BÀI TẬP MATLAB CUỐI KỲ

7/31/2019 BI TP MATLAB CUI K

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BI T P MATLAB CU I K

CT THI T K NG C KB ROTO L NG SC

Ph n I:T o giao di n guide


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H NG D N S D NG CH NG TRNH

Chng trnh thit k ng c khng ng b ba pha roto lng scT giao din ban u ta n nt ENTER sau giao din th 2 hin ra talm vic giao din ny .Trong chng trnh chia lm 6 nhm cng vic

c th nh sau :

Chng 1: Xc nh cc kch thc ch yuChng 2: Thit k li thp StatoChng 3: Thit k li thp RtoChng 4:Tnh ton mch tChng 5:Tnh ton tham s ng c tn hao v c tnh lm vicChng 6: Tnh ton c tnh khi ng

i vi tng chng, ta kch vo nt nhp vo nt nhp d liu chng th chng trnh s hin ra file nhp d liu, bn s nhp d liu vo.Kt thc vic nhp d liu bn phi lu v ng file ny li. tip theo takch vo nt tnh chng trnh s chy v thc hin tnh ton.Chng trnh chy xong ta kch vo nt xem kt qu, lc trn mnhnh xut hin file kt qu.Cc chng khc chng khc thc hin tng tCch nhp d liu tra theo ti liu thit k my in, theo li hng dntrong chng trnh.


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Cu l nh :function varargout = DC_KDB(varargin)% DC_KDB M-file for DC_KDB.fig% DC_KDB, by itself, creates a new DC_KDB or raises the existing% singleton*.%% H = DC_KDB returns the handle to a new DC_KDB or the handle to% the existing singleton*.%% DC_KDB('CALLBACK',hObject,eventData,handles,...) calls the local% function named CALLBACK in DC_KDB.M with the given input arguments.%% DC_KDB('Property','Value',...) creates a new DC_KDB or raises the% existing singleton*. Starting from the left, property value pairs are% applied to the GUI before DC_KDB_OpeningFunction gets called. An% unrecognized property name or invalid value makes property application

% stop. All inputs are passed to DC_KDB_OpeningFcn via varargin.%% *See GUI Options on GUIDE's Tools menu. Choose "GUI allows only one% instance to run (singleton)".%% See also: GUIDE, GUIDATA, GUIHANDLES% Edit the above text to modify the response to help DC_KDB% Last Modified by GUIDE v2.5 05-Mar-2012 12:19:15% Begin initialization code - DO NOT EDITgui_Singleton = 1;

gui_State = struct('gui_Name', mfilename, ...'gui_Singleton', gui_Singleton, ...'gui_OpeningFcn', @DC_KDB_OpeningFcn, ...'gui_OutputFcn', @DC_KDB_OutputFcn, ...'gui_LayoutFcn', [] , ...'gui_Callback', []);

if nargin & isstr(varargin{1})gui_State.gui_Callback = str2func(varargin{1});

endif nargout

[varargout{1:nargout}] = gui_mainfcn(gui_State, varargin{:});else

gui_mainfcn(gui_State, varargin{:});

end% End initialization code - DO NOT EDIT

% --- Executes just before DC_KDB is made visible.function DC_KDB_OpeningFcn(hObject, eventdata, handles, varargin)% This function has no output args, see OutputFcn.% hObject handle to figure% eventdata reserved - to be defined in a future version of MATLAB% handles structure with handles and user data (see GUIDATA)% varargin command line arguments to DC_KDB (see VARARGIN)


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% Choose default command line output for DC_KDBhandles.output = hObject;% Update handles structureguidata(hObject, handles);% UIWAIT makes DC_KDB wait for user response (see UIRESUME)

% uiwait(handles.figure1);

% --- Outputs from this function are returned to the command line.function varargout = DC_KDB_OutputFcn(hObject, eventdata, handles)% varargout cell array for returning output args (see VARARGOUT);% hObject handle to figure% eventdata reserved - to be defined in a future version of MATLAB% handles structure with handles and user data (see GUIDATA)% Get default command line output from handles structurevarargout{1} = handles.output;

