Hello, I am working on a matlab code for school and it is using a while loop. The problem with this is I want the while loop to run until the variable doesn't equal itself.

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Jack
Jack on 22 Oct 2024
Commented: VBBV on 22 Oct 2024
So the code won't change the value of R for the while loop which I do not understand. Along with this, in the while loop it has a big if statment with two varriables unkown. A randomly selected value for R is imput and needs to run until it R converges and gives the output of that R and the Cop# that whent with what PT you went with.
Here is the long code with the while loop portion at the top and me trying to get the output for R at the bottom.
% Inputs
D = input('What is the Duct Diameter in mm? \n') ; % Input that asks for the Duct Diameter in mm
d = input('What is the Orifice Diameter in mm? \n') ;
PD = input('What is the Pressure Difference in " W.C.? \n') ;
BP = input('What is the Barometric Pressure in Pa? \n') ;
T = input('What is the Ambient Temperature in °C? \n') ;
R = input('What is the initial guess for the Reynolds number? \n') ;
PT = input('What kind of Pressure Tap is being used? Enter 1 for Corner Tap, 2 for Flange Tap, 3 for D-D/2 Tap or 4 for an ISA-1932 Nozzle. \n') ; % Input that asks what type of tap or nozzle is being used
B = d/D ;
fprintf('The diameter ratio is %.4f ',B)
A = ((19000*B)/(R)).^0.8; % Area that will change based on B and R resulting in mm^2
Rc = 287 ;
A0 = pi*(d/2)^2 ;
fprintf('The orfice area is %.4f mm^2 ',A0)
P = PD*248.64107 ;
fprintf('The pressure difference is %.4f Pa ',P)
mu = -0.245*T^2 + 474.4*T +1.75*10^5 ;
fprintf('The absolute viscosity is %.4f kg/m s ',mu)
p = BP/(Rc*(T + 273.15)) ;
fprintf('The air density is %.4f kg/m^3 ',p)
while R ~= R
R = R + 0.1 ;
if PT == 1 % If PT is equal to 1 then a Corner Tap is being used
L1 = 0; % L1 is given in the textbook based on the tap used (page 25 ASME)
L2 = 0; % L2 is given in the textbook based on the tap used (page 25 ASME)
m = ((2*(L2))/(1 - B)); % m is given based on the L values and the Diameter Ratio
if (D >= 71.12)
Cop1 = 0.5961 + (0.0261*(B.^2)) - (0.216*(B.^8)) + 0.000521*((((10.^6)*B)/R).^0.7) + ((0.0188 + 0.0063*A)*(B.^3.5)*(((10.^6)/R).^0.3)) + (0.043 + 0.080*(exp(-10*(L1))) - 0.123*(exp((-7)*L1)))*(1 - 0.11*A)*((B.^4)/(1-B.^4)) - 0.031*((m) - (0.8*(m).^(1.1)))*B.^1.3 ; % Cop1 calculates the Discharge Coefficcient for a Corner Tap with a Diameter more than 71.12 mm
mr1 = Cop1*A0/(sqrt(1 - B^4))*sqrt(2*P*p) ;
fprintf('The mass flow rate is %.4f kg/s ', mr1)
qv1 = mr1/p ;
fprintf('The volumetric flow rate is %.4f m^3/s ',qv1)
if (B >= .1) && (B < .2) % If statement used to caculate uncertainty
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop1, (0.7 - B)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
elseif (B >= .2) && (B < .6) % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f with an uncertainty of 0.50%% \n', Cop1) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
elseif (B >= .6) && (B < .75) % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop1, ((1.6667*B) - .5)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
end
elseif (D < 71.12)
Cop1 = 0.5961 + (0.0261*(B.^2)) - (0.216*(B.^8)) + 0.000521*((((10.^6)*B)/R).^0.7) + ((0.0188 + 0.0063*A)*(B.^3.5)*(((10.^6)/R).^0.3)) + (0.043 + 0.080*(exp(-10*(L1))) - 0.123*(exp((-7)*L1)))*(1 - 0.11*A)*((B.^4)/(1-B.^4)) - 0.031*((m) - (0.8*(m).^(1.1)))*B.^1.3 + 0.011*(0.75 - B)*(2.8 - D/25.4) ; % Cop1 calculates the Discharge Coefficcient for a Corner Tap with a Diameter less than 71.12 mm
