Saturday, March 21, 2020

Dimensional Analysis Revisited


The use of dimensions or units in physics may be used to check the homogeneity of physical equations, or obtain a useful empirical equation from basic measurements, observations.

Example:

Consider a volume V of a liquid flowing per second through a pipe under steady pressure as shown below:
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Fig. 1: The fluid flow through a uniform pipe


It is reasonable to assume, in the first instance, that V (volume flow per unit time) depends upon:

a)     The coefficient of viscosity, h, of  the liquid
b)     The radius, r, of the pipe
c)      DP/L the pressure gradient arising from the flo

Then we may write: V = k hx ry (DP/L)z


Where k denotes a constant, and x, y and z are indices whose values are to be found.

We know:

V has dimensions L3 T - 1

h has dimensions M L-1 T -1

r has dimension L

DP has dimensions M L-1 T - 2

So, DP/L has dimensions: M L-2 T - 2

Then, equating dimensions:

L3 T – 1  = [M L-1 T -1]x [L]y [M L-2 T - 2]z

We next equate indices for M, L, and T on both sides:

For M:  0 = x + z

For L: z = -x + y – 2z

For T: -1 = -x – 2z

Next, solve for the indices:

a)     x = -z
b)     -1 = -(-z)- 2z = -z or z = 1
c)      -1 = -x -2z
d)     3 = 1 + y -2 or y = 4

Finally:  V = k h- 1 r 4 (DP/L)1

Or: V =  k DP r 4 /h L


We note here that k cannot be obtained by the method of dimensions, but a fuller analysis would reveal k = p/8, so:

V = p DP r 4 /8 h L    (Poiselle’s formula)

Problems:

1)      Use the method of dimensions to obtain a formula for the force experienced by a sphere of radius r moving at velocity v through a fluid of viscosity h.  (Hint: Set out the force as F = k hx v y r )

2) Repeat the exercise above to obtain the period (T) for a pendulum’s swing.


  1. Errors & Uncertainties in Measurements

In general, there are two types of errors with which we will deal in physics:

1) Systematic: an error repeated due to a fault in the instrument of the procedure.

2)     Random: an error which arises due to intrusion of some random factor, or the necessity to estimate and the limits of instrumental sensitivity.

An illustration of a systematic error is illustrated below for an ammeter.  We observe that in this case, the meter is incorrectly calibrated since the real zero is displaced from the value read by the ammeter, by and amount 0.2 A. Thus when the meter reads 1.2 A it is actually 1.0 A.
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Fig. 2: Improperly calibrated ammeter.

In this case, X = 1.0 A represents the “true value” for the measurement, and the reading (x) is registered as:

x = X + e

Then:  + e  =  x – X = 1.2A – 1.0A =  + 0.2A

The nature of a random error is illustrated in Fig. 3.:
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Fig. 3: A random error arising from multiple deviating readings.

In this case, reading of the ammeter yields a series of slightly different readings at different times, say: x1, x2, x3, and x4. In this case, x denotes the “true value” and we have:

x1 = x – c1

x2 = x – c2

x3 = x – c3

where c1, c2, and c3 are “correction factors” of the observed value to the true value, x. In such an instance, we generally reduce the effects of these random different values by taking the average value x by:

x = (x1 + x2 + x3 +…..xn)/ n

Which should be closer to the true value x, than any one of the other observed values.

In general the associated error will be:

e = (x – x )

So the fractional error is:  f = e/ x

And the percentage error is: p = (e/x) x 100%

Usually: f < < 1 and p < < 100%

Example:

As illustrated below, a measurement of 2.5 cm is made using a scale rule graduated in millimeters (mm or 0.1 cm) What would the possible error (p.e.) be in this case, and how would we write the length of the object?
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Fig. 4.: Example illustrating possible error of an object

The possible error in this case is the equivalent of the limit of accuracy for the scale or 0.1 cm. Then:

p.e. =  + 0.1 cm

The length of the object is written: L = 2.5 + 0.1 cm

The percentage possible error (p.p.e.) is:

(+ 0.1 cm/ 2.5 cm) x 100% =  + 4%

Example (2):

For the same measurement a vernier caliper may find:

L = 2.46 + 0.01 cm

Since its accuracy is a factor 10 enhanced over the scale.

The possible error is then: p.e. = + 0.01 cm

And:  the percentage possible error is:

p.p.e. = (+ 0.01 cm/ 2.46 cm) x 100% =  + 0.4%


 3. Combining Errors of Measurements

It is possible and indeed, frequently necessary, to combine errors. We consider several cases below:

Case (1): The error sum Q = a + b

Then: The total p.e. in Q = p.e. in a + p.e. in b

(Also, for absolute errors: DQ = Da + D b)

Let: a = 5.1 + 0.1 cm  and b = 3.2 + 0.1 cm 

Then Q = (5.1 + 0.1 cm ) + (3.2 + 0.1 cm ) = 8.3 + 0.2 cm 

In addition:

Q max = (5.2 + 3.3) cm = 8.5 cm,   and

Q min = (5.0 + 3.1) cm = 8.1 cm

Case (2): The error difference Q = a – b

If Q = a – b then the same rule applies, in terms of addition of possible errors. Also: Da =  D b  = + e)

Case (3) Product or quotients and fractional errors.

If a quantity x is measured and recorded as x’ units, then let: (x’ – x) denote the error in x or x’ = x + e. But we may also write:

x’ = x(1 + f)

where f = e/x is the fractional error in x. Note also that: (x’/ x) = 1 + f.

For a sum or difference: Q = a + b = (a + e1) + (b + e2)

And: f = (e1 + e2)/ (a + b)

Also:

f1 = e1/ a and f2 = e2/b

Therefore: f = f1 + f2

Or f = (af1/ a) + bf2/b)

Error in a product, viz, Q = ab

Then: Q = a (1 + f1) x b (1 + f2)

Or: Q = ab(1 + f1 + f2)

Error in a quotient, viz. Q = a/b:

Then: Q = a(1 + f1)/ b(1 + f2)

Q = a/b [ (1 + f1)(1 – f2 + …) = a/b [1 + f1 – f2]


Problems for the ambitious reader :

1) Two lengths ℓ1 and ℓ2 are measured as (10.0 + 0.1 cm ) and (9.0+ 0.1 cm ).  Find the p.e. for each length. Express the sum (ℓ1 + ℓ2) with errors-uncertainties. Thence or otherwise obtain the fractional error derived from the sum.

2) The length of a sheet of paper is measured as: ℓ = 11.0 + 0.1 cm, and its width as w =  8.5 + 0.1 cm. Find the Area A = ℓ x w in terms of the associated possible errors. Find the fractional error associated with the area.

3) A glass thermometer is scaled in tenths of a degree. If a student reads the lab temperature as 30.5 o C, express the possible error in the measurement.

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