- Engine volume (Vd) = 107 cc =0 ,000107 m3
- Maximum pressure (Pmax) = 40 atm = 4053 kPa
- Maximum temperature (Tmax) = 900o C
- Compression ratio (rc) = 9 : 1
- Angular velocity of engine (N) = 3000 rpm
- Volumetric efficiency (ηv) = 0,93
- Circle/cycle (n) = 2
- Specific heat ratio (k) = 1,35
- Material of channel = Comercial steel
- Diameter of turn channel (d) = 19 mm
- Difference of channel height (Δz) = 19 cm
For pressure drop at turn channel, it will be calculated manually. While for pressure drop at changing diameter, it will be calculated using Matlab. It’s done because at turn the diameter of the channel is uniform until it only needs one calculation, while at changing diameter it needs calculating with certain interval considering to Reynold Number and friction factor is function of diameter.
It can be defined, from the data above, supporting parameter to get pressure drop at turn channel from point A to B along 410 mm with 19 mm diameter, which are:
- Pressure of exhaust gas when exiting from engine valves
- Temperature of exhaust gas when exiting from engine valves
For standard analysis, air property is used to determine the exhaust gas, then from temperature of exhaust gas we can get
- ρ = 0,51092 kg/m3
- υ = 64,71748.10-6 m2/s
- Mass flow of the exhaust is equal to air plus fuel, at Tatmosfer = 30oC so ρatmosfer = 1,1774 kg/m3
- Average flow velocity
- Reynold Number
- Friction factor, because of turbulent so we use equation for turbulence flow, with e = 0,046 and diameter in millimeter (mm)
To get value of f we will use Matlab with M-file script as follows:
% Finding Friction Factor
syms f %Declaring f as symbolic variable
ff=1/sqrt(f)+(2*log10(((.046/19)/3.7)+(2.51/(6332.01*sqrt(f))));
%Making the friction factor equation by placing it at same space, so ff=0
f=solve(ff); %Finding value of f
f=double(f) %Changing f to be numeric variable
So we can get friction factor value, f = 0,0379
Until pressure at exit of the exhaust is
To get pressure drop at channel from point B to C, and also point C to D, we need to know the radius equation of the channel. Using Matlab, the M-file script is:
%% Finding Radius Equation of BC and CD
xBC=[0 230]; %Distance of point B-C
dBC=[19 84.7]; %Diameter of point B-C
yBC=dBC/2; %Radius of point B-C
pBC=polyfit(xBC,yBC,1) %Finding radius equation of point B-C
xCD=[0 390]; % Distance of point C-D
dCD=[84.7 100.2]; % Diameter of point C-D
yCD=dCD/2; % Radius of point C-D
pCD=polyfit(xCD,yCD,1) % Finding radius equation of point C-D
And we get radius equation of channel BC: pBC = 0,1428X + 9,5; and for channel CD: pCD = 0,0199X + 42,35
Below is M-file script for channel BC:
%% Finding Pressure Drop
%% Data
Vd=.000107; %Engine volume (m^3)
d=2*[.1428 9.5]*1e-3;%diameter of straight channel (m)
t=20; %Divided interval
l=linspace(0,230,t);%Channel length for each dividing point (cm)
dl=polyval(d,l); %Evaluating diameter value for each length l
N=3000; % Angular velocity of engine (rpm)
E=.93; %Volumetric efficiency
n=2; % Circle/cycle
z=0; % Difference of channel height at point B to C
e=.046; %Surface roughness (mm)
g=9.81; %Gravity acceleration
rhoA=1.1774; %Density of room air at temperature 30o Celsius
%% Memory Allocation
m=zeros(1,t);
V=zeros(1,t);
R=zeros(1,t);
dP=zeros(1,t);
P=zeros(1,t);
%% Data of Air Property for T=420.4 Celsius Constant
miu=64.71748-6; %Kinematics viscosity of gas (m^2/s)
rho=.51092; %Density gas (kg/m^3)
%% Supporting Data
P=208613.86; %Exhaust Pressure(Pa)
m=(E*rhoA*Vd*(N/60)/n)*(16/15);%Mass flow (kg/s)
V=m./(rho*(pi*dl.^2/4)); %Average flow velocity (m/s)
R=V.*dl./miu; %Reynold Number
%% Friction Factor
if R<=2000;
f=64./R;
else syms f
ff=1/sqrt(f)+(2*log10(((e./(dl.*1e3))./3.7)+(2.51./(R.*sqrt(f)))));
