Transcript or 0.089 cal/(cm 2 -sec) - West Virginia University

tom.h.wilson
[email protected]
Dept. Geology and Geography
West Virginia University
Tom Wilson, Department of Geology and Geography
Any questions about integration problems 1-4?
Tom Wilson, Department of Geology and Geography
Let’s continue on with these (the last set of
general integrals for you to solve)
Tom Wilson, Department of Geology and Geography
Another set
Tom Wilson, Department of Geology and Geography
Last set
Tom Wilson, Department of Geology and Geography
A brief introduction to integration by
substitution
Take a look at this one for a minute.
24
3
3
t
8

3
t
dt

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.
Tom Wilson, Department of Geology and Geography
24
3
3
t
8

3
t
dt In this problem, we can see that

d  8  3t 3 
dt
 9t 2
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
u  8  3t 3 , where
du
 9t 3 and du  9t 3dt
dt
Tom Wilson, Department of Geology and Geography
We also see that
du
 3t 3 dt
3
So we substitute and redefine the integral
u  8  3t
3
into
du
 3t 3 dt
3
24
3
3
t
8

3
t
dt

to get
Tom Wilson, Department of Geology and Geography
u
1
4
du 1 1 4
  u du
3 3
1 14
Now you have the much simpler integral  u du
3
to evaluate.
You just need to use the
power rule on this.
What would you get?
1 4 54
 u
3 5
Substitute the expression
defined as u(t) back in
Tom Wilson, Department of Geology and Geography
Our result …
5
4
3
4
8

3
t


15
Integration by substitution helps structure
the process of finding a solution.
Differentiate to verify.
Tom Wilson, Department of Geology and Geography
Try another one
xdx
 (7 x 2  3)5 
Let u =
Tom Wilson, Department of Geology and Geography
(7 x 2  3)
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?
This problem is solved using a simple equation
referred to as the heat conduction equation.
T
qx   K
x
See http://serc.carleton.edu/NAGTWorkshops/geophysics/activities/18913.html for additional discussion
Tom Wilson, Department of Geology and Geography
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-lbs
1 calorie/sec = 0.004284 kW
=4.1868 Watts
Tom Wilson, Department of Geology and Geography
Relating to the units
or =3.086 ft-lbs/second
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.
Tom Wilson, Department of Geology and Geography
To solve this problem, we use the heat
conduction equation: qx=-KT/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
Tom Wilson, Department of Geology and Geography
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!
Tom Wilson, Department of Geology and Geography
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
X=0 km
X=40 km
T  0.5 x  30 x  10
2
Tom Wilson, Department of Geology and Geography
o
What is the derivative
Using the conduction equation in differential form
dT
qx   K
dx
You see you have to take a derivative
to determine heat flow.
Tom Wilson, Department of Geology and Geography
Calculate the temperature gradient
dT d
 0.5 x 2  30 x  10 

dx dx
3
o
Given K 6 x10 cal / (cm  sec C )
106 cal / (cm2  sec)
Calculate qx at x=0 and 40km.
dT
qx   K
dx
Tom Wilson, Department of Geology and Geography
and that 1 heat flow unit =
and substitute for x to get q
You will get qx=-1.8 hfu at x=0
X=40 km
X=0 km
and qx=0.6 hfu at x=40
Heat flows out both ends of the sill.
Tom Wilson, Department of Geology and Geography
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?
heat generation rate ~  M E
Typical radiogenic heat production  for granite is
~2x10-13 cal/(gm-sec) and that for basalt – about 2x1014. 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?
Tom Wilson, Department of Geology and Geography
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.
Tom Wilson, Department of Geology and Geography
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).
Tom Wilson, Department of Geology and Geography
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
y kW
Q
20 km3
Also review total natural strain discussion and
the integration of discontinuous functions.
Tom Wilson, Department of Geology and Geography
Return to example 9.7 where your task is to find the
cross sectional area of the sand body
Tom Wilson, Department of Geology and Geography
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?
Tom Wilson, Department of Geology and Geography
In this exercise we repeat Wlatham’s analysis
using a 5th order polynomial
  ax
x
0
5
 bx  cx  dx  ex  f dx 
4
3
2
6
5
x
ax
bx
cx
What
is
this


....
integral?
6
5
4
0
For the 5th order polynomial you derive
you’ll have 6 terms including the constant
Tom Wilson, Department of Geology and Geography
4
Recall computation set up using the 2nd order
polynomial as an example
  ax
x
0
2
 bx  c dx 
x
ax3 bx 2 cx


3
2
1 0
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
Tom Wilson, Department of Geology and Geography
6
5
4
3
2
x
ax bx cx dx ex




 fx
6
5
4
3
2
0
For the 5th order polynomial you derive
you’ll have 6 terms.
(Upper limit of x)6
(Upper limit of x)5
Etc.
Enter
values
here
Tom Wilson, Department of Geology and Geography
Enter
values
here
6
5
4
3
2
x
ax bx cx dx ex




 fx
6
5
4
3
2
0
For the 5th order polynomial you derive
you’ll have 6 terms.
Tom Wilson, Department of Geology and Geography
Problem will be due next Tuesday
We’ll give you an
extra day on this, so
bring questions to
class on Thursday.
Tom Wilson, Department of Geology and Geography
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