Do the math or have the computer do it for you.

The simulation results don't lie.

simulation for one cylinder of a 617 engine.

No boost Work W_net= 2.22517 kJ

roughly 7.25psi of boost Work W_net= 2.49142 kJ

A 11.965% increase in power for the exact same energy added. Of course these are done with an ideal gas (Note I tried a simulation using data for regular air but the simulator swallowed its navel) and an ideal diesel cycle, but the trends will hold true to a real engine as far as an increase in power.

If you lost power because a sensor wasn't getting the right data all bets are off. That can play total havoc on a system. I re-ran the simulation with higher boost simulating what I think the IP would do if it thought it had no boost. It results in a power loss due to distortion of the diesel cycle.

roughly 7.25psi of boost screwed IP sensor W_net= 2.13984 kJ

or a 3.835% loss in power.

This is why we have a sensor that runs to the IP from the intake manifold to adjust for boost. Note that the same energy was added to system in all three cases. Keep those banjo bolts clean .

I have attached the revised simulation data set showing the three cases.

Attached Files

Boost-simulation-4a-diesel-cycle-revised.txt (18.2 KB, 96 views)

As far as the turbo verses supercharger goes. The reason I brought up the trubo compounded Wright turbo cyclone was to show that the loss of power due to the increased back pressure was insignifigant to the power extracted by the turbine. A supercharger draws all the power it uses from the crank an thus the output power, a turbo draws most of its power from energy normally wasted and blown out the tail pipe. All boost levels equal, a turbo charged engine has less parasitics and makes more usable power.

OK, I spent a few minutes trying to understand your model and I have a few questions:

- Is this modeled as a Diesel Cycle or an Air Standard Duel Cycle (which is the actual cycle for modern diesel engines)?

- I assume (if this is a Diesel Cycle model) that state 1-2 is constant S compression, state 2-3 is constant pressure combustion (during 1/3 of the power stroke), state 3-4 is constant S power stroke (with no combustion) and state 4-1 includes the exhaust and intake strokes. Is this correct?

- How did you (or the model) establish the temperatures at state 3 (6500 C or 11700 F) and state 4 (4580 C or 8270 F)? These values are not realistic. In an accurate model, the state 4 temp would represent the exhaust gas temperature, which should be around 1000 F or about 540 C.

Am I reading this correctly?

Its an ideal diesel cycle. I suppose I could do a Duel cycle. I t will be harder to keep the energy added the same for boost vs no boost. I don't have time now, mabey tuesday night.

Yes, that is correct. Though with the higher boost the combustion takes place in a shorter portion of the power stroke to keep the pressure level the same. In the real world the IP boost sensor would allow the IP to compensate in the same way. The same energy is added in each case.


The model is calculating temp based on pressure, volume, and mass of air. The models can be tweaked to do the same simulation with less energy added to show more realistic values. The same trend will emerge, more power from more boost however.

OK, that helps me understand a little more of what you've done, but I don't really have enough info to find the major problem in the model yet. I suspect the high temperatures being predicted are causing some issues. Ideal gas laws don't really work at those pressures and temperatures, anyway. CO2, in particular, does not act anything like an ideal gas at higher pressures. The model does seem to be generating a strange Cp value (about 1.2) under the high temperature conditions. That may be causing it to over-predict the net work output. In addition, the model does not account for the work required to expel the exhaust gases against the turbocharger back-pressure, but I'm not sure that is enough difference to explain the error. Maybe running the model with a duel cycle and realistic temperatures will produce a more reasonable result. I was just wondering what the model predicts if you run it with no heat input from combustion, does it still show any net work being generated due to the boost? This is interesting, but I'm not sure how much time you really want to spend doing this.

BTW, I just noticed something else. Your boost cases assume the inlet air temperature is 25 C, just like your non-boost case. Unless you have a 100% efficient intercooler, that is not a valid assumption.

Well, after further though the temperatures at stage 4 may not be to ridiculous. This is the temp before the exhaust valve opens. When the exhaust valve opens the temp and pressure will drop signifigantly due to the gases expanding into the manifold.

I was trying to show that boost affected power all things being equal. If you want me to add energy to the system..... Okay....


############################################################################################
# Copy (select text by dragging the mouse & pressing Ctrl_c) and paste (Ctrl_v) the following TEST-code onto a file
# to save for later use as a baseline case. To load this solution in a new session launch the appropriate daemon,
# Test>Daemons>Systems>Closed>Process>Specific>PowerCycle>Ideal
# Paste the TEST-code on this panel, and click the Load and Super-Calculate buttons to recover the solution.
############################################################################################
#
#--------------------Start of TEST-Codes-----------------------------------------------------------------------------

States {
State-1: Air;
Given: { p1= 151.325 kPa; T1= 75.0 deg-C; Vel1= 0.0 m/s; z1= 0.0 m; Vol1= 0.6288893 L; }

State-2: Air;
Given: { s2= "s1" kJ/kg.K; Vel2= 0.0 m/s; z2= 0.0 m; m2= "m1" kg; Vol2= 0.02925067 L; }

