Heat Transfer Test 3

Concept:

At stationary medium

Qconduction = Qconvection

Nusselt number = hD/K

Calculation:

Now,

\({\rm{Q_{conduction}}} = \frac{{{T_s} - {T_\infty }}}{{\frac{1}{{4\pi k}}\left( {\frac{1}{{{r_1}}} - \frac{1}{{{r_2}}}} \right)}} = \frac{{{T_s} - {T_\infty }}}{{\frac{1}{{4\pi k}}\left( {\frac{1}{{{r_1}}} - \frac{1}{\infty }} \right)}}\) (since infinite surrounding)

∴ Q_conduction = 4π kr (T – T_∞)

Now,

Qconvection = hA_sΔT

∴ Q_convection = 4π hr² (T – T_∞)

Qconduction = Qconvection

⇒ h = k/r

\(Nu = \frac{{hD}}{k} = \frac{k}{r} \times \frac{{2r}}{k} = 2\)

∴ p = 2

Concept:

For turbulent flow

Nusselt number (Nu), Reynolds number (Re) and Prandtl number(Pr) are related as

\(Nu\;\;\alpha \;{\left( {Re} \right)^{0.8}}{\left( {Pr} \right)^n}\) 

Where Pr = constant

Therefore,

\(Nu\;\alpha \;{\left( {Re} \right)^{0.8}}\) 

\(Nu = \frac{{hD}}{K}\;and\;Re = \frac{{\rho vD}}{\mu }\) 

Hence, \(h\;\alpha \;{V^{0.8}}{D^{ - 0.2}}\)

And mass flow rate is given by, ρAv

Calculation:

h ∝ V^0.8 × D^-0.2

Case 1:

Let V₁ = V₁    D₁ = D

\({\dot m_1} = \rho AV = \rho \frac{\pi }{4}{D^2}V\)

Case 2:

ṁ₂ = 4ṁ₁

\({\dot m_2} = \rho \left( {\frac{x}{4}{{\left( {2D} \right)}^2}} \right){V_2}\)

\(4\rho \frac{\pi }{4}{D^2}{V_2}\) 

∴ D₂ = 2D, V₂ = V

\(\frac{{{h_2}}}{{{h_1}}} = {\left( {\frac{{{V_2}}}{{{V_1}}}} \right)^{0.8}} \times {\left( {\frac{{{p_2}}}{{{D_1}}}} \right)^{ - 0.2}} = {\left( 1 \right)^{0.8}} \times {\left( 2 \right)^{ - 0.2}} = 0.87\)

Concept:

Corrections using forces convection internal flow:

1. For a laminar flow through a circular pipe and pipe surface maintained at the constant temperature boundary condition

\(N{u_1} = \frac{{hD}}{{{k_{fluid}}}} = 3.66\)

2. For a laminar flow through a circular pipe and pipe surface maintained at the constant heat flux boundary conditions

\(N{u_2} = \frac{{hD}}{{{k_{fluid}}}} = 4.36\)

\(\frac{{N{u_1}}}{{N{u_2}}} = \frac{{3.66}}{{4.36}}\)

Concept:

For laminar boundary layer, the hydrodynamic boundary layer and its relation with thermal boundary layer thickness δ_t is given as:

\(\delta = \frac{{5x}}{{\sqrt {Re} }}\;and\;\frac{\delta }{{{\delta _t}}} = \;P{r^{\frac{1}{3}}}\)

While for turbulent flow:

\(\delta = \frac{{0.37x}}{{R{e^{\frac{1}{5}}}}}\;and\;\delta \cong {\delta _t}\)

Calculation:

μ = 4.5 × 10^-4 Pa.s, k = 0.65 W/mK, C_p = 4200 J/kg K, ρ = 1000 kg/m³, V = 10 m/s, x = 2 m

\(Re = \frac{{\rho Vx}}{\mu } = \frac{{1000 \times 10 \times 2}}{{4.5 \times {{10}^{ - 4}}}} = 4.44 \times {10^7}\) ⇒ Turbulent region

\(\delta = \frac{{0.37x}}{{R{e^{\frac{1}{5}}}}} = \frac{{0.37 \times 2}}{{{{(4.44 \times {{10}^7})}^{\frac{1}{5}}}}} = 0.0218\;m = 21.86\;mm\)

\(\delta \cong {\delta _t} = 21.86\;mm\)

Concept:

Temperature gradient at wall \( = {\left( {\frac{{dT}}{{dy}}} \right)_{y = 0\;}}\)

\({\rm{Nusselt\;number\;}} = {\rm{\;}}\frac{{{{\left( {\frac{{dT}}{{dy}}} \right)}_{y = 0\;}}}}{{\Delta T/L}}\)

\({\rm{Heat\;flux\;}} = {\rm{\;}}\frac{q}{A} = \;h\Delta T\)

Calculation:

\({\rm{Heat\;flux}}\left( {\frac{q}{A}} \right) = h{\rm{\Delta }}T\)

