What's new

Reviewer CHEGG LINK UNLOCK PASOK KAYO DITO ACTIVE TODAY JUNE 28 2021 ( MONDAY )

Status
Not open for further replies.

Unkown User

Eternal Poster
Joined
Apr 28, 2017
Posts
585
Solutions
57
Reaction
415
Points
312
Paayos nalang nung link para maunlock ko agad pagclick ko.

You do not have permission to view the full content of this post. Log in or register now.

If need nyong magpaask tutor may iooffer ako sainyong account affordable lang for one time use aabot lang nang 3-7 days average pero pag sinwerte ka pwede umabot nang 1month at magrenew pa, Kung madami kayong iaask pwede nyong ubusin yung 20 questions sa chegg after macancel nung account kaya sulit na sulit. DM nyo lang ako. If need nyo lang naman.

1624504510322.png


1624504481423.png
 

Attachments

Question: Five Bus System Homework Assignment Thanks to Prof Chen-Ching Liu and Ph.D. Candidate Madeleine G...​


bookmark.png
(1 bookmark)
flags.png
37c%2F37c8dcae-adc6-413f-b485-e573faba62aa%2Fimage.jpg

6ea%2F6ea09a16-1432-4ef0-a25f-577818b6d69e%2Fimage.jpg

the program that is needed to do the work is power world
Show transcribed image text

Expert Answer​


  • 02_img-avatar-gry-40x40.png
    Anonymousanswered this
    Was this answer helpful?
    Thumbs up inactive