State-3: Air;
Given: { p3= 10070.8955 kPa; Vel3= 0.0 m/s; z3= 0.0 m; m3= "m2" kg; }

State-4: Air;
Given: { s4= "s3" kJ/kg.K; Vel4= 0.0 m/s; z4= 0.0 m; m4= "m3" kg; Vol4= 0.6288893 L; }
}

Analysis {
Process-A: b-State = State-1; f-State = State-2;
Given: { Q= 0.0 kJ; T_B= 25.0 deg-C; }

Process-B: b-State = State-2; f-State = State-3;
Given: { Q= 5.3297 kJ; T_B= 25.0 deg-C; }

Process-C: b-State = State-3; f-State = State-4;
Given: { Q= 0.0 kJ; T_B= 25.0 deg-C; }

Process-D: b-State = State-4; f-State = State-1;
Given: { T_B= 25.0 deg-C; }
}


#----------------------End of TEST-Codes---------------------------------------------------------------------------------
#
# Solution Time: Dec 5, 2005 1:12:10 PM;
# Daemon Path: Daemon Path: Systems>Closed>Process>Specific>PowerCycle>Ideal Gas; v-7.0bsW;
#
#--------------Spreadsheet Friendly Property Table (tab-delimited, 999999 signifies unknown value)---------------------
#
# State p(kPa) T(K) v(m3/kg) u(kJ/kg) h(kJ/kg) s(kJ/kg)
# 1 151.32 348.2 0.6603 -49.53 50.38 6.924
# 2 10070.9 1077.7 0.0307 528.47 837.74 6.924
# 3 10070.9 3694.0 0.1053 2983.74 4043.86 8.426
# 4 920.31 2117.3 0.6603 1490.94 2098.58 8.426

#
#**********************Detailed Output*************************************************************************************
#
# States
#
# State-1: Air > Ideal Gas;
# Given: p1= 151.325 kPa; T1= 75.0 deg-C; Vel1= 0.0 m/s;
# z1= 0.0 m; Vol1= 0.6288893 L;
# Calculated: rho1= 1.514547 kg/m^3; v1= 0.6602634 m^3/kg; u1= -49.53362 kJ/kg;
# h1= 50.380745 kJ/kg; s1= 6.9239554 kJ/kg.K; e1= -49.53362 kJ/kg;
# j1= 50.380745 kJ/kg; m1= 9.5248246E-4 kg; MM1= 28.97 kg/kmol;
# R1= 0.28698653 kJ/kg.K; c_p1= 1.0111803 kJ/kg.K;
#
# State-2: Air > Ideal Gas;
# Given: s2= "s1" kJ/kg.K; Vel2= 0.0 m/s; z2= 0.0 m;
# m2= "m1" kg; Vol2= 0.02925067 L;
# Calculated: p2= 10070.8955 kPa; T2= 804.51904 deg-C; rho2= 32.562756 kg/m^3;
# v2= 0.030709932 m^3/kg; u2= 528.4663 kJ/kg; h2= 837.7428 kJ/kg;
# e2= 528.4663 kJ/kg; j2= 837.7428 kJ/kg; MM2= 28.97 kg/kmol;
# R2= 0.28698653 kJ/kg.K; c_p2= 1.1510557 kJ/kg.K;
#
# State-3: Air > Ideal Gas;
# Given: p3= 10070.8955 kPa; Vel3= 0.0 m/s; z3= 0.0 m;
# m3= "m2" kg;
# Calculated: T3= 3420.8247 deg-C; rho3= 9.49976 kg/m^3; v3= 0.10526582 m^3/kg;
# u3= 2983.738 kJ/kg; h3= 4043.859 kJ/kg; s3= 8.426421 kJ/kg.K;
# e3= 2983.738 kJ/kg; j3= 4043.859 kJ/kg; Vol3= 1.00263846E-4 m^3;
# MM3= 28.97 kg/kmol; R3= 0.28698653 kJ/kg.K; c_p3= 1.2338063 kJ/kg.K;
#
#
# State-4: Air > Ideal Gas;
# Given: s4= "s3" kJ/kg.K; Vel4= 0.0 m/s; z4= 0.0 m;
# m4= "m3" kg; Vol4= 0.6288893 L;
# Calculated: p4= 920.30676 kPa; T4= 1844.1788 deg-C; rho4= 1.514547 kg/m^3;
# v4= 0.6602634 m^3/kg; u4= 1490.9385 kJ/kg; h4= 2098.5833 kJ/kg;
# e4= 1490.9385 kJ/kg; j4= 2098.5833 kJ/kg; MM4= 28.97 kg/kmol;
# R4= 0.28698653 kJ/kg.K; c_p4= 1.2338063 kJ/kg.K;
#
# Analysis
#
# Process-A: b-State = State-1; f-State = State-2;
# Given: Q= 0.0 kJ; T_B= 25.0 deg-C;
# Calculated: W_B= -0.5505348 kJ; S_gen= -0.0 kJ/K; Delta_E= 0.5505348 kJ; Delta_S= -0.0 kJ/K;
#
#
# Process-B: b-State = State-2; f-State = State-3;
# Given: Q= 5.3297 kJ; T_B= 25.0 deg-C;
# Calculated: W_B= 2.9910967 kJ; S_gen= -0.016444828 kJ/K; Delta_E= 2.3386033 kJ; Delta_S= 0.0014310727 kJ/K;
#
#
# Process-C: b-State = State-3; f-State = State-4;
# Given: Q= 0.0 kJ; T_B= 25.0 deg-C;
# Calculated: W_B= 1.4218653 kJ; S_gen= -0.0 kJ/K; Delta_E= -1.4218653 kJ; Delta_S= -0.0 kJ/K;
#
#
# Process-D: b-State = State-4; f-State = State-1;
# Given: T_B= 25.0 deg-C;
# Calculated: Q= -1.4672726 kJ; W_B= 0.0 kJ; S_gen= 0.003490184 kJ/K; Delta_E= -1.4672726 kJ;
# Delta_S= -0.0014310727 kJ/K;
#
# Cycle:
#
# Calculated: T_max= 3420.8247 deg-C; T_min= 75.0 deg-C; p_max= 10070.8955 kPa;
# p_min= 151.325 kPa; Q_in= 5.3297 kJ; Q_out= 1.46727 kJ;
# W_in= 0.55053 kJ; W_out= 4.41296 kJ; Q_net= 3.86243 kJ;
# W_net= 3.86243 kJ; eta_th= 72.46988 %; MEP= 6441.2583 kPa;
#
# **************************************************************************************************** *************************************
# TO USE THIS PANEL AS A CALCULATOR, ENTER AN EXPRESSION IN MICROSOFT EXCEL FORMAT, AND PRESS ENTER.
# PROPERTIES OF CALCULATED STATES (p1, s2, mdot3, etc.) CAN FORM PART OF THE EXPRESSION. EX: =Sin30+ln(5)*exp(-2.3),
# =2+10^2-3.4*(10/2+4*(2+3)), =sqrt(atan(.5)+alog2), =u1+p1*v1; =s2-s1, =p1*v1^1.4/v2^1.4, =mdot1*(h9-h1).
#

