Energy Systems (Solved Numerical Problems)

Wind at 1 standard atmospheric pressure and \({15^0}C\) has velocity of 15 m/s, calculate

(i) the total power density in the wind stream

(ii) the maximum obtainable power density

(iii) a reasonably obtainable power density

(iv) total power

(v) torque and axial thrust

Given: turbine diameter = 120 m, and turbine operating speed = 40 rpm at maximum efficiency.

Propeller type wind turbine is considered. (AMIE Winter 2023)

Solution

For air, the value of gas constant is R = 0.287 kJ/kg.K
1 atm = 1.01325 x 105 Pa

Air density \(\rho  = \frac{P}{{RT}} = \frac{{1.01325x{{10}^5}}}{{287}}(288) = 1.226\,kg/{m^3}\)

Total Power

\({P_{total}} = \rho A{V_1}^3/2\)

Power density

\(\begin{array}{l}\frac{{{P_{total}}}}{A} = \frac{1}{2}\rho {V_1}^3\\ = \frac{1}{2}(1.226){(15)^3}\\ = 2068.87{\mkern 1mu} W/{m^2}\end{array}\)

Maximum power density

\(\begin{array}{l}\frac{{{P_{\max }}}}{A} = \frac{8}{{27}}\rho A{V^3}_1\\ = \frac{8}{{27}}(1.226){(15)^3}\\ = 1226{\mkern 1mu} W/{m^2}\end{array}\)

Assuming efficiency = 35%

\(\begin{array}{l}\frac{P}{A} = n\left( {\frac{{{P_{total}}}}{A}} \right) = 0.35{\mkern 1mu} x{\mkern 1mu} 2068.87\\ = 724{\mkern 1mu} W/{m^2}\end{array}\)

Torque at maximum efficiency

\(\begin{array}{l}{T_{\max }} = \frac{2}{{27}}.\frac{{\rho D{V_1}^3}}{N}\\ = \frac{2}{{27}}.\frac{{1.226(120){{(15)}^3}}}{{40/60}} = 55170\,N\end{array}\)

Maximum axial thrust

\(\begin{array}{l}{F_{x,\max }} = \frac{\pi }{9}\rho {D^2}{V_1}^2\\ = \frac{\pi }{9}(1.226\,x\,{120^2}\,x\,{15^2})\\ = 1385870\,N\end{array}\)

Calculate the angle made by beam radiation with normal to a fiat plate collector on December 1, at 9.00 AM. solar time for a location at 28° 35' N. The collector is tilted at an angle of latitude plus 10°, with the horizontal and is pointing due south. (AMIE Winter 2023)

Solution

As collector is pointing towards south, \(\gamma  = 0\)

For December 1

\(\begin{array}{l}n = 365 - 30 = 335\\\delta  = 23.45\sin \left[ {\frac{{360}}{{365}}(284 + n)} \right]\\ = 23.45\left[ {\frac{{360}}{{365}}(284 + 335)} \right]\\ =  - {22.11^0}\end{array}\)

At 0900, hour angle 

\(\omega  = (9 - 12)\,x\,15 =  - {45^0}\)

\(\begin{array}{l}\beta  = latitude + {10^0}\\ = {28^0}35' + {10^0}0'\\ = {38^0}35' = {38.58^0}\end{array}\)

Now

\(\cos {\theta _T} = \sin \delta \sin (\lambda  - \beta ) + \cos \delta \cos \omega \cos (\lambda  - \beta )\)

Hence,

\(\begin{array}{l}\cos {\theta _T} = \sin ( - 22.11)\sin (28.58 - 38.58) + \cos (28.58 - 38.58)\cos ( - {22.11^0})\cos (45)\\ = 0.7104\end{array}\)

Hence, \(\theta  = {\cos ^{ - 1}}(0.45486) = {62.94^0}\)

---

The study material for AMIE/BTech/Junior Engineer exams is available at https://amiestudycircle.com