tom.h.wilson Dept. Geology and Geography West Virginia University Tom Wilson, Department of Geology and Geography

Transcription

1 tom.h.wilson Dept. Geology and Geography West Virginia University

2 Graduation! Mark your calendars.

3 For the day Strain Integration of discontinuous functions Acceleration due to gravity (derivative and integral examples) Return to Excel problem 9.7 for review and final questions. This problem will be due next class.

4 L i L e, the elongation Li Lf S 1 e, the stretch = L i Positive or negative depending on relative difference of L = L f -L i. The total natural strain,, is the sum of an infinite number of infinitely small extensions L f In our example, this gives us the definite integral L L i f dl L ln L L L i f ln( L ) ln( L) Lf ln Li f i ln( S) Where S is the Stretch

5 A side bar on math relationships Strain (or elongation) (e), stretch (S) and total natural strain () Elongation L Lf Li Lf e 1 S1 L L L i i i Total natural strain L L i f dl L ln( S) ln(1 e) expressed as a series expansion of ln(1+e) 3 4 e e e e ln(1 e) The six term approximation is accurate out to 5 decimal places! 3 4 e e e e

6 Comparison of finite elongation vs. total natural strain Initial line has a length of 10 L f e

7 Some additional integration perspectives Section 9.7 Integrating discontinuous functions M E 0 r s () r dv where (r) is hard to define analytically. Approximate the average densities 11,000 kg/m 3 4,500 kg/m 3

8 Assume Earth s mass can be approximated by two major subdivisions Approximate the average densities 11,000 kg/m 3 4,500 kg/m 3 M shell or sphere 4r i V i i r 1) an inner core of average density, i, and ) an outer shell (mantle and crust) represented by another average density, o.

9 What do we get when we integrate the surface area over r? We can simplify the problem and still obtain a useful result. Approximate the average densities 4,500 kg/m 3 11,000 kg/m 3 What s the surface area of a sphere? M R 4 r dr 0 Mass of shell r thick = area of shell x density x r M 4r r shell i i Below we sum all shells r thick x density to get mass of the larger volume N M 4r r i1 i i

10 Actually a pretty good approximation We can simplify the problem and still obtain a useful result. Approximate the average densities 11,000 kg/m 3 4,500 kg/m 3 These are thick shells so we do have to integrate them r.11000dr 4 r.4500dr r r r r M R 4 r dr r C The result 6.0 x 10 4 kg is close to the generally accepted value of 5.97 x 10 4 kg.

11 Showing the evaluation of the definite integral Mass of the core Mass of the mantle r r ( ) (0) 6000 ( ) 6000 ( ) In the above, units of the radii are converted from km to m to be units consistent! M E ~ ( ) (0) 6000 ( ) 6000 ( ) M ~ E 1.94x x x x10 Mass of the core Mass of the mantle The approximation is very close to the accepted value of 5.97x10 4

12 We could use integration of discontinuous functions to simplify the computation of lithostatic pressure Simplify this problem by breaking it down into a series of constant density steps. Sv () z gdz v Could become S gz gz

13 How does the core contribute to the acceleration of gravity measured at the Earth s surface? Mcore r dr 4 3 r r Considering only the core, we find it s mass is x10 4 kg (about 1/3 rd the total mass of the earth. Mass of core ~ 1.94 x 10 4 kg. g (acceleration) is proportional to the mass of an object.

14 On the surface at ~6371km (6,371,000m) from the center of the earth, the acceleration g g GM core r For reference, the radius of the core is about twice that of the moon which has an average density of only 3.34 gm/cm 3 The core is about 900km beneath your feet, We have to keep units consistent and use G=6.673x10-11 m 3 /(kg-sec ) M= x10 4 kg And r=6,371,000 m The contribution to the total acceleration of ~9.8 m/s due to the core is g core ~3.19m/s.

15 What is g due to the mantle? What is the acceleration of gravity produced by the Earth s mantle? M mantle r dr g GM mantle Not surprisingly it turns out to be the remaining /3rds ~6.6 m/s We can show through triple integration over the volume of a shell that it gravitates as though all its mass is concentrated at the center of the shell or center of the Earth. That is why we can use the simple formula above to evaluate the acceleration due to gravity of the mantle shell. r

16 The crust? The mass of the crust is ~.8x10 kg (with an average density of about.8gm/cm 3 ). If we use an average radius of km, we find that the acceleration of gravity due to the crust alone is only 0.05m/s! g GM crust r On such an earth it would 63.5 seconds to fall 100 meters and you would land with a velocity of 3.16m/s. This corresponds to the velocity you would land jumping from a height of ½ meter.

17 Gradient in the acceleration of gravity at the Earth s surface dg dr d GM dr r What is this derivative? m/ s What is the rate of change of g with increased r in? m

18 More on volume integrals GM dm g G r r We usually look for some symmetry to help simplify our problem. For the sphere or spherical shell this integration simplifies to the foregoing formula.

