[Solved] Heat Transfer MCQ [Free PDF] - Objective Question Answer for Heat Transfer Quiz - Download Now! (2024)

1. absorptivity = 1
2. reflectivity = 1
3. transmissivity + absorptivity = 1
4. absorptivity + reflectivity = 1

Concept:

When thermal radiation falls onto an object,

• The radiation will be absorbed by the surface of the object, causing its temperature to change
• The radiation will be reflected from the surface of the body, causing no temperature change
• The radiation will pass completely through the object, causing no temperature change

Absorptivity (α)is a measure of how much of the radiation is absorbed by the body

Reflectivity (ρ)is a measure of how much is radiation is reflected

Transmissivity (τ)is a measure of how much radiation passes through the object.

Each of these parameters is a numberthat ranges from 0 to 1.

For any given wavelength (λ), αλ+ ρλ+ τλ= 1. Objects may have different values of each of these parameters at different wavelengths.

Now,

For an opaque body, transmissivity is 0 since no wavelength is transmitted through it.

∴For an opaque body sum of absorptivity and reflectivity is 1

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1. Prandtl numberand Grashof number
2. Stanton number and Prandtl number
3. Peclet number and Stanton number
4. Prandtl number and Reynolds number

Explanation:

In convection studies, it is common practice to nondimensional the governing equations and combines the variables which group together into dimensionless numbers in order to reduce the number of total variables.

Nusselt number is a parameter to nondimensional the heat transfer coefficient.

Non-dimensional groupings

• Nusselt No. Nu = hLc/k = (convection heat transfer strength)/ (conduction heat transfer strength)
• Prandtl No. Pr = n/a = (momentum diffusivity)/ (thermal diffusivity)
• Reynolds No. Re = Ux/ν = (inertia force)/ (viscous force)

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

Grashof number, Gr, as 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)

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1. 5600 W/m²
2. 6000 W/m²
3. 5200 W/m²
4. 5300 W/m²

Concept:

To determine the rate of heat loss per unit area through a composite wall, we use the concept of thermal resistance in series.

Calculation:

Given:

• Thickness of silica brick, \( t_1 = 119 \, \text{mm} = 0.119 \, \text{m} \)
• Thickness of magnesite brick, \( t_2 = 240 \, \text{mm} = 0.24 \, \text{m} \)
• Temperature at the inside surface of silica brick, \( T_1 = 780 \, ^\circ \text{C} \)
• Temperature at the outside surface of magnesite brick, \( T_2 = 102 \, ^\circ \text{C} \)
• Thermal conductivity of silica brick, \( k_1 = 1.7 \, \text{W/m} \cdot ^\circ \text{C} \)
• Thermal conductivity of magnesite brick, \( k_2 = 6 \, \text{W/m} \cdot ^\circ \text{C} \)
• Contact thermal resistance, \( R_c = 0.003 \, ^\circ \text{C/W} \)

Calculation:

First, calculate the thermal resistance of the silica brick layer:

\( R_1 = \frac{t_1}{k_1} = \frac{0.119}{1.7} \, \text{m}^2 \cdot ^\circ \text{C/W} = 0.07 \, ^\circ \text{C/W} \)

Next, calculate the thermal resistance of the magnesite brick layer:

\( R_2 = \frac{t_2}{k_2} = \frac{0.24}{6} \, \text{m}^2 \cdot ^\circ \text{C/W} = 0.04 \, ^\circ \text{C/W} \)

Then, calculate the total thermal resistance:

\( R_{\text{total}} = R_1 + R_2 + R_c = 0.07 + 0.04 + 0.003 = 0.113 \, ^\circ \text{C/W} \)

Finally, calculate the rate of heat loss per unit area:

\( q = \frac{\Delta T}{R_{\text{total}}} = \frac{T_1 - T_2}{R_{\text{total}}} = \frac{780 - 102}{0.113} \, \text{W/m}^2 = \frac{678}{0.113} \, \text{W/m}^2 \approx 6000 \, \text{W/m}^2 \)

Therefore, the rate of heat loss per unit area of the wall is:

\( \text{6000 W/m}^2 \)

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1. can increase or decrease
2. remains constant
3. increases
4. decreases

Concept:

The capacity of an evaporator is directly related to the heat transfer rate, which depends on the overall heat transfer coefficient, the temperature difference, and the surface area of the evaporator.

Heat Transfer Equation:

The heat transfer rate ( Q ) in an evaporator can be described by the equation:

\( Q = U \cdot A \cdot \Delta T \)

Where:

\( U \) = Overall heat transfer coefficient

\( A \) = Area of the evaporator surface

\( \Delta T \) = Temperature difference between the hot and cold fluids

Given Conditions:

For the given overall heat transfer coefficient and temperature difference, if the area of the evaporator surface increases, the heat transfer rate Q will increase proportionally because Q is directly proportional to the area A .