    0
    Thumbs down inactive

    0
    27 answers
    . MAIN MENU
    =========
    1 LOAD THE PSF 8 DATA VERIFICATIO
    2 IMPORTING UTILITIES 9 LIMIT CHECKS
    3 INPUT DATA PROCESSING 10 SENSITIVITY FACTORS
    4 STUDY PREPARATION UTILITIES 11 EXPORTING UTILITIES
    5 POWER FLOW SOLUTION 12 CLOSE THE PSF
    6 NETWORK REDUCTION 13 QUIT
    7 POWER FLOW SOLUTION REPORTING
    ENTER MENU CHOICE: 5
    * POWER FLOW SOLUTION *
    =============================
    1 AUTOMATIC SOLUTION
    2 FAST DECOUPLED (XB)
    3 NEWTON RAPHSON
    4 FAST DECOUPLED (BX)
    5 LOCALIZED SOLUTION (FD XB)
    6 DC POWER FLOW SOLUTION
    7 POWER FLOW SOLUTION OPTIONS
    ITEM #, EXIT 3
    * START OF NEWTON RAPHSON ITERATIONS *
    ITERATION : 1 NEWTON RAPHSON UNSOLVED ABSOLUTE ERROR
    -------------- BUS WITH LARGEST ERROR ------------ BUSES SUMMATION
    P(P.U. MW) 0.000375 ( 30 BUS 3138. ) 0 0.0017
    V(RADIANS) 0.000179 ( 70 BUS 7138. ) 8 0.0016
    Q(P.U. MVAR) 0.001873 ( 120 BUS 12230. ) 0 0.0078
    V(P.U. KV) 0.000210 ( 30 BUS 3138. ) 3 0.0009
    ITERATION : 2 NEWTON RAPHSON UNSOLVED ABSOLUTE ERROR
    -------------- BUS WITH LARGEST ERROR ------------ BUSES SUMMATION
    P(P.U. MW) 0.000006 ( 160 BUS 16230. ) 0 0.0000
    Q(P.U. MVAR) -0.000008 ( 170 BUS 17230. ) 0 0.0000
    * POWER FLOW SOLUTION IS REACHED IN 2 ITERATIONS *
    2. MAIN MENU
    =========
    1 LOAD THE PSF 8 DATA VERIFICATION
    2 IMPORTING UTILITIES 9 LIMIT CHECKS
    3 INPUT DATA PROCESSING 10 SENSITIVITY FACTORS
    4 STUDY PREPARATION UTILITIES 11 EXPORTING UTILITIES
    5 POWER FLOW SOLUTION 12 CLOSE THE PSF
    6 NETWORK REDUCTION 13 QUIT
    7 POWER FLOW SOLUTION REPORTING
    ENTER MENU CHOICE: 9
    ** LOAD FLOW SOLUTION LIMIT CHECK **
    ====================================
    1 OVERLOADED LINES OR TRANSFORMERS
    2 CONTROL PARAMETERS VIOLATION
    3 BUS VOLTAGE OUTSIDE SPECIFIED BAND
    4 LINES ACROSS WHICH THE ANGLE EXCEEDS THE SPECIFIED VALUE
    5 CHECK FOR ALL LIMITS AND VIOLATIONS
    6 SORT OPTIONS
    ITEM #, SPECIFY SUBSYSTEM, EXIT 1
    ALL LINES TRANSFORMERS: A
    * LOAD FLOW SOLUTION LIMIT CHECK *
    1 OVERLOADED BRANCHES RATING GROUP = 1 % LOADING = 100.0
    <----- FROM BUS -----> <------ TO BUS ------> MVA MVA PERCENT
    NUMBER NAME NUMBER NAME CKT LOADING RATING LOADING
    ====== ================ ====== ================ ======== ======= ====== =======
    60 BUS 6138. 100 BUS 10138. 1 175.4 175.0 100.22
    3. ** LOAD FLOW SOLUTION LIMIT CHECK **
    ====================================
    1 OVERLOADED LINES OR TRANSFORMERS
    2 CONTROL PARAMETERS VIOLATION
    3 BUS VOLTAGE OUTSIDE SPECIFIED BAND
    4 LINES ACROSS WHICH THE ANGLE EXCEEDS THE SPECIFIED VALUE
    5 CHECK FOR ALL LIMITS AND VIOLATIONS
    6 SORT OPTIONS
    ITEM #, SPECIFY SUBSYSTEM, EXIT 3
    HI VOLTAGE LIMIT LO VOLTAGE LIMIT: L
    ENTER VALUE IN PU: .99
    * LOAD FLOW SOLUTION LIMIT CHECK *
    4 BUSES WITH V < 0.990
    <-------- BUS --------> <-- VOLTAGE -->
    NUMBER NAME TYPE BASE KV MAG ANGLE AREA ZONE OWNER ID
    ====== ================ ==== ======= ======= ======= ==== ==== ================
    30 BUS 3138. 1 138.00 0.9868 -26.02 10 1 BLANK
    40 BUS 4138. 1 138.00 0.9888 -20.27 10 1 BLANK
    90 BUS 9138. 1 138.00 0.9824 -15.83 10 1 BLANK
    120 BUS 12230. 1 230.00 0.9876 -4.73 10 1 BLANK
