tom wilson, department of geology and geography tom.h.wilson [email protected] dept. geology...

Tom Wilson, Department of Geology and Geography

[email protected].

eduDept. Geology and

GeographyWest Virginia University

Any questions about integration problems 1-4?


Let’s continue on with these (the last set of general integrals for you to solve)


Another set


Last set


A brief introduction to integration by substitution


2 343 8 3t t dt

Take a look at this one for a minute.

A lot of the problems on the in-class worksheet are similar to this. To simplify these kinds of problems, you look for something that, when differentiated, supplies another term in the integrand.


In this problem, we can see that

3

28 3

9d t

tdt

which within a factor of 1/3rd equals the leading term in the

integrand – the 3t2. So the idea is that we define a new variable

3

3 3

8 3 , where

9 and 9

u t

dut du t dt

dt

2 343 8 3t t dt


2 343 8 3t t dt

We also see that

333

dut dt

So we substitute and redefine the integral 33

3

dut dt38 3u t

into

to get 1 14 4

1

3 3

duu u du


14

1

3u duNow you have the much simpler

integral to evaluate.You just need to use the

power rule on this.

What would you get? 5

41 4

3 5u

Substitute the expression defined as

u(t) back in


5

3 448 3

15t

Our result …

Integration by substitution helps structure the process of finding a

solution.

Differentiate to verify.

Try another one


2 5(7 3)

xdx

x

Let u = 2(7 3)x

Let’s consider some heat flow problems as a companion discussion to the example in Chapter 9


Consider heat conduction through a thick glass window given two possible inside temperatures 65oF and 72.2oF and an outside temperature of 32oF. In terms of degrees C this corresponds to temperatures of 18.33OC, 22.33oC and 0oC. How much energy do you save?

See http://serc.carleton.edu/NAGTWorkshops/geophysics/activities/18913.html for additional discussion

This problem is solved using a simple equation referred to as the heat conduction equation.

x

Tq K

x


We consider this problem in terms of the heat flow over the course of the day, where heat flow (qx) is expressed in various units representing heat per unit area per time: for example, calories/(m2-s).

A qx of 1 cal/(cm2-s)=41.67 kW/m2

kW-s=737.5622 ft-lbs1 calorie/sec = 0.004284 kW

=4.1868 Watts

Relating to the units


If you lift about 3 pounds one foot in one second, then you’ve expended 1 calorie (thermomechanical) of energy.

Nutritional calories are about 1000 thermomechanical calories. So you would have burnt only 1/1000th a nutritional calorie!

You can expend 1 nutritional calorie by carrying 100 lbs up 30 feet.

or =3.086 ft-lbs/second

To solve this problem, we use the heat conduction equation: qx=-KT/x


K (thermal conductivity) 2x10-3cal/(cm-sec-oC)

Assume x=0.5cm

Then qx=0.07 cal/(cm2-sec) or 0.089 cal/(cm2-sec)

If the window has an area of 2m2

Then the net heat flowing across the window is 733 or 896 cal/sec

The lower temperature saves you 163 cal/sec

80,000 cal/sec


163 cal/sec corresponds to 1.41x107 cal/day

There are 860420.650 cal per kWh so that this corresponds to about 16.3 kWh/day. A kWh goes for about 6.64 cents so $1.08/day.

note this estimate depends on an accurate estimate of K (thermal conductiviry). Other values

are possible, but, as you can see, it can add up!

The second question concerns a hot sill


A hot sill intruded during Mesozoic time is now characterized by temperature from east-to-west that varies as

20.5 30 10oT x x

X=0 kmX=40 km

What is the derivative


x

dTq K

dx

You see you have to take a derivative to determine heat flow.

Using the conduction equation in differential form

Calculate the temperature gradient


20.5 30 10dT d

x xdx dx

Given K 36 10 / ( sec )ox cal cm C and that 1 heat flow unit = 6 210 / ( sec)cal cm Calculate qx at x=0 and 40km.

x

dTq K

dx

and substitute for x to get q


You will get qx=-1.8 hfu at x=0

and qx=0.6 hfu at x=40

X=0 kmX=40 km

Heat flows out both ends of the sill.

Another simple example : assume the mantle and core have the same heat production rate as the crust


What is the heat flow produced by the Earth in this case?Is it a good assumption?

Typical radiogenic heat production for granite is ~2x10-13 cal/(gm-sec) and that for basalt – about 2x10-

14. We use an average of about 1x10-13 for this problem. Given that the mass of the Earth is about 6x1027 grams we get a heat generation rate of about 6.6x1014 cal/sec.

What is the heat flow per cm2?

~ Eheat generation rate M

Heat flow per unit area …


To answer that, we need the total area of the Earth’s surface in cm2.

The surface area of the Earth is about 5.1 x 10 18 cm2. which gives us a heat generation rate /cm2 of about 12.9 x10-5 cal/(cm2-sec) or 129 hfu.

The global average heat flow is about 1.5 hfu.

We would have to conclude that the earth does not get much radiogenic heating from the mantle and core.

For Thursday


Review problem 9.8.

In this problem, we assume that heat is generated in the crust at the rate of 1kW/km3 and that heat generation is confined to the crust (we’ve confirmed this is a pretty good assumption).

Radiogenic heating decreases with depth until, as we suspected, below the crust there is very little heat generated through radioactive decay (the thermal gradient or direction of heat flow is from hot to cold or vertically upward).

Problem 9.8


Heat generation rate in this problem is defined as a function distance from the base of the crust. Waltham uses y for this variable and expresses heat generation rate (Q) as

320

y kWQ

km

Also review total natural strain discussion and the integration of discontinuous functions.

Return to example 9.7 where your task is to find the cross sectional area of the sand body


Waltham presents the results from a 4th order polynomial approximation of the sand bar thickness


t = -2.857E-12x4 + 1.303E-08x3 - 2.173E-05x2 + 1.423E-02x - 7.784E-02

Recall how to compute the cross sectional area?

In this exercise we repeat Wlatham’s analysis using a 5th order polynomial


5 4 3 2

0

xax bx cx dx ex f dx

For the 5th order polynomial you derive you’ll have 6 terms including the constant

What is this integral?

6 5 4

0

....6 5 4

xax bx cx

Recall computation set up using the 2nd order polynomial as an example


3 2

03 2 1

xax bx cx

2

0

xax bx c dx

The integral has factors a/3, b/2 and c.

The values a, b, and c come from the best-fit process. You already know what those numbers are from the trendline.

You also know the limits of integration:

0 to 2000meters


For the 5th order polynomial you derive you’ll have 6 terms.

6 5 4 3 2

06 5 4 3 2

xax bx cx dx ex

fx

Enter values here

Enter values here

(Upper limit of x)6

(Upper limit of x)5

Etc.


For the 5th order polynomial you derive you’ll have 6 terms.

6 5 4 3 2

06 5 4 3 2

xax bx cx dx ex

fx

Problem will be due next Tuesday


We’ll give you an extra day on this, so bring questions

to class on Thursday.


1. Hand the integral worksheets in before leaving or put in my mailbox sometime today.

2. Finish up problem 9.7 for next Tuesday

3. We’ll review other problems in the text this Thursday (e.g. 9.8 & discussions of topics illustrated in figures 9.4, 9.5 & 9.6).

4. Also look over problems 9.9 and 9.10 and bring questions to class this Thursday.

tom wilson, department of geology and geography tom.h.wilson [email protected] dept. geology...

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