Homework 5

Due at 11:59pm on Thursday, May 14th. Submit to link. It is fine to submit scanned or a picture of handwritten answers, but we must be able to read it. Indicate your final answer clearly, show your work. Each problem is 10 points (all subproblems weighted equally).

Problem 1

The lift force exerted on each blade of a three-blade wind turbine is \(1000 \text{ N}\). The center of mass of the blade is \(20 \text{ m}\) from the hub. Find the total torque generated by the turbine and the total power generated if the blades are turning at \(35 \text{ rpm}\) (revolutions per minute).

Solutions.

The torque produced by a single blade is:

\[\tau_{\text{blade}} = F \cdot r = (1000 \text{ N})(20 \text{ m}) = 20{,}000 \text{N}\cdot \text{m}\] \[\tau_{\text{total}} = 3 \cdot \tau_{\text{blade}} = 60{,}000 \text{N}\cdot \text{m}\]

The angular velocity is given by converting \(35 \text{ rpm}\) to radians per second:

\[\omega = 35 \frac{\text{rev}}{\text{min}} \cdot \frac{2\pi \text{ rad}}{1 \text{ rev}} \cdot \frac{1 \text{ min}}{60 \text{ s}} = 3.665 \text{ rad/s}\]

Power is given by

\[P = \tau_{\text{total}} \cdot \omega = 220 \text{ kW}\]

Problem 2

The wind speed at a site is uncertain. Suppose we can predict the mean speed to be \(10 \text{ m/s}\) and the deviation to be \(\pm 10 \%\). That is, wind speed is in the range of \([9,11] \text{ m/s}\). Using this information, how well can be predict wind power? That is, what is the % deviation from the mean?

Solutions.

Since wind power is proportional the cube of wind speed, the forcasted power is proportional to value in the inverval

\([9^3,11^3]=[729,1331]\) The average (center point of the interval) is \(1030\), and the deviation is \(301\), or roughly \(30 \%\) error.

Problem 3

Complete the following steps. Suppose for a given area, the wind speed before the turbine is \(w_u\) and the wind speed after the turbine is \(w_d\).

1. To compute the power that is extracted by the turbine, we need to know the wind speed at the turbine. Making a continuity assumption, suppose that the speed is at the turbine is the average of the upwind and the downwind speed. That is, the wind speed at the turbine, \(w\), is \(w=\frac{1}{2} (w_u+w_d)\).

2. The power extracted by the turbine is also given by the difference in power in upwind airmass and the power in the downwind airmass.

3. We assume air is incompressible, that is, the same amount or air is moving upwind, through the turbine, and downwind.

4. Combining the last two terms, it is now possible to write the power extracted by the turbine as

   \[P_{\text{turbine}}=P_u \cdot C_p,\]

   where \(C_p\) is in terms of the ratio \(\gamma=\frac{w_d}{w_u}\).

5. Now maximize \(C_p\) by varying \(\gamma\). This leads the to Betz ratio, which states \(C_p \leq 0.593\).

Note, this is not a trivial derivation and you may use the Internet or AI tools if needed. Although it’s worthwhile to give it a try.

Solutions.

Step 1: Wind speed at the turbine

By the continuity assumption, the speed at the turbine is the average of the upwind and downwind speeds:

\[w = \frac{1}{2}(w_u + w_d)\]

Step 2: Power extracted = upwind power − downwind power

The kinetic-energy flux (power) carried by an airmass moving at speed \(v\) with mass flow rate \(\dot{m}\) is \(\tfrac{1}{2}\dot{m}v^2\). Therefore:

\[P_{\text{turbine}} = \tfrac{1}{2}\dot{m}_u w_u^2 - \tfrac{1}{2}\dot{m}_d w_d^2\]

Step 3: Incompressibility (conservation of mass)

The same mass per unit time passes through every cross section, and the mass flow at the turbine of area \(A\) is \(\dot{m} = \rho A w\). Thus:

\[\dot{m}_u = \dot{m}_d = \dot{m} = \rho A w\]

