[quote=Steamer;21459]If you are considering a career in the engine department this will be a good way to start your education.

A modern slow-speed 2-stroke marine diesel rejects around 5 to 8 percent of the energy available in the fuel through jacket water cooling. That means if a unit of fuel contains 100 BTUs, somewhere between 5 and 8 of those BTUs will be removed from the cylinder by the cooling water.

Now, your assignment, should you choose to accept it, is to determine the number of BTUs required to produce a volume of steam that will fill the cylinder at a pressure equal to the firing pressure average (you can use BMEP since that figure is widely published for any given engine) and then figure out if there is (1): enough heat to evaporate the required amount of water, and (2): if cooling that cylinder through the “steampower” stroke will still keep it hot enough to prevent condensation in the cylinder which will kill the engine in very short order.

I think you will find that there really is no free lunch in the energy conversion business, and that is the underlying business of a marine engineer. It is also what makes this job so fascinating, once you get beyond the mechanics of keeping the plant online it becomes art and then it is really fun.

Look up: Large 2-stroke marine engine heat balance

Saturated steam tables (red flag word there)

Waste heat recovery

Combined cycle powerplants

Good luck and have fun[/quote]

Well, here’s my attempt:

Assuming a BMEP of 10 bar, a cylinder size of 100 mL, and an operating temperature of 110 degC.

We need to vaporize enough water to fill a 100 mL cyinder at 10 bar at 110 degC.

Now, according to the ideal gas eqn, the number of moles needed to fill 100 mL at 10 bar, at an operating temp of 110 degC is (values will be converted to atmospheres, liters, and Kelvins):

n=PV/TR

n=(9.87 atm * .1 L)/(.0821 L-atm/mol-degK * 383K)

n=0.0314 mol

So we need enough heat to vaporize 0.0314 mols of water, and get it to a temperature of 110 degC. The molar mass of water is 18.015 g/mol, so we need .566 g of water

The specific heat of water is 4.184 J/g/degC. Assuming the water is at 0C, we need to add

4.184 J/g/degC * .566 g * 100 degC= 23.68 J of heat

Then we need to add 2.257 J/g/degC to vaporize it, or 0.1277 J of heat.

Then we need to add more heat to bring it up to 110 degC, or

2.080 J/g/degC * .566 g * 10 degC = 11.77 J

Total heat needed is 35.58 J

And I did figure out that since 100 mL of steam can be produced from .566 mL of water, the water will expand 176.68 times in this enviroment.

Now, I’m stuck at answering the other questions, and I doubt this answer is correct.