{4}\)

∴ τ = 24.525 Pa

Concept:

In generalised equation, density (ρ) is not constant and the flow is three dimensional

\(\frac{{D\rho }}{{Dt}} + \;\rho \left( {\nabla .\vec V} \right) = 0\)

After expansion,

\(\frac{{\partial \rho }}{{\partial t}} + u\frac{{\partial \left( \rho \right)}}{{\partial x}} + \rho \frac{{\partial \left( u \right)}}{{\partial y}} + v\frac{{\partial \left( \rho \right)}}{{\partial y}} + \rho \frac{{\partial \left( v \right)}}{{\partial y}} + w\frac{{\partial \left( \rho \right)}}{{\partial z}} + \rho \frac{{\partial \left( w \right)}}{{\partial z}} = 0\)

\(\frac{{\partial \rho }}{{\partial t}} + \frac{{\partial \left( {\rho u} \right)}}{{\partial x}} + \frac{{\partial \left( {\rho v} \right)}}{{\partial y}} + \frac{{\partial \left( {\rho w} \right)}}{{\partial z}} = 0\)

\(\frac{{\partial \rho }}{{\partial t}} + \;\nabla .\left( {\rho \vec V} \right) = 0\)

Concept:

C_d = C_c × C_v

Where, C_d = Coefficient of discharge, C_c = Coefficient of vena contracta, C_v = Coefficient of velocity

Calculation:

C_d = C_c × C_v

0.58 = 0.61 × C_v

∴ C_v = 0.95

Concept:

θ = momentum thickness

\({\rm{\theta \;}} = \mathop \smallint \limits_0^\delta \frac{u}{U}\left[ {1 - \frac{u}{U}} \right]dy\)

Calculation:

\(\theta = \mathop \smallint \limits_0^\delta \frac{u}{U}\left[ {1 - \frac{u}{U}} \right]dy\)

\(\theta = \;\mathop \smallint \limits_0^\delta \frac{y}{{2\delta }}\left[ {1 - \frac{y}{{2\delta }}} \right]dy\)

\(\theta = \;\mathop \smallint \limits_0^\delta \left[ {\frac{y}{{2\delta }} - \frac{{{y^2}}}{{4{\delta ^2}}}} \right]dy\)

\(\theta = \left[ {\frac{{{y^2}}}{{4\delta }} - \frac{{{y^3}}}{{12{\delta ^2}}}} \right]_0^\delta \)

\(\theta = \frac{\delta }{4} - \frac{\delta }{{12}}\)

\(\therefore \frac{\theta }{\delta } = \frac{1}{6}\)

Explanation:

Reynold’s model law is applicable for:

(i) Pipe flow

(ii) Resistance experienced by sub-marines, airplanes, etc.

Froude model law is applicable for:

(i) free surface flows such as spillways, channels, etc.

(ii) Sea waves

Euler model law is applicable for:

Where pressure forces are dominant like flow taking place in closed pipe, cavitation phenomenon etc.

Weber model is applicable for:

(i) Capillary rise in narrow passage

(ii) Capillary movement of water in soil

Concept:

Reynolds number will be same for both model and prototype

Calculation:

Given:

\(\frac{{{L_M}}}{{{L_P}}} = \frac{1}{6}\)

\({V_p} = 60 \times \frac{5}{{18}}\;\frac{m}{s}\)

F_M = 250 N

\(Re = \;\frac{{\rho {V_M}{L_M}}}{\mu } = \frac{{\rho {V_P}{L_P}}}{\mu }\; - - - \left( 1 \right)\)

Now,

From (1)

\({V_M} = \frac{{{L_P}}}{{{L_M}}}{V_P} = 100\;\frac{m}{s}\)

Inertial force = ρL²V²

\(\frac{{{F_P}}}{{{F_M}}} = \frac{{{\rho _P}}}{{{\rho _M}}}{\left( {\frac{{{L_P}}}{{{L_M}}}} \right)^2}{\left( {\frac{{{V_P}}}{{{V_M}}}} \right)^2}\;\; - - - \left( 2 \right)\)

Density is same for model and prototype (Given)

From (2),

F_P = 250 N

Power on the prototype

P = F_PV_P

∴ P = 4.167 kW

Concept:

The Reynolds number for flow over plate can be given as:

\({R_e} = \frac{{\rho UX}}{\mu }\)

Where

X is the distance from leading edge.

