All questions of Heat Transfer for Mechanical Engineering Exam

Can you explain the answer of this question below:
For a given heat flow and for the same thickness, the temperature drop across the material will be maximum for  
  • A:
    Copper
  • B:
    Steel
  • C:
    Glass-wool
  • D:
    Refractory brick
The answer is c.

Swara Dasgupta answered
Q = -kA (dT/dx)
Qdx/A = -kdT
therefore, kdT = constant or dT ∞ 1/k

The one which has minimum thermal conductivity, will give maximum temperature drop. So, Glass-wool will have maximum temperature drop.

A thermocouple in a thermo-well measures the temperature of hot gas flowing through the pipe. For the most accurate measurement of temperature, the thermo-well should be made of:  
  • a)
    Steel
  • b)
    Brass
  • c)
    Copper
  • d)
    Aluminium
Correct answer is option 'A'. Can you explain this answer?

Bijoy Kapoor answered
(tl - ta)/(t0-ta) = (1/coshmL)
Therefor Error = tl-ta = ((t0-ta)/cosh(mL))
Where m = √(hp/KAc)
t0 = temperature of pipe wall 
ta = temperature of gas or air flowing through pipe
tl = temperature recorded by the thermometer at the bottom of the wall so to have better thermo well K should be minimum 

Material - K(W/m2-K)
Steel - 16-20
Brass - 100
Copper - 380
Aluminimum - 225
So steel will be the better material for thermo well

In descending order of magnitude, the thermal conductivity of
a. Pure iron,                                                                            
b. Liquid water,
c. Saturated water vapour, and
d. Pure aluminium can be arranged as
  • a)
    a b c d
  • b)
    b c a d
  • c)
    d a b c
  • d)
    d c b a
Correct answer is option 'C'. Can you explain this answer?

Thermal Conductivity of Different Substances

Thermal conductivity is the property of a material that determines how quickly it can transfer heat through it. It is usually measured in watts per meter per kelvin (W/m.K). In this question, we are given four substances, and we need to arrange them in descending order of thermal conductivity.

a. Pure Iron

Pure iron is a good conductor of heat, and its thermal conductivity is around 80 W/m.K. This means that it can transfer heat quite quickly through it.

b. Liquid Water

Liquid water has a thermal conductivity of around 0.6 W/m.K. This is much lower than that of pure iron, which means that water is not a good conductor of heat.

c. Saturated Water Vapour

Saturated water vapor has a thermal conductivity of around 0.03 W/m.K. This is much lower than that of liquid water, which means that water vapor is an even worse conductor of heat.

d. Pure Aluminium

Pure aluminium is an excellent conductor of heat, and its thermal conductivity is around 237 W/m.K. This is much higher than that of pure iron, which means that aluminium can transfer heat even more quickly than iron.

Arrangement of Thermal Conductivity in Descending Order

Based on the above information, we can arrange the substances in descending order of thermal conductivity as follows:

1. Pure Aluminium - 237 W/m.K
2. Pure Iron - 80 W/m.K
3. Liquid Water - 0.6 W/m.K
4. Saturated Water Vapour - 0.03 W/m.K

Therefore, option C, i.e., a-b-c-d, is the correct answer.

A steel ball of mass 1kg and specific heat 0.4 kJ/kg is at a temperature of 60°C. It is dropped into 1kg water at 20°C. The final steady state temperature of water is:  
  • a)
    23.5°C
  • b)
    300°C
  • c)
    35°C
  • d)
    40°C
Correct answer is option 'A'. Can you explain this answer?

Sinjini Bose answered
Given:
Mass of steel ball, m = 1 kg
Specific heat of steel, C_s = 0.4 kJ/kg
Initial temperature of steel ball, T_s1 = 60°C
Mass of water, M_w = 1 kg
Initial temperature of water, T_w1 = 20°C
Final steady state temperature of water, T_w2 = ?

To find:
Final steady state temperature of water

Solution:
1. Heat lost by steel ball = Heat gained by water
i.e., m × C_s × (T_s1 - T_w2) = M_w × C_w × (T_w2 - T_w1)
where, C_w is the specific heat of water (taken as 4.18 kJ/kg°C)

2. On solving the above equation, we get:
T_w2 = (m × C_s × T_s1 + M_w × C_w × T_w1) / (m × C_s + M_w × C_w)

Substituting the given values, we get:
T_w2 = (1 × 0.4 × 60 + 1 × 4.18 × 20) / (1 × 0.4 + 1 × 4.18)
T_w2 = 23.5°C

Therefore, the final steady state temperature of water is 23.5°C.

