Heat Transfer Test 2

Concept:

The efficiency of a fin is defined as the ratio of the actual heat transferred by the fin to the maximum heat transferable by fin, if entire fin area were at base temperature, for infinitely long fin:

\(\eta = \frac{{{Q_{fin}}}}{{{Q_{max}}}} = \frac{{\sqrt {Phk{A_c}\left( {{t_o} - {t_a}} \right)} }}{{hPL\left( {{t_o} - {t_a}} \right)}} = \sqrt {\frac{{k{A_c}}}{{hP}}}.\frac{1}{L}\)

\(\begin{array}{l} \eta = \frac{1}{{mL}}\\ \eta \propto \frac{1}{{\sqrt h }} \end{array}\)

As, h₁ = 2h₂

⇒ η₁ < η₂

Effectiveness of the fin is the ratio of the fin heat transfer rate to the heat transfer rate that would exist without fin.

\({\epsilon_{fin}} = \frac{{{Q_{with\;fin}}}}{{{Q_{without\;fin}}}} = \frac{{\sqrt {Phk{A_c}} \;\left( {{t_o} - {t_a}} \right)}}{{h{A_c}\left( {{t_o} - {t_a}} \right)}} = \sqrt {\frac{{Pk}}{{h{A_c}}}}\)

\({\epsilon_{fin}} \propto \frac{1}{{\sqrt h }},{h_1} = 2{h_2} \Rightarrow {\epsilon_1} < {\epsilon_2}\)

Calculation:

\(\begin{array}{l} \eta \propto \frac{1}{{\sqrt h }}\;\; and\;\; \epsilon \propto \frac{1}{{\sqrt h }} \end{array}\)

\({h_1} = 2{h_2} \Rightarrow {\epsilon_1} < {\epsilon_2} \; and \; \; {\eta_1} < {\eta_2}\)

Explanation:

Heat energy lost by radiation = Electrical energy to be dissipated

σϵAT⁴ = 75

ϵ × 5.67 × 10^-8 × 0.0225 × (560)⁴ = 75

ϵ = 0.598

Now,

For gray surface, ϵ = α = 0.598

In this case,

α + ρ = 1

Concept:

\(\frac{{{\delta _h}}}{{{\delta _t}}} = \;P{r^{1/3}}\)

Here,

δ_h = Hydrodynamic boundary layer thickness, δ_t = Thermal boundary layer thickness

Calculation:

Given:

Pr = 0.7, x = 5 mm

\({\rm{Kinematic\;viscosity\;}}\left( \gamma \right) = 30 \times {10^{ - 6}}\frac{{{m^2}}}{s}\)

Velocity u() = 10 m/s

\({\delta _h} = \frac{{5x}}{{\sqrt {R{e_x}} }}\)

\(R{e_x} = \frac{{ux}}{\gamma } = \;\frac{{10 \times 5\times 10^-3}}{{30 \times {{10}^{ - 6}}}} = 1666.66\)

\({\delta _h} = \frac{{5x}}{{\sqrt {R{e_x}} }} = \;\frac{{5\left( {5 } \right)}}{{\sqrt {1666.66} }}\)

∴ δ_h = 0.6123 mm

Now,

\(\frac{{{\delta _h}}}{{{\delta _t}}} = \;P{r^{1/3}}\)

\(\frac{{0.6123 }}{{{\delta _t}}} = \;{\left( {0.7} \right)^{1/3}}\)

Concept:

For maximum heat dissipation, the thickness of insulation should be critical thickness.

The critical thickness for wire is given as, \(r_c=\frac{k}{h}\)

Calculation:

Given:

k = 0.1 W/mK, h = 100 W/m2K

\({r_c} = \frac{k}{h} = \frac{{0.1}}{{100}} = 0.001~m = 1~mm\)

Thickness of insulation = rc - r

⇒ 1 – 0.5 = 0.5 mm

Concept:

Heat exchanger effectiveness:

\(\epsilon= \frac{{{{\rm{Q}}_{{\rm{actual}}}}}}{{{{\rm{Q}}_{{\rm{max}}}}}}\)

