Engineering Questions with Answers - Multiple Choice Questions

# Applied Chemistry MCQ – Dullong’s Formula

1 - Question

Calculate the net calorific value of a coal sample having the following composition:
C=80%, H=5%, O=4%, N=3%, S=3.5% and ash=5%.
a) 7251.8cal/g
b) 7780.5cal/g
c) 7621.5cal/g
d) 7830.75cal/g

Explanation: Apply the dulong’s formula that is: HCV = 1/100[8080C + 34500(H-O/8) + 22400S], here the C, S, O, H are the percentages of carbon, sulphur, oxygen and hydrogen. So, substitute all the given values in the formula and calculate so that you will get HCV (or) GCV as 8094.9cal/g and then apply the formula NCV=(GCV-0.09H*587), her 587cal/g is the latent heat of steam then you will get NCV=7830.75cal/g.

2 - Question

A coal has the following composition by weight:
C=90%, O=4%, N=1%, S=0.5% and ash=5%.The NCV of the fuel was found to be 8480cal/g. Calculate the percentage of hydrogen and HCV of the fuel.
a) H=4.21%, HCV=8621.80cal/g
b) H=4.521%, HCV=8221.80cal/g
c) H=4.686%, HCV=8727.37cal/g
d) H=4.1%, HCV=8221.37cal/g

Explanation: Apply the dulong’s formula that is: HCV = 1/100[8080C + 34500(H-O/8) + 22400S], here the C, S, O, H are the percentages of carbon, sulphur, oxygen and hydrogen. So, substitute all the given values in the formula and calculate so that you will get HCV. As we don’t know the value of H, you will get HCV = [7110.7+345H]Cal/g. Let it be equation -1 and then we know that NCV = (GCV-0.09H*587), here 587cal/g is the latent heat of steam. NCV is given then you will get GCV = [8480+52.83H] Cal/g and let it be equation-2. So equate both the equations to get the value of H. you will get H=4.686 and substitute in equation-1 to get the value of HCV=8727.37cal/g.

3 - Question

C=70%, O=10%, N=1%, S=5% and ash=4%.The NCV of the fuel was found to be 9210cal/g. percentage of hydrogen be x and HCV of the fuel be y. Find out y/x.
a) 747.7
b) 768
c) 777
d) 676.9

Explanation: Apply the dulong’s formula that is: HCV = 1/100[8080C + 34500(H-O/8) + 22400S], here the C, S, O, H are the percentages of carbon, sulphur, oxygen and hydrogen. So, substitute all the given values in the formula and calculate so that you will get HCV. As we don’t know the value of H, you will get HCV=[5336.75+345H]Cal/g. Let it be eauatin-1 and then we know that NCV = (GCV-0.09H*587), here 587cal/g is the latent heat of steam. NCV is given then you will get GCV=[9210+52.83H]Cal/g and let it be equation-2. So equate both the equations to get the value of H. you will get H=13.25% and let it be x and substitute in equation-1 to get the value of HCV=9908cal/g. Now let be y and divide y with x to get 747.7.

4 - Question

A formula giving the gross heating value of coal in terms of the weight fractions of carbon, hydrogen, oxygen and sulphur from the _________
a) ultimate analysis
b) proximate analysis
c) distillation
d) filtration

Explanation: The percentages of the coal components are given by ultimate analysis and the proximate analysis is used to know the percentages of volatile matter, moisture, etc.

5 - Question

The dulong’s formula is not applicable for __________
a) solid fuel
b) gaseous fuel
c) liquid fuel
d) any fuel

Explanation: For calculating the components of coal and the petroleum the dulong’s formula is applied. So, the dulong’s formula is applied for both solids and liquids and not for gaseous.

6 - Question

Modification of the dulong’s formula can be done by considering the ___________
a) latent heat
b) heat
c) fuel state
d) liquid fuels

Explanation: The modification of the formula is done by considering the latent heat. It may be of water or vapour or steam accordingly. It represents the heating value of the fuels.

7 - Question

Calculate the HCV of the coal from the given data:
Weight of the fuel burnt=0.92
Initial temperature=120C
Final temperature=19.20C
Weight of the water in calorimeter is 1458g
Water equivalent of calorimeter = 14g.
a) 11520k.cal/m3
b) 11560k.cal/m3
c) 11000k.cal/m3
d) 11590k.cal/m3

Explanation: Use the formula HCV = [(W + w)(t2-t1)]/x, where W = Weight of the water in calorimeter, w = Water equivalent of calorimeter, t2 = Final temperature, t1 = Initial temperature, x = Weight of the fuel burnt. By substituting, you will get 11520k.cal/m3 as the final answer.

8 - Question

While calculating HCV, if we need to apply the fuse wire, acid and cooling corrections then what is the formula for HCV?
a) HCV = [(W + w)(t2-t1 + cooling correction)-(acid correction + fuse correction)]/weight of fuel
b) HCV = [(W + w)(t2+t1 + cooling correction)-(acid correction + fuse correction)]/weight of fuel
c) HCV = [(W + w)(t2-t1 + acid correction)-(cooling correction + fuse correction)]/weight of fuel
d) HCV = [(W + w)(t2+t1 + acid correction)-(cooling correction + fuse correction)]/weight of fuel