mr1 = Cop1*A0/(sqrt(1 - B^4))*sqrt(2*P*p) ;
fprintf('The mass flow rate is %.4f kg/s ', mr1)
qv1 = mr1/p ;
fprintf('The volumetric flow rate is %.4f m^3/s ',qv1)
if (B >= .1) && (B < .2) % If statement used to caculate uncertainty
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop1, (0.7 - B) + 0.9*(0.75 - B)*(2.8 - D/25.4)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
elseif (B >= .2) && (B < .6) % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f with an uncertainty of %.2f % \n', Cop1, 0.5 + 0.9*(0.75 - B)*(2.8 - D/25.4)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
elseif (B >= .6) && (B < .75) % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop1, ((1.6667*B) - 0.5) + 0.9*(0.75 - B)*(2.8 - D/25.4)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
end
end
elseif PT == 3 % If PT is equal to 3 then a D or D/2 tap is being used
L1 = 1; % L1 is given in the textbook based on the tap used (page 25 ASME)
L2 = .47; % L2 is given in the textbook based on the tap used (page 25 ASME)
m = ((2*(L2))/(1 - B)); % m is given based on the L values and the Diameter Ratio
if (D >= 71.12)
Cop3 = 0.5961 + (0.0261*(B.^2)) - (0.216*(B.^8)) + 0.000521*((((10.^6)*B)/R).^0.7) + ((0.0188 + 0.0063*A)*(B.^3.5)*(((10.^6)/R).^0.3)) + (0.043 + 0.080*(exp(-10*(L1))) - 0.123*(exp((-7)*L1)))*(1 - 0.11*A)*((B.^4)/(1-B.^4)) - 0.031*((m) - (0.8*(m).^(1.1)))*B.^1.3 ; % Cop3 calculates the discharge coefficcient for D or D/2 taps with a Diameter greater than 71.12 mm
mr3 = Cop3*A0/(sqrt(1 - B^4))*sqrt(2*P*p) ;
fprintf('The mass flow rate is %.4f kg/s ', mr3)
qv3 = mr3/p ;
fprintf('The volumetric flow rate is %.4f m^3/s ',qv3)
if (B >= .1) && (B < .2) % If statement used to caculate uncertainty
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop3, (0.7 - B)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
elseif (B >= .2) && (B < .6) % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f with an uncertainty of 0.50%% \n', Cop3) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
elseif (B >= .6) && (B < .75) % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop3, ((1.6667*B) - .5)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
end
elseif (D < 71.12)
Cop3 = 0.5961 + (0.0261*(B.^2)) - (0.216*(B.^8)) + 0.000521*((((10.^6)*B)/R).^0.7) + ((0.0188 + 0.0063*A)*(B.^3.5)*(((10.^6)/R).^0.3)) + (0.043 + 0.080*(exp(-10*(L1))) - 0.123*(exp((-7)*L1)))*(1 - 0.11*A)*((B.^4)/(1-B.^4)) - 0.031*((m) - (0.8*(m).^(1.1)))*B.^1.3 + 0.011*(0.75 - B)*(2.8 - D/25.4) ; % Cop3 calculates the Discharge Coefficcient for D or D/2 taps with a Diameter less than 71.12 mm
mr3 = Cop3*A0/(sqrt(1 - B^4))*sqrt(2*P*p) ;
fprintf('The mass flow rate is %.4f kg/s ', mr3)
qv3 = mr3/p ;
fprintf('The volumetric flow rate is %.4f m^3/s ',qv3)
if (B >= .1) && (B < .2) % If statement used to caculate uncertainty
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop3, (0.7 - B) + 0.9*(0.75 - B)*(2.8 - D/25.4)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
elseif (B >= .2) && (B < .6) % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f with an uncertainty of %.2f % \n', Cop3, 0.5 + 0.9*(0.75 - B)*(2.8 - D/25.4)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
elseif (B >= .6) && (B < .75) % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop3, ((1.6667*B) - .5) + 0.9*(0.75 - B)*(2.8 - D/25.4)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
end
end
elseif PT == 2 % If PT is equal to 2 then a Flange Tap is being used
