Fins MCQ Quiz in বাংলা - Objective Question with Answer for Fins - বিনামূল্যে ডাউনলোড করুন [PDF]

Explanation:

Heat flow mainly depends on three factors

• area of the surface
• temperature difference and
• the convective heat transfer coefficient.

The rate of heat transfer can be increased by enhancing any one of these factors.

• Fins are thus used whenever the available surface area is found insufficient to transfer the required quantity of heat with the available temperature gradient and heat transfer coefficient.
• In the case of fins, the direction of heat transfer by convection is perpendicular to the direction of conduction heat flow.
• Fins are the extended surface protruding from a surface or body, and they are meant for increasing the heat transfer rate between the surface and the surrounding fluid by increasing the heat transfer area.

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

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

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

Explanation:-

When T_s and T_o are fixed, two ways to increase the rate of heat transfer are,

• To increase the convection heat transfer coefficient h.
• To increase the surface area A_s by attaching to the surface extended surfaces called fins made of highly conductive materials such as aluminum.

As it is clear from the diagram, the efficiency of the triangular fin and parabolical fin is greater than that of the straight fin because it provides more surface area near the base.

Fins with triangular and parabolic profiles contain less material and are more efficient than the ones with rectangular profiles.

Important Points

• The thermal conductivity k of the fin should be as high as possible. Use aluminum, copper, and iron.
• The ratio of the perimeter to the cross-sectional area of the fin p/Ac should be as high as possible. Use slender pin fins.  
• Low convection heat transfer coefficient h. Place fins on the gas (air) side.

Concept:

For a long fin,

\({{{Q}}_{{{fin}}}} = \left( {\sqrt {{{hpk}}{{{A}}_{{c}}}} {{\;}}} \right){{{\theta }}_0}\)

\({{{Q}}_{{{without\;fin}}}} = {{h}}{{{A}}_{{c}}}{{\rm{\theta }}_0}\)

Where,

A_c = Cross-section area of the fin, p = Perimeter of the fin, h = Convective heat transfer coefficient, k = Thermal conductivity of the fin material,

θ₀ = Temperature difference

The efectivness of the fin,

\(\Rightarrow ε= \frac{{{{{Q}}_{{{fin}}}}}}{{{{{Q}}_{{{without\;fin}}}}}}\)

\(\Rightarrow {{ε }} = \frac{{\left( {\sqrt {{{hPK}}{{{A}}_{{c}}}} {{\;}}} \right){{{\theta }}_0}}}{{{{h}}{{{A}}_{{c}}}{{{\theta }}_0}}}\)

\({\rm{ε }} = \sqrt {\frac{{PK}}{{h{A_c}}}} \;\;\;\;\; \ldots \left( 1 \right)\)

Calculation:

Given:

h₁ = 116 W/m²°C; k = 64 W/m²°C; ϵ₁ = 8.5, h₂ = 29 W/m²°C, d = 40 mm.

\({\rm{ε }} = \sqrt {\frac{{kP}}{{h{A_c}}}}\)

\({\rm{ε }} = \sqrt {\frac{{4\;\times\;k\;\times\;\pi d}}{{h\;\times\;{\pi\;\times\;d^2}}}}=\sqrt{\frac{4k}{hd}}\)

\({\rm{ε_1 }} =\sqrt{\frac{4\;\times\;64}{116\;\times\;0.04}}=7.427\)

Now keeping all the parameters the same, except heat transfer coefficient the new relation between effectiveness and heat transfer coefficient is:

\(\frac{ε_2}{ε_1}=\sqrt{\frac{h_1}{h_2}}\)

\(\frac{ε_2}{7.427}=\sqrt{\frac{116}{29}}=2\)

ε₂ = 14.855

Fins are the projections protruding from hot surface into the ambient fluid and they are meant for increasing the heat transfer by increasing the heat transfer area.

In order to find the temperature distribution along the fin surface, let us consider a small differential element of fin at a distance x from the root of the fin

Let qx is the heat conducted to the element then,

\({q_x} = \; - kA\frac{{dT}}{{dx}}\)

and qx+dx is the heat conducted out of the element then,

\({q_{x + dx}} = \;{q_x} + \left( {\frac{{\partial {q_x}}}{{\partial x}}} \right)dx\)

Heat convected from the surface of the element to the ambient fluid = \(h\; \times p \times dx\;\left( {T - {T_\infty }} \right)\)

Now from the conservation of energy

Heat conducted into the element = Heat conducted out of the element + Heat convected from the surface of the element to the fluid

\({q_x} = {q_{x + dx}} + hpdx\left( {T - {T_\infty }} \right)\)

\({q_x} = {q_x} + \left( {\frac{{\partial {q_x}}}{{\partial x}}} \right)dx + hpdx\left( {T - {T_\infty }} \right)\)

\(0 = \frac{\partial }{{\partial x}}\left( { - kA\frac{{dT}}{{dx}}} \right)dx + \;hpdx\left( {T - {T_\infty }} \right)\)

\(\frac{{{d^2}T}}{{d{x^2}}} - \frac{{hp}}{{kA}}\left( {T - {T_\infty }} \right) = 0\)

Let T – T∞ = θ = f(x)

\(\frac{{dT}}{{dx}} = \frac{{d\theta }}{{dx}}\)

\(\frac{{{d^2}T}}{{d{x^2}}} = \frac{{{d^2}\theta }}{{d{x^2}}}\;\) and Putting \({m^2} = \frac{{hP}}{{kA}}\) 

\(\frac{{{d^2}\theta }}{{d{x^2}}} - {m^2}\theta = 0\)

The solutuion for above differential equation is given by

\(\theta = {C_1}{e^{ - mx}} + {C_2}{e^{mx}}\)