Ph n 2:Cc b c tnh ton Ch ng 1:Tnh ton kch th c ch y u

Ch ng 2:Thi t k l i thp stato

Ch ng 3:Thi t k l i thp roto

Ch ng 4:Tnh ton m ch t Ch ng 5:Tnh ton tham s ng c ,t n hao v c

tnh lm vi c

Ch ng 6:Tham s kh i ng

Phn 3: Cc cu l nh v hm dng trong ch ng trnh


5/29

Ch ng 1 :ph n nh p d li u

%% CHUONG TRINH NHAP DU LIEU CHO NHOM CONG VIEC THU NHAT:% XAC DINH CAC KICH THUOC CHU YEU% Giao Vien Huong Dan :NGUYEN VAN TAN% Sinh Vien thuc Hien :NHOM LOP:092

%% --------------------------------------------------------------

Pdm=11; 15 %cong suat dinh mucU1=220; %dien ap day dinh mucf1=50; %Tan son=1500; %toc do dong bohs=0.825; %hieu suatcosphi=0.88;%chieu cao tam truc h=160 tra bang BIV trang 601Dn=27.2; %chon the chieu cao tam truc h=160 theo B10.3kD=0.64; % theo B10.2kd=0.92; %he so day quan

ks=1.11;anphas=0.7; %he so cung cuc tuA=330; %H10.3aBs=0.765; %H10.3akE=0.975 ; %H10.2

Ph n tnh ton:

% CHUONG TRINH TINH TOAN CHO PHAN THU NHAT

% Giao vien huong dan:NGUYEN VAN TAN% Thuc hien:NHOM LOP:09D2%------------------------------------------------------------------------------------% XAC DINH CAC KICH THUOC CHU YEUp=60*f1/n;D=kD*Dn;Po=kE*Pdm/(hs*cosphi);ls=6.1*10^7*Po/(anphas*ks*kd*A*Bs*D^2*n);l1=ls;l2=ls;%Buoc cuc

T=3.1415*D/(2*p);lamda=ls/T;I1=Pdm*10^3/(3*U1*hs*cosphi);

K t qu : XAC DINHKICHTHUOCCHUYEU:

1. So doicuc:p =


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Ch ng 2 :Ph n nh p d li u

% CHUONG TRINH NHAP DU LIEU CHO NHOM CONG VIEC THU HAI:% THIET KE LOAI THEP STATO% Giao Vien Huong Dan :NGUYEN VAN TAN% Sinh Vien thuc Hien :NHOM LOP:09D2

%% --------------------------------------------------------------

%chonq1=4;%chon so mach nhanh song songa1=2;%theo hinh H10.4 chon tich so AJAJ=1830;% chon so soi ghep song songn1=2;%-------------------------------------------------------------------%SAU KHI CHON DUOC CAC THONG SO TREN TA THUC HIEN VIEC LUU ROI THUC

%HIEN TINH DE XEM KET QUA CUA TIET DIEN SO BO ROI CAN CU VAO KQ DO TRA%duong kinh day va duong kinh cach dien tra BVI va thuc hien nhap cac du lieu%tiep theod=1.16;dcd=1.27;%chon buoc day quany=10;%chon la ton silic day 0.5 mm cac la ton co phu son cach dien tra bang B2.2kc=0.95;Bz1=1.75 %tr B10.5b tra theo chieu cao tam truc va so doi trucBg1=1.55%tra B10.5a tra theo chieu cao tam truc va so doi truc%thuong chonh41=0.5;

% tra bang VIII-1 chon chieu day cach dien ranh va nemc=0.4;co=0.5;

ph n tnh ton:

% CHUONG TRINH TINH TOAN CHO PHAN THU HAI% Giao vien huong dan:NGUYEN VAN TAN% Thuc hien:NHOM LOP:09D2%----------------------------------------------------------------------------

