I have no suggestions for your problem, but...

Each piston makes power every 2 revolutions. The engine makes power twice per revolution, or every 180 deg. on the crankshaft. The camshafts turn once for every two turns of the crankshaft.

 

An ideal Diesel cycle has a maximum cycle temperature of 2000*C and a cutoff ratio of 1.2. The state of the air at the beginning of the compression is P1 = 95kPa and T1 = 15*c. The cycle is executed in a four-stroke, eight-cylinder engine with a cylinder bore of 10cm and a piston stroke of 12cm. The minimum volume enclosed in the cylinder is 5 percent of the maximum cylinder volume. Determine the power produced by this engine when it is operated at 1600 rpm. Use constant specific heats at room temperature. Ans: 105 kW

So,

P2 = P3
V4 = V1 = Vmax = pi * r^2 * h = pi * .05^2 * .12 = 9.42E-4 cubic meters

m = P1V1/RT1 = 0.0010832292 kg

V2 = 0.5 * V1 = 4.7123E-5 cubic meters

V3 = Rc V2 = 1.2 * V2 = 5.65E-5 cubic meters

1-2 isentropic
T2 = T1 (V1/V2)^k-1 = 288 * 20^1.4-1 = 954.6

T3 = 2273K

Qin = m(h3-h2) = mCp(T3-T2)
Qin = 0.00108 kg * (1.005) * (2273 - 955)K
Qin = 1.44 kJ

3 - 4 isentropic
T4 = T3 (V3/V4=V1)^k-1 = 2273 * (5.66E-5 / 9.42E-4)^.4 = 738 K

Qout = m(u4- u1) = mCv(T4-T1)
Qout = 0.00108 kg * .718 * (738 - 288)K
Qout = 0.350

Wnet = Qin - Qout
Wnet = 1.44 - .350 kJ
Wnet = 1.09 kJ

Wtotal = Wnet * RPS/2 * cylinders
Wtotal = 1.09 * 26.6/2 * 8 cylinders
Wtotal = 116 kJ/s = 116 kW

 

I have a t-score of 1.36 in this class. I'm pretty sweet. h:

 

That's what I get too :dunno:

 

i should done stayed in engineering school so i can heal peoples

 

holy crap that looks confusing as hell

 

Where does the purple go? :nervous:

 

math makes me :loco:

 

I got 116 as well ;o

 

dot. too bad I'm pretty fawked in there.