\( \Rightarrow {\rm{\Delta T}} = \frac{1}{h} \times \frac{q}{A}\)

\({\rm{Also}},{\left. {\frac{{dT}}{{dy}}} \right|_{y = 0}} = Nu\;\frac{{{\rm{\Delta }}T}}{L}\)

\({\left. {\frac{{dT}}{{dy}}} \right|_{y = 0}} = Nu \times \frac{1}{L} \times \frac{1}{h} \times \frac{q}{A}\)

Now,

Nu = 18 (given)

\(\therefore \frac{q}{A} = 21,\;h = 30\frac{W}{{{m^3}k}},\;L = 0.02\;m\)

\({\left. {\frac{{dT}}{{dy}}} \right|_{y = 0}} = 18 \times \frac{1}{{0.2}} \times \frac{1}{{30}} \times 21\)

Concept:

\({\rm{Reynold\;Number\;}}\left( {{\rm{Re}}} \right) = \frac{{\rho vD}}{\mu }\)

For constant heat flux, Nusselt number (Nu) = 4.36 and Nu = hd/k

Heat flux (Q/A) = h(∆T)

Calculation:

Given:

ν = 10^-5 m²/s, D = 0.01 m, L = 15 m, V = 0.5 m/s

T_wi = 298 K = 25°C, T_wo = 330 K = 57°C

Now,

\(Re = \frac{{VD}}{\nu } = \frac{{0.5 \times 0.01}}{{{{10}^{ - 5}}}} = 500 < 2300\;\left( {laminar} \right)\)

For constant heat flux in laminar flow

Nu = 4.36

\(\therefore \frac{{hd}}{k} = 4.36\)

∴ h = 4.36 × 0.6/0.01

∴ h = 261.6 W/m²°C

Heat flux = h (ΔT)

\({\rm{\Delta T}} = \frac{{3000}}{{261.6}} = 11.467\)

\({{\rm{T}}_{\rm{s}}} = 11.467 + 57 = 68.46^\circ C\)

Concept:

Nu = F(Re, Pr)

For Turbulent flow,

Nu ∝ (Re)^0.8(Pr)ⁿ

For laminar flow,

Nu ∝ (Re)^0.5(Pr)^0.33

Calculation:

For a heated plate with air flowing over it,

Given:

Re = 10⁶ > 5 × 10⁵

∴ Flow is turbulent

For turbulent flow

Nu ∝ (Re)^0.8(Pr)ⁿ

\(\frac{{h \cdot x}}{{{k_{fluid}}}}\; \propto \;{\left( {\frac{{V \cdot x}}{\nu }} \right)^{0.8}}\)

\(\therefore h\; \propto \;\frac{{{x^{0.8}}}}{x}\)

Concept:

For flow over a flat plate:

\({R_e} = \frac{{Ux}}{\nu }\) 

R_e < 5 × 10⁵ ⇒ Boundary layer is laminar

R_e > 5 × 10⁵ ⇒ Boundary layer is turbulent

For laminar flow:

Boundary layer thickness: \(\delta = \frac{{5x}}{{\sqrt {{R_{{e_x}}}} }}\)

Local Nusselt number: \(N{u_x} = \frac{{{h_x}x}}{k} = 0.332{\left( {Pr} \right)^{\frac{1}{3}}}{\left( {{R_{{e_x}}}} \right)^{1/2}}\)

Average Nusselt number: \({\overline {Nu} _L} = \frac{{\bar hL}}{K} = 0.664{\left( {Pr} \right)^{\frac{1}{3}}}{\left( {{R_{{e_L}}}} \right)^{1/2}}\)

Thermal Boundary layer thickness: \({\delta _t} = \frac{\delta }{{{{\left( {Pr} \right)}^{\frac{1}{3}}}}}\)

For turbulent flow:

\(N{u_x} = 0.0288\;R_{{e_x}}^{0.8}{\left( {Pr} \right)^{1/3}}\) 

\({\overline {Nu} _L} = 0.036{\left( {{R_{{e_L}}}} \right)^{0.8}}{\left( {Pr} \right)^{1/3}}\) 

Calculation:

At x = 0.5 m

\({R_{{e_x}}} = \frac{{Ux}}{\nu } = \frac{{5 \times 0.5}}{{20.76 \times {{10}^{ - 6}}}} = 1.2 \times {10^5} < 5 \times {10^5}\) 

So the flow is laminar:

\(N{u_x} = 0.332{\left( {Pr} \right)^{\frac{1}{3}}}{\left( {{R_{{e_x}}}} \right)^{1/2}}\) 

\(N{u_x} = 0.332{\left( {0.697} \right)^{\frac{1}{3}}}{\left( {1.2 \times {{10}^5}} \right)^{1/2}}\) 

Nu_x = 101.9

\(N{u_x} = \frac{{{h_x}.x}}{K} \Rightarrow {h_x} = \frac{{N{u_x} \times k}}{x} = \frac{{101.9 \times 0.03}}{{0.5}} = 6.1\) 

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