I don't think so, the melting point of carbon steel is below 2800 F (about 1500 C). If you want an engine to operate, the peak temperature in the cylinders needs to be significantly less.

hmmm... thats strange.. power went up... I think that there is an optimal cutoff ratio as far as efficiency. I'm on the high side.

Its not the peak temp, its the average temp. well mabey not the average but the metal temp stays below peak

Your code is using "average" temperature (at any given time) for the ideal gas calculations. In any event, the exhaust gas has to flow through the exhaust valves and seats without melting them. The numbers in your code are not even close to reality. If you want to try to draw any valid conclusions you have to use reasonable inputs.

The trend should hold valid as far as a power increase. If the gasses were always below the melting point of the metal engines wouldn't need cooling systems.

Hold on, before you crunch any more numbers. There is no trend here, I'm just trying to help you understand why your model is generating incorrect results.

You have a computer model telling you that you can generate more power output with 50% more air and the same amount of fuel (energy input), resulting in significantly lower combustion temperatures. Those results make no sense. Lower temperatures are not going to give you higher efficiencies or more power. Part of the problem is that you have modeled the incorrect cycle. You are also assuming ridiculously high temperatures throughout the cycle (at which the ideal gas approximation is not valid). Your model is also telling you that you have a pressure of 240 psi and a temperature of 8000 F remaining in the cylinder at the end of the power stroke. These values are just silly. This model might produce meaningful results if you get the input values closer to reality.

If you use ideal gas assumptions to model a duel cycle and limit the energy input such that the state 4 temperature is in the range of 1000F and the pressure is close to atmospheric you may get useful results. Hopefully that will generate more reasonable state 2 temperatures as well. Once that generates reasonable results, you can model the affects of increasing boost. What you will find is that adding the same energy input to 50% more air will lower the state 3 temperature considerably. You will now have an engine running with a lower temperature difference between the inlet air and the combustion temperature. This, of course, will result in lower efficiency and lower (or effectively the same) power output, per the second law of thermodynamics. There is really no need to go through this exercise, but if want to see the numbers that should be the correct approach.

Not quite true, The energy is being put into heating more mass. This is why the temps are lower. By heating more air you get more expansion pressure. This is where the efficiency comes from.

The model assumes no energy is lost or absorbed by the cylinder, piston, or head. If engines were made out of perfect insulators these temps wouldn't be that far off. Unless you can tell me a way of modeling a cooling system these are as good as this is going to get. I might be able to do a duel cycle it will just take time. I also might be able to get the real air simulation to work.


The biggest thing is the cooling systems effect. That radiator is big for a reason. Regardless of the numbers that come out the trends should hold true. more boost = more power.