Page 237 - Handbook of Energy Engineering Calculations

volume in the dry exhaust the nitrogen content is: N = (76.9/100)(15) =
2
11.53 lb (5.2 kg) N per lb (0.454 kg) of fuel oil. Converting to moles, 11.53
2
lb (5.2 kg) N /28 lb (12.7 kg) fuel per hour = 0.412 mole N per lb (0.454 kg)
2
2
of fuel oil.

2. Compute the weight of N in the exhaust
2
Use the relation, percentage of CO in the exhaust gases = (CO )/(N + CO )
2
2
2
2
in moles. Substituting, (13.7)/(100) = (CO )/(N + 0.412). Solving for CO ,
2
2
2
we find CO = 0.0654 mole.
2
Now, since mole percent is equal to volume percent, for 9 percent CO in
2
the exhaust gases, 0.09 = (CO )/(CO + N ) = 0.0654/(0.0654 + N ). Solving
2
2
2
2
for N , we find N = 0.661 mole. The weight of N therefore = 0.661 × 28 =
2
2
2
18.5 lb (8.399 kg).
3. Calculate the amount of air required for combustion
The air required for combustion is found from (N ) = 18.5/0.769, where
2
0.769 = percent N in air, expressed as a decimal. Solving, N = 24.057 lb
2
2
(10.92 kg) per lb (0.454 kg) of fuel oil.

4. Find the weight of the actual air charge drawn into the cylinder
Specific volume of the air at 60°F (15.6°C) and 29.8 in (75.7 cm) Hg is 13.03
3
3
ft (0.368 m ) per lb. Thus, the actual charge drawn into the cylinder = (lb of
3
air per lb of fuel)(specific volume of the air, ft /lb)(fuel consumption,
3
3
lb/h)/3600 s/h. Or 24.1(13.02)(28)/3600 = 2.44 ft (0.69 m ) per second.
5. Compute the volumetric efficiency of this engine

Volumetric efficiency is defined as the ratio of the actual air charge drawn
into the cylinder divided by the piston displacement. The piston displacement
for one cylinder of this engine is (bore area) (stroke length)(1 cylinder)/1728
3
3
2
in /ft · Solving, piston displacement = 0.785(4.25) (6)(1)/1728 = 0.0492 ft 3
3
(0.00139 m ).
The number of suction strokes per minute = rpm/2. The volume displaced
per second by the engine = (piston displacement per cylinder)(number of

cylinders)(rpm/2)/60 s/min. Substituting, engine displacement = 0.0492(6)