19 Some details on the computation of the acceleration produced by a very long horizontal cylinder This could be a cave passage or tunnel. dm dv dxdydz G G G r r r Point of observation Surface r+dr r m dx

20 In this example, we can let the cross sectional area = dydz G dxdydz r G Adx r Again, we are interested in the vertical component of g, so G Adx cos r r+dr r Point of observation m dx

21 Zoom in on the little element dx r+dr r cos rd dx Area = R rd with cos =, we have dv rd rd dx & dv R cos cos R

22 Note that the only variable left is and the limits of integration would be from -/ to / g v G Ad cos m Point of observation A / g cos v G d m / m r+dr r=m/cos dx=rd/cos

23 Picking up the thread at this point A / g cos v G d m / What do you get? g g v v A G m A G m sin R gv G m

24 A look at text problems 9.9 & Due next Tuesday Given - r r e 1z r p r i. As part of a discrete sum of disks, what is the volume of each individual disk assuming each has thickness z. ii. What is the approximate volume of the Earth expressed as a discrete sum of disks, z thick iii. Transform the discrete sum into an integral with limits of integration extending from r p to +r p. iv. Given that r e and r p are 6378km and 6357km, respectively, what is the Earth s volume. v. Lastly, estimate the volume of the earth assuming it is a sphere and use radius equal to the average radius (average of r e and r p ) and compare z

25 Problem 9.9 Volume of the earth an oblate spheroid for which r r e 1 z In this equation r varies from r e, at the equator, to r=0 at the poles. z represents distance along the earth s rotation axis and varies from r p to r p. The equatorial radius is given as 6378km and the polar radius, as 6357km. r p

26 Volume as a summation of disks in the limit that z goes to 0 Area=r r p r rp e E e 1z r z The development up to the point is presented in the back of the text V r dz p r p Show steps taking you from the above expression to that below in part iii) V r r r 3 E e p p Combine and show calculations in steps iv) and v). Don t just write down answers from the back.

27 Problem 9.10 t t exp( x / X ) 0 In this problem, you use the bottomset bed thickness relationship to determine the cross-sectional area of the bottomset bed. The following questions are posed. i. Consider the components of a discrete sum approximation. The discrete sum would be the sum of vertical rectangles x wide and t i high. What is the area of such a triangle? ii. What is the approximate area of the bottomset bed expressed as a discrete sum of these vertical rectangles, x thick?

28 Problem 9.10 (continued) iii. Transform the discrete sum into an integral and evaluate. x X t t 0 oe dx Show details of this integration. iv. Lastly, if the the present-day rate of sediment supply is 10 m /yr, X=5km and t o = 1m, estimate the time taken to form the bed assuming the sediment supply rate has not changed through time. Be units consistent and show your calculations.

29 With regard to 9.9 and 9.10 Go beyond the text answers as noted! As with all the problems assigned from the text present the details leading to results specified in the Appendix. i.e. show steps in iii of problem 9.9. Show the details of calculations leading to part iv. Note that part v has not been provided: calculate that and compare to the result obtained in iv. In 9.10 show details of solutions leading to iii and iv.

30 Let s return to Excel problem 9.7 Remember you can format trendline and trendline labels y = E-1 x E-08x E-05x E-0x E-0 You need to format the label to get 3-4 decimal place precision in the coefficients.

31 Questions on problem 9.7? 0 x a x a x a x a x a x a dx For the 5 th order polynomial you derive you ll have 6 terms including the constant a x ax a x What is this 4 the derivative of?... x 0 List of coefficients times factors x 6 x 5 etc Coefficients come from ExcelTrendline equation

32 Lets break it down for the 4 th order polynomial Note that these are just the constants you obtain from the best fit 4 th order polynomial

33 The constants in the best fit polynomial a 4 = a 3 a 0

34 This example uses the 4 th order polynomial Note coefficients in your equation a 4 a 3 a 0 x 4 = x 3 = etc Just type in or cut and paste in the coefficients These are the factors you obtain from integration

35 This example uses the 4 th order polynomial factors a x a x a x a x a x These are just entered as text to remind you what they are.

36 This example uses the 4 th order polynomial factors a x a x a x a x a x a 4 /5= a 3 /4= etc Multiply coefficients (a s) at left by 1/5(or 0.), 1/4 (0.5), 1/3 (0.333) etc.

37 This is a definite integral. Each term is evaluated at 000 and 0 a x a x a x a x a x Note that for the definite integral, when x=0, all the terms are a3x 1 a0x a x a x a x

38 For the 4 th order polynomial a x a x a x a x a x These are the x terms evaluated at x= = = etc

39 For the 4 th order polynomial a4 5 a3 4 a 3 a1 a0 1 These are the scaled terms in the polynomial ax ax 4 ax 3 ax ax E4 notice how precision will affect result

40 For the 4 th order polynomial These are the scaled terms in the polynomial ax 4 5 ax 3 4 ax 3 ax 1 ax a x a x a x a x a x Sum them up to get the area

41 Follow directions as noted

42 Putting it all together Obtaining a trendline or a best fit line was an important task we learned earlier in the semester. When you add trendline you get a list of trendline options (see next slide).

