1. Which of the following order of petrol knocking is true?
a) aromatics > branched paraffins > olefins > cyclo paraffins > aromatics
b) straight chain paraffins > olefins > branched paraffins > cyclo paraffins > straight chain paraffins
c) straight chain paraffins > branched paraffins > olefins > cyclo paraffins > aromatics
d) straight chain paraffins > branched paraffins > olefins > aromatics > cyclo paraffins
Explanation: The order of petrol knocking is given by straight chain paraffins > branched paraffins > olefins>cyclo paraffins > aromatics. As we know that aromatics are the best anti knock agents so, the petrol knock for them is very low.
2. In fixed bed catalytic cracking, the catalyst is reactivated after every __________
a) 8-10hrs
b) 20-24hrs
c) 5-7hrs
d) 12-15hrs
Explanation: In fixed-bed catalytic cracking, the catalyst is reactivated after every 8-10hrs and continuous by burning the deposited carbon.
3. In vapour phase thermal cracking, the temperature achieved is ___________
a) 200-400oC
b) 600-650oC
c) 500-820oC
d) 700-800oC
Explanation: In vapour phase thermal cracking, the temperature achieved is 600-650oC and in fixed-bed catalytic cracking, the temperature is about 425-250oC.
4. In moving bed catalytic cracking, the catalyst is in the form of ____________
a) fine powder
b) pallets
c) liquid form
d) gaseous form
Explanation: In moving bed catalytic cracking, the catalyst is made into a fine powder and charged into the catalyst chamber. So, they move freely like fluids.
5. The gasoline vapours are purified by __________
a) thermal cracking
b) catalytic cracking
c) stabilization
d) knocking
Explanation: The stabilization is the process in which the gasoline vapours are purified. Thermal cracking and catalytic cracking are the processes to purify the petrol.
6. The pressure applied in vapour phase cracking is __________
a) 10-20kg/cm2
b) 100kg/cm2
c) 50kg/cm2
d) 80kg/cm2
Explanation: The pressure applied in the vapour phase cracking is very small about 10-20kg/cm2.The oils will be very easily vaporized.
7. The pressure applied in fixed-bed catalytic cracking is __________
a) 10kg/cm2
b) 50kg/cm2
c) 100kg/cm2
d) 1.5kg/cm2
Explanation: The pressure of 1.5kg/cm2 applied in the fixed bed catalytic cracking and during this process 40% of charge is converted into gasoline.
8. 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
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.
9. 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
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).
10. 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
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.