\({{\rm{Q}}_{{\rm{actul}}}} = {{\rm{\dot m}}_{\rm{h}}}{{\rm{c}}_{{\rm{ph}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{h}}2}}} \right) = {{\rm{\dot m}}_{\rm{c}}}{{\rm{c}}_{{\rm{pc}}}}\left( {{{\rm{T}}_{{\rm{c}}2}} - {{\rm{T}}_{{\rm{c}}1}}} \right)\)

\({{\rm{Q}}_{{\rm{max}}}} = {{\rm{C}}_{{\rm{min}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{h}}2}}} \right)\)

\(\epsilon= \frac{{{{\rm{Q}}_{{\rm{actual}}}}}}{{{{\rm{Q}}_{{\rm{max}}}}}} = \frac{{{{\rm{C}}_{\rm{h}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{h}}2}}} \right)}}{{{{\rm{C}}_{{\rm{min}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{h}}2}}} \right)}} = \frac{{{{\rm{C}}_{\rm{c}}}\left( {{{\rm{T}}_{{\rm{c}}2}} - {{\rm{T}}_{{\rm{c}}1}}} \right)}}{{{{\rm{C}}_{{\rm{min}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{c}}1}}} \right)}}\)

\({{\rm{\dot m}}_{\rm{n}}}{{\rm{C}}_{{\rm{pn}}}} = {{\rm{\dot m}}_{\rm{c}}}{{\rm{C}}_{{\rm{pc}}}} \Rightarrow {{\rm{C}}_{\rm{h}}} = {{\rm{C}}_{\rm{c}}} = {{\rm{C}}_{{\rm{min}}}} = {{\rm{C}}_{{\rm{max}}}}\)

\({\rm{\eta }} = \frac{{{{\rm{C}}_{\rm{h}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{h}}2}}} \right)}}{{{{\rm{C}}_{{\rm{min}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{C}}1}}} \right)}} = \frac{{{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{h}}2}}}}{{{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{c}}1}}}}\)

\({\rm{\eta }} = \frac{{{{\rm{C}}_{\rm{h}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{h}}2}}} \right)}}{{{{\rm{C}}_{{\rm{min}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{C}}1}}} \right)}} = \frac{{{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{h}}2}}}}{{{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{c}}1}}}}\)

Calculation:

\(\epsilon =\frac{\left( {{T}_{ce}}-{{T}_{ci}} \right)}{\left( {{T}_{hi}}-{{T}_{ci}} \right)}\)

\(0.25=\frac{{{T}_{ce}}-40}{200-40}\)

T_ce - 40 = 40

Concept:

Heat loss through the convection is given by

Q_loss = h × A × (∆T)

Nusselt number \(Nu = \frac{{h{l_c}}}{k}\)

Where l­_c is the characteristic length, in case of heater it is equal to the diameter of the sphere

Calculation:

Given, Nu = 31.4, k = 0.026 W/mK, D = 12 mm ⇒ r = 6 × 10-3 m , Surface temperature T_s = 27°C and atmospheric temperature T_∞ = 77°C

\(Nu = \frac{{h{l_c}}}{k}\)

⇒ \(h = \frac{{Nu\times{k}}}{l_c}\)

\(h = \frac{{31.4 \times 0.026}}{{0.012}} = 68.03\;W/{m^2}K\;\)

Surface area of sphere A = 4 × π × r² = 4 × 3.14 × (6 × 10^-3)²

Q_loss = 68.03 × 4 × 3.14 × (6 × 10^-3)² × (77 – 27) = 1.54 J/s

Explanation:

Given:

ρ = 2500 kg/m³, C_p = 0.52 kJ/kg k, Q = 1000 kW/m³, a = 240 W

\({\rm{Volume}} = \frac{4}{3}\pi {r^3} = \frac{4}{3}\pi \times {\left( {0.2} \right)^3} = 0.03351\;{m^3}\)

Now,

Heat received + total heat generated = Rate of change of internal energy

\(Q + q \times Volume = \rho \;{C_p}V\frac{{\partial T}}{{\partial t}}\)

\(\frac{{\partial T}}{{\partial t}} = \frac{{Q + q \times Volume}}{{\rho {C_p}V}}\)

