C=70%, O=10%, N=1%, S=5% and ash=4%.The NCV of the fuel was found

c-70-o-10-n-1-s-5-and-ash-4-the-ncv-of-the-fuel-was-found

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

I had been asked this question in an online quiz.

This interesting question is from Dullong’s Formula topic in chapter Fuel Technologies of Applied Chemistry

Correct option is A. 747.7

For explanation I would say: 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.