Combining with Step 2:

\[P_{\text{turbine}} = \tfrac{1}{2}\rho A w\,(w_u^2 - w_d^2)\]

Step 4: Express as \(P_{\text{turbine}} = P_u \, C_p\)

Substitute \(w = \tfrac{1}{2}(w_u+w_d)\) and factor \(w_u^2 - w_d^2 = (w_u+w_d)(w_u-w_d)\):

\[P_{\text{turbine}} = \tfrac{1}{4}\rho A\,(w_u+w_d)^2 (w_u-w_d)\]

The power available in the undisturbed upwind airmass crossing area \(A\) is:

\[P_u = \tfrac{1}{2}\rho A\,w_u^3\]

Form the ratio and let \(\gamma = w_d/w_u\):

\[C_p \;=\; \frac{P_{\text{turbine}}}{P_u} \;=\; \frac{\tfrac{1}{4}\rho A\,(w_u+w_d)^2(w_u-w_d)}{\tfrac{1}{2}\rho A\,w_u^3} \;=\; \tfrac{1}{2}(1+\gamma)^2(1-\gamma)\]

Step 5: Maximize \(C_p\) — the Betz limit

Differentiate with respect to \(\gamma\):

\[\frac{dC_p}{d\gamma} = \tfrac{1}{2}\!\left[\,2(1+\gamma)(1-\gamma) - (1+\gamma)^2\,\right] = \tfrac{1}{2}(1+\gamma)\bigl[2(1-\gamma) - (1+\gamma)\bigr]\] \[\frac{dC_p}{d\gamma} = \tfrac{1}{2}(1+\gamma)(1 - 3\gamma)\]

Setting \(dC_p/d\gamma = 0\) gives:

\[\gamma = \tfrac{1}{3} \quad\Longleftrightarrow\quad w_d = \tfrac{1}{3}w_u\]

There is another solution to \(dC_p/d\gamma = 0\) at \(\gamma = -1\), but this is not a physical solution.

So \(C_p^{\max} = \tfrac{1}{2}\!\left(1+\tfrac{1}{3}\right)^{\!2}\!\left(1-\tfrac{1}{3}\right) = \tfrac{1}{2}\cdot\tfrac{16}{9}\cdot\tfrac{2}{3} = \tfrac{16}{27}\)

\[C_p \;\leq\; \tfrac{16}{27} \;\approx\; 0.593\]

This is the Betz limit: no wind turbine operating on this momentum/energy principle can extract more than about \(59.3\%\) of the kinetic energy in the upstream wind, and the optimum occurs when the downwind speed is reduced to one-third of the upwind speed.

Problem 4

Each U-235 fission releases approximately \(3.20 \times 10^{-11} \text{ J}\) of energy. The heat of combustion of natural gas (primarily methane) is approximately \(55 \text{ MJ/kg}\).

Find the amount of energy \(10 \text{ kg}\)$ of \(\mbox{U}^{235}\) can produce. How much natural gas is needed to produce the same amount of energy (assuming perfect efficiency)? Hint: the molar mass of \(\mbox{U}^{235}\) is \(M = 235 \text{ g/mol}\) and Avogadro’s number is \(N_A = 6.022 \times 10^{23} \text{ mol}^{-1}\).

Solutions.

Convert mass to moles, then to atoms:

\[n = \frac{m}{M} = \frac{10{,}000 \text{ g}}{235 \text{ g/mol}} \approx 42.55 \text{ mol}\] \[N = n \cdot N_A = (42.55 \text{ mol})(6.022 \times 10^{23} \text{ mol}^{-1}) \approx 2.563 \times 10^{25} \text{ atoms}\]

Each fission releases \(3.20 \times 10^{-11} \text{ J}\), so:

\[E_{\text{U}} = N \cdot E_{\text{fission}} = (2.563 \times 10^{25})(3.20 \times 10^{-11} \text{ J})\] \[E_{\text{U}} \approx 8.20 \times 10^{14} \text{ J}\]

Using the heat of combustion \(h_c = 55 \text{ MJ/kg} = 5.5 \times 10^{7} \text{ J/kg}\):

\[m_{\text{gas}} = \frac{E_{\text{U}}}{h_c} = \frac{8.20 \times 10^{14} \text{ J}}{5.5 \times 10^{7} \text{ J/kg}}\] \[m_{\text{gas}} \approx 1.49 \times 10^{7} \text{ kg} \;\approx\; 1.49 \times 10^{4} \text{ metric tons}\]

Problem 5

We use three phase power for transmission systems. In a few short sentences, explain why we do not use four phase or higher number of phases?