\(\frac{{\partial T}}{{\partial t}} = \frac{{240 + 100 \times {{10}^3} \times 0.03351}}{{2500 \times 520 \times 0.03351}}\)

Concept:

For a lumped parameter problem, the time-temperature dependence is given by:

\({\rm{ln}}\left[ {\frac{{{\rm{T}} - {T_\infty }}}{{{T_i} - {T_\infty }}}} \right] = - \frac{{hA}}{{mc}}t = - \frac{{hA}}{{\rho V{\rm{c}}}}t\)

\(\frac{{{\rm{T}} - {T_\infty }}}{{{T_i} - {T_\infty }}} = {e^{ - \frac{{hA}}{{\rho V{\rm{c}}}}t}} = {e^{ - \frac{1}{\tau }t}}\)

\(\frac{{T - {T_\infty }}}{{{T_0} - {T_\infty }}} = {e^{ - \frac{t}{\tau }}}\)

Where, T = final temperature, T_∞ = air temperature, T_i = initial temperature, t = time, τ = time constant = ρVc/hA

Calculation:

T_i = 25°C, T_∞ = 215°C, T = 165°C

ρ = 8750 kg/m³, c = 380 J/kg°C, k (thermocouple) = 28 W/m°C and h = 145 W/m²°C

For a sphere:

\(\frac{V}{A} = \frac{{\frac{4}{3}\pi {R^3}}}{{4\pi {R^2}}} = \frac{R}{3} = \frac{D}{6}\)

Time constant:

\(\tau = \frac{{\rho Vc}}{{hA}} = \frac{{\rho cD}}{{6h}} = \frac{{8750 \times 380 \times 0.0025}}{{6 \times 145}} = 9.554\)

Putting the values in the equation:

\(\frac{{T - {T_\infty }}}{{{T_0} - {T_\infty }}} = {e^{ - \frac{t}{\tau }}}\)

\(\frac{{165 - 215}}{{25 - 215}} = {e^{ - \frac{t}{{9.554}}}}\)

Concept:

Efficiency of fin is calculated as

\(\eta = \frac{{{Q_{fin}}}}{{{Q_{max}}}}\)

\({Q_{max}} = h{A_s}\left( {{T_b} - {T_a}} \right)\)

A_s = Surface area of fin = πDL, T_b = Base temperature, T_a = Ambient temperature

Effectiveness of fin is calculated as

\(\varepsilon = \;\frac{{{Q_{fin}}}}{{{Q_{no\;fin}}}}\)

\({Q_{no\;fin}} = h{A_c}\left( {{T_b} - {T_a}} \right)\)

\({A_c} = Cross - sectional\;area\;of\;fin = \;\frac{{\pi {D^2}}}{4}\)

Calculation:

Given:

D = 5 mm, L = 100 mm, K= 400 Wm^-1K^-1, h = 40 Wm^-2K^-1,

Now,

Perimeter (P) = πD = 0.0157 m

\({A_c} = \;\frac{{\pi {D^2}}}{4} = 1.9634\; \times {10^{ - 5}}{m^2}\)

\(m = \;\sqrt {\frac{{hP}}{{K{A_C}}}} = 8.944\;{m^{ - 1}}\)

mL = 0.894

Now,

\(\eta = \;\frac{{{Q_{fin}}}}{{{Q_{max}}}} = \frac{{\sqrt {hPk{A_c}} \left( {{T_b} - {T_a}} \right){\rm{tanh}}\left( {mL} \right)}}{{h{A_s}\left( {{T_b} - {T_a}} \right)}} = \;\frac{{{\rm{tanh}}\left( {mL} \right)}}{{mL}} = 79.93\% \)

\(\varepsilon = \;\frac{{{Q_{fin}}}}{{{Q_{no\;fin}}}} = \frac{{\sqrt {hPk{A_c}} \left( {{T_b} - {T_a}} \right)\tanh \left( {mL} \right)}}{{h{A_c}\left( {{T_b} - {T_a}} \right)}}\)

Concept:

Biot number = hL/K

If biot > 0.1

Then lumped system analysis is not applicable and we use Heisler charts.

For Heisler charts Bi = hR/K

Fourier number = αt/L²

\({\theta _0} = \;\frac{{T - {T_\infty }}}{{{T_i} - \;{T_\infty }}}\)

T_i, T_∞ = initial temperature and surrounding temperature respectively.

