Homework 4

Due at 11:59pm on Friday, May 1st. Submit to https://canvas.uw.edu/courses/1883403/assignments/11402300. 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

Consider the simulated load profile shown here.

Find the following:

a) Peak load

b) Minimum Load

c) Average load

d) Peak to average ratio

e) Total energy consumed

Solutions.

\[P_{peak} = 1000 \text{ W} = 1.0 \text{ kW}\]

\[P_{min} = 200 \text{ W} = 0.2 \text{ kW}\]

Total energy over 24 hours:

\[E_{total} = (200 \times 5) + (400 \times 7) + (600 \times 6) + (800 \times 5) + (1000 \times 1) = 12{,}400 \text{ Wh}\] \[P_{avg} = \frac{12{,}400 \text{ Wh}}{24 \text{ h}} = 516.67 \text{ W}\]

\[\frac{P_{peak}}{P_{avg}} = \frac{1000}{516.67} = 1.94\]

\[\text{Total Energy} = 12{,}400 \text{ Wh} = 12.4 \text{ kWh}\]

Problem 2

AAssuming an energy rate of $0.25 \ \$/kWh$, a daily peak demand charge of $0.75 \ \$/kW$, calculate how much a consumer would be charged for the load profile shown above.

Solutions.

Energy charge

\[C_{energy} = E \times r_{energy} = 12.4 \text{ kWh} \times 0.25 \ \$/\text{kWh} = \$3.10\]

Peak demand charge

\[C_{demand} = P_{peak} \times r_{demand} = 1.0 \text{ kW} \times 0.75 \ \$/\text{kW} = \$0.75\]

Total daily charge

\[C_{total} = C_{energy} + C_{demand} = \$3.10 + \$0.75 = \$3.85\]

Problem 3

Some electric utilities offer their residential consumers a tariff where the rate is lower during off-peak hours. Calculate how much a consumer would be charged for the load profile shown above, assuming an on-peak rate of $0.30 \ \$/kWh$, an off-peak rate of $0.20 \ \$/kWh$, a daily peak demand charge of $0.75 \ \$/kW$. The off-peak hours are hours 1 to 6, the rest of the hours are on-peak.

Solutions.

Off-peak hours (1–6):

\[E_{off} = (200+200+200+400+400+600) \text{ Wh} = 2.0 \text{ kWh}\]

On-peak hours (0, 7–23):

\[E_{on} = E_{total} - E_{off} = 12.4 - 2.0 = 10.4 \text{ kWh}\]

On-peak energy charge

\[C_{on} = E_{on} \times r_{on} = 10.4 \text{ kWh} \times 0.30 \ \$/\text{kWh} = \$3.12\]

Off-peak energy charge

\[C_{off} = E_{off} \times r_{off} = 2.0 \text{ kWh} \times 0.20 \ \$/\text{kWh} = \$0.40\]

Peak demand charge

\[C_{demand} = P_{peak} \times r_{demand} = 1.0 \text{ kW} \times 0.75 \ \$/\text{kW} = \$0.75\]

Total daily charge

\[C_{total} = C_{on} + C_{off} + C_{demand} = \$3.12 + \$0.40 + \$0.75 = \$4.27\]

Problem 4

The state of California has a peak load of about $50$ GW ($50{,}000$ MW). Suppose that you own a generator that is only turned on when the load is above $40$ GW. The generator has a capacity of $10$ MW, and we assume that if it is used, then it is operating at its full capacity. Suppose that for $120$ hours out of the year the load in California is higher than $40$ GW.

Suppose the generator has a fixed cost (typically include land lost, labor, maintenance, etc) of $\$1{,}000{,}000$ per year and a variable cost (e.g., fuel cost) of $\$10/\text{MWh}$. What is price that the generator has to charge in $\$/\text{MWh}$ to break even? Note that all of the fixed cost has to be recovered from when the generator is running.

Solution.

