Dimensionless Numbers MCQ Quiz in मल्याळम - Objective Question with Answer for Dimensionless Numbers - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

\({N_u} = \frac{{hD}}{K}\)

Where D is characteristic dimension of body.

For vertical plate case D = L

Rayleigh Number (R_a) = Gr.Pr

\( = \frac{{g\beta {\rm{\Delta }}T{L^3}}}{{{v^2}}} \times Pr\) 

Where Gr = Grashoff Number, Pr = Prandtl Number

Calculation:

Nu = K(Ra)^1/4

\(\frac{{hL}}{K} = K{\left[ {\frac{{g\beta {\rm{\Delta }}T{L^3}}}{{{v^2}}} \times Pr} \right]^{1/4}}\) 

\(so,\;h \propto \frac{1}{{{L^{1/4}}}}\) 

\(\frac{{{h_1}}}{{{h_2}}} = {\left( {\frac{{{L_2}}}{{{L_1}}}} \right)^{1/4}}\) 

As L₂ = 16L₁

_{\(\left( {\frac{{{h_1}}}{{{h_2}}}} \right) = {\left( {16} \right)^{1/4}} = 2\)}

Explanation:

The relationship between the thermal boundary layer and the hydrodynamic boundary layer is given by Prandtl number 

Prandtl Number: It is defined as the ratio of momentum diffusivity to thermal diffusivity.

\(Pr = \frac{\nu }{\alpha } = \frac{{momentum\;diffusivity}}{{Thermal\;diffusivty}} = \frac{{\frac{\mu }{\rho }}}{{\frac{k}{{{c_p}\rho }}}} = \frac{{\mu {c_p}}}{k}\)

The relationship between the two is given by equation

\(\frac{{{\delta }}}{\delta_t } = P_r^{ \frac{1}{3}}\)

\(\frac{{{\delta_t}}}{\delta } = P_r^{ \frac{-1}{3}}\)

δ = the thickness of the hydrodynamic boundary layer; the region of flow where the velocity is less than 99% of the far-field velocity.

δT = the thickness of the thermal boundary layer; the region of flow where the local temperature nearly reaches the value (99%) of the bulk flow temperature.

∴ n = -1/3.

Additional Information

• If Pr > 1 the momentum or hydrodynamic boundary layer will increase more compared to the thermal boundary layer.  
• If Pr < 1 the thermal boundary layer will increase more compared to the momentum or hydrodynamic boundary layer.
• If Pr = 1 The the thermal boundary layer and momentum or hydrodynamic boundary layer will increase at the same rate.

If the velocity and thermal boundary layers coincide then Pr = 1. 

Explanation:

Thermal diffusivity:

Thermal diffusivity of material is given as \(α = \frac{k}{{ρ c}}\)

where k is thermal conductivity in W/m-k, ρ is density in kg/m3 and c specific heat capacity in J/kg-K

It is the property of the material. The larger the value of α, the faster heat will diffuse through the material. A high value of α could result either from a high value of thermal conductivity or low value of thermal heat capacity ρc. Thermal diffusivity α has units of square meters per second.

unit of \(\alpha =\frac{\frac{W}{m-K}}{\frac{kg}{m^3}\frac{J}{kg-K}}=\frac{m^2}{s}\)as Watt is J/s.

Concept:

Biot number provides a way to compare the conduction resistance within a solid body to the convection resistance external to that body (offered by the surrounding fluid) for heat transfer:

\(Bi = \frac{{{\rm{conduction\;resistance}}}}{{{\rm{convection\;resistance}}}} = \frac{{\left( {\frac{L}{{kA}}} \right)}}{{\left( {\frac{1}{{hA}}} \right)}} = \frac{{hl}}{k}\)

Biot number, Bi is the ratio of resistance within the fin (resistance to conduction) and the resistance at the surface (resistance to convection). Fins are designed for biot number less than 1

Explanation:

d = 0.01 m, V = 3 m/s, ρ = 865 kg/m³, μ = 0.0216 Pa-s, k = 0.14 W/m °C, C_p = 1780 J/kg

Now,

\(Reynold's\;number\;({R_e}) = \frac{{\rho .V.d}}{\mu }\)

\(\therefore {R_e} = \frac{{865 \times 3 \times 0.01}}{{0.0216}}\)

∴ R_e = 1201.38

∵ R_e < 2000 ⇒ Laminar flow

For laminar flow forced convection with constant wall temperature

Nu = 3.66

\(\frac{{h.d}}{{{k_f}}} = 3.66\)

\(h = \frac{{3.66 \times 0.14}}{{0.01}}\)

∴ h = 51.24 W/m² °C

Concept:

\({\rm{Mach\;Number}} = \frac{{Inertia\;force}}{{elastic\;force}} = \frac{{Inertia\;force}}{{Compressible\;force}}\)

Incompressible flow

• M < 0.3
• From the above point, it is clear that the compressibility effect can be treated as negligible when the Mach number is up to 0.2

Froude number

Froude number is the ratio of inertial force to the gravitation force.​​

\(\text{Froude }{ number}=\text{ }\!\!~\!\!\text{ }{\frac{\text{Inertia force }\!\!~\!\!\text{ }}{\text{Gravitational force }\!\!~\!\!\text{ }}}\)

Froude number has the following applications:

• Used in cases of river flows, open-channel flows, spillways, surface wave motion created by boats
• It can be used for flow classification

Skin friction coefficient

Skin friction drag is a component of profile drag that occurs differently depending on the type of flow over the lifting body

The local Skin friction coefficient for the boundary layer over a flat plate is given as

\({C_{fx}} = \frac{{0.664}}{{\sqrt {R{e_x}} }}\)

Reynolds number

Reynolds number is a dimensionless formula that is used to differentiate laminar flow from the turbulent flow.