T= temperature at time‘t’

After finding out θ₀ we can calculate the temperature ‘T’

Calculation:

L = 1m, R = 0.075 m, α = 3.95 × 10^-6 m²/s, k = 14.9 W/mK

C = 477 J/kgK, ρ = 7900 kg/m³

T_i = 450°C, T_∞ = 150°C

h = 85 W/m²k, t = 1500 s

\(Bi = h\frac{{R/2}}{k} = 0.214 > 0.1\)

(Lumped analysis not applicable)  

For Heisler chart,

Bi = hR/k = 0.428

\(\frac{1}{{{B_i}}} = 2.337\)

\({F_0} = \frac{{\alpha t}}{{{L^2}}} = \frac{{\alpha t}}{{{R^2}}} = 1.053\)

∴ θ₀ = 0.49

\(0.49 = \frac{{T - {T_\infty }}}{{{T_i} - {T_\infty }}}\)

T = T_∞ + 0.49 (T_i – T_∞)

∴ T = 297°C

Concept:

Heat transfer due to radiation, \(Q = \sigma A\left( \epsilon {T_1^4 - \alpha T_2^4} \right)\)

Heat transfer due to convection, = \(hA\Delta T\) 

Calculation:

d = 0.3 m, T_s = 3.30 K = T₁, T_∞ = 298K = T₂, E = 0.89, α = 0.92

σ = 5.67 × 10^-8 W/m² k⁴

Q₁ = σA ( T₁⁴E - α T₂⁴ )

∴ Q₁ = 187.078 A

Q₂ = hA ΔT

Q₂ = 4.5 A (330 - 298)

∴ Q₂ = 144 A

\({\rm{\% \;of\;radiation\;}} = \frac{{{\theta _1}}}{{{\theta _1} + {\theta _2}}} \times 100\)

\({\rm{\% \;of\;radiation\;}} = \frac{{187.078}}{{331.078}} \times 100\)

∴ % of radiation = 56.5057     

Concept:

Just as the absence of a free stream velocity requires use of a mean velocity to describe an internal flow, the absence of a fixed free stream temperature necessitates using a mean temperature.

The mean (or bulk) temperature of a fluid at a given cross section is defined in terms of the thermal energy transported by the fluid it moves past the cross section.

The bulk mean temperature of internal flow is given by the expression:

\({T_M} = \frac{2}{{{U_M}{R^2}}}\mathop \smallint \nolimits_0^R T\left( r \right)U\left( r \right)r\;dr\)

Calculation:

Given:

As, U(r) = A, so the mean velocity U_M = A

\(T\left( r \right) - {T_W} = B\left[ {1 - {{\left( {\frac{r}{R}} \right)}^2}} \right]\)

\({T_M} = 2\frac{A}{{{\rm{A}}{R^2}}}\mathop \smallint \nolimits_0^R \left\{ {{T_W}\; + B\left[ {1 - {{\left( {\frac{r}{R}} \right)}^2}} \right]} \right\}r\;dr\)

\({T_M} = \frac{2}{{{R^2}}}\mathop \smallint \nolimits_0^R \left\{ {{T_W}r\; + B\left[ {r - \frac{1}{{{R^2}}}{r^3}} \right]} \right\}\;dr\)

\({T_M} = \frac{2}{{{R^2}}}\left[ {\left\{ {{T_W}\frac{{{r^2}}}{2}\; + B\left[ {\frac{{{r^2}}}{2} - \frac{1}{{{R^2}}}\frac{{{r^4}}}{4}\;} \right]} \right\}} \right]_0^R\)

\({T_M} = \frac{2}{{{R^2}}}\left\{ {{T_W}\frac{{{R^2}}}{2}\; + B\left[ {\frac{{{R^2}}}{2} - \frac{{{R^2}}}{4}\;} \right]} \right\} = \;\frac{2}{{{R^2}}}\left\{ {{T_W}\frac{{{R^2}}}{2}\; + B\left( {\frac{{{R^2}}}{4}} \right)} \right\} = {T_W} + \frac{B}{2}\)

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:

Rayleigh number (Ra) is given by,

\(Ra = Gr.\Pr = \frac{{g\beta \Delta T{L^3}Pr}}{{{\vartheta ^2}}}\;\)

Where, \(\beta = \frac{1}{{{T_f}}}\) and T_f = Mean temperature

Nusselt number, Nu = \(0.60{\left( {Ra} \right)^{0.25}}\) = hL/K

We will calculate ‘h’ from this

And then calculate the heat dissipation as, Q = 2hA∆T

Calculation:

R_a = Gr – Pr \( = g\frac{{\beta {\rm{\Delta }}T\;{L^3}}}{{{V^2}}}Pr\)

β = 1/T_f, and

∵  \({T_f} = \frac{{293 + 413}}{2} = 353\;K\; = {\rm{\;}}80^\circ {\rm{C}}\;\)

Now,

\(Ra = \frac{{9.81 \times \frac{1}{{353}} \times 120 \times {{\left( {0.35} \right)}^3}}}{{{{\left( {41.09 \times {{10}^{ - 6}}} \right)}^2}}} \times 0.723\)

Ra= 61227382.83

N_u = 0.60 (Ra) ^0.25 = 53.07 = hL/k

\(h = 53.07 \times \frac{{0.05}}{{0.35}}\)

∴ h = 7.58 W/m²K

Now,

Q = 2hA ΔT

\({\rm{m}} = \sqrt {\frac{{{\rm{hp}}}}{{{\rm{k}}{{\rm{A}}_{\rm{c}}}}}} \)

Concept:

Heat dissipation from an infinitely long fin:

\({{\rm{Q}}_{{\rm{long\;fin}}}} = - {\rm{k}}{{\rm{A}}_{\rm{c}}}{\left. {\frac{{{\rm{dT}}}}{{{\rm{dx}}}}} \right|_{{\rm{x}} = 0}} = \sqrt {{\rm{hpk}}{{\rm{A}}_{\rm{c}}}} {\rm{\;}}\left( {{{\rm{T}}_{\rm{b}}} - {{\rm{T}}_\infty }} \right)\)

\(\frac{{{\rm{T}} - {{\rm{T}}_\infty }}}{{{{\rm{T}}_{\rm{o}}} - {{\rm{T}}_\infty }}} = {{\rm{e}}^{ - {\rm{mx}}}}\)

Heat dissipation form a fin insulated at the flip negligible heat loss fin tip.

\(\frac{{\rm{\theta }}}{{{{\rm{\theta }}_{\rm{o}}}}} = \frac{{{\rm{T}} - {{\rm{T}}_\infty }}}{{{{\rm{T}}_{\rm{o}}} - {{\rm{T}}_\infty }}} = \frac{{\cos {\rm{h}}\left( {{\rm{m}}\left( {{\rm{l}} - {\rm{x}}} \right)} \right)}}{{\cos {\rm{h}}\left( {{\rm{ml}}} \right)}}\)

\({{\rm{Q}}_{{\rm{insulated\;fin}}}} = - {\rm{k}}{{\rm{A}}_{\rm{c}}}{\left. {\frac{{{\rm{dT}}}}{{{\rm{dx}}}}} \right|_{{\rm{x}} = 0}} = \sqrt {{\rm{hPk}}{{\rm{A}}_{\rm{c}}}} \left( {{{\rm{T}}_{\rm{o}}} - {{\rm{T}}_\infty }} \right)\tan {\rm{h}}\left( {{\rm{mL}}} \right)\)

Calculation:

A long rod applies that it is an infinity long fin

Given: D = 20 mm = 0.02 m, T_b = 110°C, T = 20°C

h = 5 W/m²K , k = 15 W/mK

P = πd = π (0.02) = 0.063 m

\(A = \frac{\pi }{4}{d^2} = \frac{\pi }{4}{\left( {0.02} \right)^2} = 3.14 \times {10^{ - 4}}{m^2}\)

\(Q = \sqrt {hPkA} \left( {{T_b} - {T_\infty }} \right)\)

\(Q = \sqrt {5 \times 0.063 \times 3.14 \times {{10}^{ - 4}} \times 15} \;\left( {110 - 20} \right)\)