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

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:

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

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

Concept

This is a case of an unsteady-state heat transfer problem.

In order to obtain the temperature after t time the formula is given by

\(\frac{{T - {T_∞ }}}{{{T_i} - {T_{∞ \;}}}} = {\rm{exp}}\left( { - \frac{{hA}}{{ρ VC}} \times τ } \right)\)

where T = temperature after time t, T_∞ = temperature of bath or environment, T_i = initial temperature, τ is the time,

h is heat transfer coefficient, A is the area of cross-section, ρ is the density of ball, V is the volume of ball, C is the heat capacity.

Calculation:

Given: 

T_∞ = 30°C, Ti = 530°C , T = 430°C ,τ = 10 sec .

\(\frac{{T - {T_∞ }}}{{{T_i} - {T_{∞ \;}}}} = {\rm{exp}}\left( { - \frac{{hA}}{{ρVC}} \times τ } \right)\)

putting the value,

\(\frac{{430 - 30}}{{530 - 30}} = {\rm{exp}}\left( { - \frac{{hA}}{{ρVC}} \times 10} \right)\)

\(\frac{{hA}}{{ρVC}}\) = 0.022314

Now temperature of the ball after the next 10 sec means the total time is 20 sec (τ = 20 sec)

Note: Time is always taken from initial.

\(\frac{{T - {T_∞ }}}{{{T_i} - {T_{∞ \;}}}} = {\rm{exp}}\left( { - \frac{{hA}}{{ρVC}} \times τ } \right)\)

\(\frac{{T - 30}}{{530 - 30}} = {\rm{exp}}\left( { - \frac{{hA}}{{ρVC}} \times 20} \right)\)

put the value of \(\frac{{hA}}{{ρVC}}\) from the above 

\(\frac{{T - 30}}{{530 - 30}} = {\rm{exp}}\left( { - 0.022314 \times 20} \right)\)

T - 30 = 320

T = 350°C.

Concept: 

Biot-number:

Biot number provides a way to compare the conduction resistance within a solid body to the convection resistance external to that body (offered by the surrounding fluid) for heat transfer:

\(Bi = \frac{{conduction\ resisitance}}{{Convection\ resisitance}} = \frac{{\left( {\frac{L_c}{{kA}}} \right)}}{{\left( {\frac{1}{{hA}}} \right)}} = \frac{{hL_c}}{k}\)

Here k = conductivity of the solid body, Lc = characteristic length of the solid body = \(\frac{V}{A_s}\)

Explanation:

The lumped heat capacity analysis is such analysis in which temperature is only function of time and not a function of space.

T ≠ f (space)

T = f (Time) only

For lumped parameter analysis

Biot Number < 0.1

\({\rm{Biot\;No}} = \frac{{{\rm{Internal\;Conductive\;Resistance\;}}\left( {{\rm{ICR}}} \right)}}{{{\rm{External\;Convective\;Resistance\;}}\left( {{\rm{ECR}}} \right)}}\)

Concept: 

• The Heisler charts are extensively used to determine the temperature distribution and heat flow rate when both conduction and convection resistances are almost of equal importance and Biot no is 1.
• Biot number is the ratio of internal conductive resistance to external convective resistance. It is used to determine whether lumped heat analysis can be applied for the given condition or not.
• For lumped heat analysis ideally Biot number should be equal to 0. But we can apply this analysis when Biot number is less than 0.1 also.
• When Biot number is greater than 0.1, the Heisler chart is used to determine the temperature variations.
• Therefore, the best suitable option is 4

Concept:

The temperature equation with time in a lumped system is given by:

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

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

Rate of heat transfer at any time t can be obtained from:

\(\dot{Q}=-mc\frac{dT}{dt}\)

\(\dot{Q}=-mc\frac{d}{dt}\left[T_{\infty}+ ({T_i} - {T_∞ })e^{- \frac{{hA}}{{\rho CV}}t} \right]\)

\(\dot{Q}=hA({T_i} - {T_∞ })e^{- \frac{{hA}}{{\rho CV}}t}\)

We know that \(\frac{hAt}{\rho c V}=Bi Fo\)

Calculation:

Given:

(Bi Fo)₁ = 1, (Bi Fo)₂ = 2

\(\dot{Q}=hA({T_i} - {T_∞ })e^{- BiFo}\)

\(\dot{Q}\propto e^{- (BiFo)}\)

\(\frac{\dot{Q}_1}{\dot{Q}_2}= \frac{e^{-{(BiFo)}_1}}{e^{-{(BiFo)}_2}}\)

\(\frac{\dot{Q}_1}{\dot{Q}_2}= \frac{e^{-1}}{e^{-2}}=e=2.718\)

Concept:

\(\rm \frac{T - T_{∞}}{T_0 - T_{∞}} = exp \left[ - \left( \frac{hA}{ρ VC} \right) × t \right]~\)

Convective heat loss = Rate of change of energy

hA(T - T∞) = -mCp\(\frac{dT}{dt}\)

Calculation:

Given:

Radius of sphere, r = 7.86 cm, T_o = 290 °C, T_∞  = 15 °C, h = 58 W/m²K, t = 22.617 minutes, ρ = 2700  kg/m³, c = 900 j/kgK, K = 205 W/mK

\(\rm \frac{T - T_{∞}}{T_0 - T_{∞}} = exp \left[ - \left( \frac{hA}{ρ VC} \right) × t \right]~\)

\(\rm \frac{T - T_{∞}}{T_0 - T_{∞}} = exp \left[ - \left( \frac{hA}{ρ VC} \right) × t \right]~\)

\(\rm \frac{T - 15}{290 - 15} = exp [-9.1 × 10^{-4} × 1357]\)

T = 95°C

For sphere at any temperature T

Convective heat loss = Rate of change of energy

hA(T - T_∞) = -mC_p\(\frac{dT}{dt}\)

9.1 × 10^-4 × (95 - 15) =  \(\frac{dT}{dt}\)

\(\frac{dT}{dt} = -(0.073)\)

Explanation:

In case of unsteady state heat conduction, the temperature keeps changing with respect to time. so in this case, storage or accumulation of energy occurs. So in this case Fourier and Biot numbers are helpful to study the unsteady state.

