Heat Transfer MCQ Quiz - Objective Question with Answer for Heat Transfer - Download Free PDF

Explanation:

Emissivity Factor for Energy Emitted by a Grey Body

• Emissivity is a measure of the efficiency of a surface in emitting thermal radiation compared to a perfect black body. A black body is an idealized physical object that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence, and emits the maximum amount of energy at any given temperature. A grey body, however, emits less energy than a black body at the same temperature.

Mathematically, the emissivity factor (ε) is defined as the ratio of the energy emitted by a grey body per unit area per unit time (E) to the energy emitted by a perfect black body per unit area per unit time (E_B) at the same temperature:

\( \varepsilon = \frac{E}{E_B} \)

• \( E \) = Emitted energy by grey body (W/m²)
• \( E_B \) = Emitted energy by black body (W/m²)

Key Points:

• For a perfect black body, emissivity is equal to 1 because it emits the maximum possible energy at any given temperature.
• For real-world objects (grey bodies), emissivity is less than 1, as they emit less energy compared to a black body.
• Emissivity is a dimensionless quantity and typically ranges between 0 and 1.

Physical Significance:

The emissivity factor is crucial in determining the thermal radiation properties of materials. It plays a significant role in applications such as:

• Thermal engineering, where heat transfer calculations involve radiation.
• Designing heat exchangers and radiative cooling systems.
• Understanding and modeling the behavior of materials in high-temperature environments.
• Developing thermal imaging systems and sensors.

Concept:

The rate of heat transfer in a heat exchanger is calculated using enthalpy change when there is no change in kinetic or potential energy. Enthalpy for air is given by:

\( h = C_p \cdot T \)

Calculation:

• Initial temperature: \( T_1 = 15^\circ C = 288~K \)
• Final temperature after heat exchanger: \( T_2 = 800^\circ C = 1073~K \)
• Specific heat: \( C_p = 1.005~kJ/kg·K \)
• Mass flow rate: \( \dot{m} = 2~kg/s \)

Heat added in exchanger:

\( \dot{Q} = \dot{m} \times C_p \times (T_2 - T_1) \)

\( \dot{Q} = 2 \times 1.005 \times (1073 - 288) = 2 \times 1.005 \times 785 = 1580~kJ/s \)

Explanation:

Newton's Formula for Convective Heat Transfer:

• Newton's formula for convective heat transfer describes the rate of heat transfer from a fluid to a metallic wall or from a metallic wall to a fluid via convection. The formula is expressed as:

Q = ha(t_s - t_f)

Where:

• Q: Heat transfer by convection in Joules per second (J/s) or Watts (W).
• h: Heat transfer or film coefficient in J/m²°C·s or W/m²°C.
• a: Surface area through which heat is transferred in square meters (m²).
• t_s: Temperature of the surface or wall in degrees Celsius (°C).
• t_f: Temperature of the fluid in degrees Celsius (°C).

Working Principle:

Heat transfer by convection occurs when a fluid (liquid or gas) comes into contact with a solid surface (e.g., a metallic wall). The temperature difference between the fluid (tf) and the solid surface (ts) drives the heat transfer process. Newton's law of cooling provides a mathematical relationship for this process, indicating that the rate of heat transfer is proportional to:

• The temperature difference (ts - tf).
• The surface area (a) exposed to heat transfer.
• The convective heat transfer coefficient (h), which depends on the properties of the fluid and the nature of the surface.

The formula Q = ha(ts - tf) accurately describes this phenomenon and is widely used in engineering calculations for convective heat transfer analysis.

Concept:

Net heat transfer by radiation occurs when a body at higher temperature radiates energy toward a cooler body or surroundings.

According to Stefan–Boltzmann Law, a perfect black body radiates heat as,

\( Q = σ T^4 \)

For real bodies, emissivity \( \varepsilon \) is introduced:

\( Q = σ \varepsilon T^4 \)

Here,

σ = 5.67 × 10^-8 W/m²-K⁴ (Stefan–Boltzmann constant), and ε is the emissivity (0 ≤ ε ≤ 1)

Calculation:

Let Body 1 be at temperature T₁ and surroundings (or Body 2) at temperature T₂ .

The net radiative heat transfer is given by:

\( Q = σ \varepsilon_1 (T_1^4 - T_2^4) \)

Correct option is: (3) 5 / 3

In series, R_eq = R₁ + R₂ + R₃

= 1 / (2KA) + 1 / (KA) + 1 / (2KA)

= 4 / (2KA)

R_eq = 2 / (KA)

In series rate of heat flow is same

(3T − T₁) / R₁ = (3T − T) / R_eq

((3T − T₁) KA) / 1 = (2T) KA / 2

⇒ 6T − 2T₁ = T

⇒ T₁ = 5T / 2         ...(1)

Now, equate heat flow rate in 3^rd section and total section

(T₂ − T) / R₃ = (3T − T) / R_eq

((T₂ − T)(2KA)) / 1 = (2T KA) / 2

⇒ 2T₂ − 2T = T

⇒ T₂ = 3T / 2         ...(2)

By equation (1) and (2)

T₁ / T₂ = (5T / 2) / (3T / 2) = 5 / 3

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° C or 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.

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,  ΔT1 will always be greater than ΔT2.

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.

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, the ratio 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, δ is hydrodynamic boundary layer thickness and δ_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 

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

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.

Concept:

Explanation:

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

Explanation:

Insulation

• It is defined as a process of preventing the flow of heat from the body by applying insulator materials to the surface which controls the 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.

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.

Explanation:

Net radiation heat exchange between two bodies is given by:

Q̇ = AF × σ × (T₁⁴ - T₂⁴)

where F, A, σ are shape factor, Area and Stefan-Boltzmann constant respectively

Now explanding  (T14 - T24)

Q̇ = AF × σ × ((T1)²)² - (T2)²)²)

Q̇ = AF × σ × (T12 - T22) × (T12 + T22)

Q̇ = AF × σ × (T₁ - T₂)(T₁ + T₂) ×  (T12 + T22)

\(\dot Q =\frac{T_1-T_2}{\frac{1}{\sigma \times AF \times(T_1+T_2)\times(T_1^2+T_2^2)}}\)

Comparing with the electrical analogy \(i=\frac VR\)

We will get thermal resistance as \(\frac{1}{{FA\sigma \left( {{T_1} + {T_2}} \right)\left( {T_1^2 + T_2^2} \right)}}\)