Heat Transfer Test 1

Concept:

In order to provide effective insulation, the radius of insulation must be greater than the critical radius of insulation as at a critical radius of insulation maximum heat loss takes place.

The critical radius of insulation for a cylindrical pipe \({r_c} = \frac{k}{h}\)

Where k is the thermal conductivity of the insulating material and h is the heat transfer coefficient of the surrounding medium.

Critical radius for sphere is given by \({r_c} = \frac{{2k}}{h}\)

Calculation:

Given, k = 0.12 W/mK, h = 4 W/m²K,

Critical radius = \({r_c} = \frac{k}{h} = \frac{{0.12}}{4} = 0.03\;m = 30\;mm\)

Concept:

Shape Factor is defined as the fraction of radiation energy leaving a surface that reaches another surface, for example, F12 means the fraction of energy leaving the surface 1 reaches surface 2.

Summation Rule: F11 + F12 = 1

Reciprocity theorem: A1 F12 = A2 F21 (A is the surface area)

Area of cube of side a = 6a²

Area of circle of diameter d = (π/4)d²

Calculation:

Since the sphere is completely enclosed by the cube therefore,

F₁₂ = 1

By Reciprocal theorem,

A₁ F₁₂ = A₂ F₂₁

\(\begin{array}{l} 4\;π {\left( {\frac{d}{2}} \right)^2} \times 1 = 6 {d^2} \times{F_{21}}\\ {F_{21}} = \frac{π }{6} \end{array}\)

Concept:

3D, steady, state heat equation without heat generation

\(\frac{{{\partial ^2}T}}{{\partial {x^2}}} + \frac{{{\partial ^2}T}}{{\partial {y^2}}} + \frac{{{\partial ^2}T}}{{\partial {z^2}}} = 0\)

\({\nabla ^2}T = 0\) ⇒ Laplace equation

Points to remember

\({\nabla ^2}T + \frac{q}{k} = 0 \Rightarrow\) Poisson equation

\({{\rm{\Delta }}^2}T = \frac{1}{\alpha }\frac{{\partial T}}{{\partial t}}\) ⇒ Fourier equation

Concept:

For a fin, effectiveness \(\epsilon =\sqrt{\frac{kP}{hA}}\)

\(\varepsilon \propto \;\frac{1}{{\sqrt h }}\; \Rightarrow \;\frac{{{\varepsilon _1}}}{{{\varepsilon _2}}} = \sqrt {\frac{{{h_2}}}{{{h_1}}}} \)

Calculation:

h₁ = 25 W/m²K, h₂ = 10 W/m²K

Concept:

\(\left( {St} \right) \cdot {\left( {Pr} \right)^{\frac{2}{3}}} = \frac{{{C_{{f_x}}}}}{2}\)

Calculation:

\({C_{{f_x}}} = 2 \cdot \left( {St} \right) \cdot {\left( {Pr} \right)^{2/3}}\)

\({C_{{f_x}}} = 2\left( {5.585 \times {{10}^{ - 3}}} \right){\left( {0.7} \right)^{2/3}}\)

Concept:

Negligible temperature gradient in bodies refer and unsteady state conduction refer to transient analysis or lumped system analysis.

As we know for such analysis the temperature distribution is

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

Where, h = convective heat transfer coefficient, k = coefficient of thermal conductivity

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

A = area of face through which heat transfer occurs, t = time in seconds

C = specific heat (J/kgK), ρ = density

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

T = temperature at time ‘t’

∴ The correct answer is exponential.

Concept:

Emissive power of a black body (ϵ = 1):

E_b = σAT⁴ W

Emissive power of a real surface (ϵ < 1):

E = ϵσAT⁴ W

Where σ = Stefan Boltzmann constant

σ = 5.67 × 10^-8 W/ m² K⁴

Calculation:

Given: ϵ = 0.8, T_b = 1075 K, T_g = ?