Reynolds number is given by \(Re = \frac{{\rho \times V \times D}}{\mu }\)

In the case of pipe flow,

• If Re ≤ 2000 then flow is laminar
• 2000 ≤ Re ≤ 4000 then flow is transitional
• Re ≥ 4000 then flow is turbulent

Nusselt number

It is the ratio of heat flow rate by convection process to the heat flow rate by the conduction process.

\(Nu = \frac{{{Q_{conv}}}}{{{Q_{cond}}}} = \frac{{hA{\rm{\Delta }}T}}{{\frac{{kA{\rm{\Delta }}T}}{L}}} = \frac{{hL}}{k}\)

Concept:

Generally, the temperature of a body varies with time as well as position. In heat conduction under steady conditions, the temperature of a body at any point does not change with time.

In heat conduction under unsteady state conditions, the temperature of a body at any point varies with time as well as position in one- dimensional and multidimensional systems. This type of heat conduction is also known as transient heat conduction.

One such phenomenon is Lumped heat analysis, in which temperature varies with time but not space.

It is given by

\(\begin{array}{l} T = \frac{{T - {T_\infty }}}{{{T_o} - {T_\infty }}} = exp\left( { - \frac{{hA}}{{\rho cV}}t} \right)\\ \frac{{hA}}{{\rho cV}}t = \frac{h}{{\rho c{L_c}}}t = \left( {\frac{{h{L_c}}}{k}} \right)\left( {\frac{k}{{\rho c}}} \right).\frac{1}{{L_c^2}}t = \left( {\frac{{h{L_c}}}{k}} \right)\left( {\frac{\alpha }{{L_c^2}}t} \right) = Bi \times Fo \end{array}\)

Thermal Diffusivity: \(\alpha = \frac{k}{{\rho C}}\)

Fourier number: \(Fo = \frac{\alpha }{{L_c^2}}t\)

So \(T = \exp \left( { - Bi \times Fo} \right)\)

∴ Biot number and Fourier number are the dimensionless numbers associated with transient heat conduction.

Concept:

\(Nu = 0.59{\left( {Gr.Pr} \right)^{\frac{1}{4}}}\)

\(Gr = \frac{{g\beta {\rm{\Delta }}T{D^3}}}{{{\nu ^2}}};\;Nu = \frac{{hD}}{k}\)

Prandtl number is the ratio of molecular momentum diffusivity to thermal diffusivity.

\(Pr = \frac{\nu }{\alpha } = \frac{{\mu {c_p}}}{k}\)

Prandtl number is an intrinsic property of a fluid. For both the pipes Pr. No. is same as the medium is same.

The pipes are located in boiler house where ambient air is stationary. Implying a case of free convection for which:

\(Nu \propto G{r^{\frac{1}{4}}} \Rightarrow hD \propto {D^{\frac{3}{4}}} \Rightarrow h \propto \frac{1}{{{D^{\frac{1}{4}}}}}\frac{{{h_1}}}{{{h_2}}} = {\left( {\frac{{{D_2}}}{{{D_1}}}} \right)^{\frac{1}{4}}}\) 

Calculation:

D₁ = 100 mm and D₂ = 300 mm

Explanation:-

Nusselt number

It is the ratio of heat flow rate by convection process to the heat flow rate by the conduction process.

\(Nu = \frac{{{Q_{conv}}}}{{{Q_{cond}}}} = \frac{{hA{\rm{\Delta }}T}}{{\frac{{kA{\rm{\Delta }}T}}{L}}} = \frac{{hL}}{k}\)

The Nusselt number is in

here Nu = Nusselt's Number

Re = Reynold's Number, Pr = Prandtl Number, Gr = Grashoff Number 

Other important dimensionless numbers are:

• Biot number  → Ratio of internal thermal resistance to boundary layer thermal resistance
• Grashof number  → Ratio of buoyancy to viscous force
• Prandtl number  → Ratio of momentum to thermal diffusivities
• Reynolds number  → Ratio of inertia force to viscous force

Concept:

\({\rm{Fourier\;number\;}}{F_0} = \frac{{\alpha t}}{{L_e^2}}\)

\({\rm{Le}} = {\rm{\;characteristic\;length\;}} = {\rm{\;}}\frac{V}{A} = \frac{a}{6}\left( {{\rm{cube}}} \right)\)

L_e = a/b (cube)

Calculation:

Given: α = 20 × 10^-6 m²/s, t = 15 s

\({l_e} = \frac{{0.1}}{6}m\)

∴ \({F_o} = \frac{{20\; \times \;{{10}^{ - 6}} \;\times\; 15}}{{{{\left( {\frac{{0.1}}{6}} \right)}^2}}} = 1.08\)