the first derivative with respect to time is obtained by carefully applying the chain rule ( surf init ) T Tinit

Transcription

1 .005 ermal Fluids Engineering I Fall`08 roblem Set 8 Solutions roblem ( ( a e -D eat equation is α t x d erfc( u du π x, 4αt te first derivative wit respect to time is obtained by carefully applying te cain rule 3 ( x x exp 4 ( surf t t π αt 4α e first derivative of te complimentary error function is [ ] exp( u For te given temperature profile of erfc ( surf ( x x exp 4 ( surf t t 4απ αt e second derivative wit respect to position x is obtained from ( x exp 4 ( surf x π αt 4αt erefore ( x x exp 4 ( surf x π αt 4αt 4 αt ( x x α exp x 4απ 4 αt 3 t ( ( α t x 3 ( surf x x x x exp ( surf t exp ( surf t 4απ 4αt 4απ 4αt wic cecks as a solution to te eat equation. 3 3 b In part (a we ave sown tat te given temperature profile satisfies te eat equation. In order tat tis temperature profile apply specifically to a semi-infe slab it must also satisfy te two boundary conditions and te ial condition. Initial Condition: I.C. ( x,0 0 x ( x,0 erfc erfc 4α 0 ( note erfc 0 ( x,0 0 First Boundary Condition: B.C. (0, t Second Boundary Condition: ( ( ( surf surf surf ( ( ( (0, t erfc 0 surf surf

2 B.C. (, t 0 ( ( (, t erfc 0 surf c e eat flux at te surface of te slab, x0, is obtain by using Fourier s aw. ( x qxt (, k k exp 4 ( surf x x π αt 4α t ( surf k x qxt (, exp 4 αt k q(0, t ( surf d Wen two semi-infe solids are brougt into termal contact, te interface temperature must be te same and te eat flux leaving one solid must be equal to te eat flux going into te second solid. In te problem statement we are told to assume tat tere is no interfacial termal resistance between te solid. For tis reason, tere cannot be a temperature difference between te two interfaces. Secondly, since tere is no work transfer at te interface, te First aw applied to a differential piece of te interface tells us tat te flux leaving one solid must be equal to te flux going into te oter solid. k surf surf ( ( k x x e negative sign is needed in te flux equation since te coordinate systems of te two slabs are different. e Find te expression for te interface temperature. k surf surf surf ( ( k x x x 0 x 0 ( surf ( surf k k πα t πα t surf k α + k k α + k α f Using te termopysical properties given in te problem, we can calculate surf, for te tree different blocks. Copper Block: surf 0.7 C Stainless Steel Block: surf.98 C VC Block: surf 30.9 C e copper block feels coldest to te touc. g Recall from part c tat te expression for te eat flux at te surface is α k q(0, t ( surf as units of lengt. is expression can be tougt of as te penetration dept, e expression i.e., te dept into te solid tat as experienced a cange in temperature from its ial condition. is expression is often used in eat transfer analysis to obtain a scale fore te time it takes an entire solid (or a part of a solid to respond to a temperature cange at its interface.

3 In te case of cooking a burger, we can use tis lengt scale to obtain an approximate value for te time it takes to cook te burger. If we assume tat te tickness of te burger is 5 mm and te termal diffusivity of te burger is tat of water, we can calculate a cooking time scale of l t 9sec min 7 πα π (.5 0 (We used alf te tickness of te burger as our lengt scale, since eat diffuses in from bot te top and bottom surfaces. roblem e rain keeps te surface of te aspalt at a constant 0 C. et s model te aspalt as a semi-infe solid, wit an ial temperature of 50 C. From equation 6.08 in te course reader (also derived in class, te eat flux into te aspalt at a given k( s i k time is: q& s ( t were α ρc We integrate wit respect to t to get te total eat transfer per unit area into te aspalt, remembering tat 30 minutes is 800 seconds : Q k 30 min 0 q ( t dt ( πα s s i 5 J/m t t 30 min And of course, since we want te eat transferred from te aspalt, we take te negative of te value above: Q J/m out roblem 3 CV CV a Applying te First aw to CV gives us an expression for Q & du dt Since te system is at steady state and tere is no work transfer, 0 A q& Q& + A( Q& Aq& A ( 0 air Q & ( air Q& W& c e efficiency of a reversible engine operating between termal reservoirs is η max