Solutions.

A three-phase system uses three wires to carry three times the power as a single phase system using two wires. A four-phase system is possible, but it would use four wires to carry four times the power as a single phase system, so there is not a efficiency gain in terms of power/wire there. At the same time, four phase systems are both more cumbersome to construct and harder to balance in operations. Note: some systems do use a poly-phase setup, but that’s on a much smaller scale.

Problem 6

A balanced Y-connected three-phase source is connected to a balanced three phase load. At the source, the rms voltage and current per phase are \(160 \angle 50^\circ \text{ V}\) \(85 \angle 70^\circ \text{ A}\), respectively.

Find:

a) The (rms) line-to-line voltage.

b) Total power delievered by the source.

Solutions

a)

For a balanced Y-connected source, the line-to-line voltage magnitude is \(\sqrt{3}\) times the phase voltage magnitude, and it leads the phase voltage by \(30^\circ\):

\[V_{LL} = \sqrt{3}\,V_\phi \angle (\theta_\phi + 30^\circ)= \sqrt{3}\,(160) \angle (50^\circ + 30^\circ) \text{ V}=277.1 \angle 80^\circ \text{ V (rms)}\]

b)

The total complex power delivered by a balanced three-phase Y-connected source is:

\[S_{\text{total}} = 3\,V_\phi\,I_\phi^{*} = 3\,(160)(85)\angle(50^\circ - 70^\circ) \text{ VA}= 38.3 - j\,13.95 \text{ kVA}\]

Problem 7

A Y-connected balanced three-phase source is feeding a balanced three-phase load. The voltage and current on phase \(a\) of the source are:

\[v(t) = 240 \cos(100\pi t + 2\pi/9) \text{ V}\] \[i(t) = 300 \cos(100\pi t + 3\pi/18) \text{ A}\]

Find:

a) The impedance of the load if it is Y-connected.

b) The impedance of the load if it is \(\Delta\)-connected.

Solutions.

Converting to phasors:

\[\mathbf{V}_a = \frac{240}{\sqrt{2}}\angle 40^\circ \approx 169.7\angle 40^\circ \text{ V}_{\text{rms}}\] \[\mathbf{I}_a = \frac{300}{\sqrt{2}}\angle 30^\circ \approx 212.1\angle 30^\circ \text{ A}_{\text{rms}}\]

a)

\[\mathbf{Z}_Y = \frac{\mathbf{V}_a}{\mathbf{I}_a} = \frac{169.7\angle 40^\circ}{212.1\angle 30^\circ} = 0.8\angle 10^\circ \ \Omega =0.788 + j\,0.139 \ \Omega.\]

b)

Line-to-line voltage and line-to-line current:

\[\mathbf{V}_{ab} = \sqrt{3}\,\mathbf{V}_a\angle(40^\circ + 30^\circ) = 169.7\sqrt{3}\angle 70^\circ = 293.9\angle 70^\circ \text{ V}_{\text{rms}}\] \[\mathbf{I}_{AB} = \frac{\mathbf{I}_a}{\sqrt{3}}\angle(30^\circ + 30^\circ) = \frac{212.1}{\sqrt{3}}\angle 60^\circ = 122.5\angle 60^\circ \text{ A}_{\text{rms}}\] \[\mathbf{Z}_\Delta = \frac{\mathbf{V}_{ab}}{\mathbf{I}_{AB}} = 2.4\angle 10^\circ \ \Omega = 2.364 + j\,0.417 \ \Omega.\]

As a sanity check, \(\mathbf{Z}_\Delta = 3\,\mathbf{Z}_Y\).