    UP DOWN SCREEN DUMP LIST TO FILE CHANGE RATING EXIT : e
    4. MAIN MENU
    =========
    1 LOAD THE PSF 8 DATA VERIFICATION
    2 IMPORTING UTILITIES 9 LIMIT CHECKS
    3 INPUT DATA PROCESSING 10 SENSITIVITY FACTORS
    4 STUDY PREPARATION UTILITIES 11 EXPORTING UTILITIES
    5 POWER FLOW SOLUTION 12 CLOSE THE PSF
    6 NETWORK REDUCTION 13 QUIT
    7 POWER FLOW SOLUTION REPORTING
    ENTER MENU CHOICE: 7
    ** LOAD FLOW SOLUTION REPORTING **
    ==================================
    1 MISMATCH SUMMARY 11 ULTC/PS SUMMARY
    2 SUBSYSTEM SUMMARY 12 SERIES COMPENSATORS
    3 BUS SHUNT DATA 13 POWER FLOW SUMMARY
    4 PLANT DATA 14 SECTIONALIZED BRANCHES
    5 MACHINE DATA 15 STATIC TAP CHANGERS / PHASE SHIFTERS
    6 AREA INTERCHANGE DATA 16 THREE WINDING TRANSFORMERS
    7 TIE LINE FLOWS 17 INTERFACE FLOWS
    8 BUS FLOWS 18 VOLTAGE PROFILE
    9 DC CONVERTERS 19 SORT OPTIONS
    10 DC FLOWS
    ITEM #, SPECIFY SUBSYSTEM, EXIT 8
    SUBSYSTEM BUSES PICK BUSES EXIT: p
    ENTER BUS LIST FILE NAME OR BUSES TO INCLUDE OR EXCLUDE
    * SOLUTION REPORTING - BUS FLOWS *
    BUS: 60 BUS 6138.
    VOLTAGE : 1.0017 PU ( 138.2 kV) -20.67
    BUS NUM NAME AREA CKT MW MVAR MVA TAP
    ======== ================ ==== ======== ======== ======== = ======== ========
    LOAD 136.00 28.00 138.85
    SW SHUNT 0.00 100.33 100.33
    TO 100 BUS 10138. 10 1 -128.63 -107.64 167.73
    TO 20 BUS 2138. 10 1 -7.37 -20.69 21.96
    UP DOWN NEXT BUS PREVIOUS BUS SCREEN DUMP LIST TO FILE EXIT : e
    SUBSYSTEM BUSES PICK BUSES EXIT: p
    ENTER BUS LIST FILE NAME OR BUSES TO INCLUDE OR EXCLUDE
    * SOLUTION REPORTING - BUS FLOWS *
    BUS: 100 BUS 10138.
    VOLTAGE : 1.0132 PU ( 139.8 kV) -16.15
    BUS NUM NAME AREA CKT MW MVAR MVA TAP
    ======== ================ ==== ======== ======== ======== = ======== ========
    LOAD 195.00 40.00 199.06
    TO 120 BUS 12230. 10 1 -234.48 60.92 242.27 1.0000LK
    TO 110 BUS 11230. 10 1 -235.99 52.21 241.70 1.0000LK
    TO 80 BUS 8138. 10 1 69.77 -3.74 69.87
    TO 60 BUS 6138. 10 1 130.95 -131.80 185.80
    TO 50 BUS 5138. 10 1 74.76 -17.59 76.80
    UP DOWN NEXT BUS PREVIOUS BUS SCREEN DUMP LIST TO FILE EXIT : e
    5. ** LOAD FLOW SOLUTION REPORTING **
    ==================================
    1 MISMATCH SUMMARY 11 ULTC/PS SUMMARY
    2 SUBSYSTEM SUMMARY 12 SERIES COMPENSATORS
    3 BUS SHUNT DATA 13 POWER FLOW SUMMARY
    4 PLANT DATA 14 SECTIONALIZED BRANCHES
    5 MACHINE DATA 15 STATIC TAP CHANGERS / PHASE SHIFTERS
    6 AREA INTERCHANGE DATA 16 THREE WINDING TRANSFORMERS
    7 TIE LINE FLOWS 17 INTERFACE FLOWS
    8 BUS FLOWS 18 VOLTAGE PROFILE
    9 DC CONVERTERS 19 SORT OPTIONS
    10 DC FLOWS
    ITEM #, SPECIFY SUBSYSTEM, EXIT 4
    * SOLUTION REPORTING - PLANT DATA *
    =======================================
    1 ALL PLANTS
    2 ON LINE PLANTS
    3 PLANTS AT VAR LIMIT WITH UNEQUAL VAR LIMITS
    4 PLANTS WITH UNSCHEDULED REACTIVE POWER
    5 PLANTS CONTROLLING A REMOTE BUS
    6 PLANTS WITH RESERVE REACTIVE POWER
    ITEM #, EXIT 1
    * SOLUTION REPORTING *
    10 PLANTS
    PLANT BUS MACHINES
    NUM NAME TYPE I/S O/S MW MVAR QMAX QMIN
    ======== ================ ==== ======== ======== ======== ======== ========
    R 10 BUS 1138. 2 4 0 134.70 41.53 280.00 -50.00
    R 20 BUS 2138. 2 4 0 187.00 30.50 140.00 -100.00
    R 70 BUS 7138. 2 3 0 165.00 91.63 280.00 0.00