L1 = (25.4/D); % L1 is given in the textbook based on the tap used (page 25 ASME)
L2 = (25.4/D); % L1 is given in the textbook based on the tap used (page 25 ASME)
m = ((2*(L2))/(1-B)); % m is given based on the L values and the Diameter Ratio
if (D >= 71.12)
Cop2 = 0.5961 + (0.0261*(B.^2)) - (0.216*(B.^8)) + 0.000521*((((10.^6)*B)/R).^0.7) + ((0.0188 + 0.0063*A)*(B.^3.5)*(((10.^6)/R).^0.3)) + (0.043+0.080*(exp(-10*(L1))) - 0.123*(exp((-7)*L1)))*(1 - 0.11*A)*((B.^4)/(1 - B.^4)) - 0.031*((m) - (0.8*(m).^(1.1)))*B.^1.3 ; % Cop2 calculates the discharge coefficcient for flange taps with a Diameter greater than 71.12 mm
mr2 = Cop2*A0/(sqrt(1 - B^4))*sqrt(2*P*p) ;
fprintf('The mass flow rate is %.4f kg/s ', mr2)
qv2 = mr2/p ;
fprintf('The volumetric flow rate is %.4f m^3/s ',qv2)
if (B >= .1) && (B < .2) % If statement used to caculate uncertainty
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop2, (0.7 - B)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
elseif (B >= .2) && (B < .6) % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f with am uncertainty of 0.50%% \n', Cop2) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
elseif (B >= .6) && (B < .75) % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop2, ((1.6667*B) - .5)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
end
elseif (D < 71.12)
Cop2 = 0.5961 + (0.0261*(B.^2)) - (0.216*(B.^8)) + 0.000521*((((10.^6)*B)/R).^0.7) + ((0.0188 + 0.0063*A)*(B.^3.5)*(((10.^6)/R).^0.3)) + (0.043 + 0.080*(exp(-10*(L1))) - 0.123*(exp((-7)*L1)))*(1 - 0.11*A)*((B.^4)/(1-B.^4)) - 0.031*((m) - (0.8*(m).^(1.1)))*B.^1.3 + 0.011*(0.75 - B)*(2.8 - D/25.4) ; % Cop2 calculates the Discharge Coefficcient for flange taps with a Diameter less than 71.12 mm
mr2 = Cop2*A0/(sqrt(1 - B^4))*sqrt(2*P*p) ;
fprintf('The mass flow rate is %.4f kg/s ', mr2)
qv2 = mr2/p ;
fprintf('The volumetric flow rate is %.4f m^3/s ',qv2)
if (B >= .1) && (B < .2) % If statement used to caculate uncertainty
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop2, (0.7 - B) + 0.9*(0.75 - B)*(2.8 - D/25.4)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
elseif (B >= .2) && (B < .6) % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f with am uncertainty of %.2f % \n', Cop2, 0.5 + 0.9*(0.75 - B)*(2.8 - D/25.4)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
elseif (B >= .6) && (B < .75) % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop2, ((1.6667*B) - .5) + 0.9*(0.75 - B)*(2.8 - D/25.4)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
end
end
elseif PT == 4 % If PT is equal to 4 than a ISA-1932 Nozzel is being used
Cop4 = 0.9900 - 0.2262*(B).^4.1 - (0.00175*B.^2 - 0.0033*B^4.15)*((10.^6)/R).^1.15 ; % Calculates the discharge coefficient if you are using a ISA-1932 Nozzle
mr4 = Cop4*A0/(sqrt(1 - B^4))*sqrt(2*P*p) ;
fprintf('The mass flow rate is %.4f kg/s ', mr4)
qv4 = mr4/p ;
fprintf('The volumetric flow rate is %.4f m^3/s ',qv4)
if B <= .6 % If statement used to caculate uncertainty
fprintf('The discharge coefficient is %.4f with an uncertainty of 0.80%% \n', Cop4) % Outputs a statement with your discharge coefficient and uncertainty based on your diameter ratio
elseif B > .6 % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop4, ((2*B) - .4)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
end
end
end
disp(R)

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Answers (2)

Walter Roberson
Walter Roberson on 22 Oct 2024
while R ~= R
That R ~= R will only be true if R is NaN.
