Undefined function 'minus' for input arguments of type 'function_handle'.

Question

Roshni Khetan 2017년 8월 29일

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https://kr.mathworks.com/matlabcentral/answers/354468-undefined-function-minus-for-input-arguments-of-type-function_handle

댓글: Walter Roberson 2017년 8월 29일

m_tot = 8; % in kg/sec
[M2,H] = B249VALVEcurve()
C_v249= [M2];
[M,H] = B251valveperfcurve()
C_v251 = [M];% a vector for opening from 1 - 100%
wInP = 400; % in kPa
k = 1;
PD_pipe2 = 0.96 % In kPa from toolbox online - assuming 3 inch dia, 20 ft length, flow rate for 5 kg/sec
PD_DIC1  = @(m_w2)(160*(m_w2 ^2) + (2.3*10^3)*m_w2 - 10^4)/1e3;
P_out_DIC1 = @(m_w2) wInP - PD_DIC1 - PD_pipe2;
rho_w =refpropm('D','T',345,'P',400,'water');
Q_w2 = @(m_w2)m_w2/rho_w;
Q_w2_hr =@(m_w2) 3600*Q_w2; % in m3/hr
m_w1 = @(m_w2) m_tot - m_w2;
PD_pipe1 = 0.96;
PD_oc = 5;
PD_IC1= @(m_w2) (590*(m_w1 ^2) + 0.42*m_w1+ 0.56)/1e3;
PD_IC2 =@(m_w2) (590*(m_w1 ^2) + 0.4*m_w1+ 0.51)/1e3;
PD_AC=  @(m_w2)(1400*(m_w1^2) + 0.37*m_w1 + 0.44)/1e3;
P_out_AC =@(m_w2) wInP - PD_IC1 - PD_pipe2 - PD_IC2 - PD_oc - PD_AC;
rho_w1 =refpropm('D','T',360,'P',280,'water'); % assuming 280, anyways rho shouldnt vary much
Q_w1 = m_tot/rho_w1;
Q_w1_hr= 3600*Q_w1; % in m3/hr
for i = 1:100
    Kv(i) = 0.865*C_v249(i); % gpm into m3/hr
    DP_B249(i) =  @(m_w2)((Q_w2_hr/Kv(i))^2)*(10^2); % Q is in m3/hr, and Press is in kPa
    DP_B249a = DP_B249(i);
    for j = 1:100
        Kv_251(j) = 0.865*C_v251(j);
        %         Kv_251(i,j) = Kv_251(j);
        %         DP_B249(i,j) = DP_B249(i) ;
        %         PD_DIC1(i,j) = PD_DIC1(i);
        DP_B251(j) = ((Q_w1_hr/Kv_251(j))^2)*(10^2); % Q is in m3/hr, and Press is in kPa
        DP_B251a = DP_B251(j);
          PD_1 = @(m_w2)PD_pipe1  +PD_oc  +PD_IC1  +PD_IC2 + PD_AC + DP_B251a ;
          PD_2 = @(m_w2)PD_pipe2 + PD_DIC1 + DP_B249a+ PD_AC+ DP_B251a ;
          fun = @(m_w2) (PD_2 - PD_1);
          m_w2 = fzero(fun,2);
          m_w2_f(k) = m_w2;
          m_w1_f(k) = m_tot - m_w2;
          B251valve_pos(k) = j; %valve position of 251 for this flow rate
          B249valve_pos(k) = i; % valve position of 249 for this flow rate
          k =  k+1;
          j = j+1;
      end
      i = i +1;
  end

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Walter Roberson 2017년 8월 29일

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fun = @(m_w2) (PD_2(m_w2) - PD_1(m_w2));

Roshni Khetan 2017년 8월 29일

편집: Walter Roberson 2017년 8월 29일

MATLAB Online에서 열기

Hi Walter! I did this but it gives me an error:

