Description

The Evaluation of a Horizontal Axis
Wind Turbine in a Power Generation
System

Deep Dayaramani
Audrey Zhao
Mechanical Engineering 146
Professor Carey

Task 1 (I)
The goal of this task was to derive a closed-form equation based on the relationship provided for the power output of the rotor, W, in terms of the related parameters: n, ρ, v 1, C L,i , K h , R, σ, r h dot, and λ. It’s then compared to the Betz power output to obtain the Betz efficiency. The detailed derivation process was shown in Appendix I, and the results for the power output and efficiency were:

Task 2 (II)
The power and efficiency equations obtained from Task 1 was programmed in order to calculate them at specific condition. Based on the values given, W was 2.0032e+06 V, and η* was 0.8644. When the angle of attack was kept at 8°, the relationship between the setup angle ξ (in degrees) and radius between r h and R (in meters) was also plotted in Fig.1. And it can be concluded that ξ increases as radius increases. Task 3 (III)
Using task 2, we designed a horizontal axis wind turbine for the top of the Berkeley hills that will provide 1.5 kW of power at mean air density = 1.18 kg/m3 and mean wind speed (v) = 6.5 m/s. Based on provided info, we found R = 5.2000 m, W = 1.5468e+03 W, Sigma = 0.3000, Eff = 0.1916. When the angle of attack was kept at 8°, the relationship between the setup angle ξ (in degrees) and radius between r h and R (in meters) was also plotted in Fig.1. And it can be concluded that ξ increases as radius increases.
Task 4 (IV)
For the first part of the problem, we assumed that the turbine generator nominally produces 12.0 V DC current and using that we determined the delivered current for the power output in task_3. We used P = VI, to find I, and we found it was 128.9 A.

For the second part of this problem, energy was first stored and then extracted from the H 2- O 2 fuel cell at the same current I = 3.5 A for an hour. The energy input was calculated by plugging in I into V cell equation, which was found to be 1.4835 V. Similarly, the extracted power was obtained from V L formula, and was measured to be 0.9765 V . Therefore, the fraction of energy extraction in the process was approximately 66%.

Graphs
Task 2:

Figure 1 . Plot of setup angle ξ with respect to radius changing between r h and R. It demonstrates a positive trend of change in both variables.
Task 3

Figure 2 . Plot of setup angle ξ with respect to radius changing between r h and R. It demonstrates a positive trend of change in both variables.

12/12/18 12:30 PM C:UsersprincDocumentsFall… ask_1_corr.m 1 of 1
function [w_tur, eta]=task_1(rou, v1, alpha, Cl, n, Kh,sigma, rh, R,omega ) lambda = omega*R/v1; rh_bar= rh/R; a=2*R/3/lambda; b=1-rh_bar; syms r; w = Cl/3*(n*rou*lambda*Kh*(v1^2)/R) * ((r^2+a^2)^(0.5) *((r^2+a^2)*(lambda*v1/3/R* (1+sigma*rh_bar/b) – sigma*r*lambda*v1/4/(R^2)/b) + log(r+(r^2+a^2)^(0.5))*(2*sigma*R^2*v1)/ (81*lambda^3*b))); w_b = 16/27*0.5*rou*v1^3*pi *r^2; %turbine work ans = subs(w,r, [rh R]); w_tur =double(ans(2)-ans(1)); %betz work ans1 = subs(w_b,r, [rh R]); w_betz = double(ans1(2)-ans1(1)); eta=w_tur/w_betz; c=rh:0.001:R; xi = alpha*pi/180 – atan(2*R./3./lambda./c); xi1=xi*180/pi; plot(c, xi1) title (‘Setup Angle vs. Radius’); xlabel(‘Radius (m)’, ‘FontWeight’,’bold’); ylabel(‘Setup Angle (degrees)’, ‘FontWeight’,’bold’); end

alpha, Cl, n, omega) %setting R, sigma, and other variables R1=linspace(1,22,51); sigma1=linspace(0,1,51); Kh1 = 0.085*R1; rh1 = 0.1*R1; %Using task_1 to find W_tur and Efficiency for all different R and sigma for i = 1:length(R1) disp(i); for j = 1:length(sigma1) [w_tur(i,j),eta(i,j)]=task_1(rou, v1, alpha, Cl, n, Kh1(i),sigma1(j), rh1(i), R1(i),omega ); end end %finding a value closest to 1500W [a,b]= find(w_tur<1550 & w_tur>=1500); for l = 1:length(a) des_2_w_tur(l) = w_tur(a(l),b(l)); des_2_eff(l) = eta(a(l),b(l)); end [p,q] = max(des_2_eff); des_2_eff = p; des_2_w_tur = des_2_w_tur(q); %finding a value of efficiency closest to 1 [c,d]= find(eta<1.00 &eta>0.98); for l = 1:length(c) des_1_w_tur(l) = w_tur(c(l),d(l)); des_1_eff(l) = eta(c(l),d(l)); end [e,f] = max(des_1_eff); des_1_eff = e; des_1_w_tur = des_1_w_tur(f); des_1_R= R1(c(f)); des_1_sigma = sigma1(d(f)); des_2_R = R1(a(q)); des_2_sigma = sigma1(b(q)); des_R = [des_1_R,des_2_R]; des_eff = [des_1_eff,des_2_eff]; des_w_tur = [des_1_w_tur,des_2_w_tur]; %choosing between two alternatives for min sigma [overall_des_sigma,index] = min([des_1_sigma,des_2_sigma]); overall_des_R= des_R(index); overall_des_w_tur = des_w_tur(index); overall_des_eff = des_eff(index); %plotting change in setup angle vs R des_ov_rh = overall_des_R*0.1; lambda = omega*overall_des_R/v1; des_ov_c = des_ov_rh:0.001:overall_des_R; des_ov_xi = alpha*pi/180 – atan(2*overall_des_R./3./lambda./des_ov_c); des_ov_xi1=des_ov_xi*180/pi; plot(des_ov_c, des_ov_xi1) xlabel(‘Radius (m)’) ylabel(‘Setup Angle (Degrees)’) title(‘Setup Angle vs Radius’)

end function [w_tur, eta]=task_1(rou, v1, alpha, Cl, n, Kh,sigma, rh, R,omega ) lambda = omega*R/v1; rh_bar= rh/R; a=2*R/3/lambda; b=1-rh_bar; syms r; w = Cl/3*(n*rou*lambda*Kh*(v1^2)/R) * ((r^2+a^2)^(0.5) *((r^2+a^2)*(lambda*v1/3/R* (1+sigma*rh_bar/b) – sigma*r*lambda*v1/4/(R^2)/b) + log(r+(r^2+a^2)^(0.5))*(2*sigma*R^2*v1)/ (81*lambda^3*b))); w_b = 16/27*0.5*rou*v1^3*pi *r^2; %turbine work ans = subs(w,r, [rh R]); w_tur =double(ans(2)-ans(1)); %betz work ans1 = subs(w_b,r, [rh R]); w_betz = double(ans1(2)-ans1(1)); eta=w_tur/w_betz;
%c=rh:0.001:R;
%xi = alpha*pi/180 – atan(2*R./3./lambda./c);
%xi1=xi*180/pi;
%plot(c, xi1)
end
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P = 1.5468e+03;
V = 12;
I = P/V;

12/12/18 12:30 PM C:UsersprincDocumentsFall 201… ask_4b.m 1 of 1
function [fract]=task_4b(I_L, I_cell, V_rev) V_L = V_rev – 0.11 – 0.041*I_L; V_cell = V_rev + 0.11 + 0.041*I_cell; fract = V_L/V_cell; end