化工应用数学

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授課教師： 郭修伯

Partial Differentiation and Partial Differential Equations Lecture 7 Chapter 8

Partial differentiation and P.D.E.s

– Problems requiring the specification of more than one independent variable. – Example, the change of temperature distribution within a system:

The differentiation process can be performed relative to an incremental change in the space variable giving a temperature gradient, or rate of temperature rise. Partial derivatives

Figure 8.1 (contour map for u)

– If x is allowed to vary whilst y remains constant then in general u will also vary and the derivate of u w.r.t. x will be the rate of change of u relative to x, or the gradient in the chosen direction : ?u ? ? ? ? ?x ? y

u is a vector along the line of greatest slope and has a magnitude equal to that slope.

u = i ?u ?u + j = grad u ?x ?y u i ? ?u = ?x

?u ? ? ?δy due to the change in y: ? ?y ? ? ?

?u ? u will change by ? ?δxdue to the change in x, and by ? ?x ? ?u ? ? ?u ? δu = ? ?δx + ? ?δy ? ?y ? ? ?x ? ? ? δu → du δx → dx δy → dy

?u ? ? ?u ? du = ? ?dx + ? ?dy ? ?y ? ? ?x ? ? ? the “total differential” of u

?u ? ? ?u ? ? ?u ? ?dx1 + ? ?dx2 + ?? + ? du = ? In general form : ? ?x ? ? ?x ? ? ?x ?dxn ? ? 1? ? 2? ? n?

Important fact concerning “partial derivative” The symbol “ ?g“ cannot be cancelled out! ? The two parts of the ratio defining a partial derivative can never be separated and considered alone. – Marked contrast to ordinary derivatives where dx, dy can be treated separately Changing the independent variables z = g ( x, y )

x = φ (u , v) y = ψ (u, v)

?g ? ? ?g ? dz = ? ?dx + ? ?dy ? ?y ? ? ?x ? ? ?

?φ ? ? ?φ ? dx = ? ?du + ? ?dv ? ?u ? ? ?v ? ? ?ψ ? ? ?ψ ? dy = ? du + ? ? ?dv ? ?u ? ? ?v ?

φ ? ? ?g ?? ?ψ ?ψ ? ? ?g ?? ?φ dz = ? ?? du + dv ? + ? ?? du + dv ? ? ?y ? ?u ?v ? ? ?? ?v ? ? ?x ?? ?u w.r.t. u ?z ? ?z ?? ?x ? ? ?z ?? ?y ? = ? ?? ? + ? ?? ? ?u ? ?x ?? ?u ? ? ?y ?? ?u ? ? ? ?z ? ?g ?? ?φ ? ? ?g ?? ?ψ ? = ? ?? ? + ? ?? ? ?u ? ?x ?? ?u ? ? ?y ?? ?u ? ? ?

?z ?? ?xn ? ?z ? ?z ?? ?x1 ? ? ?z ?? ?x2 ? ?? ?+? ?? ? + ?? + ? =? In general form : ? ?x ?? ?u ? ? ?x ?? ?u ? ? ? ?x ?? ?u ? ?? ?u1 ? 1 ?? 1 ? ? 2 ?? 1 ? ? n ?? 1 ?

Independent variables not truly independent Vapour composition is a function of temperature, pressure and liquid composition: y = f (T , P, x)

However, boiling temperature is a function of pressure and liquid composision: T = g ( P, x ) Therefore

?f ? ? ?f ? ? ?f ? dy = ? ?dx + ? ?dT + ? ?dP ? ?x ? ? ?T ? ? ?P ? ?g ? ? ?g ? dT = ? ?dx + ? ?dP ? ?x ? ? ?P ?

?f ? ? ?f ?? ?g ?? ?? ?f ? ? ?f ?? ?g ?? dy = ?? ? + ? ?? ?? dx + ?? ? + ? ?? ?? dP ?? ?x ? ? ?T ?? ?x ?? ?? ?P ? ? ?T ?? ?P ?? Temperature increment of a fluid: ?T ? ? ?T ? ? ?T ? ? ?T ? ?dy + ? dT = ? dx + ? dz + ? ? ? ?dt ? ?y ? ? ?x ? ? ?z ? ? ?t ? ? ? dT ? ?T ? dx ? ?T ? dy ? ?T ? dz ? ?T ? =? ? +? ? ?y ? dt + ? ?z ? dt + ? ?t ? ? dt ? ?x ? dt ? ? ? ? ? ? Total time derivative A special case when the path of a fluid element is traversed T ? ?T ? ? ?T ? ? ?T ? ? ?T ? ? ?v + ? + u ? ?T = ? ?u + ? ?w + ? ? ? ?t ? ?x ? ? ?y ? ? ?z ? ? ?t ?

DT Dt Substantive derivate of element of fluid ?T compare partial derivate of element of space ?t Formulating P.D.Es

Identify independent variables ? Define the control volume ? Allowing one independent variable to vary at a time ? Apply relevant conservation law Unsteady-state heat conduction in one dimension δx x Rate of heat input at distance x and time t: T Rate of heat input at distance x and time t + δt: Considering the thermal equilibrium of a slice of the wall between a plane at distance x from the heated surface and a parallel plane at x+δx from the same surface gives the following balance. k ?T ?x k

T ? ? ?T ? + ?? k ?δt ?x ?t ? ?x ?