Conclusion:

If the area of the evaporator surface increases, then the capacity of the evaporator:

3) Increases

This is because a larger surface area allows for more heat to be transferred, thus increasing the evaporator's capacity to absorb heat.

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1. Low thermal conductivity
2. Low density
3. High permeability
4. High temperature resistance

Heat Insulating Materials:

Heat insulating materials are essential for providing protection against extreme temperatures, whether hot or cold. These materials help in reducing heat transfer and maintaining desired temperatures within a space.

Key Characteristics:

• Pore Structure: Closed and air-filled pores are the most effective heat insulating materials. The presence of these pores in the material significantly reduces the transfer of heat, making them ideal for insulation.

• Thermal Conductivity: The coefficient of heat conductivity for these materials does not exceed 0.18 kcal/m·hr·°C. Lower thermal conductivity values indicate better insulating properties.

• Air-Filled Pores: The low heat conductivity of insulating materials is largely due to their air-filled pores, which act as barriers to heat transfer.

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Heat Transfer Question 6

1. 210° F
2. 212° F
3. 208° F
4. 214° F

Heat Transfer Question 6 Detailed Solution

The correct answer is 212° F.

• At 1 atmosphere of pressure (sea level), water boils at 100° C (212° F).
• When a liquid is heated, it eventually reaches a temperature at which the vapor pressure is large enough that bubbles form inside the body of the liquid. This temperature is called the boiling point.
  • Once the liquid starts to boil, the temperature remains constant until all of the liquid has been converted to a gas.

Important Points

• The boiling point of water depends on the atmospheric pressure, which changes according to elevation.
  • Water boils at a lower temperature as you gain altitude (e.g., going higher on a mountain).
  • Water boils at a higher temperature if you increase atmospheric pressure (coming back down to sea level or going below it).
• The boiling point of water also depends on the purity of the water.
  • Water that contains impurities (such as salted water) boils at a higher temperature than pure water. This phenomenon is called boiling point elevation.
  • It is one of the colligative properties of matter.

Key Points

• Liquids have a characteristic temperature at which they turn into solids, known as their freezing point.
  • Water freezes at 32° F or 0° Cor 273.15 Kelvin.
• Pure, crystalline solids have a characteristic melting point, the temperature at which the solid melts to become a liquid.
• In theory, the melting point of a solid should be the same as the freezing point of the liquid.

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Heat Transfer Question 7

1. increases
2. decreases
3. remains constant
4. increases up to a certain temperature and then decreases

Heat Transfer Question 7 Detailed Solution

Explanation:

Gases transfer heat by the collision of molecules.

As the temperature increases, the kinetic energy of molecules of gases also increases and eventually collision between molecules also increases which increases the thermal conductivity of gases.

∴ As temperature increases the thermal conductivity of gases increases.

For liquid and solids, generally as the temperature increases, the thermal conductivity decreases.

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Heat Transfer Question 8

1. a condenser
2. an evaporator
3. a counter flow heat exchanger
4. a parallel flow heat exchanger

Heat Transfer Question 8 Detailed Solution

Explanation:

In case of the counter-flow heat exchanger when the heat capacities of both the fluids are the same.

i.e.ṁ_hc_h = ṁ_cc_c

Q = ṁ_hc_h(T_h1 – T_h2) = ṁ_cc_c(T_c2 – T_c1)

⇒ (T_h1 – T_h2) = (T_c2 – T_c1)

⇒ (T_h1 – T_c2) = (T_h2 – T_c1)

⇒ ΔT₁ = ΔT₂

For parallel flow heat exchanger,ΔT1will always be greater thanΔT2.

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Heat Transfer Question 9

1. Sodium
2. Potassium
3. Lead
4. Mercury

Heat Transfer Question 9 Detailed Solution

Thermal conductivity of different metals in liquid state is given below

Sodium (Na) – 140 W/m-K

Potassium (K) – 100 W/m-K

Lithium (Li) – 85 W/m-K

Tin (Sn) – 64 W/m-K

Lead (Pb) – 36 W/m-K

Mercury ( Hg) – 8 W/m-K

So out of given options Sodium has highest thermal conductivity.

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Heat Transfer Question 10

1. δ = δ_T
2. δ > δ_T
3. δ < δ_T
4. δ = 0 but δ_T ≠ 0

Heat Transfer Question 10 Detailed Solution

Concept:

Prandtl number Pr is defined as the ratio of momentum diffusivity to thermal diffusivity.

\(Pr = \frac{{\mu {C_p}}}{K} = \frac{{\left( {\frac{\mu }{\rho }} \right)}}{{\left( {\frac{K}{{\rho {C_p}}}} \right)}}\)

\(Pr = \frac{\nu }{\alpha } = \frac{{momentum\;diffusivity}}{{thermal\;diffusivity}}\)

In another way, we can define Prandtl number as, theratio of the rate that viscous forces penetrate the material to the rate that thermal energy penetrates the material.

