What happens when fuel is burned , what are compounds
released??
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Answer / engr. muhammad omer farooqi
End products of complete combustion are always carbon dioxide (CO2) and water (H2O). However incomplete combustion may lead to carbon monoxide (CO). Also this is true for 100% pure fuel which is not the case since every hydrocarbon has got some impurity which also shows itself in final products.
| Is This Answer Correct ? | 12 Yes | 0 No |
Answer / lafangey
along with poisonous gases,water also forms.....coz any fuel(hyrdrocarbon) has H in its molecules so it reacts with O2 and form water.
| Is This Answer Correct ? | 10 Yes | 0 No |
Answer / engineer muhammad amin
Carbon di oxide is major compound exhausted but some other
like carbon mono oxide and even small particales if present
in fuel may tends to suspended in air.The other major
compounds are nitrogen sulfur but all these creates the
pollution so now a days many precautions at=re taken into
account before manufacturing these types of fuels and well
pre treated before the final separation.
| Is This Answer Correct ? | 9 Yes | 2 No |
CHEMICAL FLUID MECHANIC - EXAMPLE 3.3 : The drag coefficient Cd = 0.05 and lift coefficient Cl = 0.4 for a levelled flow aircraft are measured. The velocity of the aircraft is v = 150 ft / s with its weight W = 2677.5 pound-force. (a) Find the value of the lift of the aircraft, L, when it is also its weight. (b) The drag of the aircraft, D = Cd M, L = Cl M. Find the value of D. (c) The power required is P = Dv. If 1 pound-force x (ft / s) = 1.356 W, find the value of P in the unit of Watt or W.
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hi all i have doe cheical engg, but i have to go for iob po interview so can anybody suggest me abt the kind of qns that can be asked.
0 Answers Indian Overseas Bank,
ENVIRONMENTAL ENGINEERING - QUESTION 22.2 : Biochemical Oxygen Demand (BOD) could be calculated using the formula BOD = (DOi - DOf) (Vb / Vs) where Vb = Volume of bottle in ml, Vs = Volume of sample in ml, DOi = Initial dissolved oxygen in mg / L, DOf = Final dissolved oxygen in mg / L. (a) By using a bottle of Vb = 300 ml with sample Vs = 30 ml, find the BOD if DOi = 8.8 mg / L and DOf = 5.9 mg / L. (b) By using a bottle Vb = 600 mL with sample Vs = 100 mL, find the BOD if DOi = 8.8 mg / L and DOf = 4.2 mg / L. (c) Find the average BOD = [ Answer of (a) + Answer of (b) ] / 2. (d) If the BOD-5 test for (a) - (c) is run on a secondary effluent using a nitrification inhibitor, find the nitrogenous BOD (NBOD) = TBOD - CBOD. Let TBOD = 45 mg / L and CBOD = Answer of (c).
In a triple effect evaporator, the heat transfer for an evaporator is calculated as q = UA (TI – TF) where TI is the initial temperature, TF is the final temperature; U and A are constants. Given that heat transfer for the first evaporator : q(1) = UA (TI – TB); second evaporator : q(2) = UA (TB – TC); third evaporator : q(3) = UA (TC – TF) where q(x) is the heat transfer function, TB is the temperature of second inlet and TC is the temperature of third inlet, prove that the overall heat transfer Q = q(1) q(2) q(3) = UA (TI – TF).
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