Applied Chemistry Questions and Answers – Fuel Technologies Problems

This set of Applied Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Fuel Technologies Problems”.

1. Calculate the GCV and NCV of a fuel from the following data. Volume of the fuel burnt at STP is 0.08m2.Weight of the water used for cooling is 24kg.Temperature of inlet and outlet water is 25oC. and 30oC respectively. Weight of water obtained by steam condensation is 0.02kg.
a) 1234.50cal/g
b) 1353.25cal/g
c) 1225.50cl/g
d) 1335.25cal/g
View Answer

Answer: b
Explanation: For calculating the GCV of the coal use the formula GCV=w (T2-T1)/V, where w=weight of water and v=volume of fuel burnt. Substitute the corresponding values and we know that NCV is given by NCV=GCV – [(M/V)587], M is the weight of the water obtained by steam.

2. Calculate the HCV and NCV of the coal for the following data.
Weight of the coal=0.8g
Water equivalent of calorimeter=460g
Weight of the water=2600g
Rise in temperature=2.42oC
Cooling corrections=0.052oC
Fuse wire corrections=10 calories
H=6% and assume latent heat of steam=600cal/g.
a) 9131.4cal/g
b) 9113.6cal/g
c) 9800cal/g
d) 9220cal/g
View Answer

Answer: a
Explanation: We know the formula for HCV and that is [(W+w) (t2-t1+cooling corrections)/x]-fuse wire corrections there W is the weight of the water and w is the water equivalent and t2 and t1 are the final and initial temperatures. Substitute the values to get the HCV and for NCV=HCV – (0.09H*latent heat of steam).

3. A sample of coal containing the following elements:
C=90%; H=5%; ash=4%; weight of the coal burnt=0.90g; weight of the water taken=600g; water equivalent of bomb calorimeter=2000g; rise in temperature=2.48oC; fuse wire correction=10cal; cooling correction=0.02oC; acid correction=50 cal. Calculate net and GCV of coal in cal/g. Assume latent heat of steam is 580cal/g.
a) HCV=7155.55cal/g; NCV=6894.55cal/g
b) HCV=6894.55cal/g; NCV=7155.55cal/g
c) HCV=7171.2cal/g; NCV=6889.2cal/g
d) HCV=6889.2cal/g; NCV=7171.2cal/g
View Answer

Answer: a
Explanation: For HCV we can use the formula HCV = (W+w) (t2-t1+cooling corrections)-(acid + fuse corrections)/weight of the fuel. So, the NCV can be given by subtracting the 0.09H*latent heat of steam from HCV.

4. Calculate the weight and volume of air required for the complete combustion of 2kg of carbon.
a) weight of air=5000g; volume of the air=17000l
b) weight of air=17000g; volume of the air=15000l
c) weight of air=5330g; volume of the air=17766.66l
d) weight of air=8220g; volume of the air=1555.45l
View Answer

Answer: c
Explanation: For calculation first the reaction is C+O2 → CO2.For this equation, 12 kg of c requires 32kg of oxygen for combustion then find for 2kg of carbon dioxide. At NTP 1g mole of oxygen occupies 22.4l then calculates for the weights of carbons that has calculated before then find volume of air by knowing that 100 parts of air contains 21 parts of oxygen.

5. The coal sample contains=80%; H=4%; O=2%; N=12%; S=2%, the remainder is ash and then calculate the volume of air required for perfect combustion of 1kg of fuel assuming STP conditions.
a) weight of oxygen required=10.65kg; volume of the oxygen=8725l
b) weight of oxygen required=11.25kg; volume of the oxygen=8500l
c) weight of oxygen required=11.25; volume of the oxygen=8200l
d) weight of oxygen required=10.65kg; volume of the oxygen=8166.6l
View Answer

Answer: d
Explanation: The total weight of oxygen required for combustion is calculated by taking every component equation. It will be 2.45kg and then calculate the volume of oxygen. At STP conditions 32g of oxygen occupies 22.4l volume. Calculate for 2450g and then find weight of oxygen required as we know 100kg of gas requires 23kg of oxygen.
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6. CO=40%, H2=12%, CO2=6%, CH2=2%, N2=50% are the components in the coal. What will be the composition of the dry flue gas, if the 50% excess air was used for burning 100m3 of gas?
a) CO2=14.9%, N2=80%, O2=5.2%
b) CO2=12.2%, N2=70.8%, O2=6.5%
c) CO2=14.9%, N2=79.9%, O2=5.2%
d) CO2=13%, N2=77%, O2=75
View Answer

Answer: c
Explanation: Form a table formulating the actual components in 100m3, combination reactions and volume required and % of dry gas. Find total volume of gas. Later calculate the volume of oxygen in excess air by calculating the volume of dry products. Add the dry products combinations to get the total products combination and then calculate the carbon dioxide, nitrogen and oxygen percentages.

7. Which of the following releases less amount of carbon dioxide per unit of energy?
a) Coal
b) Oil
c) LPG
d) Petrol
View Answer

Answer: b
Explanation: Oil releases very less amount of carbon dioxide when compared to coal, LPG and petrol but it also has very small calorific values when compared to those fuels and less efficient.

8. Which of the following is not the constitute of CNG?
a) Ethane
b) Propane
c) Isobutane
d) CO
View Answer

Answer: c
Explanation: Isobutane is also called as LPG is not the constitute of the CNG. Ethane, propane and other gases like nitrogen, CO are the constituents of the CNG.

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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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