--------% THIET KE LOAI THEP STATOdl_nhap_phan_1;ct_tinh_phan_1;dl_nhap_phan_2;Z1=6*p*q1;t1=3.1415*D/Z1ur1=A*t1*a1/I1ur1=round(ur1)W1=p*q1*ur1/a1Jo=AJ/A


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So1=I1/(a1*n1*Jo)t0=Z1/(2*p)% t0 buoc cucbeta=y/t0ky=sin(beta*pi/2)anpha=p*2*pi/Z1kr=(sin(q1*anpha/2))/(q1*sin(anpha/2))kdq=ky*kr%17 tu thong

phi=(kE*U1)/(4*ks*kdq*f1*W1)%17 mat do tu thong khe ho khong khiBdeta=phi*10^4/(anphas*T*l1)%18 so bo dinh chieu rong cua rangbz1=Bdeta*l1*t1/(Bz1*l1*kc)%19 so bo dinh chieu cao gong statohg10=phi*10^4/(2*Bg1*l1*kc)%20 kich thuoc ranh va cach dienb41=1.5+dcdhr1=((Dn-D)/2-hg10)*10d1= (pi*(D*10+2*hr1)-Z1*bz1*10)/(Z1+3.14)d2=(pi*(D*10+2*h41)-Z1*10*bz1)/(Z1-3.14)hr12=hr1-d1/3%dien tich ranh tru nemSro=pi*(d1^2+d2^2)/8 + (d1+d2)*(hr12-d1/2)/2%dien tich ranh cach dienScd= (pi*d2/2+2*hr12+(d1+d2))*c + pi*d1*co/2%dien tich co ich cua ranh;Sr=Sro-Scd%he so lap day ranhkday=ur1*n1*dcd^2/Sr%21 be rong rang starobz1o=pi*(D+2*h41*0.1+d1*0.1)/Z1-d1*0.1bz11o=pi*(D+2*(h41*0.1+hr12*0.1))/Z1-d2*0.1bz10=(bz1o+bz11o)/2;%22 chieu cao gonghg1=(Dn-D)/2 - hr1*0.1 + d2*0.1/6

%23 khe ho khong khideta=0.25+D*10/1000

K t qu : THIET KELOAITHEPSTATO:

9. So ranhstato:Z1 =

48.010.Buoc ranh stato:t1 =1.13911. So thanhdantacdungcuamotranh:


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ur1 =34.012. So vongdaynoitiepcuamotpha:W1 =

136.0Tiet diendaysobo:So1 =0.987014. Buoc tuongdoi:beta =0.833He sobuocngan:

ky =0.966He soquanrai:kr =0.958He sodayquan:

kdq =0.92516. Tu thongkhehokhongkhi:phi =0.00768017. Mat dotuthongkhehokhongkhi:Bdeta =0.76618. So bodinhchieurongcuarang:bz1 =0.52519. So bodinhchieucaogongstato:


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hg10 =2.27620. Kich thuocranh:b41 =

2.770

h41 =0.50

d1 =8.986

d2 =6.644

hr1 =26.201

hr12 =23.206

Dien tichranhtrunemSranhtn =195.288Dien tichcachdienScd =36.049Dien tichcoichcuaranhSr =159.239He solapdayranhkday =0.68921. Be rongrangstato:


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bz10 =0.546Chieu caogongstato:hg1 =

2.38722.Khe ho khongkhi:deta =0.424


% CHUONG TRINH NHAP DU LIEU CHO NHOM CONG VIEC THU BA:% THIET KE LOAI THEP ROTO

% Giao vien huong dan:NGUYEN VAN TAN% Thuc hien:NHOM LOP:09D2%% --------------------------------------------------------------

Z2=56 %SO ranh roto,chon theo bang 10.5Bz2=1.75 %tu cam trong rang rotokc2=0.95 %he so ep chat loai sat rotokI=0.9 %J2=3 %mat do dong dien trong thanh danJv=2.5 %mat do dong dien trong vanh ngan mach