Rate of heat output at distance x + δx and time t: ? k L Rate of heat output at distance x + δx and time t + δt: k

T ? ? ?T ? + ?? k ?δx ?x ?x ? ?x ? T ? ? ?T ? ? ? ?T ? ? ?T ? ? + ?? k + ?? k ?δt + ?? k ?δt ? ?x ?t ? ?x ? ?x ? ?x ?t ? ?x ? ? ?T 1 ? ? ?T ? ? Average heat input during the time interval δ t is ?? k + ?k ?δt ?δt ? ?x 2 ?t ? ?x ? ? ? Average heat output during the time interval δ t is ?T ? ? ?T ? 1 ? ? ?T ? ? ?T ? ? ? ?k + ?? k ?k + ?? k ?δx + ?δx ?δt ?δt ? ? ?x ?x ? ?x ? 2 ?t ? ?x ?x ? ?x ? ? ? ?

Heat content of the element at time t is Heat content of the element at time t + δ t is Accumulation of heat in time δ t is

ρ C pT δ x ρ C p ?T + ?

? ? ?T ? ?t ? ? ?δ t ? δ x ? ? ?T ? ρC p ? ?δ t δ x ? ?t ?

Conservation law ? ? ? ? ?T ? ?T ? 1 ?2 ? ? ?T ? ? ?? k ?δx + ?? k ?δxδt ?δt = ρC p ? ?δtδx ? ?x ? 2 ?t?x ? ?x ? ? ?t ? ? ?x ? ? assuming k is constant

2T 1 ? 3T ? ?T ? k 2 + k 2 δt = ρC p ? ? ?x 2 ?x ?t ? ?t ? 2T 1 ? 3T ? ?T ? k 2 + k 2 δt = ρC p ? ? ?x 2 ?x ?t ? ?t ? δt → 0

k ? 2T ?T = ρC p ?x 2 ?t

k is the thermal diffusivity ρC p three dimensions T k 2 ?T= ρC p ?t Mass transfer example

A spray column is to be used for extracting one component from a binary mixture which forms the rising continuous phase. In order to estimate the transfer coefficient it is desired to study the detailed concentration distribution around an individual droplet of the spray. (using the spherical polar coordinate) During the droplet’s fall through the column, the droplet moves into contact with liquid of stronger composition so that allowance must be made for the time variation of the system. The concentration will be a function of the radial coordinate (r) and the angular coordinate (θ) θ θ r r θ r θ r

Area of face AB is Area of face AD is Volume of element is

2πr sin θ (rδθ ) 2πr sin θ (δr ) 2πr sin θ (δr )(rδθ )

Material is transferred across each surface of the element by two mechanisms: Bulk flow and molecular diffusion

Input rate across AB Input rate across AD

c ? ? ? uc ? D ?2πr sin θ (rδθ ) ?r ? ? D ?c ? ? vc ? ? ?2πr sin θ (δr ) r ?θ ? ?

?c ? ? ?? ?c ? ? ? uc ? D ?2πr sin θ (rδθ ) + ?? uc ? D ?2πr sin θ (rδθ )?δr ?r ? ?r ?? ?r ? ? ? ? D ?c ? ? ?? D ?c ? ? Output rate across BC ? vc ? ?2πr sin θ (δr ) + ?? vc ? r ?θ ?2πr sin θ (δr )?δθ r ?θ ? ?θ ?? ? ? ? Accumulation rate (δr )(rδθ ) ?c 2πr sin θ ?t

Output rate across CD Conservation Law: input - output = accumulation

? ?? ?c ? ? ? ?? uc ? D ?2πr sin θ (rδθ )?δr ? ?r ?? ?r ? ?θ ? = 2πr sin θ (δr )(rδθ ) ?c ?t

? D ?c ? ?? vc ? r ?θ ?2πr sin θ (δr )?δθ ? ?? ?

? ?? ?c ? ? ? ?? uc ? D ?2πr sin θ (rδθ )?δr ? ?r ?? ?r ? ?θ ? = 2πr sin θ (δr )(rδθ ) ?c ?t

? D ?c ? ?? vc ? r ?θ ?2πr sin θ (δr )?δθ ? ?? ? Dividing throughout by the volume

?c 1 ? ?? 2 ?c ?? 1 ? ?? D ?c ? sin θ ? =? 2 r uc ? r 2 D ?? ? vc ? ? ? ? ?t r ?r ?? ?r ?? r sin θ ?θ ?? r ?θ ? ? ? ? The continuity equation z y x

Input rate through ABCD ρuδyδz ρvδxδz

C D δz A B δy δx E H G

Input rate through ADHE Input rate through ABFE Output rate through EFGH Output rate through BCGF Output rate through CDHG ρwδxδy ρuδyδz +

(ρuδyδz )δx ?x ? ρvδxδz + ( ρvδxδz )δy ?y ? ρwδxδy + ( ρwδxδy )δz ?z F

Conservation Law: input - output = accumulation

? ? ?ρ ? (ρuδyδz )δx ? (ρvδxδz )δy ? ( ρwδxδy )δz = δxδyδz ?x ?y ?z ?t

ρ ? ? ? + ( ρu ) + ( ρv ) + (ρw) = 0 ?t ?x ?y ?z

ρ ? ? ? + (ρu ) + (ρv ) + (ρw) = 0 ?t ?x ?y ?z ?ρ ?u ?v ?w ?ρ ?ρ ?ρ +ρ +ρ +ρ +u +v +w =0 ?t ?x ?y ?z ?x ?y ?z

Dρ ? ?ρ ? ? ?ρ ? ? ?ρ ? ? ?ρ ? ? ? + w? ? + ? ? = u ? ? + v? ? Dt ? ?x ? ? ?y ? ? ?z ? ? ?t ?