    R 130 BUS 13230. 3 3 0 337.53 170.64 540.00 -300.00
    R 150 BUS 15230. 2 6 0 185.00 40.36 310.00 -55.00
    R 160 BUS 16230. 2 1 0 155.00 106.41 280.00 -55.00
    R 180 BUS 18230. 2 1 0 400.00 100.73 200.00 -90.00
    R 210 BUS 21230. 2 1 0 400.00 25.25 200.00 -90.00
    R 220 BUS 22230. 2 6 0 300.00 -80.55 696.00 -560.00
    R 230 BUS 23230. 2 3 0 660.00 33.34 310.00 -125.00
    UP DOWN OTHER SCREEN SCREEN DUMP LIST TO FILE EXIT :e
    CODES: 1.) CODE FOR TAKING BUS DATA
    (busdata.m)
    % Returns Initial Bus datas of the system... function busdt = busdatas() % Type.... %
    1 - Slack Bus.. % 2 - PV Bus.. % 3 - PQ Bus.. % |Bus | Type | Vsp | theta | PGi | QGi | PLi | QLi |
    busdat14 = [1 1 1.00 0 0 0 100 0 ; 2 2 1.01 0 250 0 200 100 ; 3 3 1.00 0 0 0 300 100 ;]; busdt = busdat14; 2.) CODE FOR TAKING LINE DATA
    (linedatas.m) % Returns Line datas of the system... function linedt = linedatas() % | From | To | R | X | B/2 | % | Bus | Bus | pu | pu | pu |
    linedat14 = [1 2 0.01 0.05 0 1 3 0.015 0.05 0 2 3 0.015 0.05 0 ]; linedt = linedat14;
    3.) CODE FOR BUILDING Y-BUS MATRIX (ybusmatrix.m)
    Program to for Admittance And Impedance Bus Formation....
    function Y = ybusmatrix() % Returns Y
    linedata = linedatas(); % Calling Linedatas..
    . fb = linedata:),1); % From bus number..
    . tb = linedata:),2); % To bus number...
    r = linedata:),3); % Resistance, R...
    x = linedata:),4); % Reactance, X..
    b = linedata:),5); % Ground Admittance
    , B/2... z = r + i*x; % z matrix..
    . y = 1./z; % To get inverse of each element..
    . b = i*b;% Make B imaginary...
    nb = max(max(fb),max(tb)); % No. of buses...
    nl = length(fb); % No. of branches..
    . Y = zeros(nb,nb); % Initialise YBus... % Formation of the Off Diagonal Elements... for
    k = 1:nl Y(fb(k),tb(k)) = Y(fb(k),tb(k)) - y(k); Y(tb(k),fb(k)) = Y(fb(k),tb(k));
    end %
    Formation of Diagonal Elements.... for m = 1:nb
    for n = 1:nl
    if fb(n) == m Y(m,m) = Y(m,m) + y(n) + b(n);
    elseif
    tb(n) == m Y(m,m) = Y(m,m) + y(n) + b(n);
    end
    end
    end
    %Y; % Bus Admittance Matrix %
    Z = inv(Y); % Bus Impedance Matrix
    4.) CODE FOR THE NEWTON-RAPHSON ITERATIONS (main.m) %
    Program for Newton-Raphson Load Flow Analysis..
    nbus=3; Y = ybusmatrix(); % Calling ybusmatrix.m to get Y-Bus Matrix..
    busd = busdatas(); % Calling busdatas..
    BMva = 100; % Base MVA..
    bus = busd:),1); % Bus Number..
    type = busd:),2); % Type of Bus 1-Slack, 2-PV, 3-PQ..
    V = busd:),3); % Specified Voltage..
    del = busd:),4); % Voltage Angle..
    Pg = busd:),5)/BMva; % PGi..
    Qg = busd:),6)/BMva; % QGi..
    Pl = busd:),7)/BMva; % PLi..
    Ql = busd:),8)/BMva; % QLi..
    P = Pg - Pl; % Pi = PGi - PLi..
    Q = Qg - Ql; % Qi = QGi - QLi..
    Psp = P; % P Specified..
    Qsp = Q; % Q Specified..
    G = real(Y); % Conductance matrix..
    B = imag(Y); % Susceptance matrix..
    pv = find(type == 2 | type == 1); % PV Buses.
    . pq = find(type == 3); % PQ Buses.
    . npv = length(pv); % No. of PV buses..
    npq = length(pq); % No. of PQ buses.. T
    ol = 1; Iter = 1;
    while (Tol > 1e-5) % Iteration starting..
    P = zeros(nbus,1);
    Q = zeros(nbus,1); % Calculate P and Q for i = 1:nbus
    for k = 1:nbus
    P(i) = P(i) + V(i)* V(k)*(G(i,k)*cos(del(i)-del(k)) + B(i,k)*sin(del(i)-del(k)));