\(\frac{δ }{{{δ _T}}} = {\left( {Pr} \right)^{1/3}}\;\)where,δis hydrodynamic boundary layer thickness andδ_Tis thermal boundary layer thickness.

Calculation:

Given:

Pr = 0.7

from,\(\frac{δ }{{{δ _T}}} = {\left( {Pr} \right)^{1/3}}\;\)=\({0.7^{\frac{1}{3}}} = 0.88 < 1\)

thus,δ <δ_T.

When P_r< 1 δ_T> δ

P_r > 1 δT< δ

P_r = 1 δ_t = δ

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Heat Transfer Question 11

1. increase
2. decrease
3. remain unaffected
4. Non of the above

Heat Transfer Question 11 Detailed Solution

Explanation:

Insulation

• It is defined as a process of preventingthe flow of heat from the body by applying insulator materials to the surface which controlsthe rate of heat transfer.
• The insulating ability of an insulator depends on various factors:
  • thickness of insulator
  • material of insulator
  • surrounding conditions
  • temperature difference
• Generally, air packets are present in porous insulating materials.
• Since water which is a more conductive material is replacing air which is a less conductive material, so the overall insulating ability of the insulator will decrease. Most insulators are porous in nature.
• If it has been about Non-porous insulators then the insulating ability will remain unaffected.

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Heat Transfer Question 12

1. 0.004-0.300
2. 1.7-13.7
3. 50.500
4. 2000-1000

Heat Transfer Question 12 Detailed Solution

Explanation:

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

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

Typical ranges of Prandtl member is listed below

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Heat Transfer Question 13

1. conduction
2. convection
3. Radiation
4. In all, heat is transferred with the same speed

Heat Transfer Question 13 Detailed Solution

Concept:

Explanation:

• As we know the radiationtravels with the speed of light, thus the rate of heat transfer is maximum in radiation in form of electromagnetic radiations

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Heat Transfer Question 14

1. Wm^-2K^-1
2. WmK^-1
3. Wm-1K-1
4. WmK

Heat Transfer Question 14 Detailed Solution

Concept:

• Thermal Conductivity:When one end of a metal rod is heated, heat flows byconduction from the hot end to the coldend. In this process, each cross-section of the rod receives some heat from the adjacent cross-section towards the hot end.

It is found that theamount of heat Qthat flows from hot to cold face during steady-state:

Or,\(Q = \frac{{KA\left( {{T_1} - {T_2}} \right)t}}{x}\)

where K = Coefficient of thermal conductivity of the material.

Rate of conduction of heat energyis given by:

\(\frac{{dQ}}{t} = \frac{{KA\left( {{T_1} - {T_2}} \right)}}{x} = KA\frac{{{\bf{\Delta }}T}}{x}\)

Calculation:

Given:

Rate of conduction of heat energy is given by:

\(\frac{{dQ}}{t} = \frac{{KA\left( {{T_1} - {T_2}} \right)}}{x} = KA\frac{{{\bf{\Delta }}T}}{x}\)

Coefficient of thermal conductivity of material will be,

\(K = \frac{{dQ \times x}}{{t \times A \times {\rm{\Delta }}T}}\)

SI unit of Q = J/s = W, A = m2, x = m and ΔT = K

\(\therefore K = \frac{{dQ \times x}}{{t \times A \times {\rm{\Delta }}T}} = \frac{{J \cdot m}}{{sec \cdot {m^2} \cdot K}} = W{m^{ - 1}}{K^{ - 1}}\)

Therefore, theSI unit of the thermal conductivityisWm-1K-1.

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Heat Transfer Question 15

1. Electric heater
2. Steam condenser
3. Melting of ice
4. Boiler

Heat Transfer Question 15 Detailed Solution

Explanation:

• There are three methods of heat transfer between the two systems. They are conduction, convection, and radiation.
• Conduction is a method of heat transfer in solids and heat transfer takes place without the movement of particles.
• Convection is a method of heat transfer in fluids (gases and liquids) and heat transfer takes place due to the movement of particles.
• Radiation is a method of heat transfer where heat is transferred from one place to another without affecting the medium of heat transfer.

Now let's see what happens in a steam boiler:

• A steam boiler is designed to absorb the maximum amount of heat released from the process of combustion.
• Heat transfer within the steam boiler is accomplished by three methods: radiation, convection, and conduction. The heating surface in the furnace area receives heat primarily by radiation.
• The remaining heating surface in the steam boiler receives heat by convection from the hot flue gases. Heat received by the heating surface travels through the metal by conduction
• Heat is then transferred from the metal to the water by convection.

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