C1 and C2 are the constants of integration and are determined by the boundary conditions

One boundary condition is at x = 0, T = T0  and θ = θ0 = T0 – T∞ the second boundary condition depends upon three different cases

Case (ii)

When the fin is finite in length and the tip is insulated

\(\frac{{T - {T_\infty }}}{{{T_0} - {T_\infty }}} = \frac{{\cosh m\left( {l - x} \right)}}{{\cosh ml}}\)

Case (ii)

Fin is infinitely long or very long fin

The temperature at the fin tip is equal to the ambient temperature i.e. at x = ∞ , T = T∞ and then θ = 0

then temperature distribution within the fin is given by

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

Case (iii)

When the fin is finite in length but losing heat through the tip

\(\frac{{T - {T_\infty }}}{{{T_0} - {T_\infty }}} = \frac{{\cosh m\left( {{l_c} - x} \right)}}{{\cosh m{l_c}}}\)

where lc is the corrected length of in

\({l_c} = l + \frac{t}{2}\;\)for rectangular fiin

\({l_c} = l + \frac{d}{4}\) for pin fin

Concept:

Heat loss in infinite long fin is,

\(Q = \sqrt {hpkA} {\theta _0}\)

Calculation:

\(Q = \sqrt {20 \times \pi \times 30 \times {{10}^{ - 3}} \times 150 \times \frac{\pi }{4} \times {{0.03}^2}} \times \left( {180-24} \right)\)

Q = π × 0.14 × 156

∴ Q = 69.77 W

If the fin is finite in length and the tip is insulated, it means it is not losing the heat through the tip,

Heat conducted into the tip = 0

\(- \;kA\;{\left( {\frac{{dT}}{{dx}}} \right)_{at\;x = L}} = 0\)

\({\left( {\frac{{dT}}{{dx}}} \right)_{at\;x = L}} = 0\) , Boundary condition

The temperature distribution within the fin is given by

\(\frac{{T - {T_\infty }}}{{{T_o} - {T_\infty }}} = \frac{{\cosh m\left( {L - x} \right)}}{{\cosh mL}}\)

Heat transfer through fin = \(\sqrt {hpkA} \left( {{T_o} - {T_\infty }} \right)\tanh mL\) Watts

Where, h is the outside fluid heat transfer coefficient, p is the perimeter of the cross-section of the fin, A is the cross-sectional area of the fin and k is the thermal conductivity of the fin, L is the length of the fin and m is the slope of differential equation, T_o is the base temperature and T_∞ is the ambient temperature

Calculation:

We have been given with m i.e. slope of the differential equation and length of the fin

Therefore for other parameters to be constant we can see from the equation of heat transfer for this fin that

\(Q \propto \tanh mL\)

∴ We will check the value of tan h ⁡mL for tall the given condition and the condition which is having higher value will have the higher heat transfer.

First option: m = 0.75, L = 3, tan h⁡ mL = 0.9780

Second option: m = 1, L = 3, tan h⁡ mL = 0.9950

Third option: m = 3, L = 0.72, tan h ⁡mL = 0.9737

Fourth option: m = 2, L = 1.2, tan h⁡ mL = 0.9836

Concept:

For long fins, the rate of heat loss from fin is given by:

\(Q=\sqrt{hpkA}\;θ_o\)

Where k = thermal conductivity, p = perimeter of the fin, h = heat transfer coefficient, A = cross-sectional area,  and θ₀ is the temperature difference.

\(Q\propto\sqrt{k}\;\) if other parameter remains constant.

Calculation:

Given:

Let the amount of heat loss be Q̇₁ when thermal conductivity is (k₁) = k

Let the amount of heat loss be Q̇₂ when thermal conductivity is (k₂) = 2k

\( \frac{{{Q_{2}}}}{{{Q_{1}}}} = \sqrt {\frac{{k_2{}}}{{k_1}}}\)

\( \frac{{{Q_{2}}}}{{{Q_{1}}}} = \sqrt {\frac{{2k{}}}{{k}}}\)

\(\frac{{{Q_{2}}}}{{{Q_{1}}}} = \sqrt {{{2{}}}{{}}}\)

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:

Fin efficiency:

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 the entire fin area were at base temperature, for an infinitely long fin.

\(\eta_{fin}=\frac{(\dot Q_{fin})_{actual}}{(\dot Q_{fin})_{maximum}}\)

where \((̇ Q_{fin})_{maximum}\) = hA_fin(T₀ - T_∞);

where A_fin = surface area of the fin = Perimeter × Length.

Case I:

Insulated tip:

\((\dot Q_{fin})_{actual}=\sqrt{hPkA_c}(T_o-T_{\infty})\tan h(mL)\)

\(\eta_{fin}=\frac{(\dot Q_{fin})_{actual}}{(\dot Q_{fin})_{maximum}}\)

\(\eta_{fin}=\frac{\sqrt{hPkA_c}(T_o-T_{\infty})\tan h(mL)}{h(pL)(T_o-T_{\infty})}=\frac{\tanh(mL)}{mL}\)

where \(m=\sqrt \frac{hP}{kA_c}\)

Case II:

Infinitely long fin:

\((\dot Q_{fin})_{actual}=\sqrt{hPkA_c}(T_o-T_{\infty})\)

\(\eta_{fin}=\frac{(\dot Q_{fin})_{actual}}{(\dot Q_{fin})_{maximum}}\)

\(\eta_{fin}=\frac{\sqrt{hPkA_c}(T_o-T_{\infty})}{h(pL)(T_o-T_{\infty})}=\frac{1}{mL}\)

where \(m=\sqrt \frac{hP}{kA_c}\)