1 Dρ ?u ?v ?w + + + =0 ρ Dt ?x ?y ?z

Continuity equation for a compressible fluid Boundary conditions O.D.E.

– boundary is defined by one particular value of the independent variable – the condition is stated in terms of the behaviour of the dependent variable at the boundary point. P.D.E.

– each boundary is still defined by giving a particular value to just one of the independent variables. – the condition is stated in terms of the behaviour of the dependent variable as a function of all of the other independent variables. Boundary conditions for P.D.E. Function specified – values of the dependent variable itself are given at all points on a particular boundary

Derivative specified

– values of the derivative of the dependent variable are given at all points on a particular boundary

Mixed conditions ? Integro-differential condition Function specified

Example 8.3.1 (time-dependent heat transfer in one dimension): The temperature is a function of both x and t. The boundaries will be defined as either fixed values of x or fixed values of t:

– at t = 0, T = f (x) – at x = 0, T = g (t)

Steady heat conduction in a cylindrical conductor of finite size: The boundaries will be defined as by keeping one of the independent variables constant: – at z = a, T = f (r, θ) – at r = r0, T = g (z, θ) Derivative specified

In some cases, (e.g., cooling of a surface and eletrical heating of a surface), the heat flow rate is known but not the surface temperature. ? The heat flow rate is related to the temperature gradient. ? Example: C G H B F A E z D

The surface at x = 0 is thermally insulated. x C G H z D B F A E

x Input rate through ADHE

T ?k δxδz ?y ?T Input rate through ABFE ?k δxδy ?z ? ?T ? ? ?T Output rate through BCGF ? k δxδz + ? ? k δxδz ?δy ? ?y ?y ? ?y ? ? ?T ? ? ?T ? Output rate through DCGH ? k δxδy + ? ? k δxδy ?δz ?z ?z ? ?z ? ?T ? ? ?T ? Output rate through EFGH ? k δyδz + ? ? k δyδz ?δx ?x ?x ? ?x ? ? ?T ? Accumulation of heat in time δ t is ρC p ? ?δ x δ y δ z ? ?t ? Heat balance gives

2T ? 2T ? 2T ? 2T ? ?T ? k δ xδ zδ y + k 2 δ xδ zδ y + k 2 δ zδ y + k 2 δ xδ zδ y = ρ C p ? ?δ x δ z δ y 2 ?y ?z ?x ?x ? ?t ? ? 2T ? 2T ? 2 T k? 2 + + 2 ? ?y ?z ?x 2 ?

?T ? ?T ? ?δ x + k = ρC p ? ?δ x ? ?x ? ?t ? ? size → 0 δx → 0 T =0 ?x

This is the required boundary condition. at x = 0 Example

A cylindrical furnace is lined with two uniform layers of insulting brick of different physical properties. What boundary conditions should be imposed at the junction between the layers? B layer 1 δr A C D a

layer 2

Due to axial symmetry, no heat will flow across the faces of the element given by θ = constant but will flow in the z direction. θ One boundary condition: T1 = T2 at r = a

T1 (aθδz ) The rate of flow of heat just inside the boundary of the first layer is ? k1 ?r

The rate of flow of heat into the element across the face CD is r=a k1

T1 (rθδz ) ? ? ? ? k1 ?T1 (rθδz )? 1 δr ? ? ?r ?r ? ?r ?2

T1 1 ? ? ?T1 ? + k1 (θδrδz ) ? r ? r =a Input across CD = ? k1 (aθδz ) ?r 2 ?r ? ?r ? B

layer 1

Input across CD = δr A C D a layer 2 θ

T1 1 ? ? ?T1 ? ? k1 (aθδz ) + k1 (θδrδz ) ? r ? r =a ?r 2 ?r ? ?r ? Output across AB = k 2 (aθδz )

T2 1 ? ? ?T ? ? k 2 (θδrδz ) ? r 2 ? r = a ?r 2 ?r ? ?r ? The heat flow ratesin the z direction Input at face z = 1 ? ? (aθδr ) (k1T1 + k 2T2 ) 2 ?z

? ? 1 ? 1 ? Output at face z + δz = ? (aθδr ) (k1T1 + k 2T2 ) + ?? (aθδr ) (k1T1 + k 2T2 )?δz 2 ?z ?z ? 2 ?z ? Accumulation within the element

1 ?T 1 ?T ρ1C p1 (aθδrδz ) 1 + ρ 2C p 2 (aθδrδz ) 2 2 ?t 2 ?t The complete heat balance on the element

1 (aθδrδz ) (ρ1C p1T1 + ρ 2C p 2T2 ) = ?t 2 ? ?T 1 ? ? ?T ?? ? k1 (aθδz ) 1 + k1 (θδrδz ) ? r 1 ?? ? ?r 2 ?r ? ?r ?? ? ? ?T 1 ? ? ?T ?? ? ?? k 2 (aθδz ) 2 ? k 2 (θδrδz ) ? r 2 ?? ?r 2 ?r ? ?r ?? ? ? ? ?1 ? ? ? + ? ? (aθδr ) (k1T1 + k 2T2 )?δz ? ?z ? ? ? ?z ? 2 dividing by aθδz δr → 0

T1 ?T2 k1 = k2 ?r ?r

This is the second boundary condition. And??

