Extracting function from contour plot

Question

I have made a contour plot via Minitab and got an image like this one:

With

• $z$ being the enthalpy $[BTU/lb]$
• $x$ being the temperature $[°F]$
• $y$ being the concentration [%]

How can I mathematically access its function $z = f(x,y)$ from just the data behind the picture?
Is there a possibility to get $z = f(x,y)$ from this via programming/software?

EDIT:

Part of my data:

x = Concentration [%]	y = Temperature [°F]	z = Enthalpy [BTU / lb]
0	32	4
5	32	-16
10	32	-32
15	32	-48
20	32	-63
25	32	-76
30	32	-90
35	32	-103
40	32	-116
45	32	-124
50	32	-128
55	32	-132
60	32	-136
65	32	-139
70	32	-136
75	32	-128
80	32	-118
85	32	-100
90	32	-75
95	32	-40
100	32	2
0	50	16
5	50	0
10	50	-18
15	50	-32
20	50	-48
25	50	-63
30	50	-78
35	50	-90
40	50	-100
45	50	-111
50	50	-118
55	50	-123
60	50	-125
65	50	-128
70	50	-125
75	50	-121
80	50	-112
85	50	-96
90	50	-70
95	50	-37
100	50	4

Picture:

@Claude:

• BTU/lb_5 = - 48,72 + 1,007 Temp - 0,000163 Temp^2
• BTU/lb_10 = - 65,25 + 1,001 Temp - 0,000287 Temp^2
• BTU/lb_15 = - 77,55 + 0,9209 Temp - 0,000163 Temp^2
• BTU/lb_20 = - 90,01 + 0,8385 Temp + 0,000043 Temp^2
• BTU/lb_25 = - 100,7 + 0,7528 Temp + 0,000226 Temp^2

Answer 1

Follows a MATHEMATICA script that I hope, will express with sufficient accuracy, the formulation needed $z = f(x,y)$

data1 = {{5, -48.72}, {10, -65.25}, {15, -77.55}, {20, -90.01}, {25,-100.7}};
data2 = {{5, 1.007}, {10, 1.001}, {15, 0.9209}, {20, 0.8385}, {25, 0.7528}};
data3 = {{5, -0.000163}, {10, -0.000287}, {15, -0.000163}, {20, 0.000043}, {25, 0.000226}};

fc1 = c10 + c11 w + c12 w^2 + c13 w^3 + c14 w^4;
fc2 = c20 + c21 w + c22 w^2 + c23 w^3 + c24 w^4;
fc3 = c30 + c31 w + c32 w^2 + c33 w^3 + c34 w^4;

c0 = NonlinearModelFit[data1, fc1, {c10, c11, c12, c13, c14}, w];
c1 = NonlinearModelFit[data2, fc2, {c20, c21, c22, c23, c24}, w];
c2 = NonlinearModelFit[data3, fc3, {c30, c31, c32, c33, c34}, w];

f[Temp_, w_] := -(c0[w] + c1[w] Temp + c2[w] Temp^2)

ContourPlot[f[Temp, w], {w, 5, 25}, {Temp, 0, 100}, Contours -> 15, ContourShading -> None]

Plot[{f[Temp, 15], f[Temp, 17], f[Temp, 20]}, {Temp, 0, 100}]

Added a python script to define $z = f(x,y)$

from scipy.optimize import curve_fit

def coefs(w,c0,c1,c2,c3,c4):
    return c0 + c1*w + c2*w ** 2 + c3*w ** 3 + c4*w ** 4

x_values  = [5., 10., 15., 20., 25.]
c0_values = [48.72, 65.25, 77.55, 90.01, 100.7]
c1_values = [-1.007, -1.001, -0.9209, -0.8385, -0.7528]
c2_values = [0.000163, 0.000287, 0.000163, -0.000043, -0.000226]

coefsc0, _ = curve_fit(coefs, x_values, c0_values)
coefsc1, _ = curve_fit(coefs, x_values, c1_values)
coefsc2, _ = curve_fit(coefs, x_values, c2_values)


def C(p, x):
    val = 0
    for i, pp in enumerate(p):
        val += pp * x**i
    return val


def f(t,w):
    return C(coefsc0,w) + (C(coefsc1,w) + C(coefsc2,w)*t)*t

print(f(0,15))

Answer 2

Being myself thermodynamicist, I think you should consider the enthalpy as a function of temeperature for a given concentration because it must be a smooth function, probably well approximated by a quadratic polynomial for any concentration.

So, let us suppose that for a given concentration $y$ we have $$z_y =a_y+b_y\, x + c_y\, x^2$$ the three coefficients are easy to obtain. Now, look how each of them varies with $y$. Suppose that the first one is linear, the second cubic and the third quadratic. In such a case, the model would be $$z=(a_0+a_1\,y)+(b_0+b_1\,y+b_2\,y^2 +b_3\,y^3)x+(c_0+c_1\,y+c_2\,y^2 )x^2$$ Now, use all the data and refit the parameters (it is just a linear regression).