1. Tom Wilson, Department of Geology and Geography tom.h.wilson tom.wilson@mail.wvu.edu Dept. Geology and Geography West Virginia University

2. Tom Wilson, Department of Geology and Geography For the 5 th order polynomial you derive you’ll have 6 terms including the constant What is this integral?

3. Tom Wilson, Department of Geology and Geography For the 5 th order polynomial you derive you’ll have 6 terms. Enter values here (Upper limit of x) 6 (Upper limit of x) 5 Etc.

4. Last set Tom Wilson, Department of Geology and Geography

5. Another example of integration by substitution Tom Wilson, Department of Geology and Geography Let u =

6. Let’s consider some heat flow problems as a companion discussion to the example in Chapter 9 Tom Wilson, Department of Geology and Geography Consider heat conduction through a thick glass window given two possible inside temperatures 65 o F and 72.2 o F and an outside temperature of 32 o F. In terms of degrees C this corresponds to temperatures of 18.33 O C, 22.33 o C and 0 o C. How much energy do you save? See http://serc.carleton.edu/NAGTWorkshops/geophysics/activities/18913.html for additional discussionhttp://serc.carleton.edu/NAGTWorkshops/geophysics/activities/18913.html This problem is solved using a simple equation referred to as the heat conduction equation.

7. Power = energy/time 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 (q x ) is expressed in various units representing heat per unit area per time: for example, calories/(m 2 -s). A q x of 1 cal/(cm 2 -s)=41.67 kW/m 2 kW-s=737.5622 ft-lbs 1 calorie/sec = 0.004284 kW =4.1868 Watts

8. Relating to the units Tom Wilson, Department of Geology and Geography 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

9. To solve this problem, we use the heat conduction equation: q x =-K  T/  x Tom Wilson, Department of Geology and Geography K (thermal conductivity)  2x10 -3 cal/(cm-sec- o C) Assume  x=0.5cm Then q x =0.07 cal/(cm 2 -sec) or 0.089 cal/(cm 2 -sec) If the window has an area of 2m 2 Then the net heat flowing across the window is 733 or 896 cal/sec The lower temperature saves you 163 cal/sec

10. 80,000 cal/sec Tom Wilson, Department of Geology and Geography 163 cal/sec corresponds to 1.41x10 7 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!

11. The second question concerns a hot sill Tom Wilson, Department of Geology and Geography 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

12. What is the derivative Tom Wilson, Department of Geology and Geography You see you have to take a derivative to determine heat flow. Using the conduction equation in differential form

13. Calculate the temperature gradient Tom Wilson, Department of Geology and Geography Given K and that 1 heat flow unit = Calculate q x at x=0 and 40km.

14. and substitute for x to get q Tom Wilson, Department of Geology and Geography You will get q x =-1.8 hfu at x=0 and q x =0.6 hfu at x=40 X=0 km X=40 km Heat flows out both ends of the sill.

15. Another simple example : assume the mantle and core have the same heat production rate as the crust Tom Wilson, Department of Geology and Geography 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 6x10 27 grams we get a heat generation rate of about 6.0x10 14 cal/sec. What is the heat flow per cm 2 ?

16. Heat flow per unit area … Tom Wilson, Department of Geology and Geography To answer that, we need the total area of the Earth’s surface in cm 2. The surface area of the Earth is about 5.1 x 10 18 cm 2. which gives us a heat generation rate /cm 2 of about 117 x10 -6 cal/(cm 2 -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.

17. Problem 9.8 Tom Wilson, Department of Geology and Geography 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 Also review total natural strain discussion and the integration of discontinuous functions.

18. Tom Wilson, Department of Geology and Geography Where z is the distance in km from 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 ii. What is the heat generated in a in a small box-shaped volume  z thick and 1km x 1km surface? Since

19. Tom Wilson, Department of Geology and Geography The heat generated will be iii & iv. Heat generated in the vertical column This is a differential quantity so there is no need to integrate In this case the sum extends over a large range of  z, so However, integration is the way to go.

20. Tom Wilson, Department of Geology and Geography v. Determining the flow rate at the surface would require evaluation of the definite integral vi. To generate 100MW of power An area about 67km on a side

21. Tom Wilson, Department of Geology and Geography LiLi LfLf The total natural strain, , is the sum of an infinite number of infinitely small extensions In our example, this gives us the definite integral Where S is the Stretch

22. Tom Wilson, Department of Geology and Geography Strain (or elongation) (e), stretch (S) and total natural strain (  ) Elongation Total natural strain  expressed as a series expansion of ln(1+e) The six term approximation is accurate out to 5 decimal places!

23. Comparison of finite elongation vs. total natural strain Tom Wilson, Department of Geology and Geography

24. Integrating discontinuous functions Tom Wilson, Department of Geology and Geography 11,000 kg/m 3 We can simplify the problem and still obtain a useful result. Approximate the average densities 4,500 kg/m 3

25. Tom Wilson, Department of Geology and Geography 11,000 kg/m 3 We can simplify the problem and still obtain a useful result. Approximate the average densities 4,500 kg/m 3 The result – 6.02 x 10 24 kg is close to the generally accepted value of 5.97 x 10 24 kg.

26. Tom Wilson, Department of Geology and Geography Volume of the earth – an oblate spheroid 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 6457km.

27. Problem 9.10 Tom Wilson, Department of Geology and Geography 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

28. Consider this simple problem (turn in next time) Tom Wilson, Department of Geology and Geography 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).

29. Gravitational acceleration can vary in response to subsurface density contrasts Tom Wilson, Department of Geology and Geography From Newton’s universal law of gravitation g gvgv Sulfide deposit r z x  

30. We will develop a solution as follows: Tom Wilson, Department of Geology and Geography Express r in terms of x and z Rewrite g in terms of x and z Express g v in terms of g and  Replace the trig function in the above with its spatial equivalent and rewrite g v. Factor z out of the denominator to obtain

31. The vertical component of the acceleration due to gravity of the sulphide deposit Tom Wilson, Department of Geology and Geography Given that G=6.6732 x 10 -11 nt-m 2 /kg 2, x=1km, z=1.7km, R deposit =0.5km and  =2gm/cm 3, calculate g v.

32. Spatial variation in the gravity anomaly over the sulphide deposit Tom Wilson, Department of Geology and Geography Note than anomaly is symmetrical across the sulphide accumulation

33. Start reviewing materials for the final! … Current to-do list Tom Wilson, Department of Geology and Geography 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