1 ? (aθδrδz ) (ρ1C p1T1 + ρ 2C p 2T2 ) = 2 ?t ? ?T1 1 ? ? ?T1 ?? + k1 (θδrδz ) ? r ?? ?? k1 (aθδz ) ?r 2 ?r ? ?r ?? ? ? ?T2 1 ? ? ?T2 ?? ? ?? k 2 (a

θδz ) ? k 2 (θδrδz ) ? r ?? ?r 2 ?r ? ?r ?? ? ? ? ?1 ? ? ? + ? ? (aθδr ) (k1T1 + k 2T2 )?δz ? ?z ? ? ? ?z ? 2

If the heat balance is taken in either layer (say layer 1) subscript 2 → 1 a→r

1 ? ? ?T1 ? ? 2T1 ρC p ?T ?r ?+ 2 = r ?r ? ?r ? ?z k ?t

Heat conduction in cylindrical polar coordinates with axial symmetry. Mixed conditions

The derivative of the dependent variable is related to the boundary value of the dependent variable by a linear equation. ? Example: surface rate of heat loss is governed by a heat transfer coefficient. T k = h(T ? T0 ) ?x

rate at which heat is removed from the surface per unit area rate at which heat is conducted to the surface internally per unit area Integro-differential boundary condition Frequently used in mass transfer

– materials crossing the boundary either enters or leaves a restricted volume and contributes to a modified driving force.

Example: a solute is to be leached from a collection of porous spheres by stirred them as a suspension in a solvent. Determine the correct boundary condition at the surface of one of the spheres.

The rate at which material diffuse to the surface of a porous sphere of radius a is:

c ? 4πa D ?r 2 a

D is an effective diffusivity and c is the concentration within the sphere. If V is the volume of solvent and C is the concentration in the bulk of the solvent: C ?c 2 V = ?4 Nπa D ?t ?r

For continuity of concentration, c = C at r = a : a

N is the number of spheres. at r = a,

4 Nπa 2 D ?c C = ?∫ a dt 0 V ?r t

Boundary condition

Initial value and boundary value problems Numer of conditions: – O.D.E.

the number of B.C. is equal to the order of the differential equation – P.D.E.

no rules, but some guild lines exist. 2T ?T α 2 = ?x ?t

Two boundary conditions are needed at fixed values of x and one at a fixed value of t.

Initial value or boundary value?

When only one condition is needed in a particular variable, it is specified at one fixed value of that variable.

– The behaviour of the dependent variable is restricted at the beginning of a range but no end is specified. The range is “open”.

When two or more conditions are needed, they can all be specified at one value of the variable, or some can be specified at one value and the rest at another value. – When conditions are given at both ends of a range of values of an independent variable, the range is “closed” by conditions at the beginning and the end of the range. – When all conditions are stated at one fixed value of the variable, the range is “open” as far as that independent variable is concerned. The range is closed for every independent variable: a boundary value problem (or, a jury problem). ? The range of any independent variable is open: an initial value problem (or, a marching problem). 1本文由ohngipeex贡献

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授課教師： 授課教師： 郭修伯 Lecture 2 實驗數據的分析 實驗精確程度

The degree of accuracy sought in any investigation should depend upon the projected use of the results, and the accuracy of the required data and calculations should be consistent with the desired accuracy in the results. It is desirable to complete the investigation and obtain the required accuracy with a minimum of time and expense. The accuracy of a number representing the value of a quantity is the degree of concordance between this number and the number that represents the true value of the quantity; it may be expressed in either absolute or relative terms. 誤差來源

Accidental errors of measurement – Such errors are inevitable in all measurements and that they result from small unavoidable errors of observation due to more or less fortuitous variation in the sensitivity of measuring instrument and the keenness of the senses of perception. （例如用了不準的A來校正B，用B的錯誤校正曲 線進行量測） Precision and constant errors

– A result may be extremely precise and at the same time highly inaccurate. – Constant errors can be detected only by performing the measurement with a number of different instruments and , if possible, by several independent methods and observers. （例如用了不準確的儀器或樣品取樣在不具代表 性的地方） Errors of Methods

– These arises as a result of approximations and assumptions made in the theoretical development of an equation used to calculate the desired result. （例如在計算時，用的錯誤的假設）

Variance and distribution of random errors If an experimental measurement is repeated a number of times, the recorded values

of the measured quantity almost invariably differ from one another. The data so obtained may be used for two purposes: – to evaluate the precision of the measurement – to obtain an estimate of the probability that the mean of the measurement differs from the true value of the measured quantity by some special amount The “scatter” of the repeated measurements of the quantity is commonly reported in terms of the “variance” or “standard deviation” of the sets of measurements. Sample variance and standard deviation Sample mean x = ∑ n k =1 xk n

Sample variance σ 2 = ∑ n k =1

( xk x )2 n

Sample standard deviation σ = σ 2

Population variance and standard deviation Population mean x = lim ∑ n k =1 xk n→ ∞ n

Population variance σ 2 =

∑ lim n→ ∞ n k =1

( xk x )2 n

Population standard deviation σ = σ 2

Population and sample Sample 1 Population

Normal frequency distribution

If an infinite data set the variation in x are random, it was first shown by Gauss that the distribution of values of x about the population mean is given by exp{ (1 / 2 )[( x x ) / σ ] 2 } f = σ 2π

– f is frequency, or probability of occurrence, of a value of magnitude x. Normal frequency distribution

The probability that a single measurement will give a value lying between x - dx/2 and x + dx/2 is ∫ fdx = ∫

exp{ (1 / 2 )[( x x ) / σ ] 2 } dx σ 2π

The probability is less than 5% that a single measurement of x will differ from x by more than twice the standard deviation, i.e. by more than ± 2σ. The range ± 2σ is frequently called the “95 percent confidence belt on x”. More about variance??