    Q(i) = Q(i) + V(i)* V(k)*(G(i,k)*sin(del(i)-del(k)) - B(i,k)*cos(del(i)-del(k)));
    end
    end
    % Calculate change from specified value
    dPa = Psp-P;
    dQa = Qsp-Q; k = 1;
    dQ = zeros(npq,1);
    for i = 1:nbus
    if type(i) == 3 dQ(k,1) = dQa(i);
    k = k+1; end
    end
    dP = dPa(2:nbus);
    M = [dP; dQ]; % Mismatch Vector % Jacobian % J1 - Derivative of Real Power Injections with Angles..
    J1 = zeros(nbus-1,nbus-1);
    for i = 1:(nbus-1)
    m = i+1;
    for k = 1:(nbus-1)
    n = k+1;
    if n == m for n = 1:nbus
    J1(i,k) = J1(i,k) + V(m)* V(n)*(-G(m,n)*sin(del(m)- del(n)) + B(m,n)*cos(del(m)-del(n)));
    end
    J1(i,k) = J1(i,k) - V(m)^2*B(m,m);
    else J1(i,k) = V(m)* V(n)*(G(m,n)*sin(del(m)-del(n)) - B(m,n)*cos(del(m)-del(n)));
    end
    end
    end
    % J2 - Derivative of Real Power Injections with V..
    J2 = zeros(nbus-1,npq);
    for i = 1:(nbus-1) m = i+1;
    for k = 1:npq n = pq(k);
    if n == m for n = 1:nbus
    J2(i,k) = J2(i,k) + V(n)*(G(m,n)*cos(del(m)-del(n)) + B(m,n)*sin(del(m)-del(n)));
    end
    J2(i,k) = J2(i,k) + V(m)*G(m,m);
    else
    J2(i,k) = V(m)*(G(m,n)*cos(del(m)-del(n)) + B(m,n)*sin(del(m)-del(n)));
    end
    end
    end
    % J3 - Derivative of Reactive Power Injections with Angles.. J3 = zeros(npq,nbus-1); for i = 1:npq m = pq(i);
    for k = 1:(nbus-1) n = k+1;
    if n == m for n = 1:nbus
    J3(i,k) = J3(i,k) + V(m)* V(n)*(G(m,n)*cos(del(m)- del(n)) + B(m,n)*sin(del(m)-del(n)));
    end
    J3(i,k) = J3(i,k) - V(m)^2*G(m,m); else J3(i,k) = V(m)* V(n)*(-G(m,n)*cos(del(m)-del(n)) - B(m,n)*sin(del(m)-del(n)));
    end
    end
    end
    % J4 - Derivative of Reactive Power Injections with V..
    J4 = zeros(npq,npq);
    for i = 1:npq
    m = pq(i);
    for k = 1:npq n = pq(k);
    if n == m
    for n = 1:nbus
    J4(i,k) = J4(i,k) + V(n)*(G(m,n)*sin(del(m)-del(n)) - B(m,n)*cos(del(m)-del(n)));
    end
    J4(i,k) = J4(i,k) - V(m)*B(m,m);
    else J4(i,k) = V(m)*(G(m,n)*sin(del(m)-del(n)) - B(m,n)*cos(del(m)-del(n)));
    end
    end
    end
    J = [J1 J2; J3 J4]; % Jacobian Matrix..
    X = inv(J)*M; % Correction Vector
    dTh = X(1:nbus-1); % Change in Voltage Angle..
    dV = X(nbus:end); % Change in Voltage Magnitude..
    % Updating State Vectors..
    del(2:nbus) = dTh + del(2:nbus); % Voltage Angle..
    k = 1;
    for i = 2:nbus
    if type(i) == 3 V(i) = dV(k) + V(i); % Voltage Magnitude..
    k = k+1;
    end
    end
    Iter = Iter + 1;
    Tol = max(abs(M)); % Tolerance..
    end
    loadflow(nbus,V,del,BMva); % Calling Loadflow.m..
    5.) CODE FOR THE PRINTING THE LOADFLOW SOLUTION (loadflow.m)
    % Program for Bus Power Injections, Line & Power flows (p.u)...
    function [Pi Qi Pg Qg Pl Ql] = loadflow(nb,V,del,BMva) Y = ybusmatrix(); % Calling Ybus program..
    lined = linedatas(); % Get linedats..
    busd = busdatas(); % Get busdatas.
    . Vm = pol2rect(V,del); % Converting polar to rectangular.
    . Del = 180/pi*del; % Bus Voltage Angles in Degree...
    fb = lined:),1); % From bus number..
    . tb = lined:),2); % To bus number...
    nl = length(fb); % No. of Branches..
    Pl = busd:),7); %
    PLi.. Ql = busd:),8); %
    QLi.. Iij = zeros(nb,nb);
    Sij = zeros(nb,nb);
    Si = zeros(nb,1); % Bus Current Injections..
    I = Y*Vm; Im = abs(I);