4 erefore te efficiency of tis Stirling engine is η SE 0. c W& Q& Combining tis result wit te expression for Q & given above, we get an expression relating te power output of te Stirling engine to te temperature of te solar collector. c c W& 0. Q 0. ( Aq A( air & & ( o maximise te work, we take te derivative of te power wit respect to and set it equal to zero. dw& d c 0. air A q& + ( ( + c Aq& A 0. ( 0 ( c cair q& c + air 339. K b e power output of te generator at tis operating condition is determined by plugging in te value of into equation (. We get W &.4 W W&.4 c e efficiency for solar to electric power conversion is η total & roblem 4 Aq 0 0 Below is one possible realisation of te Ericsson cycle. (You may ave oter designs tat run te same termodynamic cycle.

5 For rocess -, te reversible constant pressure cooling at, W dv mr mr dv ( V V mr( mr( (In te above equation, we used te fact tat te pressure stays constant over te process -, and is equal to From te st aw, Q W + ( U U W + mcv ( W + mcv ( Q mr( + mcv ( mc p ( is eat transfer is negative, wic means tat eat leaves te ideal gas and goes into te regenerator. For rocess -3, te reversible isotermal eat transfer at, e analysis proceeds in te same way as for rocess 4-. In tis case, W Q 3 3 mr Since <, te eat transfer and work transfer are negative eat leaves te system and work enters te system. For rocess 3-4, te reversible constant pressure eating at, e analysis proceeds in te same way as for rocess -. In tis case, W mr( and Q mc ( p Note tat te eat and work transfers in tis case are equal and opposite to te eat and work transfers in process -. All te eat tat was transferred to te regenerator in process - is transferred back to te gas in process 3-4. o find te cycle efficiency, we need to find te net work output and te eat input. e net work output is W net W4 + W + W 3 + W3 4 W net mr( mr + mr( + mr + mr( e eat input occurs in process 4-. (eat is rejected to te low temperature reservoir in process -3. As sown previously, te eat transfers in processes - and 3-4 are equal and opposite: tere is no net eat transfer to te regenerator. Q in Q4 mr

6 e efficiency of te cycle is defined as In tis case, it is calculated to be W η Q net mr( η mr in is is te same as tat of a Carnot engine. is is not surprising, since any reversible engine operating between te same fixed temperature reservoirs and will ave te same efficiency. d e eat absorbed at is Q in J / cycle e eat rejected at is Q out Q J / cycle 00 e net work transfer is W net ( J / cycle e termal efficiency of te cycle is roblem 5 a Since System (as sown in te figure below is reversible and adiabatic, te entropy of te system does not cange from te ial state to te final state (by te Second aw. erefore, we ave A, VA, B, VB, S + A + S B ma cv + R mb cv + R 0 A, VA, B, VB, System System We are told tat te final volumes of A and B are te same. Note tat all times, te pressures in A and B must be same, because of te frictionless, massless piston. Since te masses of gases in A and B are also te same, te final temperatures of A and B must also be te same (tis just follows from te ideal gas law. et tis final temperature be We are not given te volumes of A and B, but we only require te volume ratios in te Second aw equation. et te ial volume of A be V A,. Since te mass and ial pressure of A and B are te same, ma,r mb,r VA, VB, VA, 50 V 550 A, B, V B, B, VA,. V A, A, A, B, erefore, te total ial volume is V A, + V B, 3.V A,. Since te final volumes are equal, and te total volume remains te same, we ave V A, V B,.6V A,. B,

7 VA, VB,.6 erefore,. 6 and V V. A, B, lugging tis back into te Second aw equation, we get m A c v + R.6 + mb cv + R Since te masses are equal, tey drop out of te equation. Combining terms, we get c v + R ( R / cv ( b e total work extracted from te cylinder is found by applying te First aw to System : U A + U B Q W ere is no eat transfer from te system (Q A and Q B are internal to te system. erefore, W ( U + U ( m c + m c ( 9. kj A B A v ( A, B v B, 8 K c If te eat engine is not allowed to communicate wit any oter termal reservoir wic is at a temperature different from te final temperature of gases A and B ten te maximum amount of work as been extracted from te cylinder consisting of gas A and B. No more positive work can be extracted from te eat engine. d e process witin System (sown in te figure is also reversible and adiabatic. If we apply te Second aw to tis system, we get S C + S C block D 0 C, C, + C block D, D, 0 D, C, D, Using tis result and applying te First aw to System, we can solve for C, : U C + U D Q W ere is no eat transfer to tis system; we are told tat te work transfer is W J C + C ( ( block ( C, C, block D, D, ( C C, + ( D, D, C 400 C, D, 300 K C,D, C, C, C,, 800C, is equation as two solutions, C, K and C, K For C, K, we get D, K For C, K, we get D, K Since we are told tat C is te colder block, D, K and C, K roblem 6 A reversible cycle executes te cycle sown, between a maximum temperature of 800 K and a minimum temperature of 600 K. is cycle is different from te two eat-reservoir cycles we discussed in class C,