The sample mean is the best estimate of the population mean. The sample variance is not the best estimate of the population variance. A better estimate is given by Sample estimate of the population variance s = 2

( xk x ) 2 ∑ k =1 n n 1

n = σ2 n 1 sample Population

Number of measurment

By how much can the mean of n measurement be expected to differ from the best value of the population mean?

Population x , σ 2 Set 1, k times x1 , s12 Set 2, k times x 2 , s2 2 ? n sets Estimate of the set of means, xm Estimate variance of the set of means, sm < si 2 2

Sample variance of the mean

The grand mean xm = ∑ n i=1 xi

(Sample猜測的population mean) n

The sample estimate of the variance of the set of means sm 2 = ∑ n i=1

(xi xm )2 n 1

(Sample猜測的population variance)

The sample estimate of the variance of the set of means may be estimated by a single set s 2 m s 2 k sample Population

Confident limits for small samples

To associate the magnitude of deviations of xi from the population mean x with the probability of the occurrence of such deviations. It is known that if the sample set contains at least 20 entries, the error introduced by the previous slide is not serious. s 2 2 2 s m k ≈ σ m

For smaller samples, however, s2/k is not an adequate estimate of σm2. Student’s t test The solution to small number of entries was first pointed out in Biometrika, Vol. VI, 1908, by W.S. Gossett, who signed his article “STUDENT”. The dimensionless quantity of particular interest in a confidence-limit analysis is called “Student’s t”:

xi x t = sm

It involves estimates obtained from a sample of finite size. Student’s t test t is the difference between the measured sample mean and the true population mean

divided by the sample estimate of the standard deviation of the population of means. “Student” derived the frequency distribution for t: t 2 ( f 1 ) / 2 f t dt = C f {( 1 + ) } dt f

– Cf is a function of f only – f is the “degree of freedom”, defined as the number of values used to calculate the means on which t is based, less the number of means so calculated. Student’s t test

The distribution funcion of t is used to calculate the probability value of the size of the sample (the degrees of freedom f). Probability calculation of this kind have been carried out over a wide range of conditions, and the results are tablulated in the handout given in the course. t

sample Population T test 範例

Two methods were used to measure a quantity. It is desired to use these data to obtain the following information:

X i1 X i2 X i3 X i4 X i5 Xi 2 Σ (X ik- X i) 2 S 1 (X i) 2 Sm 95% confidence limits Procedure 1 55.3 56.9 55.8 57.3 57.7 56.6 4.1 1.0 0.2 55.3 = X 1 = 57.9 Procedure 2 52.6 54.3 58.0 52.7 60.0 55.5 43.9 11 2.2 51.3 = X 2 = 59.9 T test 範例（續） 範例（

The confidence limits to be assigned to the results of procedures 1 and 2 The significance of the difference between the mean values of the results of procedures 1 and 2 The “best value” to be assigned to the sample analysis The confidence limits of the best value

T test 範例（續） 範例（

Confidence limits – The sample mean of procedure 1 is easily found to be 56.6 – The external probability limit on t will be arbitrarily set at 0.05, corresponding to 95 percent confidence limits. – Five measured values are used and one mean is calculated, degree of freedom: f = 5 - 1 = 4 – From the “t table” for f = 4, values of t lying outside ± 2.776 only are 0.05 probable. – The 95 percent confidence limits on t are: s ∑ ( x x ) /( k 1 ) = 4 . 1 / 4 = 0 . 20 = – s 2 2 2 1 1i 1 m ,1

2 . 776 ≤ t ≤ 2 . 776 k k 5

– The 95 percent confidence limits on x1 are then

x 1 x 1 = ± 2 . 776 × s m ,1 = ± 2 . 776 × 55 . 3 ≤ x 1 ≤ 57 . 9 0 . 20 = ± 1 . 3

T test 範例（續） 範例（

The significant of the difference – The mean value obtained by procedure 1 is 56.6 with 95 percent limits of 55 . 3 ≤ x 1 ≤ 57 . 9 – The mean value obtained by procedure 2 is 55.5 with 95 percent limits of 51 . 3 ≤ x 2 ≤ 59 . 9 – The sample

means are different, which might be taken to indicate a systematic difference or bias between the two methods of analysis. – The 95 percent confidence limits analysis shows that the mean of sample 2 is included with the confidence limits of sample 1, and vice versa. – It may be concluded that the difference between the two means has no statitical significance. T test 範例（續） 範例（

The best value? – Since the difference between the mean value obtained by procedure 1 and that by procedure 2 is not statistically significant, the best value to be assigned to the sample analysis is a weighted combination of the two mean values. – If the difference between the means has been significant, it would have been concluded that one or both of the procedures were affected by non-random factors (errors of method or bias). In this case, the best value cannot be estimated. T test 範例（續） 範例（