I want the while loop to run until the variable doesn't equal itself.
Possibly you want to store the previous version, into something like R_old, and test R ~= R_old and inside the loop,
R_old = R;
R = something that updates R
VBBV
VBBV on 22 Oct 2024 
    while R ~= Rc

You probably need to change the while loop condition like above. Given the input value for Reynolds number you need to check it against its critical value which i presume as Rc

  2 Comments
VBBV
VBBV on 22 Oct 2024

You may also add another condition in case user inputs R value exactly as 287.

if isequal(R, Rc)
  fprintf('message')
else 
  while R~=Rc 
    ...
  end
end
VBBV
VBBV on 22 Oct 2024
Ran in:
% Inputs
D = 1050; %input('What is the Duct Diameter in mm? \n') ; % Input that asks for the Duct Diameter in mm
d = 50; %input('What is the Orifice Diameter in mm? \n') ;
PD = 1.2; %input('What is the Pressure Difference in " W.C.? \n') ;
BP = 101000;%input('What is the Barometric Pressure in Pa? \n') ;
T = 25; %input('What is the Ambient Temperature in °C? \n') ;
R = 100; %input('What is the initial guess for the Reynolds number? \n') ;
PT = 1; %input('What kind of Pressure Tap is being used? Enter 1 for Corner Tap, 2 for Flange Tap, 3 for D-D/2 Tap or 4 for an ISA-1932 Nozzle. \n') ; % Input that asks what type of tap or nozzle is being used
B = d/D ;
fprintf('The diameter ratio is %.4f ',B)
The diameter ratio is 0.0476
A = ((19000*B)/(R)).^0.8; % Area that will change based on B and R resulting in mm^2
Rc = 287 ;
A0 = pi*(d/2)^2 ;
fprintf('The orfice area is %.4f mm^2 ',A0)
The orfice area is 1963.4954 mm^2
P = PD*248.64107 ;
fprintf('The pressure difference is %.4f Pa ',P)
The pressure difference is 298.3693 Pa
mu = -0.245*T^2 + 474.4*T +1.75*10^5 ;
fprintf('The absolute viscosity is %.4f kg/m s ',mu)
The absolute viscosity is 186706.8750 kg/m s
p = BP/(Rc*(T + 273.15)) ;
fprintf('The air density is %.4f kg/m^3 ',p)
The air density is 1.1803 kg/m^3
k = 1;
while R ~= Rc & R <= Rc
R = R + 10 ; % for convenience stp size is increased
fprintf('iteration no %d ', k) % record iteration
if PT == 1 % If PT is equal to 1 then a Corner Tap is being used
L1 = 0; % L1 is given in the textbook based on the tap used (page 25 ASME)
L2 = 0; % L2 is given in the textbook based on the tap used (page 25 ASME)
m = ((2*(L2))/(1 - B)); % m is given based on the L values and the Diameter Ratio
if (D >= 71.12)
Cop1 = 0.5961 + (0.0261*(B.^2)) - (0.216*(B.^8)) + 0.000521*((((10.^6)*B)/R).^0.7) + ((0.0188 + 0.0063*A)*(B.^3.5)*(((10.^6)/R).^0.3)) + (0.043 + 0.080*(exp(-10*(L1))) - 0.123*(exp((-7)*L1)))*(1 - 0.11*A)*((B.^4)/(1-B.^4)) - 0.031*((m) - (0.8*(m).^(1.1)))*B.^1.3 ; % Cop1 calculates the Discharge Coefficcient for a Corner Tap with a Diameter more than 71.12 mm
mr1 = Cop1*A0/(sqrt(1 - B^4))*sqrt(2*P*p) ;
fprintf('The mass flow rate is %.4f kg/s ', mr1)
qv1 = mr1/p ;
fprintf('The volumetric flow rate is %.4f m^3/s ',qv1)
if (B >= .1) && (B < .2) % If statement used to caculate uncertainty
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop1, (0.7 - B)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
