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Geomath

Geomath. Geology 351 -. Wrap up chapter 9 - integral calculus. tom.h.wilson tom.wilson@mail.wvu.edu. Dept. Geology and Geography West Virginia University. Question 9.7: review and questions. What is this integral?.

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Geomath

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  1. Geomath Geology 351 - Wrap up chapter 9 - integral calculus tom.h.wilson tom.wilson@mail.wvu.edu Dept. Geology and Geography West Virginia University Tom Wilson, Department of Geology and Geography

  2. Question 9.7: review and questions What is this integral? For the 5th order polynomial you derive you’ll have 6 terms including the constant Tom Wilson, Department of Geology and Geography

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

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

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

  6. 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. See http://serc.carleton.edu/NAGTWorkshops/geophysics/activities/18913.html for additional discussion Tom Wilson, Department of Geology and Geography

  7. 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

  8. 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

  9. 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

  10. 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

  11. 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 Tom Wilson, Department of Geology and Geography

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

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

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

  15. 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 6.0x1014 cal/sec. What is the heat flow per cm2? Tom Wilson, Department of Geology and Geography

  16. 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

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

  18. 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 Tom Wilson, Department of Geology and Geography

  19. The heat generated will be 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. Tom Wilson, Department of Geology and Geography

  20. 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 Tom Wilson, Department of Geology and Geography

  21. Li Strain Lf 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 Tom Wilson, Department of Geology and Geography

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

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

  24. We can simplify the problem and still obtain a useful result. Approximate the average densities 11,000 kg/m3 4,500 kg/m3 Integrating discontinuous functions Tom Wilson, Department of Geology and Geography

  25. We can simplify the problem and still obtain a useful result. Approximate the average densities 11,000 kg/m3 4,500 kg/m3 The result – 6.02 x 1024kg is close to the generally accepted value of 5.97 x 1024kg. Tom Wilson, Department of Geology and Geography

  26. Problems 9.9 and 9.10 Volume of the earth – an oblate spheroid 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

  27. Problem 9.10 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

  28. 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

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

  30. 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

  31. The vertical component of the acceleration due to gravity of the sulphide deposit 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

  32. 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

  33. Start reviewing materials for the final!… Current to-do list • Finish up problem 9.7 for next Tuesday • Finish up the gravity computation for Tuesday • We’ll follow up with questions on 9.9 and 9.10 in class next Tuesday • Start reviewing class materials. The following week will be a final review week Tom Wilson, Department of Geology and Geography

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