Index exceeds matrix dimensions for this line: m_w2 = fzero(fun,2);

m_tot = 8; % in kg/sec RUN THE MODEL FOR DIFFERENT m_tot
[M2,H] = B249VALVEcurve()
C_v249= [M2];
[M,H] = B251valveperfcurve()
C_v251 = [M];% a vector for opening from 1 - 100%
wInP = 400; % in kPa
k = 1;
PD_pipe2 = 0.96 % In kPa from toolbox online - assuming 3 inch dia, 20 ft length, flow rate for 5 kg/sec
PD_DIC1  = @(m_w2)(160*(m_w2 ^2) + (2.3*10^3)*m_w2 - 10^4)/1e3;
rho_w =refpropm('D','T',345,'P',400,'water');
Q_w2 = @(m_w2)m_w2/rho_w;
Q_w2_hr =@(m_w2) 3600*Q_w2(m_w2); % in m3/hr
m_w1 = @(m_w2) m_tot - m_w2;
PD_pipe1 = 0.96;
PD_oc = 5;
PD_IC1= @(m_w2) (590*(m_w1(m_w2) ^2) + 0.42*m_w1(m_w2)+ 0.56)/1e3;
PD_IC2 =@(m_w2) (590*(m_w1(m_w2) ^2) + 0.4*m_w1(m_w2)+ 0.51)/1e3;
PD_AC=  @(m_w2)(1400*(m_w1(m_w2)^2) + 0.37*m_w1(m_w2) + 0.44)/1e3;
rho_w1 =refpropm('D','T',360,'P',280,'water'); % assuming 280, anyways rho shouldnt vary much
Q_w1 = m_tot/rho_w1;
Q_w1_hr= 3600*Q_w1; % in m3/hr
for i = 1:100
    Kv(i) = 0.865*C_v249(i); % gpm into m3/hr
    DP_B249(i) =  @(m_w2)((Q_w2_hr(m_w2)/Kv(i))^2)*(10^2); % Q is in m3/hr, and Press is in kPa
    DP_B249a = DP_B249(i);
    for j = 1:100
        Kv_251(j) = 0.865*C_v251(j);
        %         Kv_251(i,j) = Kv_251(j);
        %         DP_B249(i,j) = DP_B249(i) ;
        %         PD_DIC1(i,j) = PD_DIC1(i);
        DP_B251(j) = ((Q_w1_hr/Kv_251(j))^2)*(10^2); % Q is in m3/hr, and Press is in kPa
        DP_B251a = DP_B251(j);
          PD_1 = @(m_w2)PD_pipe1  +PD_oc  +PD_IC1(m_w2)  +PD_IC2(m_w2) + PD_AC + DP_B251a ;
          PD_2 = @(m_w2)PD_pipe2 + PD_DIC1(m_w2) + DP_B249a(m_w2)+ PD_AC+ DP_B251a ;
          fun = @(m_w2) (PD_2(m_w2)- PD_1(m_w2));
          m_w2 = fzero(@ (m_w2) fun(m_w2),4);
          m_w2_f(k) = m_w2;
          m_w1_f(k) = m_tot - m_w2;
          B251valve_pos(k) = j; %valve position of 251 for this flow rate
          B249valve_pos(k) = i; % valve position of 249 for this flow rate
          k =  k+1;
          j = j+1;
      end
      i = i +1;
  end

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Answer 1

Star Strider 2017년 8월 29일

1
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이 답변에 대한 바로 가기 링크

https://kr.mathworks.com/matlabcentral/answers/354468-undefined-function-minus-for-input-arguments-of-type-function_handle#answer_279582

MATLAB Online에서 열기

My guess is that this assignment (and others like it in your code) are throwing the error:

P_out_AC =@(m_w2) wInP - PD_IC1 - PD_pipe2 - PD_IC2 - PD_oc - PD_AC;

The solution is to provide arguments to your other functions in that assignment:

P_out_AC =@(m_w2) wInP - PD_IC1(m_w2) - PD_pipe2 - PD_IC2(m_w2) - PD_oc - PD_AC(m_w2);

Now they return values, and should work to calculate ‘P_out_AC’.