    Ia = angle(I); %Line Current Flows..
    for m = 1:nl
    p = fb(m);
    q = tb(m);
    Iij(p,q) = -(Vm(p) - Vm(q))*Y(p,q);%
    Y(m,n) = -y(m,n).. Iij(q,p) = -Iij(p,q);
    end Iij = sparse(Iij);
    Iijm = abs(Iij);
    Iija = angle(Iij);
    % Line Power Flows..
    for m = 1:nb
    for n = 1:nb
    if m ~= n Sij(m,n) = Vm(m)*conj(Iij(m,n))*BMva;
    end
    end
    end
    Sij = sparse(Sij);
    Pij = real(Sij);
    Qij = imag(Sij); % Line Losses..
    Lij = zeros(nl,1); for m = 1:nl
    p = fb(m); q = tb(m); Lij(m) = Sij(p,q) + Sij(q,p); end Page | 19 Lpij = real(Lij); Lqij = imag(Lij); % Bus Power Injections.. for i = 1:nb for k = 1:nb Si(i) = Si(i) + conj(Vm(i))* Vm(k)*Y(i,k)*BMva; end end Pi = real(Si); Qi = -imag(Si); Pg = Pi+Pl; Qg = Qi+Ql; disp('##################################################################### ####################'); disp('--------------------------------------------------------------------- --------------------'); disp(' Newton Raphson Loadflow Analysis '); disp('--------------------------------------------------------------------- --------------------'); disp('| Bus | V | Angle | Injection | Generation | Load |'); disp('| No | pu | Degree | MW | MVar | MW | Mvar | MW | MVar | '); for m = 1:nb disp('----------------------------------------------------------------- ------------------------'); fprintf('%3g', m); fprintf(' %8.4f', V(m)); fprintf(' %8.4f', Del(m)); fprintf(' %8.3f', Pi(m)); fprintf(' %8.3f', Qi(m)); fprintf(' %8.3f', Pg(m)); fprintf(' %8.3f', Qg(m)); fprintf(' %8.3f', Pl(m)); fprintf(' %8.3f', Ql(m)); fprintf('\n'); end disp('--------------------------------------------------------------------- --------------------'); fprintf(' Total ');fprintf(' %8.3f', sum(Pi)); fprintf(' %8.3f', sum(Qi)); fprintf(' %8.3f', sum(Pi+Pl)); fprintf(' %8.3f', sum(Qi+Ql)); fprintf(' %8.3f', sum(Pl)); fprintf(' %8.3f', sum(Ql)); fprintf('\n'); disp('--------------------------------------------------------------------- --------------------'); disp('##################################################################### ####################'); disp('--------------------------------------------------------------------- ----------------'); disp(' Line FLow and Losses '); disp('--------------------------------------------------------------------- ----------------'); disp('|From|To | P | Q | From| To | P | Q | Line Loss |'); disp('|Bus |Bus| MW | MVar | Bus | Bus| MW | MVar | MW | MVar |'); for m = 1:nl p = fb(m); q = tb(m); disp('----------------------------------------------------------------- --------------------'); fprintf('%4g', p); fprintf('%4g', q); fprintf(' %8.3f', full(Pij(p,q))); fprintf(' %8.3f', full(Qij(p,q))); Page | 20 fprintf(' %4g', q); fprintf('%4g', p); fprintf(' %8.3f', full(Pij(q,p))); fprintf(' %8.3f', full(Qij(q,p))); fprintf(' %8.3f', Lpij(m)); fprintf(' %8.3f', Lqij(m)); fprintf('\n'); end disp('--------------------------------------------------------------------- ----------------'); fprintf(' Total Loss '); fprintf(' %8.3f', sum(Lpij)); fprintf(' %8.3f', sum(Lqij)); fprintf('\n'); disp('--------------------------------------------------------------------- ----------------'); disp('##################################################################### ################');
    You do not have permission to view the full content of this post. Log in or register now.
 

