Finiding a value at which a function is zero



In [1]:

    
import numpy as np
from scipy.optimize import minimize, minimize_scalar, newton



In [2]:

    
# assuming currents 
I1 = 1000
I2 = 1000

# bars dimensions
b1 = 25.4e-3
a1 = 0.25*25.4e-3
b2 = 25e-3
a2 = 6e-3

# temperatures
T1 = 84.9
Ta = 50

# temperature resistance change coeff
alpha = 3.9e-3

def f(x):
    return np.sqrt((2*b1 + 2*a1)/(2*b2+2*a2)) * ((T1-Ta)/(x-Ta))**0.61 \
* np.sqrt((a1*b1*(1+alpha*(x-20)))/(a2*b2*(1+alpha*(T1-20)))) - I1/I2

Trying to use minimize_scalar function



In [3]:

    
result = minimize_scalar(f)



In [4]:

    
result.success









    Out[4]:





True



In [5]:

    
T2 = result.x
T2









    Out[5]:





49.999999988366476



In [6]:

    
f(T2)









    Out[6]:





(-202715.36714150239+563060.81555927359j)

The result above is a really strange result. Most probably some constaints are missing.

Trying to use Newton method



In [7]:

    
# Newton method used to find zero; x0 is a starting point
res = newton(f, x0=90)



In [8]:

    
res









    Out[8]:





88.077777089701527



In [9]:

    
f(res)









    Out[9]:





0.0



In [10]:

    
# Excel goal seek result
f(88.035)









    Out[10]:





0.00061994046218716292

Trying approach with the function



In [11]:

    
def findTemperature(I1, I2, a1, b1, a2, b2, T1, Ta, T20):
    """Function used to calculate the temperature of a bar in the new conditions.
    
    Inputs
    ---------------
    I1: float
        Tested current in Amps
    I2: float
        New current value in Amps
    a1: float
        Tested bar thickness in meters
    b1: float
        Tested bar width in meters
    a2: float
        New bar thickness in meters
    b2: float
        New bar width in meters
    T1: float
        Absolute temperature of a tested bar in deg C
    Ta: float
        Ambient temperature in deg C
    T20: float
        initial value of the calculated temperature in deg C
        starting point for the Newton method
    
    Returns
    -----------------
    T2: float
        Absolute temperature of a bar in new conditions in deg C"""
    
    # temperature resistance change coeff
    alpha = 3.9e-3

    def f(x):
        return np.sqrt((2*b1 + 2*a1)/(2*b2+2*a2)) * ((T1-Ta)/(x-Ta))**0.61 \
    * np.sqrt((a1*b1*(1+alpha*(x-20)))/(a2*b2*(1+alpha*(T1-20)))) - I1/I2
    
    T2 = newton(f, x0=T20)
    return T2



In [12]:

    
T2 = findTemperature(I1=1000, I2=1000, b1 = 25.4e-3, a1 = 0.25*25.4e-3, b2 = 25e-3, a2 = 6e-3, T1=84.9, Ta=50, T20=90)



In [13]:

    
T2









    Out[13]:





88.077777089701527



In [14]:

    
T22 = findTemperature(I1=1000, I2=800, b1 = 25.4e-3, a1 = 0.25*25.4e-3, b2 = 25e-3, a2 = 6e-3, T1=84.9, Ta=50, T20=90)
T22









    Out[14]:





75.574979923285781



In [15]:

    
T23 = findTemperature(I1=1000, I2=1500, b1 = 25.4e-3, a1 = 0.25*25.4e-3, b2 = 25e-3, a2 = 6e-3, T1=84.9, Ta=50, T20=90)
T23









    Out[15]:





132.16715870829645



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