The best value obtainable from a series of sets of measurements exhibiting statistically equivalent n means is given by

the number of measurements in the ith set Population variance x best value = ∑W

i =1 n i =1 i xi i ∑W Wi = (k )i σ 2 i

(k )i 1 = 2 2 s m ,i si x best value = ∑W

i =1 n i =1 n i xi i ∑W =

56 . 6 / 0 . 2 + 55 . 5 / 2 . 2 = 56 . 6 1 / 0 .2 + 1 / 2 .2 T test 範例（續） 範例（

The confidence limits of the best value – the variance of the best value is

needed: σ

2 m , best value = 1

∑ [( k ) i =1 n i /σ 2 i 1 ]

∑ (1 / s i =1 n =

2 m ,i )

1 = 0 . 18 1 / 0 .2 + 1 / 2 .2

– the degree of freedom: f = 10 - 2 = 8 – the t limits are ± 2.3 – Consequetly, the 95 percent confidence limits on the best value are ± 2 .3 ×

0 . 18 = ± 1 . 0

∴ 55 . 6 ≤ x ≤ 57 . 6 Other test ( L test) L test:

– A calculation to determine the probability that the samples represent normal populations exhibiting the same population variance σ2, but without regard to the population means

Other test (F test) F test:

– In a single set of data, two types of error are possible - random errors and errors of method or bias. The magnitude of the random errors is estimated by the with-sample or error variance. However, from the single set of data, no estimate of the error due to method is possible. Suppose that other sets of data are available which are likewise subject to both random and method errors. For each data set, the within-sample or error variance may be calculated. – The statistic F is the ratio of the variance which contains both random and method error to the variance which includes random errors only. – The magnitude of F is a measure of the importance of errors of method which differ from one data set to the next. 最小平方法 (least squares)

Recall: A best straight line as the one for which the sum of the squares of the error terms is a minimum. Y = a + b ( x x ) The best measure of the precision with

which the points fit the line is the variance of estimate: se ( yi ) = 2 ∑ (Y i =1 n i

yi ) 2

degrees of freedom (2 = a, b) n2 2 2

The estimate of the error variance of Yi is

1 ( xi x )2 se (Yi ) = se [a + b( xi x )] = se (a) + ( xi x ) se (b) = se ( yi )[ + n ] ni ( xi x )2 ∑ 2 2 2 2 i =1

The confidence limits of Yi is ( ± t ) se2(Yi ) 1本文由eithelal贡献

ppt文档可能在WAP端浏览体验不佳。建议您优先选择TXT，或下载源文件到本机查看。 化工應用數學

授課教師： 授課教師： 郭修伯 Lecture 1 授課內容

實驗數據處理與分析 應用數學表達物理現象 常微分方程式 傅立葉分析 偏微分方程式 數值解析 最佳化設計 給分標準 – 期中考 30% – 期末考 40% – 作業 30% 教科書 及 參考書

教科書： 科書： – Mathematical Methods in Chemical Engineering (Author: V.G. Jenson and G.V. Jeffreys; Second Edition; Publisher: Academic Press) 參考書： 參考書： – Partial Differential Equations and Boundary-Value Problems with Applications (Author: M.A. Pinsky; Publisher: McGraw-Hill) – Applied Mathematics in Chemical Engineering (Author: H. S. Mickley, T. K. Sherwood, and C. E. Reed) – Advanced Engineering Mathematics (Author: P.V. O’Neil; Publisher: Wadsworth) 應用數學 Constructing a mathematical model to describe, understand, and predict behaviour of a physical process or system.

Identify the main components of the system Observe the behaviour of the system Attempt to relate the variables by means of one or more equations 應用數學

Experiment – How to analyse experimental results? – Is it accurate? Mathematics – How to build mathemitical models to describe physical problems? – How to solve the mathematical equations? 實驗

Why do experiments?

– test theoretical predictions by comparison with experiment – develop empirical formulae

Data analysis

– Some of the data points are often inaccurate and methods must be found for eliciting reliable information with reasonable certainty. – Statistics – Presenting the results graphically

Purpose: to extract maximum information

Chapter 10 Treatment of experimental results 實驗結果繪圖

選用何種座標？ 選用何種座標？ – 有因次式 – 無因次式 選用何種繪圖紙？ 選用何種繪圖紙？ – Linear graph paper – Semi-logarithmic graph paper – Logarithmic graph paper – Triangular graph paer 無因次群 If a particular piece of equipment compares with other work on similar equipment, the appropriate dimensionless groups should be used along the axes. If the dimensionless groups are chosen correctly, each group should only contain one principal variable and the rest of its constituent parts should be parameters. 目的在比較scale不同，但進行相似實驗的結果 繪圖

Axes: avoid plotting a dimensionless group against a dimensional variable. Many types of graph paper can be used to present experimental data. The most desirable shape for a curve is a straight line. 繪圖紙

Linear graph paper – y = a + bx Semi-logarithmic graph paper

– useful for the cases involving the approach to steadystate conditions. The dependent variable is a decaying exponential function of the independent variable. – y = ke bx ln y = ln k + bx

– The gradient, b, must be determined from the linear measurement of lny and not from reading the scales which gives values of y 繪圖紙

Logarithmic graph paper – covers a wide range – Dimensional analysis frequently indicates an empirical equation of the form y = Cx n

y and x are dimensionless groups ln y = ln C + n ln x Y = a + bX 1 y 3 1 2 ln y 3 2 x ln x 繪圖紙

