Air Compressor Model Analysis

[Donahue Maverick]

I have an interest in storing, for later use, energy from excess shaft power developed by a wind turbine. Currently I am assuming that the compressor is running in a steady fashion, continually drawing 300K, 1.0 atm. air into the compression chamber and compressing it to 20 atms. before shuttling it to an adiabatic accumulator at the discharge pressure of 20 atms. For continuous operation at 1850 RPM, how would I compute the the piston work rate [kW] and the heat transfer rate [kW]? I am neglecting the fins on storage cylinder due to the fact that in my initial analysis the energy due to HT is significantly lower than the work.

Found/Given:
n = 1.4;
Compressor piston diameter: 5 cm (single piston)
Compressor cylinder volume when piston is at bottom dead center: 150 cc. [0.000015 m^3]
Storage tank (accumulator): Cylindrical tank with a diameter of 90 cm and a length of 300 cm.
Compressor speed: 1850 RPM
I would also like to determine the time to charge the storage tank [sec] from 2 atm. to 20 atm.

This is the start of my analysis coded in MATLAB - any tweaks/suggestions would be appreciated.

% Heat Transfer Variables
omega = 1850; %rpm
D = 0.09; %m
L = 0.3; %m
V_piston_open = (D^2)*0.25*pi*L; %m^3

% Compression Variables
Cp = 1.00; %kJ/kgK @ 300K
Cv = 0.718; %kJ/kgK @ 300K
k = Cp/Cv;
R = 8.314 / 28.97; %kJ/kg*K

%% Thermodynamic Processes
P_3 = 20 * 101325; %Pa

% 1 - 2: Air Intake (Isobaric Process)
T_1 = 300;%K
P_1 = 101325; %Pa
V_1 = 1.7*10^-5; %m^3
u_1 = 214; %kJ/kg @ 300K

T_2 = T_1;
P_2 = P_1;
u_2 = u_1;

% 2 - 3: Isentropic Compression
T_3 = T_2 * (P_3/P_2)^((k - 1)/k);
V_2 = 0.00015; %m^3
h_3 = 713; %kJ/kg @ 700K
u_3 = 512; %kJ/kg @ 700K

h_1 = u_1 + P_1 * V_1;

V_3 = V_2 * ((P_2/P_3)^(1/k)); %Questionable
h_2 = u_1 + P_1 * V_2;

% 3 - 4: Discharge of Compressed Air
P_4 = P_3;
u_4 = 364; %kJ/kg
h_4 = 762; %kJ/kg
V_4 = (h_4 - u_4)/P_4;
T_4 = T_3 * (V_3/V_4)^(k-1);

% Net Work
W_12 = P_1*(V_2 - V_1);
% W_12 = h_2 - h_1
W_23 = (P_3*V_3 - P_2*V_2)/(((1 - k)));
%W_23 = h_3 - h_2;
W_34 = P_4*(V_4 - V_3);
%W_34 = h_4 - h_3;
W_41 = 0;

W_tot_piston = W_12 + W_34;

% Net Heat Transfer
Q_12 = u_2 - u_1 + W_12;
Q_23 = u_3 - u_2 + W_23;
Q_34 = u_4 - u_3 + W_34;
Q_41 = u_1 - u_4;

Q_tot = Q_41 + Q_12 + Q_23 + Q_34;

% Mass Flow Rate

m = (-Q_tot + W_tot_piston)/(h_4 - h_1);

W_tot_cycle = m * (u_4 - u_1 + P_4*V_4 - P_1*V_1 + Q_34);
Q_tot_cycle = m * Q_tot;
W_tot_piston2 = m * W_tot_piston;

% Time to charge the storing tank
V_tank = pi * D * .25 * D * L;
w = (omega / 60) * 2 * pi;
n = (P_2 * V_2) / (R * 28.97 * T_2);

 

[trurle]

Seems you come to design starting from the wrong side - starting from the engineering details instead of specifications.
In particular, your gas cylinder would lose ~200 kW in nearly-adiabatic process without insulation, while compressor seems to be about 5 kW (therefore your accumulator is isothermic, not adiabatic)
You need to add good thermal isolation, likely in form of ~40cm of glass wool.

The powers above could be calculated using following rules:
* Heat transfer - radiative is by Stefan-Boltzman law, convective as 20 W/m2K, conductive from specs of insulator material
* Piston work - integrate pressure under piston by the travel of piston

Overall, such design is generally inferior to the pumped water storage, exactly because of air compressibility and resulting thermal losses.