Since the black and grey body are emitting the same amount of radiation:

E_b = E_g

σT_b⁴ = ϵσT_g⁴

\({\left( {\frac{{{T_g}}}{{{T_b}}}} \right)^4} = \frac{1}{\epsilon} = \frac{1}{{0.8}} = 1.25\)

\(\frac{{{T_g}}}{{{T_b}}} = {\left( {1.25} \right)^{\frac{1}{4}}} = 1.0574\)

Concept:

• Fourier number is defined as the ratio of heat conducted to heat stored in the system.
• Also, it is defined as the ratio of operating time to diffusion time.
• \(Fourier\;No = \frac{{Operating\;time}}{{diffusion\;time}}\)

\({F_o} = \frac{t}{{\left( {\frac{{L_c^2}}{\alpha }} \right)}}\)

Where,

T = operating time, L_c = Characteristic length, α = thermal diffusivity \(\left( {\frac{{{m^2}}}{{sec}}} \right)\)

Concept:

Making the resistance circuit

\(\begin{array}{l} {R_{th}} = \frac{{{L_1}}}{{{k_1}{A_1}}} + \frac{{{L_2}}}{{{k_2}{A_2}}} + \frac{{{L_3}}}{{{k_3}{A_3}}}\\ Q = \frac{{{\rm{\Delta }}T}}{{{R_{th}}}}\\ \Rightarrow Q = \frac{{{t_1} - {t_2}}}{{\frac{{{L_1}}}{{{k_1}{A_1}}} + \frac{{{L_2}}}{{{k_2}{A_2}}} + \frac{{{L_3}}}{{{k_3}{A_3}}}}} \end{array}\)

Which one has minimum thermal conductivity that will give maximum temperature drop.

Concept:

Effectiveness for parallel flow heat exchanger,

\(e = \;\frac{{1 - NTU\left( {1\; + \;C} \right)}}{{1 + C}}\)

\(NTU = \frac{{UA}}{{{C_{min}}}}\)

Calculation:

C_r = 0.420 = C_min/C_max

U = 800 W/m²K

ε = 0.52

C_min = 800 W/K

\(\varepsilon = \frac{{1 - {e^{ - NTU\left( {1 + C} \right)}}}}{{1 + C}} = 0.52\)

⇒ 1 – e^{-NTU (1.42)} = 1.42 × 0.52

\({\rm{NTU\;}} = {\rm{\;}}0.94{\rm{\;}} = \frac{{UA}}{{{C_{min}}}}\)

Concept:

The heat transfer from the base of the fin is expressed as :

\(Q = - k{\left( {\frac{{dT}}{{dx}}} \right)_{x = 0}}\)

Calculation:

T = 6x² - 5x + 3

\(\frac{{dT}}{{dx}} = 12x - 5\)

\({\left( {\frac{{dT}}{{dx}}} \right)_{x = 0}} = - 5\)

Concept:

First we will calculate the Reynolds number to determine the type of flow

\(Re = \frac{{\rho vD}}{\mu }\)

If the flow is turbulent (which will come in this case) we use the following

\({\rm{Nusselt\;number\;}}\left( {Nu} \right) = 0.023{\left( {Re} \right)^{0.8}}{\left( {Pr} \right)^{0.4}}\)

Nu = h_id/k

Overall heat transfer coefficient,

\(\frac{1}{U} = \frac{1}{{{h_i}}} + \frac{1}{{{h_o}}}\)

 Calculation:

ρ = 970 kg/m³, D = 20 mm, V = 2m/s, h_o = 12000 W/m²k

μ = 4 × 10^-4 kg/m-s, Pr = 2.54

\(Re = \frac{{\rho VD}}{\mu } = \frac{{970 \times 2 \times 0.02}}{{4 \times {{10}^{ - 4}}}} = 97000\)

Re > 2300 ⇒ Turbulent flow

Using, Nu = 0.023 (Re)^0.8 (Pr)^0.4

\(\frac{{{h_i}d}}{k} = 0.023\;{\left( {97000} \right)^{0.5}}{\left( {2.54} \right)^{0.4}}\)

\({h_i} = 325.895 \times \frac{{0.345}}{{0.02}}\)

\({h_i} = 5621.70\;W/{m^2}k\)

\({U^{ - 1}} = \frac{1}{{{h_i}}} + \frac{1}{{{h_o}}} = 2.612 \times {10^{ - 4}}\)

Concept:

For an adiabatic/insulated tip condition, the heat transfer from the fin is given as:

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

\(m = \sqrt {\frac{{hP}}{{kA}}} \)

Calculation:

L = 40 cm = 0.4 m, D = 0.002 m, k = 140 W/mK, T_b = 50°C, T_∞ = 25°C, h = 1000 W/m²K

\(\frac{P}{A} = \frac{{{\rm{\pi D}}}}{{\frac{{{\rm{\pi }}{{\rm{D}}^2}}}{4}}} = \frac{4}{D}\)

Fin parameter:

\(m = \sqrt {\frac{{hP}}{{kA}}} = \;\sqrt {\frac{{4h}}{{kD}}} = \sqrt {\frac{{4 \times 1000}}{{140 \times 0.002}}} = \;119.5\;{m^{ - 1}}\)

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

\(Q = \sqrt {1000 \times \pi \times 0.002 \times 140 \times \frac{{\pi \times {{\left( {0.002} \right)}^2}}}{4}} \left( {50 - 25} \right)\tanh \left( {119.5 \times 0.4} \right)\)

Concept:

Average heat transfer coefficient is given as

\(\bar h = \frac{1}{L}\mathop \smallint \limits_0^L {h_x}dx\)

Calculation:

Average heat transfer coefficient

\(\bar h = \frac{1}{L}\;\mathop \smallint \limits_0^L A{x^{ - 0.2}}dx\)

\(\bar h = \frac{1}{L}\left[ {\frac{{A{x^{0.8}}}}{{0.8}}} \right]_0^L\)

h̅ = 1.25 AL^-0.2

Now,

Local heat transfer coefficient

h_X = Ax^-0.2

at x = L,

h_x=L = AL^-0.2

\(\frac{{\bar h}}{{{h_{x = L}}}} = \frac{{1.25\;A{L^{ - 0.2}}}}{{A{L^{ - 0.2}}}}\)

Concept:

Net radiation heat exchange between two infinitely long concentric cylindrical surfaces:

\({Q_{1 - 2}}=\frac{{\sigma \left( {T_1^4 - T_2^4} \right){A_1}}}{{\frac{1}{{{\epsilon_1}}} + \frac{{{A_1}}}{{{A_2}}}\left( {\frac{1}{{{\epsilon_2}}} - 1} \right)}}W\)

When a small body is kept in a large enclosure:

\({A_1} \ll {A_2} \Rightarrow \frac{{{A_1}}}{{{A_2}}} \approx 0\)

Q_{1 - 2} = σ ϵ₁ A₁ (T₁⁴ - T₂⁴)

For steady state condition of the filament:

Q = σ ϵ_f A_f (T_f⁴ - T_∞⁴)

Calculation:

Given:

Q = 100 W, T = 3000° C = 3273 K, L = 250 mm = 0.25 m

A_f = π DL = 025 π D mm²

Now,

Q = σ ϵ_f A_f T_f⁴

100 = 5.67 × 10^-8 × 1 × 0.25 π D × (3273)⁴ = 5110.4 × 10³D

D = 0.02 × 10^-3 m

∴ D = 0.02 mm

Concept:

For condenser [C = 0]

Effectiveness (ϵ) = 1 – e^-NTU

T_h1 = 110°C, T_c1 = 20°C, T_c2 = 75°C

Also,

\(ϵ = \frac{{{T_{c2}} - {T_{c1}}}}{{{T_{h1}} - {T_{c1}}}} = \frac{{75 - 20}}{{110 - 20}} = 0.61111\)

∴  We have,

0.61111 = 1 – e^-NTU

∴ NTU₁ = 0.9445

\(NT{U_1} = \frac{{UA}}{{{C_{min}}}} = 0.9445\)

By doubling flow rate,

\(NT{U_2} = \frac{{UA}}{{2{C_{min}}}} = \frac{{0.9445}}{2} = 0.47225\)

\(\therefore ϵ = 1 - {e^{ - NT{U_2}}} = 0.3764\)

\(\therefore ϵ = 0.3764 = \frac{{{T_{c2}} - {T_{c1}}}}{{{T_{h1}} - {T_{c1}}}}\)

\(0.3764 = \frac{{75 - 20}}{{{T_{h1}} - 20}}\)

Concept:

Radiosity = leaving energy from surface.

Radiosity = ρ × G + ϵσ Ts4

Calculation:

Given,

ϵ = 0.8, G = 240 W/m², T_s = 0°C = 273 K

For opaque body, τ = 0

α + ρ = 1

using kirchoff’s law,

ϵ = α

∴ ρ = (1 - ϵ)

Radiosity = leaving energy from surface.

