Chapter 5, Problem 5.73P

To determine

i.

The relationship in dimensionless form.

Answer to Problem 5.73P

The Pi terms are $f\left(n,\frac{\Omega D}{v}\right)=\frac{P}{D^2\rho v^3}$

Explanation of Solution

Given:

Windmill parameter:

Diameter = 50 cm

Power developed = 2.7 kW

Speed of rotation = 4800 rpm

Concept Used:

$\begin{array}{l}P=f(D,\rho ,v,\Omega ,n)\\ k=5\\ j=3(M,L,T)\\ Numberof{\displaystyle \prod terms}=k-j\end{array}$

Calculation:

$\begin{array}{l}P=f(D,\rho ,v,\Omega ,n)\\ k=5\\ j=3(M,L,T)\\ Numberof{\displaystyle \prod terms}=k-j\\ Numberof{\displaystyle \prod terms}=5-3\\ Numberof{\displaystyle \prod terms}=2\end{array}$

The dimensions of each variable are,

$\begin{array}{l}\left\{P\right\}=\left\{M{L}^2{T}^{-3}\right\}\\ \left\{D\right\}=\left\{L\right\}\\ \left\{\rho \right\}=\left\{M{L}^{-3}\right\}\\ \left\{v\right\}=\left\{L{T}^{-1}\right\}\\ \left\{\Omega \right\}=\left\{T^{-1}\right\}\\ \left\{n\right\}=1\\ Nowtaking(D,\rho ,v)asrepeating\mathrm{var}iables,weget\\ f\left(n,\frac{\Omega D}{v}\right)=\frac{P}{D^2\rho v^3}\end{array}$

Conclusion:

The Pi terms are $f\left(n,\frac{\Omega D}{v}\right)=\frac{P}{D^2\rho v^3}$.

To determine

ii.

The power developed by a geometrical and dynamically similar prototype with a diameter of 5 m.

Answer to Problem 5.73P

The power developed is 5.989 kW

Explanation of Solution

Given:

Geometric parameter of the prototype:

Diameter = 5 m

Wind speed = 12 m/s

Altitude = 2000 m

Concept Used:

$f\left(n,\frac{\Omega D}{v}\right)=\frac{P}{D^2\rho v^3}=cons\tan t$

Calculation:

For geometric similarity $n\left(numberofblades\right)=constant$

For Dynamic similarity:

$\begin{array}{l}\left(\frac{\Omega D}{v}\right)=\frac{P}{D^2\rho v^3}=cons\tan t\\ \frac{P_1}{D_1{}^2{\rho }_1{v}_1{}^3}=\frac{P_2}{D_2{}^2{\rho }_2{v}_2{}^3}\\ P_2=\frac{P_1*D_2{}^2{\rho }_2{v}_2{}^3}{D_1{}^2{\rho }_1{v}_1{}^3}\\ Thedensityofairat2000maltitudeis1.0067kg/m^3\\ Thedensityofairatsealevelaltitudeis1.2255kg/m^3\\ P_2=\frac{2700*5^2*1.0067*{12}^3}{{0.5}^2*1.2255*{40}^3}\\ P_2=5989W\Rightarrow 5.989kW\end{array}$

Conclusion:

The power developed is 5.989 kW.

To determine

iii.

The rotation rate by a geometrical and dynamically similar prototype with a diameter of 5 m.

Answer to Problem 5.73P

The rotation rate is 144 rev/min

Explanation of Solution

Given:

Geometric parameter of the prototype:

Diameter = 5 m

Wind speed = 12 m/s

Altitude = 2000 m

Concept Used:

$f\left(n,\frac{\Omega D}{v}\right)=\frac{P}{D^2\rho v^3}=cons\tan t$

Calculation:

For geometric similarity $n\left(numberofblades\right)=constant$

For Dynamic similarity:

$\begin{array}{l}\left(\frac{\Omega D}{v}\right)=cons\tan t\\ \frac{{\Omega }_1{D}_1}{v_1}=\frac{{\Omega }_2{D}_2}{v_2}\\ {\Omega }_2=\frac{{\Omega }_1{D}_1}{v_1}*\frac{v_2}{D_2}\\ {\Omega }_2=\frac{4800*0.5}{40}*\frac{12}{5}\\ {\Omega }_2=144rev/\min \end{array}$

Conclusion:

The rotation rate is 144 rev/min.