Transcript and total natural strain - IIS Windows Server

tom.h.wilson
[email protected]
Dept. Geology and Geography
West Virginia University
Tom Wilson, Department of Geology and Geography
  ax
x
0
5
 bx4  cx3  dx2  ex  f dx 
For the 5th order polynomial you derive
you’ll have 6 terms including the constant
Tom Wilson, Department of Geology and Geography
x
6
5
4
ax
What bx
is this cx


....
integral?
6
5
4
0
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
Last set
Tom Wilson, Department of Geology and Geography
Another example of integration by substitution
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
Power = energy/time
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 6.0x1014 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
117 x10-6 cal/(cm2-sec) or 117 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
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
z kW Where z is the distance in km from
Q
20 km3
the base of the Earth’s crust
i. Determine the heat generation rate at 0,
10, 20, and 30 km from the base of the
crust
0 kW
kW
0 3
20 km
km
10 kW
kW

0.5
20 km3
km3
ii. What is the heat generated in a
in a small box-shaped volume z
thick and 1km x 1km surface? Since
V  Az
Tom Wilson, Department of Geology and Geography
V  zkm3
The heat generated will be
QV  Qz kW
This is a differential quantity so
there is no need to integrate
iii & iv. Heat generated in the vertical
column
n
In this case the sum extends
q   Qi z
over a large range of z, so
i 1
However, integration is
the way to go.
Tom Wilson, Department of Geology and Geography
q
z
0

z
dz
20
1 z
zdz
20 0
v. Determining the flow rate at the
surface would require evaluation of
z
the definite integral q  0 Qdz
30
z2
q
 22.5kW
40 0
vi. To generate 100MW of power
100, 000kW
 4444km 2
22.5 kW 2
km
An area about 67km on a side
Tom Wilson, Department of Geology and Geography
Li
e
L
, the elongation
Li
Lf
S  1  e, the stretch
The total natural strain, , is the
sum of an infinite number of
infinitely small extensions
In our example,
this gives us the
definite integral
 
Lf
Li
 ln L L f
L
i
 ln( L f )  ln( Li )
L 
 ln  f 
 Li 
Tom Wilson, Department of Geology and Geography
dL
L
  ln( S )
Where S is the Stretch
Strain (or elongation) (e), stretch (S) and total natural strain ()
e
Elongation
Total natural strain
 
Lf
Li
 expressed as a series
expansion of ln(1+e)
L L f  Li L f


1  S 1
Li
Li
Li
dL
L
e 2 e3 e 4
  e     ...
2 3 4
The six term approximation is
accurate out to 5 decimal places!
Tom Wilson, Department of Geology and Geography
  ln( S )  ln(1  e)
Comparison of finite elongation vs. total natural strain
Tom Wilson, Department of Geology and Geography
Integrating discontinuous functions
We can simplify the problem and still obtain a useful result.
Approximate the average densities
11,000
kg/m3
4,500
kg/m3
N
M   Area of sphere x r x density of shell
i 1
Area of sphere x r = volume of shell
N
M   4 ri 2 i r
i 1
Tom Wilson, Department of Geology and Geography
R
M   4 r 2  dr
0
We can simplify the problem and still obtain a useful result.
Approximate the average densities
11,000
kg/m3
4,500
kg/m3
R
M   4 r 2  dr
0
4 3
 r  C
3

3480
0
4 r 2 .11000dr  
6371
3480
3480
4
  r 3 1
3
0
3480
44000 3

r
3
0
Tom Wilson, Department of Geology and Geography
4 r 2 .4500dr
6371
4
  r 3 2
3
3480
6371
18000 3

 r 2
3
3480
The result – 6.02 x
1024kg is close to the
generally accepted
value of 5.97 x 1024kg.
Volume of the earth – an oblate spheroid
  2 
r  r 1   z 2  
  rp  
2
2
e
In this equation r varies from re, at the
equator, to r=0 at the poles. z represents
distance along the earth’s rotation axis
and varies from –rp to rp.
The equatorial radius is given as 6378km
and the polar radius, as 6457km.
Tom Wilson, Department of Geology and Geography
Problem 9.10
t  t0 exp( x / X )
In this problem, we return to the
thickness/distance relationship for the
bottomset bed. Have a look before next
Tuesday.
Problems 9.9 and 9.10 will be due next
Tuesday after next
Tom Wilson, Department of Geology and Geography
Consider this simple problem (turn in next time)
A gravity meter measures the vertical component
of the gravitational field .
Assume that you are searching for a sulphide
deposit of roughly spherical dimensions buried at a
depth z beneath the surface (see figure next page).
Tom Wilson, Department of Geology and Geography
Gravitational acceleration can vary in response
to subsurface density contrasts
GM
g 2
r
From Newton’s universal law of gravitation
gv 
g
x

r
z
Sulfide
deposit
Tom Wilson, Department of Geology and Geography
We will develop a solution as follows:
•Express r in terms of x and z
• Rewrite g in terms of x and z
• Express gv in terms of g and 
• Replace the trig function in the above with its
spatial equivalent and rewrite gv.
• Factor z out of the denominator to obtain
Tom Wilson, Department of Geology and Geography
The vertical component of the acceleration due
to gravity of the sulphide deposit
gv 
GM

2 x
z  2  1
z

2
3
2
Given that G=6.6732 x 10-11nt-m2/kg2, x=1km,
z=1.7km, Rdeposit=0.5km and =2gm/cm3,
calculate gv.
Tom Wilson, Department of Geology and Geography
Spatial variation in the gravity anomaly over
the sulphide deposit
Note than anomaly is
symmetrical across the
sulphide accumulation
Tom Wilson, Department of Geology and Geography
Start reviewing materials for the final!
… Current to-do list
1. Finish up problem 9.7 for next Tuesday
2. Finish up the gravity computation for
Tuesday
3. We’ll follow up with questions on 9.9 and
9.10 in class next Tuesday
4. Start reviewing class materials. The
following week will be a final review week
Tom Wilson, Department of Geology and Geography