Applied Mathematics for Engineering

1. The Spherical Cow

A physicist, an engineer, and a psychologist are called in as consultants to a dairy farm whose production has been below par. Each is given time to inspect the details of the operation before making a report.

The size of the stalls for the cattle should be decreased. Efficiency could be improved if the cows were more closely packed, with a net allotment of 275 cubic feet per cow. Also, the diameter of the milking tubes should be increased by 4 percent to allow for a greater average flow rate during the milking periods.

— The engineer

The inside of the barn should be painted green. This is a more mellow color than brown and should help induce greater milk flow. Also, more trees should be planted in the fields to add diversity to the scenery for the cattle during grazing, to reduce boredom.

— The psychologist

Assume the cow is a sphere …

— The physicist

Before doing anything else, abstract out all irrelevant details! (Krauss, 2007)

Before attempting a question, make sure you understand the fundamental concept. Draw sketches where necessary. Make sure you derive your equations correctly, then substitute the values. This is useful in preparing a general algorithm and reducing approximation errors.
Your answers should be in SI units and scientific notation. Use the following form \(p.qrs \times 10^t\) units. Where \(p,q,r,s\) is a number between \(0-9\), \(t\) is a signed integer, and units is the relevant SI units. This is useful in establishing a quick objective meaning of size/scale.

Example 1

Given that the weight of a normal cow is \(x\), what is the weight of a super cow thrice as big?

If I wanted to make leather shoes out of the super cow’s hide, how much more hide would it yield than the normal cow?

In comparison to the normal cow, how much pressure is the super cow’s skin experiencing?

Figure 1. From: the brutal realities of leather.

Solution:

Let the weight of the normal cow \(x = W_{nc} = mg\)

\(W_{nc} = mg\)
\(m = \rho \times v\)

A cow has an irregular shape that would be complicated to calculate its volume. For simplicity, let’s approximate that shape into a sphere.

\(v = \frac{4\pi}{3} r^3\)
\(W_{nc} = \rho \times \frac{4\pi}{3} r^3 \times g\)

The weight of a super cow, \(W_{sc}\), thrice as big:

\(W_{sc} = \rho \times \frac{4\pi}{3} (3r)^3 \times g\)
\(W_{sc} = 27 \times \rho \times \frac{4\pi}{3} r^3 \times g\)

A super cow thrice as big as the normal cow would weigh 27 times more than the normal cow.

Having formulated the necessary mathematical equations that describe the problem at hand, we can translate the equations into a computer program. We shall use Python programming. Install python to your computer using Anaconda https://docs.anaconda.com/anaconda/install/.

Solution:

 1# import the libraries you intend to use
 2import numpy as np
 3
 4# translate the main equation into a function
 5def cow_weight(r):
 6    cw = rho * ((4*np.pi)/3)*r**3 * g
 7    return cw
 8
 9# define the variables
10rho = 1850 # kg/m^3 bone?[https://physics.nist.gov/cgi-bin/Star/compos.pl?matno=119] (assumed constant)
11g = 9.8 # m/m^2 (assumed constant) 
12r1 = 0.6 # m (normal cow radius)
13r2 = 3*r1 # m (super cow radius - thrice)
14
15# get answers to your questions
16Wnc = cow_weight(r1)
17Wsc = cow_weight(r2)
18ratio = Wsc/Wnc
19
20# print answers (IN SCIENTIFIC NOTATION!) 
21print("Normal Cow Weight ~\t %.2E kg.m/s^2" %Wnc) # (= N) confirm units
22print("Super Cow Weight ~\t %.2E N" %Wsc)
23print("Super Cow Weighs ~%.2E more than the Normal Cow" %ratio)

Normal Cow Weight ~	 1.64E+04 kg.m/s^2
Super Cow Weight ~	 4.43E+05 N
Super Cow Weighs ~2.70E+01 more than the Normal Cow

Scientific notation gives you a better visualization of size. Following the same method and taking the hide surface area to be \(A = 4\pi r^2\), and exerted pressure to be \(P = \frac{F}{A}\), we can deduce that a cow thrice as big weighs 27 times as much and holds 9 times as much skin and 3 times as much pressure due to its weight. Here, the critical scaling factor is the radius.

1.1. Modelling Burning Candles

Example 2

Formulate general mathematical equations to predict how long it takes for candles to burn out.

Figure 2. From: which candle will burn out first?

Solution:

Modelling a burning candle requires knowledge of the chemistry of combustion and fluid dynamics. However, we can use simple mathematical reasoning to arrive at a gain useful preliminary insights about this problem.

First, identify key variables of interest while neglecting other parameters that do not play a key role or do not affect the question at hand. This is the art of applying mathematics to formulate approximate models about a certain natural phenomenon.