Attachments

Question: The molar constant-pressure molar heat capacity of oxygen gas from 300 K to 1200 K is given by C...​


bookmark.png
flags.png
Calculate molar entropy

dd4%2Fdd44b401-105f-4af0-ab00-bd1722563dbf%2Fimage.jpg
Show transcribed image text

Expert Answer​


  • 02_img-avatar-gry-40x40.png
    editedanswered this
    Was this answer helpful?
    Thumbs up inactive

    1
    Thumbs down inactive

    0
    321 answers
    Substituting the temperatures in K in the given relation for molar constant pressure ,molar heat capacity of oxygen gas we get-
    Cp,m = 25.72 + ( 12.98 x 10-3 K-1 ) 300 - ( 38.62 x 10-7 K-2 ) (1200)2
    -------- = 24.05 K-1
    Now we know that change in entropy of a gas between two temperatures is given by-
    Delta S = Cp,m [ ln Tf / Ti ] : where Tf & Ti represent final &initial temperatures in Kelvin
    ------------ = 24.05 ln [1200/300 ]
    ------------ = 33.34 JK-1
    You do not have permission to view the full content of this post. Log in or register now.
  • 1624544621962.png
 

Attachments

Question: 1.Creating the CellPhone Class Wireless Solutions, Inc. is a business that sells cell phones and ...​


bookmark.png
flags.png
I need help writing this in Python 3, please and thank you!!
2Fae7b0023-5ed1-4d5a-a76f-e7dfd742cba7%2FphpBngjxW.png
Show transcribed image text

Expert Answer​


  • 02_img-avatar-gry-40x40.png
    Anonymousanswered this
    Was this answer helpful?
    Thumbs up inactive

    2
    Thumbs down inactive

    0
    71,819 answers
    class CellPhone:
    def init(self, man, model, price):
    self.manufact = man
    self.model = model
    self.retail_price = price

    def set_manufact(self, man):
    self.manufact = man

    def set_model(self, model):
    self.model = model

    def set_retail_price(self, price):
    self.retail_price = price

    def get_manufact(self):
    return self.manufact

    def get_model(self):
    return self.model

    def get_retail_price(self):
    return self.retail_price

    man = input('Enter the manufacturer: ')
    model = input('Enter the model number: ')
    price = float(input('Enter the retail price: '))
    phone = CellPhone(man, model, price)
    print('Here is the data that you have entered')
    print('Manufacturer: ' + phone.get_manufact())
    print('Model Number: ' + phone.get_model())
    print('Retail Price: $' + '%.2f' % phone.get_retail_price())

    2Fcb9a73b1-6ea1-456d-8e6d-3417e4140bb4%2FphpGXG1am.png
    You do not have permission to view the full content of this post. Log in or register now.
  • 1624545780663.png
 

Attachments

Question: Problem 3: A 13.2-kV single-phase generator supplies power to a load through a transmission line....​


bookmark.png
(1 bookmark)
flags.png
2Fe3c31ee8-bdbd-4a4a-a1a6-6b5911573afb%2FphpA8QsYq.png
Show transcribed image text

Expert Answer​


  • 02_img-avatar-gry-40x40.png
    srinivasaraoanswered this
    Was this answer helpful?
    Thumbs up inactive

    1
    Thumbs down inactive

    0
    1,541 answers
    2F4893a4ea-8f02-46c0-b0a2-ca8205a1ec99%2FphpyDCzZC.png2Fac21d778-f6de-4263-ad91-395860c8067d%2FphpyUha1g.png2Fb3455930-ec2e-4439-a8ee-ef4aea2b4fcf%2FphpExO8pU.png
    You do not have permission to view the full content of this post. Log in or register now.
 

Attachments

Status
Not open for further replies.
Back
Top