Latest Heat Transfer MCQ Objective Questions
Heat Transfer Question 1:
For an opaque body
- absorptivity = 1
- reflectivity = 1
- transmissivity + absorptivity = 1
- absorptivity + reflectivity = 1
Answer (Detailed Solution Below)
Option 4 : absorptivity + reflectivity = 1
Heat Transfer Question 1 Detailed Solution
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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Heat Transfer Question 2:
In freeconvection, Nusselt number is the function of
- Prandtl numberand Grashof number
- Stanton number and Prandtl number
- Peclet number and Stanton number
- Prandtl number and Reynolds number
Answer (Detailed Solution Below)
Option 1 : Prandtl numberand Grashof number
Heat Transfer Question 2 Detailed Solution
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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Heat Transfer Question 3:
A wall of a furnace is made up of inside layer of silica brick 119 mm thick covered with a layer of magnesite brick 240 mm thick. The temperatures at the inside surface of silica brick wall and outside surface of magnesite brick wall are 780 °C and 102 °C respectively. The contact thermal resistance between the two walls at the interface is 0.003 °C/W per unit wall area. Thermalconductivitiesofsilica and magnesite bricksare1.7 W/m °C and 6 W/m °C. The rate of heat loss per unit area of the wall is
- 5600 W/m^{2}
- 6000 W/m^{2}
- 5200 W/m^{2}
- 5300 W/m^{2}
Answer (Detailed Solution Below)
Option 2 : 6000 W/m^{2}
Heat Transfer Question 3 Detailed Solution
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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Heat Transfer Question 4:
For the given overall heat transfer coefficient and temperature difference, if the area of evaporator surface increases, then the capacity of evaporator:
- can increase or decrease
- remains constant
- increases
- decreases
Answer (Detailed Solution Below)
Option 3 : increases
Heat Transfer Question 4 Detailed Solution
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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Heat Transfer Question 5:
Which of the following is a requirement for thermal insulation in materials?
- Low thermal conductivity
- Low density
- High permeability
- High temperature resistance
Answer (Detailed Solution Below)
Option 1 : Low thermal conductivity
Heat Transfer Question 5 Detailed Solution
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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Top Heat Transfer MCQ Objective Questions
Heat Transfer Question 6
Download Solution PDFBoiling point of water at sea level is ______.
- 210° F
- 212° F
- 208° F
- 214° F
Answer (Detailed Solution Below)
Option 2 : 212° F
Heat Transfer Question 6 Detailed Solution
Download Solution PDFThe 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
Download Solution PDFAs the temperature increases, the thermal conductivity of a gas
- increases
- decreases
- remains constant
- increases up to a certain temperature and then decreases
Answer (Detailed Solution Below)
Option 1 : increases
Heat Transfer Question 7 Detailed Solution
Download Solution PDFExplanation:
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
Download Solution PDFIn a heat exchanger, it is observed that ΔT_{1} = ΔT_{2}, where ΔT_{1} is the temperature difference between the two single phase fluid streams at one end and ΔT_{2} is the temperature difference at the other end. This heat exchanger is
- a condenser
- an evaporator
- a counter flow heat exchanger
- a parallel flow heat exchanger
Answer (Detailed Solution Below)
Option 3 : a counter flow heat exchanger
Heat Transfer Question 8 Detailed Solution
Download Solution PDFExplanation:
In case of the counter-flow heat exchanger when the heat capacities of both the fluids are the same.
i.e.ṁ_{h}c_{h} = ṁ_{c}c_{c}
Q = ṁ_{h}c_{h}(T_{h1} – T_{h2}) = ṁ_{c}c_{c}(T_{c2} – T_{c1})
⇒ (T_{h1} – T_{h2}) = (T_{c2} – T_{c1})
⇒ (T_{h1} – T_{c2}) = (T_{h2} – T_{c1})
⇒ ΔT_{1} = ΔT_{2}
_{}
For parallel flow heat exchanger,ΔT1will always be greater thanΔT2.
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Heat Transfer Question 9
Download Solution PDFLiquid metal having highest thermal conductivity is of _______.
- Sodium
- Potassium
- Lead
- Mercury
Answer (Detailed Solution Below)
Option 1 : Sodium
Heat Transfer Question 9 Detailed Solution
Download Solution PDFThermal 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
Download Solution PDFIn the laminar flow of air (Pr = 0.7) over a heated plate if δ and δ_{T} denote, respectively, the hydrodynamic and thermal boundary layer thicknesses, then
- δ = δ_{T}
- δ > δ_{T}
- δ < δ_{T}
- δ = 0 but δ_{T} ≠ 0
Answer (Detailed Solution Below)
Option 3 : δ < δ_{T}
Heat Transfer Question 10 Detailed Solution
Download Solution PDFConcept:
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_{T}is 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
Download Solution PDFThe insulating ability of an insulator with the presence of moisture would
- increase
- decrease
- remain unaffected
- Non of the above
Answer (Detailed Solution Below)
Option 2 : decrease
Heat Transfer Question 11 Detailed Solution
Download Solution PDFExplanation:
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
Download Solution PDFThe typical range of Prandtl number for water is
- 0.004-0.300
- 1.7-13.7
- 50.500
- 2000-1000
Answer (Detailed Solution Below)
Option 2 : 1.7-13.7
Heat Transfer Question 12 Detailed Solution
Download Solution PDFExplanation:
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
Fluid | Pr |
Liquid metals | 0.004 – 0.030 |
Gases | 0.7 – 1.0 |
Water | 1.7 – 13.7 |
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Heat Transfer Question 13
Download Solution PDFIn which process, the rate of transfer of heat is maximum:
- conduction
- convection
- Radiation
- In all, heat is transferred with the same speed
Answer (Detailed Solution Below)
Option 3 : Radiation
Heat Transfer Question 13 Detailed Solution
Download Solution PDFConcept:
Three modes of Transmission of heat/heat flow | |||
Sl. No | Conduction | Convection | Radiation |
1 | Heat dissipates from one place to another bymolecular vibration | Heat is transfer from one place to another by thetransfer of molecules | It transfers heat in the form ofelectromagnetic wave |
2 | Conduction is relevant tosolid only. | Convection happens inliquid or gases | It can heat any form of material. |
3 | Need mediumto transfer heat | Need mediumto transfer heat | No need for medium |
4 | Good Conductor- The objects which transfer heat easily. Ex-metals, human bodyetc Bad Conductor- The objects which do not transfer heat easily. Ex-Wood, Air,etc. Thermal Insulator- No heat is transferred by any means. Ex-Abonite, asbestosetc. | When molecules are heated they headed upward and upper molecules go downward and this cyclic process continues. Boiling of fluid | Heat travels in terms of energy packets or waves The heat absorbed by the body gain energy Radiations of Sun |
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
Download Solution PDFThe SI unit of Thermal Conductivity is
- Wm^{-2}K^{-1}
- WmK^{-1}
- Wm-1K-1
- WmK
Answer (Detailed Solution Below)
Option 3 : Wm-1K-1
Heat Transfer Question 14 Detailed Solution
Download Solution PDFConcept:
- 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
Download Solution PDFHeat is transferred by all three modes of transfer, viz, conduction, convection and radiation in
- Electric heater
- Steam condenser
- Melting of ice
- Boiler
Answer (Detailed Solution Below)
Option 4 : Boiler
Heat Transfer Question 15 Detailed Solution
Download Solution PDFExplanation:
- 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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