Heat Transfer Test 2

Concept:

Lumped system analysis is determined by a parameter known as Biot number which is given as,

\(Bi = \frac{{hL}}{k}\)

Where, h = convective heat transfer coefficient

k = coefficient of thermal conductivity

L = characteristic length of the body equal to volume upon area.

Also,

Convective resistance = 1/hA

Conductive resistance = L/kA

If Biot number is less than 0.1 then we use lumped system analysis.

Biot number can also be written as,

\(\frac{{Internal\;conductive\;resistance}}{{External\;convective\;resitance\;}} = hL/K\) = Bi

Therefore, option C is correct.

Concept:

Fourier number is given by,

\(Fo = \frac{{\alpha t}}{{\;{L^2}}}\)

Where, α = Thermal diffusivity, t = time

\({\rm{\;L}} = {\rm{\;characteristic\;length}} = \frac{{Volume}}{{Area}}\)

Calculation:

Substituting the given values,

We get,

L = a/6 = 0.05/6

t = 20 s

\(Fo = {\rm{\;}}42.2\; \times {10^{ - 6}} \times \frac{{20 \times {6^2}}}{{{{0.05}^2}}}\)

∴ F_o = 12.15 

Concept:

In lumped system analysis the temperature distribution is given by,

\(\frac{{T - {T_\infty }}}{{{T_i} - \;{T_\infty }}} = {e^{ - \frac{{hAt}}{{\rho Vc}}}}\;\)             ---(1)

Where, \(T =\) temperature at time‘t’

\(\frac{{hA}}{{\rho Vc}} = \frac{1}{{Time\;constant}}\)

\({T_i},\;{T_\infty }\) = initial temperature and surrounding temperature respectively.

Calculation:

Comparing eq.1 with given relation we get,

\(\frac{{T - \;\alpha }}{\beta } = {e^{ - \gamma \tau }}\)

\(\gamma = \frac{{hA}}{{\rho Vc}}\)

\({\rm{Time\;constant}} = \frac{{\rho Vc}}{{hA}}\)

\(\therefore {\rm{Time\;constant}} = \frac{1}{\gamma }\)

Concept:

The time required to reach temperature difference between body and surrounding to 36.7% of initial temperature difference is called as thermal time constant.

Thermal time constant \(\left( {{\tau ^*}} \right)\) is given as,

\({\tau ^*} = \frac{{\rho C{L_c}}}{h}\)

Calculation:

For sphere,

\({L_c} = \frac{{Volume}}{{surface\;area}} = \frac{R}{3}\)

\({L_c} = \frac{{30}}{3} = 10\;mm\)

\({\tau ^*} = \frac{{7800\left( {\frac{{Kg}}{{{m^3}}}} \right) \times 600\left( {\frac{J}{{KgK}}} \right) \times 0.01\left( m \right)}}{{20\left( {\frac{W}{{{m^2}K}}} \right)}}\)

\({\tau ^*} = 2340\;sec\)

• Also we can find \({\tau ^*}\) using,

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

\(0.367 = {e^{\frac{{ - t}}{{{\tau ^*}}}}}\)

\(\ln \left( {0.367} \right) = \frac{{ - t}}{{{\tau ^*}}}\)

\(- 1 = \;\frac{{ - t}}{{{\tau ^*}}}\)

\({\tau ^*} = t = \frac{{\rho C{L_c}}}{h}\)

∴ τ^* = 2340 sec

Concept:

Lumped system analysis is applicable if Bi ≤ 0.1

Biot Number, \({B_i} = \frac{{h{L_c}}}{k}\)

Characteristic length:

\({L_c} = \frac{V}{{{A_s}}}\)

Calculation:

Given: D = 1 mm, k = 35 W/m°C, ρ = 8500 kg/m³, C_p = 320 J/kg°C, h = 210 W/m²°C

For sphere:

\({L_c} = \frac{V}{{{A_c}}} = \frac{{\frac{1}{6}\pi {D^3}}}{{\pi {D^2}}} = \frac{D}{6} = \frac{1}{6}mm = \frac{{0.001}}{6}m\)

L_c = 1.67 × 10^-4m

\({B_i} = \frac{{h{L_c}}}{K} = \frac{{210 \times 1.67 \times {{10}^{ - 4}}}}{{35}} = 0.001 < 0.1\)

In order to read 99% of the initial temperature difference (T_i - T_∞):

\(\frac{{T\left( t \right) - {T_\infty }}}{{{T_i} - {T_\infty }}} = 0.01\)

\(\frac{{T - {T_\infty }}}{{{T_i} - {T_\infty }}} = {e^{ - \frac{{hA}}{{\rho CV}}t}}\)

\(\ln \left( {\frac{{T - {T_\infty }}}{{{T_i} - {T_\infty }}}} \right) = - \frac{{hA}}{{\rho CV}}t = \frac{h}{{\rho C{L_c}}}t\)

\(\ln \left( {0.01} \right) = - \frac{{210}}{{8500 \times 320 \times 1.67 \times {{10}^{ - 4}}}}t\)

-4.6051 = -0.46231 t

t = 9.96 s ≈ 10 s

Therefore, we must wait for at least 10 S for the temperature of the thermocouple junction to approach within 1 percent of the initial gas temperature difference.

Hint:

For example, when T_i = 0°C and T_∞ = 100°C, a thermocouple is considered to have read 99 percent of this applied temperature difference when its reading indicates T(t) = 99°C

or T(t) = Ti + 0.99(T∞ - Ti)

T(t) = Ti - 0.99(Ti - T∞) = 0.01 Ti + 0.99 T∞

T(t) = 0.01 Ti - 0.01 T∞ + T∞ = 0.01 (Ti - T∞) + T∞

T(t) - T∞ = 0.01 (Ti - T∞)

Concept:

Biot number = hL/k

If Biot number > 0.1

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

We find θ₀ corresponding to 1/Bi and Fourier number from Heisler chart.

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

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

T= temperature at time‘t’

Fourier number = αt/L²

α = thermal diffusivity, L = characteristic length, t = time (in seconds)

Calculation:

2l = 10 cm ⇒ L = 0.05 m

α = 1.2 × 10^-5 m²/s

k = 43 W/mK, T_i = 250°C, T_∞ = 45°C

h = 700 W/m²K, T = 100°C

\(Bi = \frac{{hL}}{k} = 0.814 > 0.1\)

(Lumped analysis not applicable)

\({\theta _0} = \frac{{T - {T_\infty }}}{{{T_i} - {T_\infty }}} = 0.268\)

\(\frac{1}{{{B_i}}}\; = 1.2289\)

\({F_0} = \frac{{\alpha t}}{{{L^2}}} = 2.4\)

\(t = 2.4 \times \frac{{{L^2}}}{\alpha } = 500\;sec.\)

Concept:

Biot Number:

\({{B}_{i}}=\frac{h{{L}_{c}}}{K}\)

The characteristic length: \({{L}_{c}}=\frac{V}{{{A}_{s}}}\)

To determine the temperature T(t) of a body a time t:

\(\frac{T\left( t \right)-{{T}_{\infty }}}{{{T}_{i}}-{{T}_{\infty }}}={{e}^{-\left[ \frac{hA}{\rho V{{C}_{p}}} \right]}}\)

Calculation:

Given: D = 0.3 cm = 0.003 m, h = 60 W/m²K, k = 8.9 W/mK,

α = k/(ρC_p) = 0.016 m²/hr = 4.44 × 10^-6 m²/s

(T(t) – T_∞) = 0.5 (T_i – T_∞)

\({{L}_{c}}=\frac{V}{{{A}_{s}}}=\frac{\frac{\pi }{4}{{D}^{2}}L}{\pi DL}=\frac{D}{4}=7.5\times {{10}^{-4}}m\)