U is the free mean velocity.

μ is the dynamic viscosity of fluid flowing.

ρ is the density of fluid flowing.

If Reynolds number is less than or equal to critical Reynold’s number than boundary layer formed over flat plate is laminar i.e.R_e ≤ R_eCr­

Calculation:

Given:

U = 1 m/s ; ρ = 1000 kg/m³ ; μ = 1 centipoise = 10^-3 N s/m²

R_{e cr} = 5 × 10⁵

Let ‘x’ is the distance from leading edge up to which Boundary layer formed is laminar,

Now,

R_ex ≤ R_{e cr}

\(\frac{{\rho Ux}}{\mu } \le 5 \times {10^5}\)

\(\frac{{{{10}^3} \times 1 \times x}}{{{{10}^{ - 3}}}} \le 5 \times {10^5}\)

x ≤ 0.5 m

∴ The maximum distance from leading edge of the plate up to which boundary layer is laminar is 0.50 m.

Concept:

\(u = \left( {\frac{{ - \partial P}}{{\partial x}}} \right)\left[ {\frac{{{R^2} - {r^2}}}{{4\mu }}} \right]\)

Calculation:

Given: D = 0.5 mm ⇒ R = 0.25 mm, L = 2m, Q = 48 cm³/min = 8 × 10^-7 m³/sec,

μ = 1.92 centipoise = 1.92 × 10^-3 Ns/m²

Now,

Q = A × V_avg

\(8 \times {10^{ - 7}} = \frac{\pi }{4} \times {\left( {0.5 \times {{10}^{ - 3}}} \right)^2} \times {V_{avg}}\)

∴ V_avg = 4.074 m/s

\(\therefore {u_{max}} = \left( {\frac{{ - \partial P}}{{\partial x}}} \right)\left[ {\frac{{{R^2}}}{{4\mu }}} \right]\)

And u_mean = V_avg u_max

\(\therefore {V_{avg}} = \left( {\frac{{ - \partial P}}{{\partial x}}} \right)\left[ {\frac{{{R^2}}}{{8\mu }}} \right]\)

\(\therefore {\rm{\Delta }}P = \frac{{{V_{avg}} \times L \times 8\mu }}{{{R^2}}}\)

∴ ΔP = 2 MPa

Points to Remember:

\({\rm{\Delta }}P = \frac{{32\;\mu \;{V_{avg}}\;L}}{{{D^2}}}\)

Concept:

\(U = \; - \frac{{\partial \emptyset }}{{\partial x}} = \; - \frac{{\partial \varphi }}{{\partial y}}\;\; - - - \left( 1 \right)\)

\(V = \; - \frac{{\partial \emptyset }}{{\partial y}} = \;\frac{{\partial \varphi }}{{\partial x}}\;\;\; - - - \left( 2 \right)\)

\(Q = \;{\varphi _2} - {\varphi _1}\;\;\; - - - \left( 3 \right)\)

Q is flow per unit width

Now,

From (1),

\(U = \; - 2x = \; - \frac{{\partial \varphi }}{{\partial y}}\; \Rightarrow \;\varphi \left( {x,y} \right) = 2xy + f\left( x \right)\; - - - \;\left( 4 \right)\)

\(V = \;2y = \;\frac{{\partial \varphi }}{{\partial x}}\; \Rightarrow \;\varphi \left( {x,y} \right) = 2xy + f\left( y \right)\;\; - - - \;\left( 5 \right)\)

At (1, 2)

U = -2 m/s

V = 4 m/s

\(Velocity = \;\sqrt {{{\left( { - 2} \right)}^2} + {4^2}} \)

∴ Velocity = 4.47 m/s

Now,

From (4) and (5)

φ(x, y) = 2xy

φ(1, 3) = 6

φ(3, 4) = 24

Q = φ(3, 4) - φ(1, 3) = 24 - 6

∴ Q = 18 units 

Concept:

For a fluid flow to be possible, it should obey the continuity equation which is given by

\(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0~\) (Incompressible, 2D)

The convective acceleration is given by

\({{\alpha }_{xC}}=u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}~~;~~{{\alpha }_{yC}}=u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}\)

The local acceleration is given by

\({{\alpha }_{xL}}=\frac{\partial u}{\partial t}~~;~~{{\alpha }_{yL}}=\frac{\partial v}{\partial t}\)

The acceleration in a direction is given by

α = α_C + α_L  

Calculation:

Given V = (2x²t + y) î - 4xyt ĵ ⇒ u = 2 x² t + y ; v = - 4 x y t ;

\(\frac{\partial \left( 2{{x}^{2}}t+y \right)}{\partial x}+\frac{\partial \left( -4xyt \right)}{\partial y}=4xt-4xt=0~~\)  (Option 1 is wrong) 

Convective acceleration in the x-direction is

\({{\alpha }_{xC}}=\left( 2{{x}^{2}}t+y \right)\frac{\partial \left( 2{{x}^{2}}t+y \right)}{\partial x}+\left( -4xyt \right)\frac{\partial \left( 2{{x}^{2}}t+y \right)}{\partial y}\)

⇒ α_xc = [(2 x² t + y) × 4 x t] + [(- 4 x y t) × 1] = 8 x³ t²

∴ α_xc at (1, 2) and t = 3 sec is =  8 × 1³ × 9 = 72 m/s² (Option 2)

Now,

Local acceleration in y-direction is

\({{\alpha }_{yL}}=\frac{\partial \left( -4xyt \right)}{\partial t}=-4xy\)

∴ α_yL = -4 × 1 × 2 = - 8 m/s² (Option 3 is wrong) 

Now,

Local acceleration in x-direction is

\({{\alpha }_{x}}=\frac{\partial \left( 2{{x}^{2}}t+y \right)}{\partial t}=2{{x}^{2}}=2\times {{1}^{2}}=2\)

∴ Acceleration in x-direction is α_x = 72 + 2 = 74 m/s² ;

Now,

Convective acceleration in y-direction is

\({{\alpha }_{yC}}=\left( 2{{x}^{2}}t+y \right)\frac{\partial \left( -4xyt \right)}{\partial x}+\left( -4xyt \right)\frac{\partial \left( -4xyt \right)}{\partial y}\)

⇒ α_yc = [(2 x² t + y) × - 4 y t] + [(- 4 x y t) × - 4 x t] = 16 x² y t² – 4 y² t – 8 x² y t² ;

⇒ α_yc = 8 x² y t² – 4 y² t

α_yc at (1, 2) and t = 3 sec is = 96 m/s² ;

∴ Acceleration in y-direction is α_y = 96 – 8 = 88 m/s²;

Total acceleration = (26² + 88²)^0.5

∴ α_total = (26² + 88²)^0.5 = 91.76 m/s² (Option 4 wrong)

Concept:

To check possibility of fluid flow, continuity equilibrium should be verified, i.e.

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

Angular velocity about z-axis,

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

Calculation:

\(u = 2x + \frac{{{y^3}}}{3} - {x^2}y\)

\(v = x{y^2} - 2y - \frac{{{x^3}}}{3}\)

\(\frac{{\partial u}}{{\partial x}} = 2 - 2xy\;\;\;;\;\frac{{\partial u}}{{\partial y}} = {y^2} - {x^2}\)

\(\frac{{\partial v}}{{\partial x}} = {y^2} - \frac{{3{x^2}}}{3} = {y^2} - {x^2}\)

\(\frac{{\partial v}}{{\partial y}} = 2xy - 2\)

For 2-D flow, continuity equation,

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

(2 – 2xy) + (2xy - 2) = 0

Hence, it represents possible case of fluid flow.

Now,

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

\({\omega _z} = \frac{1}{2}\left[ {\left( {{y^2} - {x^2}} \right) - \left( {{y^2} - {x^2}} \right)} \right] = 0\)