Question

2. Experiments are carried out on a plastic extrusion die to determine the relationship between the mass flow rate m and the pressure difference P. We expect the relationship to be of the form m = APN, where A and n are constants. The measurements yield the mass flow rate m for different pressure differences P as: 12.8 15.5 17.5 19.8 m (kg/h) P (atm) 22.0 30.0 10.0 15.0 20.0 25.0 Obtain a best fit to these data and determine the coefficients A and n. Is this a good best fit, or should we consider other functional relationships?

please help for thermal design subject

Answer 1

Taking natural logarithm to both sides of the equation we get

$m=AP^n$

or, In m In A + nln P

$or,~z=c+bx$

where z = ln m, c = ln A, b = n and x = ln P. Converting the given data, we get

m	P	z = ln m	x = ln P
12.8	10	2.549445	2.302585
15.5	15	2.74084	2.70805
17.5	20	2.862201	2.995732
19.8	25	2.985682	3.218876
22	30	3.091042	3.401197

Now we have the case of a linear fit. In our case, we have N = 5 datasets. Using standard equations, the value of b and c can be determined as follows :-

6 N i=1 zili N N N i=117

Using Excel to calculate the terms in this equation, we get

Στη , = 41.991 [=1

Σ t, = 14.626 I=1

ΣΗ = 14.229 =1

$\sum_{I=1}^nx_i^2=43.539$

Putting these values in the equation we get

6 N i=1 zili N N N i=117

$or,~b=\frac{41.991-\frac{14.626*14.229}{5}}{43.539-\frac{14.626^2}{5}}$

$or,~b=\frac{41.991-41.623}{43.539-42.784}$

$or,~b=0.487$

Using this the value of c can be caluclated as

Σ C 1=1 Ν

$or,~c=\frac{14.229-(0.487)*14.626}{5}$

or, c = 1.421

Thus the linear fit is

$or,~z=1.421+0.487x$

From our prvious assumption,

$c=ln~A$

$or,~A=e^c=e^{1.421}=4.142$

Thus the exponential equation fit to the data is

$m=4.142P^{0.487}$

The value of R^2 will determine how good a fit this is. We can determine it from the linear model. I have used Excel for the calculations to avoid human error. All formulae used are written in the headings.

A B C D E F G H K L M N 1 2 P z-zm m 12.8 (z'-zm)^2 0.092102 3 10 4 15.5 15 0.011241 R^2 = J11/H11 0.998157 z = In m x= In P 2.549445 2.302585 2.74084 2.70805 2.862201 2.995732 2.985682 3.218876 3.091042 3.401197 z' = 1.421+0.487x 2.54235894 2.739820448 2.879921617 2.988592527 3.077383125 (z-zm)^2 0.0878511 0.0110254 0.0002676 0.0195552 0.0601232 z'-zm -0.303483153 -0.106021645 0.034079524 0.142750433 0.231541032 5 -0.296396922 -0.105002069 0.016358788 0.139839844 0.24520036 17.5 20 6 19.8 25 0.001161 0.020378 0.053611 7 22 30 8 9 mean of z = zm = 2.845842 sum (z-zm)^2 0.1788226 sum (z'-zm)^2 0.178493 10 11 12

Thus we see that the value of R^2 is fairly close to 1. Although the given data would fit several other types of polynomial and transcendental equations, this fit is quite good for the given data set.

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