%KICH THUOC ROTO VA VANH NGAN MACHhr2=15h12=8.5d10=6.5d20=4.5b42=1h42=1

ph n tnh ton:

% CHUONG TRINH TINH TOAN CHO PHAN THU BA% Giao vien huong dan:NGUYEN VAN TAN% Thuc hien:NHOM LOP:09D2

%------------------------------------------------------------------------------------% THIET KE LOAI THEP ROTOdl_nhap_phan_1;ct_tinh_phan_1;dl_nhap_phan_2;ct_tinh_phan_2;dl_nhap_phan_3;%25Duong kinh ngoai rotoDo=D-2*deta*0.1%26Buoc rang ro to


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t2=pi*Do/Z2%27So bo dinh chieu rong rotobz2o=Bdeta*l2*t2/(Bz2*l2*kc2)%28Duong kinh truc rotoDt=0.3*D%29dong dien trong thanh danItd=kI*I1*6*W1*kdq/Z2%30Dong dien trong vanh ngan mach

Iv=Itd/(2*sin(pi*p/Z2))%31Tiet dien thanh danStd=Itd/J2%32Tiet dien vanh ngan machSv=Iv/Jv%33kich thuoc vanh ngan macha=hr2b=Sv/aDv=D*10-(a+1)%35Dien tich ranh rotoSr2=pi*(d10^2+d20^2)/8+(d10+d20)*h12/2%36Be rong roto o 1/3 chieu caobz2=pi*(Do-2*h42*0.1 -4*(h12*0.1+d20*0.1)/3)/Z2-0.1*d20%37chieu cao gong rotohg2=(Do-Dt)/2-hr2*0.1+d20*.1/6%38Do nghieng ranh rotobn=t1

K t qu : THIET KELOAITHEPROTO:

24. So ranhroto:Z2 =56.0

25.Duong kinh ngoairoto:Do =17.3226.Buoc rang roto:t2 =0.97227. So bodinhchieurongroto

bz2o =0.439978Tu camtrongrangroto:Bz2 =1.750


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28. Duong kinhtrucroto:Dt =5.22229.Dong dien trongthanhdanroto:

Itd =270.3030.Dong dien trongvanhnganmach:Iv =1207.08431. Tiet dienthanhdanbangnhomstd =90.10038832. Tiet dienvanhnganmach:Sv =482.83379333. Kich thuocmachrotovavanhnganmach:hr2 =15.0

h12 =8.5

b42 =1.0

h42 =1.0

d10 =6.5

d20 =4.5


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34. Dien tichranhroto:Sr2 =71.29435.Be rong rango1/3chieucaorang:

bz2 =0.41336.Chieu cao gong:hg2 =4.62538. Buoc nghienranh:bn =1.139


% CHUONG TRINH NHAP DU LIEU CHO NHOM CONG VIEC THU TU:% THIET KE LOAI THEP ROTO% Giao vien huong dan:NGUYEN VAN TAN% Thuc hien:NHOM LOP:09D2%% --------------------------------------------------------------

Hz1=17.9 %cuong do tu truong tren rang stato tra theo bang V-6Hz2= 28 %cuong do tu truong trong rang roto tra theo bang V-6Hg1=8.3 %cuong do tu truong o tren gong stato tra theo bang V-9Hg2=1.8 %cuong do tu truong o tren rang roto tra theo bang V-9

ph n tnh ton:% CHUONG TRINH TINH TOAN CHO PHAN THU BA% Giao vien huong dan:NGUYEN VAN TAN% Thuc hien:NHOM LOP:09D2%------------------------------------------------------------------------------------% TINH TOAN MACH TUdl_nhap_phan_1;ct_tinh_phan_1;dl_nhap_phan_2;


