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tutorial-1/solution01-1.ipynb
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tutorial-1/solution01-2.ipynb
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%% Cell type:markdown id: tags:
# Energy System Modelling - Solutions to Tutorial II.1
%% Cell type:markdown id: tags:
## Imports
%% Cell type:code id: tags:
``` python
import numpy as np
import matplotlib.pyplot as plt
import scipy.integrate as integrate
from scipy.optimize import minimize
%matplotlib inline
```
%% Cell type:markdown id: tags:
## Parameters
%% Cell type:code id: tags:
``` python
A_w = 0.4
A_s = 0.75
A_l = 0.1
T = 1
phi = 0
```
%% Cell type:markdown id: tags:
## Functions
%% Cell type:code id: tags:
``` python
def w(t, phi=0):
return 1 + A_w * np.cos(2*t*np.pi/T-phi)
```
%% Cell type:code id: tags:
``` python
def s(t):
return 1 - A_s * np.cos(2*t*np.pi/T)
```
%% Cell type:code id: tags:
``` python
def l(t):
return 1 + A_l * np.cos(2*t*np.pi/T)
```
%% Cell type:code id: tags:
``` python
def c(gamma):
return 1 - gamma
```
%% Cell type:code id: tags:
``` python
def mismatch_a(alpha, t, phi=phi):
return alpha*w(t, phi) + (1-alpha)*s(t)-l(t)
```
%% Cell type:code id: tags:
``` python
def mismatch_c(alpha, gamma, t):
return gamma*(alpha*w(t) + (1-alpha)*s(t))+c(t)-l(t)
```
%% Cell type:code id: tags:
``` python
def objective_a(alpha):
return 1/T * integrate.quad(lambda t: mismatch_a(alpha,t,phi=phi)**2, 0, T)[0]
```
%% Cell type:code id: tags:
``` python
def objective_c(alpha):
return 1/T * integrate.quad(lambda t: mismatch_c(alpha,gamma,t)**2, 0, T)[0]
```
%% Cell type:markdown id: tags:
## Plots
%% Cell type:code id: tags:
``` python
x = np.arange(0,1,0.01)
```
%% Cell type:markdown id: tags:
Availability time series
%% Cell type:code id: tags:
``` python
wind = plt.plot(x, w(x), label='wind')
solar = plt.plot(x, s(x), label='solar')
load = plt.plot(x, l(x), label='load')
plt.legend()
plt.show()
```
%% Output
%% Cell type:markdown id: tags:
Mismatch
%% Cell type:code id: tags:
``` python
alphas = [0, 0.25, 0.5, 0.75, 1]
for a in alphas:
plt.plot(x, mismatch_a(a,x),label='wind share = '+str(a))
plt.legend()
plt.show()
```
%% Output
%% Cell type:markdown id: tags:
## Solution
%% Cell type:markdown id: tags:
***
**(a) What is the seasonal optimal mix $\alpha$, which minimizes**
$$\langle\left[ \alpha W(\cdot) + (1-\alpha) S(\cdot) - L(\cdot) \right]^2 \rangle = \frac1T \int_0^T \left[ \alpha W(t) + (1-\alpha) S(t) - L(t) \right]^2 \,\mathrm d t$$
%% Cell type:code id: tags:
``` python
alpha = minimize(objective_a, x0=0, method='nelder-mead', options={'xtol': 1e-8, 'disp': True}).x[0]
alpha
```
%% Output
Optimization terminated successfully.
Current function value: 0.000000
Iterations: 37
Function evaluations: 74
0.739130432128907
0.73913043212890694
%% Cell type:markdown id: tags:
***
**(b) How does the optimal mix change if we replace $A_L \to -A_L$?**
%% Cell type:code id: tags:
``` python
A_l *= (-1)
```
%% Cell type:code id: tags:
``` python
alpha = minimize(objective_a, x0=0, method='nelder-mead', options={'xtol': 1e-8, 'disp': True}).x[0]
alpha
```
%% Output
Optimization terminated successfully.
Current function value: 0.000000
Iterations: 36
Function evaluations: 72
0.5652173919677739
0.56521739196777387
%% Cell type:code id: tags:
``` python
A_l *= (-1)
```
%% Cell type:markdown id: tags:
***
**(c) Now assume that there is a seasonal shift in the wind signal
$$ W(t) = 1 + A_W \cos \left( \omega t - \phi \right).$$
Express the optimal mix $\alpha$ as a function of $\phi$.**
%% Cell type:code id: tags:
``` python
phis = np.arange(0,2*np.pi,np.pi/100)
alphas = []
for p in phis:
phi = p
bnds = [(0, 1)]
alpha = minimize(objective_a, x0=0, method='TNC', bounds=bnds, options={'xtol': 1e-8}).x[0]
alphas.append(alpha)
plt.plot(phis, alphas)
```
%% Output
[<matplotlib.lines.Line2D at 0x7f6a03f0f5c0>]
[<matplotlib.lines.Line2D at 0x7f2c3ef2d3c8>]
%% Cell type:markdown id: tags:
***
**(d) A constant conventional power source $C(t) = 1 - \gamma$ is now introduced. The mismatch then becomes
$$\Delta(t) = \gamma \left[ \alpha W(t) + (1-\alpha) S(t) \right] + C(t) - L(t)$$
Analogously to (a), find the optimal mix $\alpha$ as a function of $0 \leq \gamma \leq 1$, which minimizes $\langle{\Delta^2}\rangle$.**
%% Cell type:code id: tags:
``` python
gammas = np.arange(0,1,0.01)
alphas = []
for g in gammas:
gamma = g
bnds = [(0, 1)]
alpha = minimize(objective_c, x0=0, method='TNC', bounds=bnds, options={'xtol': 1e-8}).x[0]
alphas.append(alpha)
plt.plot(gammas, alphas)
```
%% Output
[<matplotlib.lines.Line2D at 0x7f6a03f4d630>]
[<matplotlib.lines.Line2D at 0x7f2c3ee857f0>]
......
