Chapter 2.8, Problem 121RP

To determine

The diameter of downstream channel and power produced by the wind mill

Answer to Problem 121RP

The diameter of downstream channel is $\underset{\_}{\text{7}\text{.7658}\text{m}}$.

The magnitude of power produced by the wind mill is $\underset{\_}{3.98\text{kW}}$.

Explanation of Solution

Write the expression the specific volume of the air.

$v=\frac{RT}{P}$ (I).

Here, specific volume of the air is v, gas constant is R, temperature of the air is T, and pressure of the air is P.

Use the continuity equation, equate velocity of air flow at rotor and wind channel downstream.

$A_1{V}_1=A_2{V}_2$ (II).

Here, area of cross section at rotor is $A_1$, velocity of wind at rotor is $V_1$, area of wind channel downstream is $A_2$, velocity of wind at downstream channel is $V_2$.

Write the expression for area in terms of diameter in equation (II).

$\pi \frac{D_1^2}{4}{V}_1=\pi \frac{D_2^2}{4}{V}_2$

$D_2=\sqrt{\frac{V_1}{V_2}}{D}_1$ (III).

Here, diameter of the rotor is $D_1$ and diameter of downstream channel is $D_2$.

Write the expression for the mass flow rate of wind mill.

$\dot{m}=\frac{A_1{V}_1}{v}$

$\dot{m}=\frac{\pi D_1^2{V}_1}{4v}$ (IV).

Here, mass flow rate is $\dot{m}$.

Write the expression for the power produced by the wind mill.

$\dot{W}=\dot{m}\left(\frac{V_1^2-V_2^2}{2}\right)$ (V).

Here, power produced by the wind mill is

Conclusion:

Substitute $0.287\text{kPa}{\text{m}}^{\text{3}}/\text{kg}\text{K}$ for R, $20°\text{C}$ for T, and 100 kPa for P to find v in equation (I).

$\begin{array}{c}v=\frac{\left(0.287\text{kPa}{\text{m}}^{\text{3}}/\text{kg}\text{K}\right)\left(20+273.15\right)\text{K}}{100\text{kPa}}\\ =0.8413\frac{{\text{m}}^{\text{3}}}{\text{kg}}\end{array}$

Thus, specific volume of the air is $0.8413{\text{m}}^{\text{3}}/\text{kg}$.

Substitute 7m for $D_1$, $8\text{m}/\text{s}$ for $V_1$, and $6.5\text{m}/\text{s}$ for $V_2$ to find $D_2$ in equation (III).

$\begin{array}{c}{D}_2=\sqrt{\frac{8\text{m}/\text{s}}{6.5\text{m}/\text{s}}}\times 7\text{m}\\ =\text{7}\text{.7658}\text{m}\end{array}$

Thus, the diameter of downstream channel is $\underset{\_}{\text{7}\text{.7658}\text{m}}$.

Substitute 7m for $D_1$, $0.8413{\text{m}}^{\text{3}}/\text{kg}$ for v, and $8\text{m}/\text{s}$ for $V_1$ to find mass flow rate in equation (IV).

$\begin{array}{c}\dot{m}=\frac{\pi {\left(7\text{m}\right)}^2\left(8\text{m}/\text{s}\right)}{4\times 0.8413{\text{m}}^{\text{3}}/\text{kg}}\\ =365.9352\frac{\text{kg}}{\text{s}}\\ =366\frac{\text{kg}}{\text{s}}\end{array}$

Thus, mass flow rate is $366\text{kg}/\text{s}$.

Substitute $366\text{kg}/\text{s}$ for $\dot{m}$, $8\text{m}/\text{s}$ for $V_1$, and $6.5\text{m}/\text{s}$ for $V_2$ to find $\dot{W}$ in equation (V).

$\begin{array}{c}\dot{W}=366\text{kg}/\text{s}\left(\frac{{\left(8\text{m}/\text{s}\right)}^2-{\left(6.5\text{m}/\text{s}\right)}^2}{2}\right)\\ =3,979.45\frac{\text{kg}{\text{m}}^{\text{2}}}{{\text{s}}^{\text{3}}}\left(\frac{\text{1}\text{kJ}/\text{kg}}{\text{1,000}{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\\ =3.979\frac{\text{kJ}}{\text{s}}\\ \approx 3.98\text{kW}\end{array}$

Thus, power produced by the wind mill is $\underset{\_}{3.98\text{kW}}$.