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ct_tinh_phan_2;dl_nhap_phan_3;dl_nhap_phan_4ct_tinh_phan_3;%39 he so khe ho khong khinuy1=(b41/deta)^2/(5+b41/deta)kdeta1=t1/(t1-nuy1*deta*0.1)

nuy2=(b42/deta)^2/(5+b42/deta)kdeta2=t2/(t2-nuy2*deta*0.1)kdeta=kdeta1*kdeta2%41 Suc tu dong khe ho khong khiFdeta=1.6*Bdeta*kdeta*deta*0.1*10^4%42mat do tu thong o rang statoBz10=Bdeta*l1*t1/(bz1*l1*kc)%44 suc tu dong tren rang statohz10=hr1-d2/3Fz1=2*hz10*Hz1*0.1%45mat do tu thong o rangBz2=Bdeta*l2*t2/(bz2*l2*kc2)%47 suc tu dongtren rane rotohz20=hr2*0.1-d2*0.1/3Fz2=2*hz20*Hz2%he so bao hoa rangkz=(Fdeta+Fz1+Fz2)/Fdeta%49 mat do tu thong tren gong stataBg1=(phi*10^4)/(2*hg1*l1*kc)

% 51chieu dai mach tu o gong statoLg1=3.14*(Dn-hg1)/(2*p)

%52 Suc tu dong om gong statoFg1=Lg1*Hg1%mat do tu thong tren gong rotoBg2=(phi*10^4)/(2*hg2*l2*kc)%chieu dai mach tu o gong rotoLg2=3.14*(Dt+hg2)/(2*p)

Fg2=Lg2*Hg2F=Fdeta+Fz1+Fz2+Fg1+Fg2%he so bao hoa toan machkm=F/Fdeta%dong tu hoaIm=p*F/(2.7*W1*kdq)%dong tu hoa phan tramImphantram=Im*100/I1

K t qu : TINH TOANMACHTU:

39. He sokhehokhongkhi:

kdeta1 =1.160

kdeta2 =1.034


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kdeta =1.19940.Dung thep kithuatdiencannguoidanghuong:41.Suc tu dongkhehokhongkhi:Fdeta =

612.46542.Mat do tuthongorangstato:Bz1 =1.75043. Cuong dotutruongtrenrangstato

Hz1 =17.90000044.Suc tu dongtrenrangStato:Fz1 =79.66845. Mat dotuthongorangroto:Bz2 =1.863

46.Cuong do tutruongtrenrangroto:Hz2 =28.0047.Suc tu dongtrenrangroto:Fz2 =71.41548. He sobaohoarang:kz =1.24668049 Mat dotuthongtrengongstato:Bg1 =1.48192050. uong dotutruongotrengongStato:


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Hg1 =8.351. Chieu daimachtuogongStato :Lg1 =

19.34452 Suc tudongotrengongstato :Fg1 =160.55353.Mat do tuthongtrengongroto:Bg2 =0.82054 . Cuong do tutruongotrengongroto:Hg2 =1.80055.Chieu dai machtuogongroto:Lg2 =7.731

56. Suc tudongtrengongroto:Fg2 =13.91493057 Tong suctudongcuamachtu:F =938.01535258He so baohoatoanmach:km =1.559 Dong dientuhoa:Im =5.690Dong dientuhoaphantram:


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Imphantram =24.788

Ch ng 5 :Ph n nh p d li u % CHUONG TRINH NHAP DU LIEU CHO NHOM CONG VIEC THU 5:% TINH TOAN THAM SO DONG CO VA TINH TON HAO% Giao vien huong dan:NGUYEN VAN TAN% Thuc hien:NHOM LOP:09D2%% --------------------------------------------------------------B=1 %theo bang 3-4rot1=0.72 %theo bang 5.3xichma1=0.0062 %theo bang 5.2arot2=1