Energy generated per year

\[E = 10 \text{ MW} \times 120 \text{ h} = 1{,}200 \text{ MWh}\] \[\text{Total Cost} = \text{Fixed Cost} + \text{Variable Cost}= \$1{,}012{,}000\]

Break-even price (revenue = cost):

\[p \times 1{,}200 \text{ MWh} = \$1{,}012{,}000\] \[p =\$843.33/\text{MWh}\]

So the generator must charge approximately $\$843.33/\text{MWh}$ to break even. Note (this is not part of the solution), these plants are called peaker plants, and their business model is to strategically bid into the market at very high prices. For reference, the average price of electricity ranges from $10 \ \$/MWh$ to $40 \ \$/MWh$.

Problem 5

A $500 \text{ cm}^2$ solar cell is operating at $35^\circ\text{C}$ where the output current is $2.5 \text{ A}$, the load voltage is $0.5 \text{ V}$ and the saturation current of the diode is $0.5 \text{ nA}$. The series resistance of the cell is $25 \text{ m}\Omega$, and the parallel resistance is $180 \text{ }\Omega$. At a given time, the solar power density is $180 \text{ W/m}^2$. Compute

a) The electric losses in the cell. That is, the losses through the resistors.

b) The irradiance efficiency and the overall efficiency.

Solutions.

a) Series resistor loss

\[P_{R_s} = I^2 R_s = (2.5)^2 (0.025) = 0.156\,\text{W}\]

Shunt resistor loss: the diode voltage $V_d = V + IR_s = 0.5625$ V appears across $R_{sh}$:

\[P_{R_{sh}} = \frac{V_D^{\,2}}{R_{sh}} = \frac{(0.5625)^2}{180}= 1.76\,\text{mW}\] \[P_{loss} = 0.156 + 0.00176 = 0.158\,\text{W}\]

b) Incident solar power

\[P_{in} = \rho \cdot A = (180)(0.05) = 9\,\text{W}\]

The current source delivers $I_s$ at the internal node voltage $V_d$ and

\[I_s = I_d + I_{sh} + I\] \[V_T = \frac{kT}{q} = 0.02655\,\text{V}\] \[I_d = I_0\left[\exp\!\left(\frac{V_D}{V_T}\right) - 1\right] = (0.5\times10^{-9})(1.585\times10^{9} - 1)=0.792\,\text{A}\] \[I_{sh} = \frac{V_D}{R_{sh}} = \frac{0.5625}{180} =3.13\,\text{mA}\] \[I_s = I_D + I_{sh} + I = 3.30\,\text{A}\] \[P_{source} = I_L \cdot V_D = (3.30)(0.5625) \approx 1.853\,\text{W}\] \[\eta_{irr} = \frac{P_{source}}{P_{in}} = 20.6\,\%\] \[P_{out} = V \cdot I = (0.5)(2.5) = 1.25\,\text{W}\] \[\eta = \frac{P_{out}}{P_{in}} = \frac{1.25}{9}= 13.9\,\%\]

Problem 6

A solar panel consists of $5$ parallel columns of PV cells. Each column has $8$ PV cells in series. Each cell produces $2.5$ W at $0.6$ V. Compute the voltage and current of the panel.

Solutions.

Current per cell:

\[I_{cell} = \frac{P_{cell}}{V_{cell}} = \frac{2.5 \text{ W}}{0.6 \text{ V}} \approx 4.167 \text{ A}\]

Series connection (8 cells per column): voltages add, current stays the same.

\[V_{column} = 8 \times 0.6 = 4.8 \text{ V}\] \[I_{column} = 4.167 \text{ A}\]

Parallel connection (5 columns): currents add, voltage stays the same.

\[V_{panel} = 4.8 \text{ V}\] \[I_{panel} = 5 \times 4.167 \approx 20.83 \text{ A}\]

Problem 7

A solar cell with a reverse saturation current of $2$ nA has a source current of $2$ A. The operating temperature of the cell is $35^\circ\text{C}$. A load draws $1$ A. Find the output voltage (i.e., the diode voltage) and the output power of the cell.

Solutions.