An oil cooler in a high performance engine has an outside surface area 0.12 m2 and a surface temperature of 65 degree Celsius. At any intermediate time air moves over the surface of the cooler at a temperature of 30 degree Celsius and gives rise to a surface coefficient equal to 45.4 W/ m 2 K. Find out the heat transfer rate?
  • a)
    238.43 W
  • b)
    190.68 W
  • c)
    543.67 W
  • d)
    675.98 W
Correct answer is option 'B'. Can you explain this answer?

Asha Basu answered
To find the heat transfer rate in the oil cooler, we can use the formula:

Q = h * A * ΔT

Where:
Q = Heat transfer rate (in watts)
h = Surface coefficient (in W/m2K)
A = Surface area (in m2)
ΔT = Temperature difference between the surface and the air (in K)

Given:
Surface area (A) = 0.12 m2
Surface temperature (T1) = 65 °C = 65 + 273.15 = 338.15 K
Air temperature (T2) = 30 °C = 30 + 273.15 = 303.15 K
Surface coefficient (h) = 45.4 W/m2K

Substituting the given values into the formula, we have:

ΔT = T1 - T2 = 338.15 K - 303.15 K = 35 K

Q = 45.4 W/m2K * 0.12 m2 * 35 K
Q = 190.68 W

Therefore, the heat transfer rate in the oil cooler is 190.68 W.

Hence, option B (190.68 W) is the correct answer.

For an opaque plane surface the irradiation, radiosity and emissive power are respectively 20, 12 and 10 W/m2.What is the emissivity of the surface?  
  • a)
    0.2
  • b)
    0.4
  • c)
    0.8
  • d)
    1.0
Correct answer is option 'C'. Can you explain this answer?

Dishani Desai answered
Given:

Irradiation = 20 W/m²
Radiosity = 12 W/m²
Emissive Power = 10 W/m²

To find: Emissivity of the surface

Formula used:

Emissivity (ε) = Emissive power (P) / Irradiation (I)

Calculation:

Substituting the given values in the formula,

ε = 10 / 20

ε = 0.5

Therefore, the emissivity of the surface is 0.5 or 50%.

Answer: Option (c) 0.8

A plane wall is 25 cm thick with an area of 1 m2, and has a thermal conductivity of 0.5 W/mK. If a temperature difference of 60°C is imposed across it, what is the heat flow?  
  • a)
    120W
  • b)
    140W
  • c)
    160W
  • d)
    180W
Correct answer is 'A'. Can you explain this answer?

We know, Q = kA(dT/dx)

Where, k = thermal conductivity = 0.5, 
A = area = 1 m^2, 
dT = temperature difference = 60 deg C, 
dx = thickness =  25 cm = 0.25 m

So, Q = 0.5 x 1 x (60/0.25) = 120 W

Hence Option (A) is correct

For Notes on Modes Of Heat Transfer click on the link given below:

Can you explain the answer of this question below:

Consider the following statements: 
The Fourier heat conduction equation Q = presumes
1. Steady-state conditions
2. Constant value of thermal conductivity.
3. Uniform temperatures at the wall surfaces
4. One-dimensional heat flow. Of these statements:

  • A:

    1, 2 and 3 are correct

  • B:

    1, 2 and 4 are correct

  • C:

    2, 3 and 4 are correct

  • D:

    1, 3 and 4 are correct

The answer is d.

Rajeev Sharma answered
Assumptions of Fourier equation:1. Steady state heat conduction.2. One directional heat flow.3. Bounding surfaces are isothermal in character that is constant and uniform temperatures are maintained at the two faces.4. Isotropic and homogeneous material and thermal conductivity ‘k’ is constant.5. Constant temperature gradient and linear temperature profile.6. No internal heat generation.

Assertion (A): Cork is a good insulator.
Reason (R): Good insulators are highly porous. 
  • a)
    Both A and R are individually true and R is the correct explanation of A
  • b)
    Both A and R individually true but R in not the correct explanation of A
  • c)
    A is true but R is false
  • d)
    A is false but R is true
Correct answer is option 'A'. Can you explain this answer?

It is the air which fills the cells in the cork that makes it an excellent thermal insulator. The same protective principle can also be found in other natural products, such as wool and feathers, and in industry, such as double glazed windows, for example. However, none of these materials have such a high level of insulation as cork.

Two spheres A and B of same material have radii 1 m and 4 m and temperature 4000 K and 2000 K respectively. Which one of the following statements is correct? The energy radiated by sphere A is:   
  • a)
    Greater than that of sphere B
  • b)
    Less than that of sphere B
  • c)
    Equal to that of sphere B
  • d)
    Equal to double that of sphere B
Correct answer is option 'C'. Can you explain this answer?