elseif (B >= .2) && (B < .6) % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f with an uncertainty of 0.50%% \n', Cop1) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
elseif (B >= .6) && (B < .75) % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop1, ((1.6667*B) - .5)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
end
elseif (D < 71.12)
Cop1 = 0.5961 + (0.0261*(B.^2)) - (0.216*(B.^8)) + 0.000521*((((10.^6)*B)/R).^0.7) + ((0.0188 + 0.0063*A)*(B.^3.5)*(((10.^6)/R).^0.3)) + (0.043 + 0.080*(exp(-10*(L1))) - 0.123*(exp((-7)*L1)))*(1 - 0.11*A)*((B.^4)/(1-B.^4)) - 0.031*((m) - (0.8*(m).^(1.1)))*B.^1.3 + 0.011*(0.75 - B)*(2.8 - D/25.4) ; % Cop1 calculates the Discharge Coefficcient for a Corner Tap with a Diameter less than 71.12 mm
mr1 = Cop1*A0/(sqrt(1 - B^4))*sqrt(2*P*p) ;
fprintf('The mass flow rate is %.4f kg/s ', mr1)
qv1 = mr1/p ;
fprintf('The volumetric flow rate is %.4f m^3/s ',qv1)
if (B >= .1) && (B < .2) % If statement used to caculate uncertainty
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop1, (0.7 - B) + 0.9*(0.75 - B)*(2.8 - D/25.4)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
elseif (B >= .2) && (B < .6) % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f with an uncertainty of %.2f % \n', Cop1, 0.5 + 0.9*(0.75 - B)*(2.8 - D/25.4)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
elseif (B >= .6) && (B < .75) % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop1, ((1.6667*B) - 0.5) + 0.9*(0.75 - B)*(2.8 - D/25.4)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
end
end
elseif PT == 3 % If PT is equal to 3 then a D or D/2 tap is being used
L1 = 1; % L1 is given in the textbook based on the tap used (page 25 ASME)
L2 = .47; % L2 is given in the textbook based on the tap used (page 25 ASME)
m = ((2*(L2))/(1 - B)); % m is given based on the L values and the Diameter Ratio
if (D >= 71.12)
Cop3 = 0.5961 + (0.0261*(B.^2)) - (0.216*(B.^8)) + 0.000521*((((10.^6)*B)/R).^0.7) + ((0.0188 + 0.0063*A)*(B.^3.5)*(((10.^6)/R).^0.3)) + (0.043 + 0.080*(exp(-10*(L1))) - 0.123*(exp((-7)*L1)))*(1 - 0.11*A)*((B.^4)/(1-B.^4)) - 0.031*((m) - (0.8*(m).^(1.1)))*B.^1.3 ; % Cop3 calculates the discharge coefficcient for D or D/2 taps with a Diameter greater than 71.12 mm
mr3 = Cop3*A0/(sqrt(1 - B^4))*sqrt(2*P*p) ;
fprintf('The mass flow rate is %.4f kg/s ', mr3)
qv3 = mr3/p ;
fprintf('The volumetric flow rate is %.4f m^3/s ',qv3)
if (B >= .1) && (B < .2) % If statement used to caculate uncertainty
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop3, (0.7 - B)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
elseif (B >= .2) && (B < .6) % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f with an uncertainty of 0.50%% \n', Cop3) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
elseif (B >= .6) && (B < .75) % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop3, ((1.6667*B) - .5)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
end
elseif (D < 71.12)
Cop3 = 0.5961 + (0.0261*(B.^2)) - (0.216*(B.^8)) + 0.000521*((((10.^6)*B)/R).^0.7) + ((0.0188 + 0.0063*A)*(B.^3.5)*(((10.^6)/R).^0.3)) + (0.043 + 0.080*(exp(-10*(L1))) - 0.123*(exp((-7)*L1)))*(1 - 0.11*A)*((B.^4)/(1-B.^4)) - 0.031*((m) - (0.8*(m).^(1.1)))*B.^1.3 + 0.011*(0.75 - B)*(2.8 - D/25.4) ; % Cop3 calculates the Discharge Coefficcient for D or D/2 taps with a Diameter less than 71.12 mm