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Roshni Khetan 2017년 8월 29일

편집: Walter Roberson 2017년 8월 29일

MATLAB Online에서 열기

Thank you for the response! I tried this but now it says:

Index exceeds matrix dimensions for this line: m_w2 = fzero(fun,2);

I dont understand why

%load flow characteristic curves
m_tot = 8; % in kg/sec RUN THE MODEL FOR DIFFERENT m_tot
[M2,H] = B249VALVEcurve()
C_v249= [M2];
[M,H] = B251valveperfcurve()
C_v251 = [M];% a vector for opening from 1 - 100%
wInP = 400; % in kPa
k = 1;
PD_pipe2 = 0.96 % In kPa from toolbox online - assuming 3 inch dia, 20 ft length, flow rate for 5 kg/sec
PD_DIC1  = @(m_w2)(160*(m_w2 ^2) + (2.3*10^3)*m_w2 - 10^4)/1e3;
rho_w =refpropm('D','T',345,'P',400,'water');
Q_w2 = @(m_w2)m_w2/rho_w;
Q_w2_hr =@(m_w2) 3600*Q_w2(m_w2); % in m3/hr
m_w1 = @(m_w2) m_tot - m_w2;
PD_pipe1 = 0.96;
PD_oc = 5;
PD_IC1= @(m_w2) (590*(m_w1(m_w2) ^2) + 0.42*m_w1(m_w2)+ 0.56)/1e3;
PD_IC2 =@(m_w2) (590*(m_w1(m_w2) ^2) + 0.4*m_w1(m_w2)+ 0.51)/1e3;
PD_AC=  @(m_w2)(1400*(m_w1(m_w2)^2) + 0.37*m_w1(m_w2) + 0.44)/1e3;
rho_w1 =refpropm('D','T',360,'P',280,'water'); % assuming 280, anyways rho shouldnt vary much
Q_w1 = m_tot/rho_w1;
Q_w1_hr= 3600*Q_w1; % in m3/hr
for i = 1:100
    Kv(i) = 0.865*C_v249(i); % gpm into m3/hr
    DP_B249(i) =  @(m_w2)((Q_w2_hr(m_w2)/Kv(i))^2)*(10^2); % Q is in m3/hr, and Press is in kPa
    DP_B249a = DP_B249(i);
    for j = 1:100
        Kv_251(j) = 0.865*C_v251(j);
        %         Kv_251(i,j) = Kv_251(j);
        %         DP_B249(i,j) = DP_B249(i) ;
        %         PD_DIC1(i,j) = PD_DIC1(i);
        DP_B251(j) = ((Q_w1_hr/Kv_251(j))^2)*(10^2); % Q is in m3/hr, and Press is in kPa
        DP_B251a = DP_B251(j);
          PD_1 = @(m_w2)PD_pipe1  +PD_oc  +PD_IC1(m_w2)  +PD_IC2(m_w2) + PD_AC + DP_B251a ;
          PD_2 = @(m_w2)PD_pipe2 + PD_DIC1(m_w2) + DP_B249a(m_w2)+ PD_AC+ DP_B251a ;
          fun = @(m_w2) (PD_2(m_w2)- PD_1(m_w2));
          m_w2 = fzero(fun,2);
          m_w2_f(k) = m_w2;
          m_w1_f(k) = m_tot - m_w2;
          B251valve_pos(k) = j; %valve position of 251 for this flow rate
          B249valve_pos(k) = i; % valve position of 249 for this flow rate
          k =  k+1;
          j = j+1;
      end
      i = i +1;
  end

Star Strider 2017년 8월 29일

You need to check the dimensions of your arrays. I cannot determine that from the information you posted.

Walter Roberson 2017년 8월 29일

MATLAB Online에서 열기

At the command line give the command

dbstop if caught error

and then run your program. If it stops for any reason other than the "index exceeds" error, give the command "dbcont". When it stops because of the "index exceeds" error, examine the size of your matrices.

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