xichma2=0.0092 %theo bang 5.2cbeta0=0.32 %tra theo hinh H6.1

ph n tnh ton:% CHUONG TRINH TINH TOAN CHO PHAN THU NAM% Giao vien huong dan:NGUYEN VAN TAN% Thuc hien:NHOM LOP:09D2%------------------------------------------------------------------------------------% TINH TOAN THAM SO DONG CO VA TINH TON HAOdl_nhap_phan_1;ct_tinh_phan_1;dl_nhap_phan_2;ct_tinh_phan_2;dl_nhap_phan_3;ct_tinh_phan_3;dl_nhap_phan_4ct_tinh_phan_4;dl_nhap_phan_5%60 chieu dai phan dau noiToy=pi*(D+hr1*0.1)*(y/Z1)%chieu rong trung binh cua boi dayld1=1.3*Toy+2*B%61chieu dai trung binh nua vong day cua day quanltb=l1+ld1%62chieu adi day quan mot pha cua StatoL1=2*ltb*W1*10^-2

%63 Dien tro tac dung cua day quan stator1=L1/(41*n1*a1*So1)%tinh thao don ci tuong doir1sao=r1*I1/U1rtd=l2*0.01/(23*Sr2)rv=(3.14*Dv*0.1)/(23*Z2*Sv)*0.01tamgiac=2*sin(3.14*p/Z2)r2=rtd+2*rv/(tamgiac^2)% he so quy doigama=4*3*(W1*kdq)^2/Z2r20=gama*r2/3


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r2sao=r20*I1/U1%he so tu dan tan ranh statokbeta0=(1+3*beta)/4kbeta=(1+3*kbeta0)/4h1=hr1-0.1*d2-2*c-coh2=-(d1/2-2*c-co)lamdar1=h1*kbeta/(3*d1)+(0.785-b41/(2*d1)+h2/d1+h41/b41)*kbeta0k41=1-(0.033*(b41*0.1)^2/(t1*deta))

lamdat1=0.9*t1*(q1*kdq)^2*rot1*k41*xichma1/(kdeta*deta)*10lamdad1=0.34*q1*(ld1-0.64*beta*T)/ls%72 he so tu dan tan statolamda1=lamdar1+lamdat1+lamdad1x1=0.158*f1*W1^2*ls*lamda1/((100)^3*p*q1)x1sao=x1*I1/U1lamdar2=(h12*(1-3.14*d10^2/(8*Sr2))^2/(3*d10)+0.66-b42/(2*d10))+h42/b42lamdat2=(0.9*t2*(Z2/(3*2*p))^2*rot2*k41*xichma2)/(kdeta*deta*0.1)%76 he so tu tan phan dau noilamdad2=((2.3*Dv*0.1)/(Z2*l2*(tamgiac)^2))*log10(4.7*Dv*0.1/(a*.1+2*b*0.1))lamdarn=0.5*lamdat2*(bn/t2)^2lamda2=lamdar2+lamdat2+lamdad2+lamdarn %saai cho nayx2=7.9*f1*l2*lamda2*10^-8x20=gama*x2x20sao=x20*I1/U1x12=(U1-Im*x1)/Imx12sao=x12*I1/U1%tinh lai KekE1=(U1-Im*x1)/U1Gz1=7.8*Z1*bz1*hz10*0.1*l1*kc*10^-3Gg1=7.8*l1*Lg1*hg1*2*p*kc*10^-3PFez1=1.8*2.5*Bz1^2*Gz1*10^-3PFeg1=1.6*2.5*Bg1^2*Gg1*10^-3P0Fe=PFez1+PFeg1B0=beta0*kdeta*Bdetapbm=0.5*2*(Z1*n/10000)^1.5*(10*B0*t1)^2Pbm=(2*p*T*l2*pbm*10^-7)*(t2-b42*0.1)/t2