Thermal voltage:

\[V_T = \frac{kT}{q} = \frac{(1.381 \times 10^{-23})(308.15)}{1.602 \times 10^{-19}} = 0.02656 \text{ V}\]

Diode current:

\[I_d = I_{source} - I_{load} = 2 - 1 = 1 \text{ A}\]

Output (diode) voltage is given by the diode equation $I_d = I_0(e^{V_d/V_T} - 1)$:

\[V_d = V_T \ln\!\left(\frac{I_d}{I_0} + 1\right) = 0.02656 \cdot \ln\!\left(\frac{1}{2 \times 10^{-9}} + 1\right)= 0.532 \text{ V}\]

Output power:

\[P_{out} = V_d \times I_{load} = 0.532 \text{ W}\]

Problem 8

A solar cell with a reverse saturation current of $2$ nA has a solar current (that is, the source current) of $1.2$ A. The operating temperature of the cell is $35^\circ\text{C}$. Find the maximum output power of the cell and the resistance of the load that achieves this power. You can solve for the maximum power by plotting the power vs voltage or taking the derivative. (Note, you won’t get this question directly in the midterm since it requires more than a calculator.)

Solutions.

Thermal voltage:

\[V_T = \frac{kT}{q} \approx 0.02656 \text{ V}\]

Cell I-V equation:

\[I = I_{sc} - I_0\!\left(e^{V/V_T} - 1\right)\]

To find the MPP, set $\dfrac{dP}{dV} = 0$ where $P = VI$. This yields:

\[\left(1 + \frac{V_{mp}}{V_T}\right) e^{V_{mp}/V_T} = \frac{I_{sc}}{I_0} + 1\]

\[\frac{I_{sc}}{I_0} + 1 = \frac{1.2}{2\times 10^{-9}} + 1 \approx 6 \times 10^{8}\]

Let $x = V_{mp}/V_T$. We solve $x + \ln(1+x) = \ln(6\times 10^8)$:

\[x = 17.305 \;\;\Longrightarrow\;\; V_{mp} = 17.305 \times 0.02656 = 0.460 \text{ V}\]

Current at this operating point:

\[I_{mp} = I_{sc} - I_0\!\left(e^{x}-1\right) = 1.135 \text{ A}\]

Maximum output power:

\[P_{max} = V_{mp} \times I_{mp} \approx 0.460 \times 1.135 \approx 0.521 \text{ W}\]

Problem 9

This is a bonus problem. The goal of having peak prices is to influence the consumer to reduce their peak load consumption. Cnsider the load profile below.

We are going to write this profile out more formally as a vector, $z = (z_1, z_2, z_3, z_4)$ where $z_i$ is the consumption at hour $i$. Peak pricing can be written as:

\[C = \alpha \sum_{i=1}^{4} z_i + \beta \max(z_1, \ldots, z_4),\]

$\alpha = 0.5 \text{ \$/kWh}$ and $\beta = 2 \text{ \$/kW}$.

Suppose the user has a budget of $\$5$. This is profile is too costly. So let’s assume that the customer would change to a profile that is closest to the original one. What is the new profile of the customer such that it is the closest one to the original load, while coming under budget? Hint: think about what closest means in a vector setting.

Solutions.

This problem requires a notion of vector norm. Given a vector $x$, the norm of $x$, written as $\Vert x\Vert$, is a way to measure the size of $x$. There are many different norms, for example, see https://en.wikipedia.org/wiki/Norm_(mathematics). In this question, we will use the 2-norm, defined as

\[\Vert x \Vert_2 = \sqrt{x_1^2 + \cdots + x_n^2}\]

The problem we want to solve is to find a vector $x$, such that $x$ is under budget and $x$ is closest to $z$ in the 2-norm. More formally, we can write this as a constrained optimization problem:

\[\min_{x} \; \Vert x - z \Vert_2^2\] \[\text{s.t.} \quad \text{cost}(x) \leq 5\]

This optimization problem turns out to be a very simple one for a computer to solve (it’s an example of a convex optimization problem). Solving it gives

\[x = (0.875,\ 0.875,\ 1.375,\ 1.375)\]