tutorial-1/tutorial01-2.ipynb
View file @ 4a0e0cb9
%% Cell type:markdown id: tags:
# Energy System Modelling - Tutorial II.1
%% Cell type:markdown id: tags:
We use approximations to seasonal variations of wind and solar power generation $W(t)$
and $S(t)$ and load $L(t)$:
$$W(t) = 1 + A_W \cos \omega t$$
$$S(t) = 1 - A_S \cos \omega t$$
$$L(t) = 1 + A_L \cos \omega t$$
The time series are normalized to $\langle{W}\rangle = \langle{S}\rangle = \langle{L}\rangle := \frac{1}{T} \int_0^T L(t)
d t = 1$, and the constants have the values
$$\omega = \frac{2\pi}{T} $$
$$T = 1 \text{ year}$$
$$ A_W = 0.4 $$ $$ A_S = 0.75 $$ $$ A_L = 0.1 $$
%% Cell type:markdown id: tags:
## Parameters
%% Cell type:code id: tags:
``` python
A_w = 0.4
A_s = 0.75
A_l = 0.1
T = 1
phi = 0
```
%% Cell type:markdown id: tags:
## Imports
%% Cell type:code id: tags:
``` python
import numpy as np
import matplotlib.pyplot as plt
import scipy.integrate as integrate
from scipy.optimize import minimize
%matplotlib inline
```
%% Cell type:markdown id: tags:
## Functions
%% Cell type:code id: tags:
``` python
def w(t, phi=0):
return 1 + A_w * np.cos(2*t*np.pi/T-phi)
```
%% Cell type:code id: tags:
``` python
def s(t):
return 1 - A_s * np.cos(2*t*np.pi/T)
```
%% Cell type:code id: tags:
``` python
def l(t):
return 1 + A_l * np.cos(2*t*np.pi/T)
```
%% Cell type:code id: tags:
``` python
def c(gamma):
return 1 - gamma
```
%% Cell type:code id: tags:
``` python
def mismatch(???):
return ???
```
%% Cell type:code id: tags:
``` python
def objective(alpha):
return 1/T * integrate.quad(lambda t: mismatch(???)**2, 0, T)[0]
```
%% Cell type:markdown id: tags:
## Plots
%% Cell type:code id: tags:
``` python
x = np.arange(0,1,0.01)
```
%% Cell type:markdown id: tags:
Availability time series as shown on the worksheet.
%% Cell type:code id: tags:
``` python
wind = plt.plot(x, w(x), label='wind')
solar = plt.plot(x, s(x), label='solar')
load = plt.plot(x, l(x), label='load')
plt.legend()
plt.show()
```
%% Output
%% Cell type:markdown id: tags:
## Problem II.1
## Problem I.2
%% Cell type:markdown id: tags:
***
**(a) What is the seasonal optimal mix $\alpha$, which minimizes**
$$\langle\left[ \alpha W(\cdot) + (1-\alpha) S(\cdot) - L(\cdot) \right]^2 \rangle = \frac1T \int_0^T \left[ \alpha W(t) + (1-\alpha) S(t) - L(t) \right]^2 \,\mathrm d t$$
%% Cell type:code id: tags:
``` python
```
%% Cell type:markdown id: tags:
***
**(b) How does the optimal mix change if we replace $A_L \to -A_L$?**
%% Cell type:code id: tags:
``` python
```
%% Cell type:markdown id: tags:
***
**(c) Now assume that there is a seasonal shift in the wind signal
$$ W(t) = 1 + A_W \cos \left( \omega t - \phi \right).$$
Express the optimal mix $\alpha$ as a function of $\phi$.**
%% Cell type:code id: tags:
``` python
```
%% Cell type:markdown id: tags:
***
**(d) A constant conventional power source $C(t) = 1 - \gamma$ is now introduced. The mismatch then becomes
$$\Delta(t) = \gamma \left[ \alpha W(t) + (1-\alpha) S(t) \right] + C(t) - L(t)$$
Analogously to (a), find the optimal mix $\alpha$ as a function of $0 \leq \gamma \leq 1$, which minimizes $\langle{\Delta^2}\rangle$.**
%% Cell type:code id: tags:
``` python
```
......
tutorial-6/solution06.ipynb
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