%87 Ton hao dap machBdm=nuy1*deta*Bz2/(2*t2)*0.1Gz2=7.8*Z2*hz20*bz2*l2*kc*10^-4Pdm=0.11*(Z1*n*10*Bdm/10000)^2*Gz2*10^-3%Tong ton hao thepPFe=P0Fe+Pbm+Pdm%ton hao coPco=(n/1000)^2*(Dn/10)^4*10^-3%90 ton hao khong taiPkotai=PFe+Pco%DAC TINH LAM VIECC1=1+x1/x12Idbx=ImIdbr=(PFe*10^3+3*Im*r1)/(3*U1)E1=U1-Im*x1Ki=6*W1*kdq/Z2I2=ItdI20=I2/KiSdm=I20*r20/E1Sm=r20/(x1/C1+x20)rns=C1^2*(r1/C1+r20/Sdm)rnsm=C1^2*(r1/C1+r20/Sm)xns=C1^2*(x1/C1+x20)Zns=sqrt(rns^2+xns^2)


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21/29

64.Dien tro tacdungcuadayquanRoto:rtd =0.00006988265 .Dien tro vanhnganmach:

rv =0.00000081266. Dien troroto :r2 =0.00010267 He soquydoi :gama =3391.4333033068. Dien trorotodaquydoi :r20 =0.11566. Tinh theo donvituongdoi:r2sao =

0.01151069He so tudantanranhstato:lamdar1 =1.213812870.He so tudantantapstato:lamdat1 =1.225871 . He so tutanphandaunoi:lamdad1 =1.275872.He so tudantanStato:lamda1 =3.7153


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23/29

83.Trong luong rangstato :Gz1 =5.1319584.Trong luong gongtustato:

Gg1 =15.7892385.Ton hao theptrongloaisatstato :PFez1 =0.07072Trong gong:PFeg1 =0.1380Trong coloaisatstato:P0Fe =0.208786.Ton hao bemattrenrangroto :Pbm =

0.0121825587.Ton hao dapmachtrenrangroto :Pdm =0.00335688.Tong ton haothep:

PFe =0.224289.Ton hao co :Pco =0.123290.Tin hao khongtai:P0 =0.347


24/29

rns =11.8975

xns =

2.1882

Zns =12.0971

I2dm0 =18.5556

cosphi20 =0.9835

sinphi20 =0.1809

I1r =18.2387

I1x =8.9208

I1dm =20.3035

cosPhi =0.8983

P1 =12.0375

Pcu1 =0.6130


25/29

Pcu2 =0.0064

Pf =

0.0602

Pt =1.0271

P2 =11.0105

hsuat =91.4679

Ch ng 6 :

Ph n nh p d li u

% CHUONG TRINH NHAP DU LIEU CHO NHOM CONG VIEC THU 6:% TINH TOAN DAC TINH KHOI DONG

% Giao vien huong dan:NGUYEN VAN TAN% Thuc hien:NHOM LOP:09D2%% --------------------------------------------------------------song=1 %h10.13phi2=0.2 %h10.13kbh=1.43 %thuoc khoang 1.3-1.45kbeta=0.89sideta=0.6 %h10.15

ph n tnh ton:% CHUONG TRINH TINH TOAN CHO PHAN THU 6% Giao vien huong dan:NGUYEN VAN TAN% Thuc hien:NHOM LOP:09D2%------------------------------------------------------------------------------------% TINH TOAN DAC TINH KHOI DONGdl_nhap_phan_1;ct_tinh_phan_1;dl_nhap_phan_2;ct_tinh_phan_2;dl_nhap_phan_3;


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ct_tinh_phan_3;dl_nhap_phan_4ct_tinh_phan_4;dl_nhap_phan_5ct_tinh_phan_5;dl_nhap_phan_6%91 BOI SO MOMEN CUC DAImmax=(I2m0/I2dm0)^2*(Sdm/Sm)