\(\frac{hA}{\rho V{{C}_{p}}}=\frac{h}{\rho {{C}_{p}}{{L}_{c}}}=\frac{h\alpha }{k{{L}_{c}}}=\frac{60\times 4.44\times {{10}^{-6}}}{8.9\times 7.5\times {{10}^{-4}}}=0.03991\)

(T(t) – T_∞) = 0.5 (T_i – T_∞)

\(\frac{T\left( t \right)-{{T}_{\infty }}}{{{T}_{i}}-{{T}_{\infty }}}={{e}^{-\left[ \frac{hA}{\rho V{{C}_{p}}} \right]t}}\)

0.5 = e^-0.03991t

ln (0.5) = -0.6931 = -0.03991t

∴ t = 17.37 sec

Concept:

Q = hA_s [T – T_∞]

Using Lumped analysis,

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

Calculation:

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

\({Q_{total}} = \mathop \smallint \limits_0^t {Q_t}\;dt\)

\({Q_{total}} = \;\mathop \smallint \limits_0^t h\;{A_s}\;\left( {{T_0} - {T_\infty }} \right){e^{ - \frac{t}{{{\tau ^*}}}}}\)

\({Q_{total}} = \left[ {h{A_s}\left( {{T_0} - {T_\infty }} \right)\frac{{{e^{ - \frac{t}{{{\tau ^*}}}}}}}{{\left( { - \frac{1}{{{\tau ^*}}}} \right)}}} \right]_0^t\;\)

As, \({\tau ^*} = \frac{{\rho c{L_c}}}{h} = \frac{{\rho CV}}{{h{A_s}}}\)

We get,

\({Q_{total}} = \; - mC\;\left( {{T_0} - {T_\infty }} \right)\left[ {{e^{ - \frac{t}{{{\tau ^*}}}}} - 1} \right]\)

Now,

\(\rho = \frac{{mass}}{{volume}}\)

\(\therefore mass = 7800 \times \frac{4}{3}\pi {\left( {0.005} \right)^3}\)

m = 4.084 × 10^-3 k

\({Q_{total}} = \; - 4.084 \times {10^{ - 3}} \times 600 \times \left( {800\; - 300} \right)\left[ {{e^{ - \frac{8}{{16}}}} - 1} \right]\)

Lumped parameter analysis:

Internal/conductive resistance is very little as compared to surface convective resistance and the temperature distribution is given by

\(\frac{{{T}_{i}}-{{T}_{\infty }}}{T-{{T}_{\infty }}~}=~{{e}^{\left( \frac{hA}{\rho V{{C}_{p}}} \right)t}}\)

\(\frac{hA}{\rho V{{C}_{p}}}t=\ln\left( \frac{{{T}_{i}}-{{T}_{\infty }}}{T-{{T}_{\infty }}~} \right)\)

where

• T_i = Initial temperature of beat at t = 0,
• T = Temperature of body at any instant ‘t’ sec
• T_∞ = Ambient fluid temperature
• h = heat transfer coefficient, ρ is the density of the metal bead

For sphere: \(\frac{V}{A} = \frac{{Volume\;of\;body}}{{surface\;area}} = \frac{{\frac{4}{3}\pi {r^3}}}{{4\pi {r^2}}} = \frac{r}{3}\) 

V/A = r/3

Calculation:

Given, C_p = 400 J/kgK, ρ = 8500 kg/m³, t = 4.35 sec, r = 0.5 mm

T_i = 100°C, T_∞ = 20°C, T = 28°C

Using equation (1) and putting the given values:

\(\ln\left( {\frac{{100 - 20}}{{28 - 20}}} \right) = \left( {\frac{{h \times 3 \times 1000}}{{0.5 \times 8500 \times 400}}} \right) \times 4.35\)

\(\Rightarrow h = \frac{{2.302 \times \;0.5 \times 8500 \times 400}}{{3000 \times 4.35}} = 299.87\;W/{m^2}K\;\)