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 = AP”, where A and n are constants. The measurements yield the mass flow rate m for different pressure differences P as: m (kg/h) 12.8 15.5 17.5 19.8 22.0 P (atm) 10.0 15.0 20.0 25.0 30.0 Obtain a best fit to these data and determine the coefficients A and n. Is this good best fit, or should we consider other functional relationships?

Answer 1

SP = 225D 5m = 1586.33 expected relationship >> M = AP" Em = 87.6 Smp 22.0 19.8 m(kg/h) 12.8 15.5 17.5 EP = 100 1865.5 25 30 20 10 15 P (atm) m = Ap (Apply natural logarithm on both sides ln en m = A + n ln p en (m) = ln A+ neap let ln(m) = Y ln A=a & ln P = x Y = a + nx (x, Yar Y are known an are to be determined) 2 х XY Y m Р = ln (m) 2.5494 Х = ln (p) 2.3026 5:3019 5.8103 12.8 10 7.42.23 7.3335 2.7081 15 2.7408 15.5 2.9957 421318.5744 8.9744 2.8622 17.5 20 3.2189 9. C105 10.3612 25 2.9857 19.8 10.5132 3.4012 11.5681 3.0910 22.0 30 Y =a+nx an arc 2 constants) un knowns. SO 2 equations are needed to dete o minc them EY=2(a + x) ΣΥ= 50 + ΠΣΧ I ne is Y=a+nx (multiply X on both sides) XY= ax + nx² I one 15 ΣΧΥ = ΑΣΧ+ΠΣ,

The two equations are EY=5a + EX ΣΧΥ = ΔΣX +ΟΣ EXY = 41.9907 EY= 14.2291 EX= 14.6265, EX = 43.5391 EY-40.6724 14.2291 = 5a + 14.6265n 41.9907 = 14.6265 a + 43.53910 a = 1.4212 and n = 0.4870 Solving for a&n gives then the equation is Y = 1.4212 + 0.4870 X en (m) = 1.4212 +0.4870 In(P) m = (64212) (p) 0:48 70 ח ..m= 4.1421p0.4870 0-4870 m= 4.1421 P :. Best fit (by least squares method for m&p is m = APP A = 4.1421 & n = 0.4870 Costelation cocfficient (for data set 1&y is) n (Ezy) - (EI)Ey) J (n=2? - (E1)) (nEy?- [Ey)") n = no of data set n = 5 for the given me data n(2 m) - (Em)(EP) J(nžm?- (Em)") (n zp - Ep!") 5(1865.5) - 87.6X100 | (5X1586.38 - 87.6%) (5 x2250 -100) = 0.999 since fis close to 1, It is a good fit.
I hope you understand. comment if you have any doubt.