Radiosity = ρ × 240 + ϵσ T_s⁴

Radiosity = (1-E) × 240 + 0.8 × 5.67 × 10^-8 × (273)⁴

Radiosity = 0.2 × 240 + 251.95

∴ Radiosity = 299.95 W/m²

Concept:

The relation between hydrodynamic boundary layer thickness (δ) and thermal boundary layer thickness (δ_T) is given as

\(\frac{\delta }{{{\delta _T}\;}} = {\left( {Pr} \right)^{1/3}}\)

The relation between Nusselt number and Prandtl number for laminar flow over a flat plate is

Nu = F (Re, Pr)

For constant temperature, boundary condition

Nu = 0.332 (Re)^1/2 (Pr)^1/3

For constant heat flux boundary condition

Nu = 0.453 (Re)^1/2 (Pr)^1/3

Calculation:

\(\frac{{{{\left( {Nu} \right)}_A}}}{{{{\left( {Nu} \right)}_B}}} = \frac{{\left( {Re} \right)_A^{1/2}\;\left( {Pr} \right)_A^{1/3\;}}}{{\left( {Re} \right)_B^{1/2}\left( {Pr} \right)_B^{1/2}}}\)

\(\frac{{{{\left( {Nu} \right)}_A}}}{{{{\left( {Nu} \right)}_B}}} = \frac{{\left( {Pr} \right)_A^{1/3}}}{{\left( {Pr} \right)_B^{1/3}}}\)

\({\left( {Nu} \right)_B} = \frac{{\left( {Pr} \right)_B^{1/3}{{\left( {Nu} \right)}_A}}}{{\left( {Pr} \right)_A^{1/3}}}\)

\({\left( {Nu} \right)_B} = \frac{{\left( {Pr} \right)_B^{1/3}{{\left( {Nu} \right)}_A}}}{{\left( {Pr} \right)_A^{1/3}}}\)

Now,

\({\left( {Pr} \right)_A} = \left( {\frac{\delta }{{\delta T}}} \right)_A^3 = {\left( {\frac{1}{2}} \right)^3} = \frac{1}{8}\)

\({\left( {Pr} \right)_B} = \left( {\frac{\delta }{{{\delta _T}}}} \right)_B^3 = {\left( 2 \right)^3}\)

\(\therefore {\left( {Pr} \right)_B} = 8\)

∴ (Pr)_B = 8

\({\left( {Nu} \right)_B} = \frac{{{{\left( 8 \right)}^{\frac{1}{3}}}}}{{{{\left( {\frac{1}{8}} \right)}^{\frac{1}{3}}}}} \times 35\)

Ans. (a) Temperature slope is higher for low conducting and lower for high conducting fluid. Thus A is for 1, B for 2. Temperature profile in C is for insulator. Temperature rise is possible only for heater and as such D is for guard heater.

Concept:

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

\({R_a} = Gr \cdot Pr\)

\(Gr = \frac{{g\beta \left( {{\rm{\Delta }}T} \right)L_c^3}}{{{\nu ^2}}}\)

\(\nu = \frac{\mu }{s}\)

Calculation:

Given: Pr = 0.7, ρ=1.21, ν = 15.52 × 10-6 m2/s, k= 0.02435 W/mK, Nu = 0.15 (Ra)1/3

Twall = 70˚C = 343 K, T_∞ = 25°C = 298 K and diameter = 70 mm 

Now,

Rayleigh number (R_a) = Gr⋅Pr

\(Gr = \frac{{g\beta \left( {{\rm{\Delta }}T} \right)L_c^3}}{{{\nu ^2}}}\)

\(\beta = \frac{1}{{\left( {\frac{{{T_s} + {T_∞ }}}{2}} \right)}} = \frac{1}{{\left( {\frac{{343 + 298}}{2}} \right)}} = 0.0031\)

ΔT = (T_s – T_∞) = (343 - 298) = 45

\({L_c} = \frac{{Area}}{{Perimeter}} = \frac{{\frac{\pi}{4}{d^2}}}{{\pi d}} = \frac{d}{4}\)

\({L_c} = \frac{{70\left( {mm} \right)}}{4}\)

∴ L_c = 0.0175 m

\(Gr = \frac{{9.81 \times \left( {0.021} \right) \times 45 \times {{\left( {0.0175} \right)}^3}}}{{{{\left( {15.52 \times {{10}^{ - 6}}} \right)}^2}}}\)

Gr = 30449.06

Now,

R_a = Gr.Pr

R_a = 30449.06 × 0.7 = 21314.34

Nu = 0.15 (R_a)^1/3

∴ Nu = 4.15

\(\frac{{h{L_c}}}{{{k_{fluid}}}} = 4.15\)

\(h = \frac{{7.8693 \times 0.02435}}{{0.0175}}\)