8 because te eat input takes place over a range of temperatures (600 K 800 K. is engine interacts wit several ig-temperature reservoirs and one low-temperature reservoir (at 600 K. 800 K rocess 600 K rocess 400 kj/k 800 kj/k (a and (c Determine te direction in wic te engine executes te cycle, and indicate te pat along wic eat is rejected. Since tis cycle is a eat engine, i.e. a power-producing system, it must take eat (and entropy from ig temperature reservoirs and dump te entropy (along wit some of te eat to te low temperature reservoir, extracting some useful work in te process. So te eat transfer in must be at a iger temperature tan te eat transfer out (in contrast to a refrigerator, were te eat transfer in is at a low temperature, and te eat transfer out is at a ig temperature. o figure out weter eat is entering or leaving a system during a reversible process, we look at te entropy. From te second law for a reversible process,, a positive eat transfer must increase te entropy of te system. A negative eat transfer decreases te entropy of te system. We can break te cycle into two processes te semi-circular arc (process and te straigt line (process. o produce power, te eat transfer must be positive during process. erefore, te entropy of te system must increase during process. Similarly, te eat transfer must be out of te system during process, and terefore, te entropy of te system must decrease during process. So te engine executes a clockwise cycle and eat is rejected in process (te straigt-line part of te cycle. (b e net work For a cycle, te st aw is e cyclic integral of eat for a reversible process is (from te nd aw, wic is just te area under te -S grap. Since tis region is semi-circular, its area can be calculated as te radius is 00. erefore, te cyclic integral of eat, work. (d e termodynamic efficiency e efficiency for an engine is defined as, were, is 683 kj, wic is also equal to te net We found W in part (b; we now ave to find Q. o do tis, we apply te nd aw to process, te eat input process. e eat input is given by, wic is te area under te curve for process. is is te sum of te areas of te semi-circle, and te rectangle between (400 kj/k to 800 kj/k and (0 K to 600 K. is total area is kj. So te efficiency is W/Q 683/ % (e e Carnot efficiency for a cycle running between 600 K and 800 K. 5%. is is iger tan te efficiency of te reversible cycle we calculated. is is not an inconsistency te cycle we ave analysed does not operate between two fixed temperature reservoirs like a Carnot cycle and tus does not ave te same efficiency as te Carnot cycle, even toug it is completely reversible.

9 e irreversible engine We now take a look at an irreversible cycle wit W irrev 0.9W rev kj. e eat input is still 3083 kj. e entropy transferred during te eat input is still te same as tat in te reversible case. We know tat Q,rev Q W rev kj. is is transferred to te (fixed low temperature reservoir at 600 K. e entropy transferred to te low temperature reservoir in te reversible cycle is 40000/ kj/k. erefore, te entropy transferred in must also be 400 kj/k. A muc easier way of seeing tis is looking at process. e entropy cange in process (te eat input process is 400 kj/k (from te cycle plot. Since tis process is reversible, all te entropy cange must be due to entropy transfer. (f and (g Calculate, and is Q,rev greater tan, less tan, or equal to Q,irrev? Applying te st aw to te reversible cycle, Q Q,rev W rev Q,rev Q - W rev kj Applying te st aw to te irreversible cycle, Q Q,irrev W irrev Q,irrev Q - W irrev kj Q,rev is smaller tan Q,irrev. is is expected less work is extracted in te irreversible cycle. is is due to entropy generation in te irreversible engine. Because te entropy being transferred to te 600 K reservoir is te sum of te entropy transferred in during process and te entropy generated, te Q required is greater in te irreversible case. o find te entropy generation in one cycle, apply te nd aw to te irreversible engine over a complete cycle. is te entropy transferred in during te eat input process, wic we calculated to be 400 kj/k. simplifies to negative lugging tese values in,, wic is -4683/ kj/k (Q leaves te system and is 0.47 kj/k