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)

Ideal fluid are those which have no viscosity and surface tension and are incompressible.

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.

The depth of flow at which the specific energy attains a minimum value is called the critical depth.

The flow at critical depth is called critical flow and the velocity of flow known as the critical velocity.

When the depth is less than the critical depth, the velocity of flow is greater than the critical velocity and hence the flow is said to be supercritical or rapid.

On the contrary, if the depth of flow is greater than the critical depth, the velocity is less than the critical velocity and hence the flow is said to be subcritical or tranquil.

∴ The velocity for which the specific energy is minimum is known as critical velocity.

Important point:

As can be observed from the graph, specific energy is minimum for a given discharge at a critical state.

As can be observed from the graph, at critical state discharge is maximum for a given specific energy.

At critical state, specific force is minimum for a given discharge.

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

In case of gases the cohesion forces are small and molecular momentum transfer predominates. With the increase in temperature, molecular momentum transfer increases and hence viscosity increases.

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

The liquid used in manometer should be of high density so as to measure even small variation in pressure.

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,

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

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

The pressure measured above the absolute zero or complete vacuum is called absolute pressure.

Absolute pressure = Gauge pressure + Local atmosphere pressure.

also Absolute pressure = Local atmosphere pressure - Negative gauge pressure.