Triangular graph paper

– In the study of liquid-liquid extraction systems where three components are present in two phases, a convenient graphical representation of the composition of a phase is needed. – The data can be presented on a triangular diagram by using the geometrical properties of a triangle. A

PE PF PD + + =1 BE CF AD P E F B D C

誤差擴大(Propagation of errors)

The absolute error in the result is the sum of the absolute errors in the constituent parts. – z= x+ y – δz = δx + δy The absolute error in the difference of two quantities is the sum of the absolute errors in those quantities. – z= x-y – δz = δx + δy The relative error in a production is the sum of the relative errors in the constituent parts. – z= xy – zr = xr + yr The relative error in a quotient is the sum of the relative errors in the constituent parts. – z= x/y – zr = xr + yr

derivation 誤差範例

Propagation of errors through a general functional relationship – z = f (x,y) – δz =

f f δx + δy x y Example:

– If z = x y n with n known and x and y determined experimentally, determine the relative error in z in terms of the relative errors in x and y. z=xyn δz =

f f δx + δy x y f ( x , y ) = xy n

Z 相對於 x 和 y 的誤差為何？ δ z = y n δ x + nxy n 1 δy

δ z = xy n δx x

+ nxy n δy y

z = xy

n δz z = δx x +n δy y

Ans: zr = xr + n yr 誤差範例

If a chemical reaction A B has a first order reaction rate constant k (s-1), the concentration of A leaving a tubular reactor of length L (m) with velocity u (m/s) is:

C = C0 e kL / u

where C0 is the initial concentration of A, diffusion has been neglect and plug flow has been assumed. How accurately must k be known and the flow rate be steady for it to be possible to design a reactor to give 94.5 - 95.5% completion? 0.95 ±0.05 C0 L C

C = C0 e kL / u f = C/C0 f =e kL / u

假設誤差的來源為 k 及 u f f δf = δ k + δu k u

L kL δ f = fδ k + 2 fδ u u u L kL δ f = fδ k + 2 fδ u u u δf

kL δk kL δu = ( ) + ( ) f u k u u f = e kL / u δf f

= (ln f ) δk k

+ ( ln f ) δu u

f = 0.95 δf = 0.005 δk k

+ δu u = 0.1

兩者的相對誤差和不能大於10％ 實驗式 (Empirical equation) The equation must be truely representative of the experimental data and it should be simple in form. The form of the equation is frequently suggested by a theoretical analysis and it is necessary only to evaluate certain contants. The general problem of fitting data by an empirical equation may be divided into two parts:

– the determination of a suitable form of equation – the evaluation of the constants

常見的實驗式型態

Equations involving more than two constants should be avoided. y = a + bx y = ax n n

y = c + ax y = ae bx y = ab x

y = a+ y = b x

x a + bx y = a + bx + cx 2 x y = +c a + bx 繪曲線(Curve fitting) How do you fit a set of experimental data for x and y? – graph fitting – method of averages – method of least squares 平均法 (Method of averages)

The best curve is the one passing through the average points. Procedure

– determine the type of curve – arrange the value of x in ascending order – divide the experimental results into groups and the number of groups must equal the number of unknown parameters – groups contain approximately equal numbers of points – substitute the “average point” from the grouped data into the equation of the chosen curve – determine the equation of the curve 平均法範例

The thermal conductivity of graphite varies with temperature according to the equation

k = k0 αT

Experimentally, it is only possible to obtain a mean conductivity over a temperature range. It is required to find the point conductivity from the mean conductivity given below.

T (° C) k m (W/m° C) 390 500 1000 1500 141 138 119 115 km is determined between 25°C and T T1 = 25 C L T2

An elementary heat balance k dT T T = km 1 2 dx L

k = k0 αT

( k0 αT ) dT = k m T1 T2 dx L

B.C. x=0, T=T1 對兩邊積分 x=L, T=T2

1 2 k0T1 + αT1 = C 2 1 2 k0T2 + αT2 = C + k m (T1 T2 ) 2 k0 0.5α (T1 + T2 ) = k m k0 0.5α (T1 + T2 ) = k m 續上頁??

平均法： 將數據分為兩組

T (° C) k m (W/m° C) 390 500 1000 1500 141 138 119 115 k0 0.5α (25 + 390) = 141 k0 0.5α (25 + 500) = 138 2 k 0 470 α = 279

k0 0.5α (25 + 1000) = 119 k0 0.5α (25 + 1500) = 115 2 k 0 1275 α = 234

k = k 0 α T = 152 . 6 + 0 . 056 T 最小平方法 (Method of least squares)

Most frequenly used method to fit the best straight line to a set of data. This method defines the best straight line as the one for which the sum of the squares of the error terms is a minimum. Determine the values of m and c which gives the least mean squares fit of the equation: y = m x + c m = c = N ∑ N 2 n

xn yn xn 2

∑ ∑x ∑ N∑ yn xn

∑x ∑ y (∑ x ) ∑ x y ∑ (∑ x ) n n 2 n n n 2 2 n xn

derivation 最小平方法範例

It has been proposed that the second order chemical reaction CO + Cl2 → COCl2

proceeds on the surface of an activated carbon catalyst after adsorption of the two reactants. Each of the three substances is adsorbed to a different, but the number of sites occupied by carbon monoxide is small compared with sites otherwise occupied. Assuming that the process is controlled by the surface reaction, which is irreversible, find the best values of the adsorption coefficients from the following experimental results.

