Question

MATLAB Program Problem D 0.02 m d 0.01 m 0 P1 3.0 kPa Air Review the tank diagram above. Air flows steadily from a tank, through a hose of diameter D-0.02 m, and exits to the atmosphere from a nozzle of diameter d-0.01 m. The pressure in the tank remains constant at 3.0 KPa and the atmospheric conditions are standard temperature and pressure. Determine the flow rate at the exit and the pressure in the hose. File Submission Save your MATLAB Code (.m) file. Use a naming convention that includes your first initial and last name and the module number N (i.e., LSmithN). Do not add punctuation or special characters. Submit the file by the posted due date. Submit your MATLAB Code (.m) file through 3.6.1 MATLAB File Upload

Answer 1

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%% Note:
% In the given problem it is mentioned P1 is 3 kPa and air is flowing out.
% This means the mentioned pressure is gauge pressure. Because If it is
% absolute pressure then as outside air pressure is 101.325 kPa, flow should
% be into the system, but it is not. Hence the pressure given is gauge.
% Therefore absolute pressure in tank is P1a=P1 + 101.325 = 104.325 kPa.
% Initial in tank at 1 velocity V1 = 0 m/s and P3 =Patm = 101.325 kPa.
% Also it is not mentioned about any head loss in the hose. Considering it
% as ideal streamlined flow from tank to hose to outside through nozzle.

% Applying Bernoulli's Equation between 1 and 2.
% P1a/Rho + V1^2/2 = P2a/Rho + V2^2/2 ..........(1)

% Applying continuity beteen 2 and 3 Q=flow rate
% Q2=Q3 .......................................(2)

% Applying Bernoulli's Equation between 2 and 3.
% P2a/Rho + V2^2/2 = P3/Rho + V3^2/2 ...........(3)

% From (1) and (3) :
% P1a/Rho + V1^2/2 = P3/Rho + V3^2/2 ...........(4)

%% Initializing the Problem

P1 = 3; % in kPa
Patm = 101.325; % in kPa
T = 273.15; % in Kelvin
P1a = 3 + Patm; % in kPa
V1 = 0; % in m/s
D = 0.02; % in m
d = 0.01; % in m
a3=pi*d^2/4; % Area of nozzle at exit in m^2
A2=pi*D^2/4; % Area of hose at exit in m^2
P3=Patm;
R = 0.287; % Gas constant for air in kJ/kgK;
Rho = Patm/(R*T); % in kg/m^3 for air.

%% Solving the problem
% From eqn(4)
V3=sqrt(2*1000*(P1a-P3)/Rho);
Q3 = V3*a3 % Flow rate in m^3/sec

% From eqn(2)
V2=Q3/A2;

% From eqn(3)
P2a = (-V2^2/2 + P3*1000/Rho + V3^2/2)*Rho/1000;
P2=P2a-Patm

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