Many variables could play a role: candle diameter, the width of the wick, composition of the candle material, atmospheric pressure, ambient temperature, and so on.

Let us make the following assumptions.

Assume the candles are placed at the same atmospheric pressure and ambient temperature. Each candle is cylindrical with a radius, \(R\), and height, \(H\).
The candles are made from the same wax material. The chemical energy density, \(E\) (energy per unit volume), is independent of the wax size and type.
The volume of a candle is hence \(\pi R^2 H\), say, \(cm^3\), and contains \(E \pi R^2 H\) joules to convert to heat and light.
All candles have the same wick and dissipate the same amount of energy per unit of time, \(T\). How fast energy is lost, \(P\), is independent of the height of the candle.
Let \(T\) be the time taken for a candle of size \(H\) to be consumed. In time \(T\), the candle consumes \(E H \pi R^2\) joules and releases \(PT\) joules.

\[PT = E H \pi R^2\]

Since a candle of height \(H\) burns in a time \(T\), the flame velocity, \(v = \frac{H}{T}\).

\[\begin{align*} v &= \frac{H}{T} = \frac{P}{\pi E} \times \frac{1}{R^2} \\ v &\sim \frac{1}{R^2} \end{align*}\]

From the above equation, we can extract the scaling information. The flame velocity depends on the inverse of the radius squared, which implies that a candle of twice the diameter would burn four times slower.

To test this prediction, we can set up candles of different sizes and measure their radius and flame velocity.

Figure 3. Candle experiments (Goriely, 2018)

From Figure 3, we can see that the theoretical prediction using simple mathematical reasoning, \(v \sim \frac{1}{R^2}\) is useful. However, the theoretical prediction seems to agree more with wider candles but not so much with thinner ones. Be that as it may, this model provides us with a basic law that we can test and improve.

Another way to model a burning candle is to consider the composition of a candle as going from solid to gas. The solid is given a value \(u = 1\) and the gas a value \(u = 0\) see Figure 4.

Figure 4. Candle burning model (Goriely, 2018)

The propagation of the flame is described by a variable \(u\) as a function of both position and time. In the extreme case, we have \(u = 1\) for the solid part, and \(u = 0\) in the gas part where the solid part is consumed. A differential equation for \(u\) can later be found using the fundamental principles of combustion and energy release.

1.2. Modelling Cooking Time

Example 3

A food supplier package instructs you to cook a \(5 kg\) turkey for \(2\frac{1}{2} hrs\) and a \(10 kg\) turkey for \(4 hrs\). Using applied mathematics techniques, come up with equations that could help us to predict the time required to cook, say, an ostrich, with a typical weight varying between \(60 kg\) and \(150 kg\).

Solution:

As seen in Figure 5 this problem could be solved by carrying out an empirical experiment. Be that as it may, we can use simple mathematical reasoning to develop useful equations.

Figure 5. Cooking time vs. mass of various poultry in a log plot. (Goriely, 2018)

To develop a mathematical model for this problem, we can start by asking: how does the cooking time, at a given fixed temperature depend on weight? If I know the cooking time for a \(5 kg\), turkey, what would be the cooking time for a \(10 kg\) or \(100 kg\) bird?

As usual, the first step is to start with useful parameters while making some reasonable assumptions. Again, this is the art of modelling.

There are many factors that we could consider. The age of the bird, its size, corn-fed, free range, and so on. If there is any hope of progress, we may have to make stringent simplifications. The resultant equations should capture the basic system. These basic equations can continually be improved to continue to get closer to the real phenomenon.

Let’s make the following assumption.

The bird is a sphere of radius \(R\) with mass \(M\) and uniform constant density \(\rho\)
The bird has an initial temperature, \(T_i\) and it is cooked at a constant oven temperature \(T_o\). That is, the surface temperature at the surface of the sphere is maintained at a constant \(T_o\).
The transfer of heat in the bird is described by a unique constant coefficient of thermal diffusivity, \(k\). This coefficient describes the ability of a material to conduct heat. A larger \(k\) means that the bird cooks faster compared to a bird with a smaller value. This \(k\) has dimensions of length squared divided by time.
The density \(\rho\) is the same for all birds. The mass \(M = \rho \times 4 \pi \frac{R^3}{3}\).

Heat is a physical process and these assumptions are based on the physical theory of heat transfer. We have identified the key variables as \(R\), \(M\), \(T_i\), \(T_o\), and \(k\). In the spirit of the spherical cow assumption, we have replaced our bird with a sphere of homogeneous material subject to a constant temperature at the surface. Surprisingly, in many cases, these simplifications provide excellent results.