Partial Pressures Cl2 0.352 0.363 0.320 0.333 0.218 0.118 0.608 Reaction Rate g mols/h gm catalyst -3 4.14 x 10 -3 4.40 x 10 -3 2.41 x 10 -3 2.45 x 10 -3 1.57 x 10 -3 3.90 x 10 -3 2.00 x 10

CO 0.406 0.396 0.310 0.287 0.253 0.610 0.179 COCl2 0.226 0.231 0.356 0.367 0.522 0.231 0.206 CO + Cl2 → COCl2 光氣

Let C = number of sites in the stage specified K = adsorption coefficient k = specific reaction rate constant p = partial pressure r = rate of reaction t = total number of sites v = vacant site a+b→c The equilibrium of the three components between the catalyst and the vapour phase can be expressed as:

The rate law of the elementary reaction C a = K a p a Cv Cb = K b pbCv

rc = kCa Cb 反應速率方程式 Number of total active sites 吸附方程式

C c = K c pc C v

Ct = Cv + Ca + Cb + Cc C a = K a p a Cv Cb = K b pbCv

We measured pa, pb, pc, and rc 消去 Ca, Cb, Cc C c = K c pc C v rc = kCa Cb

Ct = Cv (1 + K b pb + K c pc ) rc = kK a K a pa pbCv 消去 Cv 2

Ct = Cv + Ca + Cb + Cc

1 + K b pb + K c pc Ct kK a K b Pa Pb r =0

1 + αxn + βyn γz n = Rn Experimental error

1 + αxn + βyn γz n = Rn

Square the left-hand of the equation and sum over the N experimental points: N +α2 ∑xn + β 2 ∑yn +γ 2 ∑zn + 2α∑xn + 2β∑yn 2γ ∑zn + 2αβ∑xn yn 2 ∑xn zn 2βγ∑yn zn = ∑Rn αγ 2 2 2 2

Differentiate the equation partially with respect to α, β, and γ equal to zero, which gives the minimum value of sum (Rn2)

α ∑ x n 2 + ∑ x n + β ∑ xn y n γ ∑ x n z n = 0 β ∑ y n 2 + ∑ y n + α ∑ xn y n γ ∑ y n z n = 0

γ ∑ z n + ∑ z n + α ∑ xn z n + β ∑ y n z n = 0 2

α = 2.61 β = 1.60 γ = 0.393

K b = 2.61 K c = 1.60 Ct kK a = 0.243 數值積分 (Numerical integration)

It is sometimes necessary to perform a calculation which involves integration. For example, the volumetric flow rate of a gas through a duct can be determined from the linear velocity distribution by evaluating a suitable integral. One way of integrating a set of data is by fitting an empirical equation to the points and then integrating the equation analytically. Polynomials are used to fit data for integration purposes. The trapezium rule The linear equation

I1 = 1 ( y2 + y1 ) × ( x2 x1 ) 2

I = I1 + I 2 = 1 1 ( y2 + y1 ) × ( x2 x1 ) + ( y3 + y2 ) × ( x3 x2 ) 2 2 1 = ( y1 + 2 y2 + y3 )x 2 1 = ( y1 + 2 y2 + y3 )( x3 x1 ) 4 I1

I = average height x width Simpson’s rule The cubic equation

y = a0 + a1 x + a2 x 2 + a3 x 3 – It passes through any three chosen points at equally spaced values of x.

1 I = average height x width I = ( y1 + 4 y2 + y3 )( x3 x1 ) 6 – If the range of integration is subdivided into equal intervals by using any odd number of ordinates (say 7), Simpson’s rule can be applied to each group of three points and the result of all integrations added together. 1 I = ( y1 + 4 y2 + 2 y3 + 4 y4 + 2 y5 + 4 y6 + y7 )( x7 x1 ) 18 Gauss’s method

It enables a polynomial of degree (2n-1) to be fitted to n points. I = ∫ ydx = 2 yaverage 1 1 1 -1 1

If the values of y at the three positions indicated by the green arrows can be determined, the shape of the quintic curve can also be determined. (n = 3) Two-points method: y1 and y2 are evaluated at x = ± 0.5773

∫ ydx = y + y Four-points method: y1, y2, y3 and y4 are evaluated at x = -0.8611, 0.3400, 0.3400, 0.8611 I = ∫ ydx = 0 . 3478 ( y + y ) + 0 . 6522 ( y + y ) Homework 1 1 2 1 1 1 4 2 3 I = 1

Summary of today’s course

無因次群的目的在於比較「相似」系統的實驗數據。 無因次群的目的在於比較「相似」系統的實驗數據。 避免無因次群對有因次群作圖 無因次群對有因次群作圖。 避免無因次群對有因次群作圖。 應用不同的繪圖紙，以利繪出直線關係式。 應用不同的繪圖紙，以利繪出直線關係式。 三角座標圖可用於顯示溶劑中三種不同溶質的濃度。 三角座標圖可用於顯

示溶劑中三種不同溶質的濃度。 實驗誤差隨著之後的運算而擴大。 實驗誤差隨著之後的運算而擴大。 可用平均法及最小平方法。 實驗值的 fitting 可用平均法及最小平方法。後者較常 使用。 使用。 數值積分可用 trapezium, Simpson’s, Gauss’s 等三個方 法。 – 似線性關係時，多用 trapezium 方法。 1

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