Understanding Blackbody Radiation
The energy radiated by a blackbody is described by the Stefan-Boltzmann Law, which states that the total energy radiated per unit surface area is proportional to the fourth power of its absolute temperature.
Stefan-Boltzmann Law
- The formula is given by:
E = σT^4
where E is the energy radiated per unit area, σ is the Stefan-Boltzmann constant, and T is the absolute temperature in Kelvin.
Calculating Energy for Sphere A and B
- Sphere A:
- Radius = 1 m
- Temperature = 4000 K
- Surface Area = 4π(1^2) = 4π m²
- Energy radiated (EA) = σ(4000^4) * 4π
- Sphere B:
- Radius = 4 m
- Temperature = 2000 K
- Surface Area = 4π(4^2) = 64π m²
- Energy radiated (EB) = σ(2000^4) * 64π
Comparing Energy Radiated
- Calculating EA and EB:
- EA = σ * 4π * (4000^4)
- EB = σ * 64π * (2000^4)
- Simplifying the comparison:
- 4000^4 = 256 * 1000^4
- 2000^4 = 16 * 1000^4
- Thus, EB becomes 64π * (16 * 1000^4) = 1024π * σ
- Final Comparison:
- EA = 1024π * σ
- EB = 1024π * σ
Conclusion
Both spheres radiate equal energy despite their different sizes and temperatures, hence the correct answer is option C: Equal to that of sphere B.

When all the conditions are identical, in the case of flow through pipeswith heat transfer, the velocity profiles will be identical for:  
  • a)
    Liquid heating and liquid cooling
  • b)
    Gas heating and gas cooling
  • c)
    Liquid heating and gas cooling
  • d)
    Heating and cooling of any fluid
Correct answer is option 'A'. Can you explain this answer?

Velocity Profiles in Flow Through Pipes with Heat Transfer

Introduction:
In the case of flow through pipes with heat transfer, the velocity profiles can vary depending on the conditions. However, when all the conditions are identical, the velocity profiles will be identical for liquid heating and liquid cooling.

Explanation:

1. Flow Through Pipes:
In flow through pipes, the velocity of the fluid can vary across the cross-section of the pipe. This variation is known as the velocity profile. The velocity profile is influenced by various factors such as fluid properties, pipe geometry, and boundary conditions.

2. Heat Transfer:
Heat transfer occurs when there is a temperature difference between the fluid and the pipe wall. This temperature difference can result in either heating or cooling of the fluid as it flows through the pipe.

3. Identical Conditions:
When all the conditions are identical, it means that the fluid properties, pipe geometry, and boundary conditions are the same for both heating and cooling scenarios. This includes factors such as pipe diameter, fluid viscosity, fluid density, and flow rate.

4. Velocity Profile:
The velocity profile in a pipe is governed by the Navier-Stokes equation, which describes the conservation of momentum. When all the conditions are identical, the velocity profile remains the same for both heating and cooling scenarios.

5. Liquid Heating and Liquid Cooling:
For liquid heating and liquid cooling, the fluid properties remain constant. The viscosity and density of the liquid do not change significantly with temperature. Therefore, the velocity profile will be the same for both heating and cooling of the liquid.

6. Gas Heating and Gas Cooling:
For gas heating and gas cooling, the fluid properties can vary significantly with temperature. The viscosity and density of the gas change with temperature, which can affect the velocity profile. Therefore, the velocity profiles will not be identical for gas heating and gas cooling.

Conclusion:
When all the conditions are identical, the velocity profiles will be identical for liquid heating and liquid cooling. This is because the fluid properties, pipe geometry, and boundary conditions remain constant for both heating and cooling scenarios. However, the velocity profiles will not be identical for gas heating and gas cooling due to the significant changes in fluid properties with temperature.

Which of the following would lead to a reduction in thermal resistance?  
1. In conduction; reduction in the thickness of the material and an increase in the thermal conductivity.
2. In convection, stirring of the fluid and cleaning the heating surface.
3. In radiation, increasing the temperature and reducing theemissivity.
Codes:
  • a)
    1, 2 and 3
  • b)
    1 and 2
  • c)
    1 and 3
  • d)
    2 and 3
Correct answer is option 'B'. Can you explain this answer?

Arshiya Dey answered
Ans. (b) 1. In conduction, heat resistance = Δ x/kA
Thus reduction in thickness and increase in area result in reduction of thermal resistance.
2. Stirring of fluid and cleaning the heating surface increases value of h, and thus reduces thermal resistance.
3. In radiation, heat flow increases with increase in temperature and reduces with reduction in emissivity. Thus thermal resistance does not decrease.
Thus 1 and 2 are correct.

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