# Engineering Chemistry Questions and Answers – Calorific Value – 2

This set of Engineering Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Calorific Value – 2”.

1. Which of the following factor does not affect the gross calorific value?
a) CO2 emissions from the fuel
b) Latent heat produced in the fuel
c) Size of the fuel
d) Moisture content of the fuel

Explanation: In calculating gross calorific value, the products of combustion are cooled down which does not depends upon the size of fuel.

2. A coal sample contains C=75.5%, H=11.25%, O=11.25, N=1%, S=1%. Caluclate gross calorific value?
a) 9568 cal/gm
b) 10000 cal/gm
c) 9799.010 cal/gm
d) 9518.893 cal/gm

Explanation: By using Dulong’s formulae –
GCV = [8080 %C + 34500(%H/1 – %O/8) + 2240 %S]/100,
Placing the values we get
GCV = 9518.893 cal/gm.

3. Calculate the net calorific value of the fuel if it’s ultimate analysis comes out to be containing C = 64%, O = 14.56%, H = 10.44%, N = 6.52%, S = 4.48%?
a) 8245.452 cal/gm
b) 7865.550 cal/gm
c) 7693.907 cal/gm
d) 8890.654 cal/gm

Explanation: NCV = [GCV – 0.09%H×587]cal/gm
For calculating GCV use Dulong’s formulae,
GCV = [8080 %C + 34500(%H/1 – %O/8) + 2240 %S]/100,
GCV = 8245.452 cal/gm
NCV = 7693.907 cal/gm.

4. How can we calculate the calorific value of a fuel at constant pressure?
a) QC.V. = QC.P. – (Δn)×R×T
b) QC.V. = QC.P. + (Δn)×R×T
c) QC.P. = QC.V. – (Δn)×R×T
d) QC.V. = QC.P. – (Δn)×R×T×S

Explanation: If there is a decrease in the number of gaseous molecules formed after reaction, then Δn will have a negative value and consequently will be higher than QC.V.
QC.P. = QC.V. – (Δn)×R×T
where QC.P. is calorific value at constant pressure, QC.V. is calorific value at constant volume, R is gas constant, T is temperature and S is entropy.

5. When does a fuel burn with high calorific intensity?
a) When solid fuel burns in high local fuel-bed temperature
b) When the fuel burns with an appreciable flame
c) When the combustion of fuel is not proper
d) When the fuel is of large size

Explanation: This is because when fuel burns without appreciable flame, the whole heat liberated is concentrated over a relatively smaller area.

6. Flaming fuels burn with lower calorific intensities then the flameless fuels.
a) True
b) False

Explanation: Since the total heat produced by the fuel is liberated over the entire area of the burning mass and the larger the area of flame the lesser is the concentration of heat.

7. On what factors does the temperature of flame depends?
a) Calorific value of gas, total gaseous products formed and pressure
b) Calorific value of gas, total gaseous products formed and specific heat
c) Calorific value of gas, total gaseous products formed and area
d) Calorific value of gas, total gaseous products formed and intensity

Explanation: The rate of heating of an object increases as the difference in temperature between the flame and the object increases which is depended on the following factors.

8. Why does the flame temperatures calculated solely are invariably higher?
a) Since the combustion takes place instantly and completely
b) Since some heat is lost as latent heat of steam produced
c) Since the dissociating gas molecules does not absorb a little heat
d) Since the specific heat of gases decreases with temperature.

Explanation: When we calculate flame temperature solely we cannot get an exact maximum temperature due to which there is a loss in latent heat of the seam produced.

9. How can fuel have good flexibility?
a) Should have high calorific value
b) Should have high moisture content
c) Should be easy to handle and control
d) Should have high cost

Explanation: Flexibility is defined as the rate of response in heat liberation with the variation in operating conditions.

Sanfoundry Global Education & Learning Series – Engineering Chemistry.

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