Energy Engineering Questions and Answers – Calorific Value of Fuels – 2

This set of Energy Engineering Interview Questions and Answers for freshers focuses on “Calorific Value of Fuels – 2”.

1. Which is the common method to relate higher calorific value to lower calorific value?
a) HCV = LCV + HV (nH2O, out / nfuel, in)
b) LCV = HCV + HV (nH2O, out / nfuel, in)
c) HCV = LCV + HV (nfuel, in/ nH2O, out)
d) LCV = HCV + HV (nfuel, in/ nH2O, out)
View Answer

Answer: a
Explanation: HV – heat vaporization of water.
nH2O, out – moles of water vaporized.
nfuel, in – number of moles of fuel combusted.
High calorific value is equal to low calorific value plus, product of heat of vaporization of water and moles of water vaporized by moles of fuel combusted.

2. Based on what basis are fuels compared?
a) Fire point value
b) High calorific value
c) Flash point value
d) Low calorific value
View Answer

Answer: d
Explanation: On basis of low calorific value the fuels are compared. Low calorific value is the amount of heat evolved when a unit weight of fuel is completely burnt and water vapor leaves with combustion products.

3. Which value is determined by bringing all products of combustion back to original pre-combustion temperature?
a) Higher calorific value
b) Low calorific value
c) Flash point value
d) Fire point value
View Answer

Answer: a
Explanation: High calorific value is determined, as all fuels contain hydrogen; they produce water vapor during combustion. When the products of combustion containing water vapor are cooled back to initial temperature, then all water vapors formed condense and evolve latent heat. This adds up to the heat liberated by burning the fuel, producing maximum amount of heat per kg of fuel. This heat is known as the higher calorific value of fuel, and it is denoted by HCV.

4. Which calorific value is same as the thermodynamic heat of combustion?
a) Net calorific value
b) Flash point value
c) Gross calorific value
d) Fire point value
View Answer

Answer: d
Explanation: Gross calorific value is as same as the thermodynamic heat of combustion since the enthalpy change for the reaction assumes a common temperature of the compounds before and after combustion, in which case the water produced by combustion is condensed to a liquid, hence yielding its latent heat of vaporization.

5. Which value is determined by subtracting the heat of vaporization of the water from the higher heating value?
a) Gross calorific value
b) Net calorific value
c) Ignition temperature
d) Fire point temperature
View Answer

Answer: b
Explanation: Net calorific value determined. In most of the combustion processes, the products of combustion cannot be cooled to its initial temperature. Thus water vapors don’t condense and hence the latent heat of water vapor is lost to the atmosphere. The resultant heat liberated by the fuel which excludes the latent heat of evaporation of water vapors is known as lower calorific value of fuel.
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6. Which formula is used to determine higher calorific value of fuel?
a) Rayleigh’s formula
b) Lamme’s equation
c) Dulongs’s formula
d) Cauchy’s formula
View Answer

Answer: c
Explanation: Higher calorific value of the fuel can be determined by using Dulongs’s formula. Let C, H, O and S represent the percentage by weight of carbon, carbon, oxygen and sulfur respectively.
HCV = 1/100 [33900 + 144000(H – (O/8)) + 9295 S] kJ/Kg
This formula gives gross heating value in terms of the weight fractions of carbon, hydrogen, oxygen and sulfur from the ultimate analysis.

7. Lower calorific value can be determined by equation:
a) LCV = HCV – m*2466
b) LCV = HCV + (m/2466)
c) LCV = HCV – (m/2466)
d) LCV = HCV + (m*2466)
View Answer

Answer: a
Explanation: Lower calorific value can be determined equation [LCV = HCV – m*2466]. The latent heat lost to the atmosphere depends on evaporation pressure and the amount of water vapors formed. Due to difficulty in measuring the evaporation pressure, it is assumed that evaporation takes place at a saturation temperature of 15°C. The latent heat corresponding to this saturation is 2466 kl/kg.
m = mass of water vapour formed per kg of fuel burnt.

8. Which fuel has higher calorific value among given fuels?
a) Natural gas
b) Gasoline
c) Diesel
d) Fuel oil
View Answer

Answer: b
Explanation: Gasoline also called as petrol, has the highest calorific value. Gasoline is a transparent petroleum derived liquid that is used primarily as a fuel in internal combustion engines. It consists of mostly of organic compounds obtained by the fractional distillation of petroleum, enhanced with variety of additives.

9. What is amount of minimum air required per kg of liquid fuel for complete combustion using carbon, oxygen, hydrogen and sulfur?
a) 1/23 [8/3 C + 8(H – (O/8)) + S]
b) 1/100 [8/3 C + 8(H – (O/8)) + S]
c) 1/100 [8/3 C + 8(H – (0/8))]
d) 1/23 [8/3 C + 8(H – (0/8))]
View Answer

Answer: a
Explanation: Let C, H, O and S represent percentage by mass of carbon(C), Hydrogen (H2), oxygen and sulfur respectively.
The mass of oxygen required for complete combustion of fuel is given by,
= 1/100 [8/3 C + 8H – O + S]
= 1/100 [8/3C + 8(H – (0/8)) + S]
As air contains 23% of oxygen by mass, minimum air required for burning one kg of liquid fuel completely is given by,
Min. air required = 1/100 [8/3 C + 8(H – (O/8)) + S] 100/23
= 1/23 [8/3 C + 8 (H – (O/8)) + S].

10. What is minimum amount of air required per m3 of gaseous fuel for complete combustion?
a) 1/21 [(H2/2) + (CO/2) + 2CH4 + 3C2H4] m3/m3 of fuel
b) 1/100 [(H2/2) + (CO/2) + 2CH4 + 3C2H4] m3/m3 of fuel
c) 1/21 [(H2/2) + (CO/2) + 3C2H4] m3/m3 of fuel
d) 1/100 [(H2/2) + (CO/2) + 3C2H4] m3/m3 of fuel
View Answer

Answer: a
Explanation: Volumetric analysis of fuels hydrogen, carbon monoxide, methane, ethane, carbon dioxide and nitrogen is done and required minimum amount of oxygen is found for one m3 of gaseous fuel:
O2 required/m3 of fuel = 1/100[(H2/2) + (CO/2) + 2CH4 + 3C2H4] m3
As atmospheric air contains 21% of O2 by volume, minimum air required t burn one m3 of gaseous fuel is given by,
Minimum volume of air required (cm3/m3 of fuel):
= 1/100 [(H2/2) + (CO/2) + 2CH4 + 3C2H4] 100/21
= 1/21 [(H2/2) + (CO/2) + 2CH4 + 3C2H4] m3/m3 of fuel.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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