Let the cooking time \(t_M\) be the earliest time at which a safe temperature is reached at all points of the mass \(M\). This time is a function of the following variables.

\[\begin{align*} t_M = f(R, T_i, T_o, k) \end{align*}\]

Using dimensional analysis, we notice that the only variable in the function that contains the dimension of time is the thermal diffusivity \(k\). Since we are interested in the time needed to cook the bird, any time in the function must be related to thermal diffusivity through its inverse.

\[\begin{align*} \frac{1}{k} &\sim \frac{t}{L^2} \\ t_M &= \frac{1}{k} f(R, T_i, T_o) \end{align*}\]

Since the quantity \(t_M\) on the left has the dimension of time, so must the quantity on the right side. So the function must have dimensions of length squared to cancel the length dependence of the thermal diffusivity. The only length available is \(R\).

\[\begin{align*} t_M = \frac{R^2}{k} f(T_i/ T_o) \end{align*}\]

The remaining unknown function \(f\) only depends on temperature and has to be dimensionless. This dimensionless number is independent of the mass of the bird.

From here we learn that if we hold constant \(k\), cooking oven, and \(T_i\), the cooking time increases with the square of the radius. But at fixed \(\rho\), the radius varies as the cube root of the mass.

\[\begin{align*} R = \left( \frac{3}{4 \pi} \times \frac{M}{\rho} \right)^{\frac{1}{3}} \end{align*}\]

We can rewrite \(t_M\) as follows.

\[\begin{align*} t_M &= C \times M^{\frac{2}{3}} \\ C &= \frac{1}{k} \left( \frac{3}{4 \pi} \times \frac{M}{\rho} \right)^{\frac{2}{3}} \times h(T_i/T_o) \end{align*}\]

According to our assumptions, the density, \(\rho\) is constant and so are the temperatures, \(T_i\) and \(T_o\). Therefore, \(C\) is constant in our model. For a \(5 kg\) and a \(10 kg\) turkey,

\[\begin{align*} t_5 &= C \times 5^{\frac{2}{3}} \\ t_{10} &= C \times 10^{\frac{2}{3}} \\ \frac{t_{10}}{t_5} &= 2^{2/3} \\ t_{10} &= 2^{2/3} t_5 \end{align*}\]

If we know that the cooking time of a \(5 kg\) turkey is about \(2 \frac{1}{2} hrs\), the cooking time \(t_{10}\) of a \(10 kg\) turkey, \(t_{10} = 2^{2/3} t_5 \approx 4 hrs\).

Following the same reasoning, we learn that to cook \(100 kg\) ostrich, \(t_{100} = t_{10} 10^{2/3} \approx 4 \times 4.6 \approx 18\frac{1}{2} hrs\)

You can use the same reasoning to find a rough estimate of how long it would take to cook \(500 kg\) stuffed camel using the same oven.

1.3. Modelling Giants of Old

Example 4

Using simple mathematical reasoning, explain why a giant 12 times larger than a normal human could not exist. Apart from its size, the giant is in all aspects identical to humans.

Figure 6. From: Giants of Old (Nephilim)

Solution:

From a thermal point of view, we can think of a mammal as a turkey in an inverted oven, where the heat source would be placed inside the turkey. We lose heat through our body surface in the same way that turkeys acquire heat through their exposed surface.

Heat gain is proportional to their surface area. Smaller mammals lose more heat than larger mammals per unit mass because their area per unit mass is larger. The greater the surface area, the greater the potential heat gain or loss through it. Consequently, a small surface area to volume ratio implies minimum heat gain and heat loss.

\[\begin{align*} \text{Surface Area} &: \text{Volume} \\ 4 \pi R^2 &: \frac{4}{3} \pi R^3 \\ 1 &: \frac{1}{3} R \\ \text{Say, } \mathbf{small} \text{ animal } R &= 1 \\ \frac{\text{Surface Area}}{\text{Volume}} &= \frac{1}{\frac{1}{3}\times 1} = \mathbf{3} \\ \text{Say, } \mathbf{big} \text{ animal } R &= 12 \\ \frac{\text{Surface Area}}{\text{Volume}} &= \frac{1}{\frac{1}{3}\times 12} = \mathbf{0.25} \\ \end{align*}\]

Since heat is exchanged through the surface of an object, the giant would cool down to the room temperature \(12^2 = 144\) times faster than an average-sized human. Such a giant with human physiology would quickly die from hypothermia, especially in a cold climate and overheating in a warm climate.

This simple scaling argument explains why the smallest mammals, such as mice must have an extremely fast metabolism, compared to larger animals, to keep their body temperature constant.

Using a simple \(\frac{w}{l}\) comparison such as in Figure 7, in his 1638 Discoursi, Galileo correctly noted that the aspect ratio, \(\frac{w}{l}\), or bones changes with size.