43 Add trendline Select polynomial and then choose order 5 And don t forget to display equation and R!

44 You will get a 6 coefficient 5 th order polynomial, which when integrated will yield Make sure you increase the precision of the coefficients by right clicking the equation label box and selecting format trendline label. a x a x a x a x a x a x Select Scientific from the drop list. 3 to 4 decimal places should do OK.

45 Assignment is due thisthursday

46 Reminder that in 9.9 and 9.10 give attention to the following! As with all the problems assigned from the text, present the details leading to results specified in the Appendix. i.e. show steps in iii of problem 9.9. Show the details of calculations leading to part iv. Show calculations for parts iv and v. In 9.10 show details of integration leading to iii and calculations yielding iv.

47 To do list For the remainder of today s class, continue to work on lab problem 9.7. The Excel version of this problem is due next class. NEXT TIME - Bring questions to class on problems 9.9 and These problems will be due next Tuesday.

48 We ll review some of these problems next time as time permits 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 per unit volume (Q) as Q y kw 0 km y is height above the base of the crust and, and we ll rewrite as 3 Q z kw 0 km 3 where z is the distance towards the surface from the base of the crust.

49 Q z kw 0 km 3 Where z is the distance in km from the base of the Earth s crust i. Determine the heat generation rate at 0, 10, 0, and 30 km from the base of the crust 0 kw 0 km kw km 10 0 kw km kw 0.5 km ii. What is the heat generated in a small box-shaped volume z thick and 1km x 1km surface? Since z V Az V zkm 3 1km

50 The heat or power generated by that crustal box would be QV Qz kw This is a differential quantity so there is no need to integrate iii & iv. Heat generated in the vertical column In this case the sum extends over a large range of z, so However, integration is the way to go. n q Q z q 1 0 i1 0 z 0 i z dz 0 z zdz

51 Evaluate the integral v. Determining the energy or heat flow rate at the surface would require evaluation of z the definite integral z q q 0.5kW Qdz vi. To generate 100MW of power 100, 000kW 4444km.5kW km An area about 67km on a side

52 More on heat flow how much energy do you save by turning the thermostat down? Consider heat conduction through a thick glass window given two possible inside temperatures 65 o F and 7. o F and an outside temperature of 3 o F. In terms of degrees C this corresponds to temperatures of O C,.33 o C and 0 o C. How much energy do you save? This problem is solved using a simple equation referred to as the heat conduction equation. See for additional discussion T qx K x

53 window Thermal gradient across the window 0.5 cm Outside temperature of 3 o 65 o or 7 o F

54 Power = energy/time We consider this problem in terms of the heat flow over the course of the day, where heat flow (q x ) is expressed in various units representing heat per unit area per time: for example, calories/(m -s). A q x of 1 cal/(cm -s)=41.67 kw/m kw-s= ft-lbs 1 calorie/sec = kw = Watts T qx K x Energy flow through a specified area in unit time

55 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.

56 To solve this problem, we use the heat conduction equation: q x =-KT/x k (thermal conductivity) x10-3 cal/(cm-sec- o C) The rate at which energy is transferred through 1 cm for each o C difference in temperature If the window is x=0.5cm thick Then q x =0.073 cal/(cm -sec) or cal/(cm -sec) If the window has an area of 1m Then the net heat flowing across the window is 733 or 933 cal/sec The lower temperature saves you 00 cal/sec T qx K x

57 Translating into $$s 163 cal/sec corresponds to 1.73x10 7 cal/day There are cal per kwh so that this corresponds to about 0 kwh/day. A kwh goes for about 7 cents so $1.40/day. note this estimate depends on an accurate estimate of K (thermal conductivity). Other values are possible, but, as you can see, it can add up!

58 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=40 km Intrusive sill T x x X=0 km o

59 What is the derivative This is an idealized problem that considers heat flow only on one direction, but it illustrates general principles associated with cooling. Discrete form T qx K x Using the conduction equation in differential form q x dt K dx Take derivative to get heat flow

60 Calculate the temperature gradient dt d 0.5x 30x 10 dx dx Given K 6 10 cal / ( cm sec) 3 o 6x10 cal / ( cm sec C) and that 1 heat flow unit = Calculate q x at x=0 and 40km. q x dt K dx

61 and substitute for x to get q You will get q x =-1.8 hfu at x=0 X=40 km X=0 km and q x =0.6 hfu at x=40 Heat flows out both ends of the sill. The sill is hotter in its interior

62 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 rate for granite is ~x10-13 cal/(gm-sec) and that for basalt about x We use an average of about 1x10-13 for this problem. Given that the mass of the Earth is about 6x10 7 grams we get a heat generation rate of about 6.0x10 14 cal/sec. What is the heat flow per cm?

63 Heat flow per unit area To answer that, we need the total area of the Earth s surface in cm. The surface area of the Earth is about 5.1 x cm. which gives us a heat generation rate /cm of about 117 x10-6 cal/(cm -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.

64 As noted above! For the remainder of today s class, continue to work on lab problem 9.7. The Excel version of this problem is due next class. NEXT TIME - Bring questions to class on problems 9.9 and These problems will be due next Tuesday.