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 Heat flow problems to accompany the example in Chapter 9 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 O C, 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. qx K x See for additional discussion T

3 Some units 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). 1 cal/s= W 1 joule = 0.74 ft-lb 1 joule =0.24 calories 1 Watt = 1 Joule/sec 1 calorie/sec = kw

4 Relating to the units 1 cal/s=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.

5 To solve this problem, we use the heat conduction equation: q x =-KT/x K (thermal conductivity) 2x10-3 cal/(cm-sec- o C) Assume x=0.5cm Then q x =0.07 cal/(cm 2 -sec) or cal/(cm 2 -sec) If the window has an area of 2m 2 Then the net heat flowing across the window is 1466 or 1786 cal/sec The lower temperature saves you 320 cal/sec

6 80,000 cal/sec 320 cal/sec corresponds to 2.82x10 7 cal/day There are cal per kwh so that this corresponds to about 32.8 kwh/day. A kwh goes for about 6.64 cents so $2.18/day. note this estimate depends on an accurate estimate of K, but, as you can see, it can add up!

7 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 X=0 km T x x o

8 What is the derivative Using the conduction equation in differential form q x dt K dx You see you have to take a derivative to determine heat flow.

9 Calculate the temperature gradient dt d 2 0.5x 30x 10 dx dx Given K 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

10 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.

11 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 2x 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 6.6x10 14 cal/sec. What is the heat flow per cm 2?

12 Heat flow per unit area 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 cm 2. which gives us a heat generation rate /cm 2 of about 12.9 x10-5 cal/(cm 2 -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.

13 Lastly consider the cooling time for the Palisades intrusive sill In this example, we consider a granitic sill with initial temperature of ~750 o C and dimensions of 25km x 25km x 0.3km. In this problem, we approximate cooling time as 2 ~ L k Where L is the thickness of the sill and k is the thermal diffusivity.

14 Cooling time The thermal diffusivity k K C p K is the thermal conductivity introduced earlier is the density of the sill, and C p is the heat capacity of the granite. As with all these problems you have to look things up in a table of constants. In this case we are given that K/C p ~0.01 cm 2 /sec.

15 Cooling time 2 4 3x10 cm 2 For this case, we have L k cm / sec 10 9x10 sec Calculate the number of seconds per year to find that this sill will cool off in about 2850 years.

16 For Thursday Review problem 9.8. In this problem, we assume that heat is generated in the crust at the rate of 1kW/km 3 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).

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 Q y kw 20 km 3 Also review total natural strain discussion and the integration of discontinuous functions.

18 Q z kw 20 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, 20, and 30 km from the base of the crust 0 kw 20 km kw km kw km kw 0.5 km ii. What is the heat generated in a in a small box-shaped volume z thick and 1km x 1km surface? Since V Az V zkm 3

19 The heat generated will 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 n q Qi z over a large range of z, so However, integration is the way to go. q z z dz 20 0 z zdz i1

20 v. Determining the flow rate at the surface would require evaluation of z the definite integral q Qdz 0 q 2 z kW vi. To generate 100MW of power 100, 000kW 4444km 22.5kW km 2 2 An area about 67km on a side

21 L i e L, the elongation L i 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 L L i f dl L ln( L ) ln( L) Lf ln Li f i ln L L L i f ln( S) L f Where S is the Stretch

22 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) e e e e The six term approximation is accurate out to 5 decimal places!

23 Comparison of finite elongation vs. total natural strain Pretty close for relatively small strains.

24 Integrating discontinuous functions 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 M N i1 Area of sphere x r x density of shell Area of sphere x r = volume of shell N M 4r r i1 2 i i 2 M R 4 r dr 0

25 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 2 M R 4 r dr r C r.11000dr 4 r.4500dr r r r r 3 3 The result 6.02 x kg is close to the generally accepted value of 5.97 x kg.

26 Volume of the earth an oblate spheroid r r 2 2 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 6457km. 2 r 2 p Solve this integral before leaving today.

27 Problem 9.10 t t exp( x / X ) 0 In this problem, we return to the thickness/distance relationship for the bottomset bed. See self assessment problem. Problems 9.9 and 9.10 will be due next Tuesday

28 We ll go over these group problems next week as part of our review

29 These provide some additional practice with finding integrals

30 Another both indefinite and definite integrals

31 Have a look before Tuesday if time permits

32 We ll also have a look at an integration by substitution approach Take a look at this one for a minute t 8 3t 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.

33 t 8 3t dt In this problem, we can see that d 8 3t dt 3 9t which within a factor of 1/3 rd equals the leading term in the integrand the 3t 2. So the idea is that we define a new variable u du dt t, where 3 3 t du t dt 2 9 and 9

34 We also see that du t dt So we substitute and redefine the integral u83t 3 into du t 8 3t dt to get 1 1 du u u du 3 3 t dt

35 Now you have the much simpler integral to evaluate. You just need to use the power rule on this u du What would you get? u Substitute the expression defined as u(t) back in 5 4

36 Our result t Integration by substitution helps structure the process of finding a solution. Differentiate to verify.

37 1. Problems 9.9 and 9.10 are due next Tuesday 2. They will be returned and discussed next Thursday during the final review session 3. Start reviewing your notes with special emphasis on the material covered since the midterm.