a0=hr2-h41si=0.067*a0kR=1+phi2rtdsi=kR*rtd%93 dien tro roto khi xet den hieu ung mat ngoair2si=rtdsi+2*rv/tamgiac^2r2si0=gama*r2silamdar2si=(h12*((1-3.14*d10^2/(8*Sr2))^2)/(3*d10)+0.66-b42/(2*d10))*song+h42/b42lamda2si=lamdar2si+lamdat2+lamdad2+lamdarnx2si=x20*lamda2si/lamda2R=r1+r2si0xnsi=x1+x2siznsi=sqrt(R^2+xnsi^2)Insi=U1/znsi%93 tham so cua dong co khi xet den ca hieu ung mat ngoai va su bao hoa%mach tu tanIbhsi=kbh*InsiFzbh=0.7*Ibhsi*ur1*(kbeta+ky*kdq*Z1/Z2)/a1Cbh=0.64+0.25*sqrt(deta/(t1+t2))Bphideta=Fzbh*10^-4/(1.6*Cbh*deta)c1=(t1-b41*0.1)/(1-sideta)tamgiaclamda1bh=((h41*0.1+0.58*0.45)*C1)/(b41*0.1*(c1+1.5*b41*0.1))lamdar1bh=lamdar1-tamgiaclamda1bhlamdat1bh=lamdat1*sidetalamda1bh=lamdar1bh+lamdat1bh+lamdad1

x1bh=x1*lamda1bh/lamda1c2=(t2-b42)*(1-sideta)tamgiaclamda2bh=h42*c2/(b42*(c2+b42))lamdar2sibh=lamdar2si-tamgiaclamda2bhlamdat2bh=lamdat2*sidetalamdarnbh=lamdarn*sidetalamda2sibh=lamdar2sibh+lamdat2bh+lamdad2+lamdarnbhx2sibh0=x20*lamda2sibh/lamda2rnsi=r1+r2si0xnsibh=(x1bh+x2sibh0)Znsibh=sqrt(rnsi^2 +xnsibh^2)Ikd=U1/Znsibhik=Ikd/I1dmx12n=x12*kmC2sihb=1+x2sibh0/x12nI2k0=Ikd/C2sihbmk=(I2k0/I2dm0)^2*Sdm*r2si0/r20

K t qu : TINH TOANDACTINHKHOIDONG

91. Boi somomencucdai:mmax =


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2.302992.Tham so cuadongcokhixetdenhieuungmatngoai :.Dien tro rotokhixetdenhieuungmatngoai:r2si =0.0001

.Dien roto quydoi:

r2si0 =0.3944. He so tudanranhrotokhixetdenhieuungmatngoai:lamdar2si =1.839776Tong hesotudanrotokhixetdenhieuungmatngoai :lamda2si =8.7284. Dien khang rotokhixetdenhieuungmatngoai :x2si =1.33986. Tong tro nganmachkhixetdenhieuungmatngoai:

rnsi =0.906895

xnsi =2.1423

znsi =2.3264. Dong dien nganmachkhichixetdenhieuungmatngoai :Insi =94.568593.Tham so cuadongcodienkhichixetdenhieuungmatngoaivasubaohoacuamachtutan:.Dong dien nganmachkhixetdenhieuungmatngoaivasubaohoacuamachtutan :


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Ibhsi =135.23291.Suc tu dongtrungbinhcuamotranhstato:Fzbh =

2664.73909.Tong he sotutanranhstatokhixetdenhesobaohoamachtutan:lamda1bh =2.89792.Dien khang statokhisxetdenbaohoamachtutan:x1bh =0.6065.He so tutanranhrotokhixetdenbaohoamachtutan:lamdar2sibh =1.8512.He so tutantaprotokhixetdenbaohoamachtutan :lamdat2bh =2.0566.He so tutan doranhnghiengrotokhixetdenbaohoamachtutan :

lamdarnbh =1.413270.Tong he sotutanrotokhixetdenbaohoatoanmachtutanvahieuungmatngoai:lamda2sibh =6.4266.Dien khang rotokhixetdenhieuungmatngoai vasubaohoacuamachtutan :x2sibh0 =0.986594.Cac tha songanmachkhixetdenhieuungmatngoaivasubaohoacuamachtutan :xnsibh =1.5930


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R =0.9069

Znsibh =

1.8330995.Dong dien khoidong :Ikd =120.0156896.Boi so dongdienkhoidong:ik =5.920797.Boi so momenkhoidong:mk =1.4645

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