mr3 = Cop3*A0/(sqrt(1 - B^4))*sqrt(2*P*p) ;
fprintf('The mass flow rate is %.4f kg/s ', mr3)
qv3 = mr3/p ;
fprintf('The volumetric flow rate is %.4f m^3/s ',qv3)
if (B >= .1) && (B < .2) % If statement used to caculate uncertainty
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop3, (0.7 - B) + 0.9*(0.75 - B)*(2.8 - D/25.4)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
elseif (B >= .2) && (B < .6) % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f with an uncertainty of %.2f % \n', Cop3, 0.5 + 0.9*(0.75 - B)*(2.8 - D/25.4)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
elseif (B >= .6) && (B < .75) % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop3, ((1.6667*B) - .5) + 0.9*(0.75 - B)*(2.8 - D/25.4)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
end
end
elseif PT == 2 % If PT is equal to 2 then a Flange Tap is being used
L1 = (25.4/D); % L1 is given in the textbook based on the tap used (page 25 ASME)
L2 = (25.4/D); % L1 is given in the textbook based on the tap used (page 25 ASME)
m = ((2*(L2))/(1-B)); % m is given based on the L values and the Diameter Ratio
if (D >= 71.12)
Cop2 = 0.5961 + (0.0261*(B.^2)) - (0.216*(B.^8)) + 0.000521*((((10.^6)*B)/R).^0.7) + ((0.0188 + 0.0063*A)*(B.^3.5)*(((10.^6)/R).^0.3)) + (0.043+0.080*(exp(-10*(L1))) - 0.123*(exp((-7)*L1)))*(1 - 0.11*A)*((B.^4)/(1 - B.^4)) - 0.031*((m) - (0.8*(m).^(1.1)))*B.^1.3 ; % Cop2 calculates the discharge coefficcient for flange taps with a Diameter greater than 71.12 mm
mr2 = Cop2*A0/(sqrt(1 - B^4))*sqrt(2*P*p) ;
fprintf('The mass flow rate is %.4f kg/s ', mr2)
qv2 = mr2/p ;
fprintf('The volumetric flow rate is %.4f m^3/s ',qv2)
if (B >= .1) && (B < .2) % If statement used to caculate uncertainty
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop2, (0.7 - B)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
elseif (B >= .2) && (B < .6) % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f with am uncertainty of 0.50%% \n', Cop2) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
elseif (B >= .6) && (B < .75) % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop2, ((1.6667*B) - .5)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
end
elseif (D < 71.12)
Cop2 = 0.5961 + (0.0261*(B.^2)) - (0.216*(B.^8)) + 0.000521*((((10.^6)*B)/R).^0.7) + ((0.0188 + 0.0063*A)*(B.^3.5)*(((10.^6)/R).^0.3)) + (0.043 + 0.080*(exp(-10*(L1))) - 0.123*(exp((-7)*L1)))*(1 - 0.11*A)*((B.^4)/(1-B.^4)) - 0.031*((m) - (0.8*(m).^(1.1)))*B.^1.3 + 0.011*(0.75 - B)*(2.8 - D/25.4) ; % Cop2 calculates the Discharge Coefficcient for flange taps with a Diameter less than 71.12 mm
mr2 = Cop2*A0/(sqrt(1 - B^4))*sqrt(2*P*p) ;
fprintf('The mass flow rate is %.4f kg/s ', mr2)
qv2 = mr2/p ;
fprintf('The volumetric flow rate is %.4f m^3/s ',qv2)
if (B >= .1) && (B < .2) % If statement used to caculate uncertainty
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop2, (0.7 - B) + 0.9*(0.75 - B)*(2.8 - D/25.4)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