for k=1:t;
f(k)=solve(ff(k));
end
f=double(f); % Friction Factor
end
%% Pressure Drop
dP=rho.*g.*(z+(f.*(l.*1e-2).*V.^2./(dl.*2*g)));
%Pressure drop for each length l and diameter dl
%% Exit Pressure (Pexit = Pexhaust - dP)
for k=1:t;
if (k+1)<=t;
k=k+1;
P(k)=P(k-1)-dP(k);
end
end
%% Pressure Drop Table
format long
disp(' Pressure Drop for each Length l and Diameter dl for channel of BC')
disp(' ')
disp(' Length Diameter Pressure Drop Exit Pressure')
disp([l' dl' dP' P'])
Program execution:
Pressure Drop for each Length l and Diameter dl for channel of BC
Length Diameter Pressure Drop Exit Pressure
1.0e+005 *
0 0.00000019000000 0 2.08613860000000
0.00012105263158 0.00000022457263 0.00012783497589 2.08601076502411
0.00024210526316 0.00000025914526 0.00012857743223 2.08588218759188
0.00036315789474 0.00000029371789 0.00010601026525 2.08577617732663
0.00048421052632 0.00000032829053 0.00008320307040 2.08569297425623
0.00060526315789 0.00000036286316 0.00006464672510 2.08562832753113
0.00072631578947 0.00000039743579 0.00005040396001 2.08557792357113
0.00084736842105 0.00000043200842 0.00003963789455 2.08553828567658
0.00096842105263 0.00000046658105 0.00003149626600 2.08550678941058
0.00108947368421 0.00000050115368 0.00002529694192 2.08548149246866
0.00121052631579 0.00000053572632 0.00002053160314 2.08546096086552
0.00133157894737 0.00000057029895 0.00001683030491 2.08544413056062
0.00145263157895 0.00000060487158 0.00001392516437 2.08543020539625
0.00157368421053 0.00000063944421 0.00001162157256 2.08541858382369
0.00169473684211 0.00000067401684 0.00000977713789 2.08540880668580
0.00181578947368 0.00000070858947 0.00000828673832 2.08540051994748
0.00193684210526 0.00000074316211 0.00000707201110 2.08539344793638
0.00205789473684 0.00000077773474 0.00000607396266 2.08538737397372
0.00217894736842 0.00000081230737 0.00000524774764 2.08538212622608
0.00230000000000 0.00000084688000 0.00000455895361 2.08537756727247
While for channel of CD, it is still used M-file script above, only changes data for diameter, channel length and exhaust pressure. For exhaust pressure, we take pressure exit at point C, which is 208537,756727247 Pa.
Program execution:
Pressure Drop for each Length l and Diameter dl for channel of CD
Length Diameter Pressure Drop Exit Pressure
1.0e+005 *
0 0.00000084700000 0 2.08537756727247
0.00020526315789 0.00000085516947 0.00000031618658 2.08537725108589
0.00041052631579 0.00000086333895 0.00000060877506 2.08537664231083
0.00061578947368 0.00000087150842 0.00000087940128 2.08537576290955
0.00082105263158 0.00000087967789 0.00000112958123 2.08537463332832
0.00102631578947 0.00000088784737 0.00000136072056 2.08537327260775
0.00123157894737 0.00000089601684 0.00000157412331 2.08537169848444
0.00143684210526 0.00000090418632 0.00000177099980 2.08536992748465
0.00164210526316 0.00000091235579 0.00000195247395 2.08536797501069
0.00184736842105 0.00000092052526 0.00000211958996 2.08536585542074
0.00205263157895 0.00000092869474 0.00000227331833 2.08536358210241
0.00225789473684 0.00000093686421 0.00000241456156 2.08536116754085
0.00246315789474 0.00000094503368 0.00000254415920 2.08535862338165
0.00266842105263 0.00000095320316 0.00000266289262 2.08535596048903
0.00287368421053 0.00000096137263 0.00000277148932 2.08535318899972
0.00307894736842 0.00000096954211 0.00000287062692 2.08535031837280
0.00328421052632 0.00000097771158 0.00000296093681 2.08534735743599
0.00348947368421 0.00000098588105 0.00000304300757 2.08534431442842
0.00369473684211 0.00000099405053 0.00000311738799 2.08534119704043
0.00390000000000 0.00000100222000 0.00000318459003 2.08533801245040
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