Fourier number is the ratio of heat conduction through medium to the rate of thermal energy storage. It is a dimension less number and it is denoted as Fr. It is also defined as ratio of operating time to diffusion time. the main significance of this number is it decides the thermal response of body.

Biot number is dimensionless number, the ratio of internal conductive resistance to external convective resistance and it is denoted as Bi. the main significance of this number is, it measures the temperature gradient of body.

If Bi < 0.1, then it indicates Lumped body.

Additional Information

Schmidt number (Sc): Ratio of molecular diffusivity of momentum to molecular diffusivity of mass transfer.

\(Sc=\frac{\mu}{\rho D}=\frac{\nu}{ D}\)

Where v is kinematic viscosity and D is diffusion coefficient

Prandtl number:

Prandtl number is the ratio of momentum diffusivity to thermal diffusivity.

\(Pr = \frac{\nu }{\alpha } = \frac{\mu }{{\frac{{\rho k}}{{\rho {C_p}}}}} = \frac{{\mu {C_p}}}{k}\)

Where α is thermal diffusivity.

Now,

\(\frac{{{S_c}}}{{Pr}} = \frac{{\frac{v}{D}}}{{\frac{v}{\alpha }}} = \frac{\alpha }{D}\)   

Fourier number:   

Fourier number is used in transient heat conduction. It is the ratio of diffusive or conductive transport rate to the quantity storage rate.

\(\begin{array}{l} {F_0}\; = \;\frac{{Conduction\;rate}}{{Thermal\;energy\;storage\;rate}}\\ {F_0}\; = \;\frac{{\alpha t}}{{{L^2}}} \end{array}\)

where \(\alpha \; = \;\frac{k}{{{\rho \times{{C_p}}}}}\) is thermal diffusivity, t = characteristic time, L = Length through which conduction occurs

Grashof number:

The dimensionless parameter which represents the natural convection effects is called the Grashof number.

Grashof number, Gr, is the ratio between the buoyancy force and the viscous force:

\(Gr = \frac{{g\beta \left( {{T_s} - {T_\infty }} \right)L_c^3}}{{{\nu ^2}}}\)

Nusselt number is a function of the Grashof number and the Prandtl number alone. Nu = f (Gr, Pr)

• Heisler charts are helpful when lumped heat capacity method fails i.e. when the biot number is more than 0.1
• Also, heisler chart solution is acceptable only when fourier number exceeds 0.2

Concept:

To calculate the temperature of a cooling stainless steel shaft in a given time, we will use the transient heat conduction solution for a finite cylinder with given parameters.

Calculation:

Given:

Diameter of the shaft, \( D = 15 \, \text{cm} = 0.15 \, \text{m} \)

Thermal conductivity, \( K = 14.9 \, \text{W/mK} \)

Density, \( \rho = 7900 \, \text{kg/m}^3 \)

Specific heat, \( C = 477 \, \text{J/kgK} \)

Thermal diffusivity, \( \alpha = 3.95 \times 10^{-6} \, \text{m}^2/\text{s} \)

Initial temperature, \( T_i = 450^\circ \text{C} \)

Ambient temperature, \( T_\infty = 150^\circ \text{C} \)

Heat transfer coefficient, \( h = 85 \, \text{W/m}^2\text{K} \)

Time, \( t = 25 \, \text{min} = 1500 \, \text{s} \)

Calculation:

First, calculate the Biot number (Bi):

\( \text{Bi} = \frac{h \cdot r_0}{K} = \frac{85 \cdot 0.075}{14.9} \approx 0.428 \)

Given:

\( \theta_0 = 0.49 \) at \( \text{Bi}^{-1} = 2.337 \)

\( \theta_0 = 0.88 \) at \( \text{Bi}^{-1} = 4.672 \)

Calculate the inverse Biot number for our case:

\( \text{Bi}^{-1} = \frac{1}{\text{Bi}} = \frac{1}{0.428} \approx 2.336 \)

Since our Bi-1 value is very close to 2.337, we can approximate \( \theta_0 \) by interpolation.

Calculate the Fourier number (Fo):

\( \text{Fo} = \frac{\alpha t}{r_0^2} = \frac{3.95 \times 10^{-6} \times 1500}{(0.075)^2} \approx 1.053 \)

Given Fo matches with the provided data, use the θ0 corresponding to Bi-1 = 2.337 directly:

\( \theta_0 \approx 0.49 \)

Calculate the final temperature \( T \):

\( \theta_0 = \frac{T - T_\infty}{T_i - T_\infty} \)

\( 0.49 = \frac{T - 150}{450 - 150} \)

\( 0.49 = \frac{T - 150}{300} \)

\( T - 150 = 0.49 \times 300 \)

\( T - 150 = 147 \)

\( T = 147 + 150 \)

\( T = 297^\circ \text{C} \)

The temperature at the end of 25 minutes after the cooling starts is approximately:

\( 297^\circ\text{C} \)