elseif (B >= .2) && (B < .6) % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f with am uncertainty of %.2f % \n', Cop2, 0.5 + 0.9*(0.75 - B)*(2.8 - D/25.4)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
elseif (B >= .6) && (B < .75) % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop2, ((1.6667*B) - .5) + 0.9*(0.75 - B)*(2.8 - D/25.4)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
end
end
elseif PT == 4 % If PT is equal to 4 than a ISA-1932 Nozzel is being used
Cop4 = 0.9900 - 0.2262*(B).^4.1 - (0.00175*B.^2 - 0.0033*B^4.15)*((10.^6)/R).^1.15 ; % Calculates the discharge coefficient if you are using a ISA-1932 Nozzle
mr4 = Cop4*A0/(sqrt(1 - B^4))*sqrt(2*P*p) ;
fprintf('The mass flow rate is %.4f kg/s ', mr4)
qv4 = mr4/p ;
fprintf('The volumetric flow rate is %.4f m^3/s ',qv4)
if B <= .6 % If statement used to caculate uncertainty
fprintf('The discharge coefficient is %.4f with an uncertainty of 0.80%% \n', Cop4) % Outputs a statement with your discharge coefficient and uncertainty based on your diameter ratio
elseif B > .6 % Elseif statment if prior statment was untrue
fprintf('The discharge coefficient is %.4f \n with an uncertainty of %.2f % \n', Cop4, ((2*B) - .4)) % Gives a statement with the calculated Discharge Coefficient and uncertainty based on the Diameter Ratio used
end
end
k=k+1;
end
iteration no 1
The mass flow rate is 32969.3195 kg/s
The volumetric flow rate is 27932.2114 m^3/s
iteration no 2
The mass flow rate is 32856.8967 kg/s
The volumetric flow rate is 27836.9647 m^3/s
iteration no 3
The mass flow rate is 32759.3527 kg/s
The volumetric flow rate is 27754.3237 m^3/s
iteration no 4
The mass flow rate is 32673.7854 kg/s
The volumetric flow rate is 27681.8295 m^3/s
iteration no 5
The mass flow rate is 32598.0164 kg/s
The volumetric flow rate is 27617.6366 m^3/s
iteration no 6
The mass flow rate is 32530.3758 kg/s
The volumetric flow rate is 27560.3302 m^3/s
iteration no 7
The mass flow rate is 32469.5608 kg/s
The volumetric flow rate is 27508.8066 m^3/s
iteration no 8
The mass flow rate is 32414.5386 kg/s
The volumetric flow rate is 27462.1908 m^3/s
iteration no 9
The mass flow rate is 32364.4792 kg/s
The volumetric flow rate is 27419.7796 m^3/s
iteration no 10
The mass flow rate is 32318.7076 kg/s
The volumetric flow rate is 27381.0011 m^3/s
iteration no 11
The mass flow rate is 32276.6683 kg/s
The volumetric flow rate is 27345.3846 m^3/s
iteration no 12
The mass flow rate is 32237.9001 kg/s
The volumetric flow rate is 27312.5394 m^3/s
iteration no 13
The mass flow rate is 32202.0163 kg/s
The volumetric flow rate is 27282.1380 m^3/s
iteration no 14
The mass flow rate is 32168.6904 kg/s
The volumetric flow rate is 27253.9037 m^3/s
iteration no 15
The mass flow rate is 32137.6444 kg/s
The volumetric flow rate is 27227.6010 m^3/s
iteration no 16
The mass flow rate is 32108.6402 kg/s
The volumetric flow rate is 27203.0281 m^3/s
iteration no 17
The mass flow rate is 32081.4725 kg/s
The volumetric flow rate is 27180.0111 m^3/s
iteration no 18
The mass flow rate is 32055.9631 kg/s
The volumetric flow rate is 27158.3991 m^3/s
iteration no 19
The mass flow rate is 32031.9569 kg/s
The volumetric flow rate is 27138.0606 m^3/s
disp(R)
290

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