Calculations for Calcium Carbide / Acet ylene Reactor Blo c k Diagram (Class Example)

No v em b er 1, 2011

1 In tro duction

Y ou ha v e inherited the w orld’s largest storehouse of calcium carbi de (CaC 2 ) from y our grandfather, who made a fortu ne starting precious metal mining companies. Y ou wish to use this free (?) source of energy to pro vide pro cess heat to a l iquid so dium r e fi nery do wn the street, as there aren’t an y more mining companies that use CaC 2 for lan terns, and what used to b e v ehicle carbide lamps are no w filamen t or LED ligh ts. The so dium refiners will pa y y ou handsomely for y ou r services and pro cess heat, as long as y ou can pro vide what they need.

2 Problem Statemen t

The so dium refiners require y ou to pro vide a steady stream of pro ces s heat to their system in order to heat up liquid so dium metal. They require y ou to heat 10 kg of so d ium from 100C to 500C. They are asking y ou to pro vide the heat e x c hanger to p erform this op e r ation.

Design, c haracte r iz e, and optimize a system to pro vide this pro cess heat in the most efficien t manner. Start b y defini ng a system, mak e educated guesses for k ey missing parame t e rs, and iterate along y our design to reduce cost, material strain, resource c on s u m p tion, and an yth ing else that could reduce y our profit margins or safet y margins.

3 Starting the Solution

Start b y constructing a blo c k diagram. An example of one that (at fi rst) solv es the problem is sho wn in Figure 1.

3.1 CaC 2 Reactor

Think ab out what c omp onen ts y ou need in order to get energy from calcium carbide in to the liquid so diu m . Since y ou ha v e an inexhaustible supply of CaC 2 , let’s start with that. Lo oking up t ypical reactions for calcium carbide, w e can find one that is used in carbide lamps (and man y other CaC 2 devices). It in v olv es a reaction with w ater to pr o duce acet ylene ( C 2 H 2 ), whic h is v ery energy dense & unstable, and lime slurry:

C aC 2 ( s ) + 2 H 2 O ( l , g ) → C 2 H 2 ( g ) + C a ( O H ) 2 ( s ) Δ H R X N =??? (1)

Let’s start b y dra win g a blo c k for the CaC 2 reactor as sho wn on the upp er left of Figure 1, where it will react with w ater to pro duce ace t ylene. W rite the reaction in Equation 1 inside the blo c k so the readers kno w what’s hap p ening inside. Y ou can also dra w an input line for CaC 2 and an output line for Ca (OH) 2 at th is p oin t, as y ou kno w they will ha v e to b e inputs and outputs, resp ectiv ely .

Let’s also assume that w e w an t gaseous w ater en tering the CaC 2 reactor, whic h will help increase th e reactivit y of this system. Dra w a line represen ting steam en tering the CaC 2 reactor. W e will finish connecting this line later, in Section 3. 6.

F i g u r e 1: B l o ck diagram for i n i t i a l setup of t h i s system. Note t h a t a nu mb e r of parameters are m i ss i n g , and i t seems t h a t t h e r e are man y more v a r i a b l e s th an e q u a t i o n s t o solv e.

3.2 Acet ylene Burner

Next, w e w an t to extract as m uc h energy as p ossible from the acet ylene gas. Acet ylene i s commonly used in high temp erature w elding torc hes as o xy-acet ylene, where flame temp eratures can e xceed 3,000C. Acet ylene burns (com b us t s ) with o xygen to pro duce w ater v ap or and carb on dio xide:

2 C 2 H 2 ( g ) + 5 O 2 ( g ) → 4 C O 2 ( g ) + 2 H 2 O ( g ) Δ H R X N =??? (2)

W e exp ect that thi s reaction will h a v e a v ery large, negativ e en thalp y of reaction, as w e kno w it is v ery exothermic. W e’ll w orry ab out finding out ho w exothermic it is later. Dra w a blo c k for the acet ylene burner, as sho wn in the upp er righ t of Figure 1. W rite the reaction in Equation 2 inside the blo c k so the readers kno w what’s happ enin g in the burner.

Y ou can no w dra w a line sho wing acet ylene flo wing out of the CaC 2 reactor an d in to the acet ylene burner. W e will w an t the acet ylene gas to b e at ro om temp erature (25C), since it is tec hnically thermo dyn am ically unstable (p ositiv e en thalp y of for m ation ).

Y ou ma y ha v e noticed that the o xygen in Equation 2 has to come f rom somewhere. Rather than supply the bu rner with p ure o xygen, whic h w ould b e exp ensiv e, y ou can just add air. Dra w a blo c k for an input of air in to the acet ylene burner. F or simplicit y’s sak e, let’s mo del the air as 80% nitrogen (N 2 ) and 20% o xygen (O 2 ). Dra w t w o sub-b o xes to indicate eac h of these gases, so that the reader kno ws ho w y ou are mo deling this mixture of gases b etter k no wn as air.

No w dra w a line lea ving the acet ylene reactor, that will con tain th e heated gaseous pro ducts (carb on dio xide and w ater v ap or) with it. This will b e what actually transfers heat in to the so dium via a heat exc hanger.

3.3 Heat Exc hanger with So dium

No w that w e ha v e a v ery hot stream of gases, let’s use it to transfer heat in to the molten so dium (Na). Dra w a b o x for the gas/Na heat exc hanger, with a dot te d line do wn th e mid dle to represen t the ph ysical separation b et w een the fluid streams, as seen on th e righ t of Figure 1. Y ou can no w dra w arro ws to represen t the flo w of Na, le ad ing in and out of the heat exc hanger. W rite do wn the k ey parameters of the Na giv en to y ou b y y our partner in industry . Let’s use a coun terflo w heat exc hanger, as those tend to b e more c ompact than comparable parallel flo w heat exc hangers. W e will mak e edu c ated guesses as to the temp eratures of the gas es en tering an d exiting the heat exc hanger so on.

3.4 Steam Reco v ery and Gas Separation

It is also desired to k eep the steam inside the cycle, as higher purit y steam w i ll result in less corrosion in the system, and require few er inputs as a whole. A condenser will w ork w ell to separate the condensable gases (w ater v ap or) from the non-condensable gases (ca r b on d io xide).

Dra w a b o x to represen t the condenser, as sho wn at the b ottom of Figure 1. Y ou c an no w dra w a line that represen ts the carb on dio xide lea ving the condenser, and sho wing liquid w ater (whic h has b een c ondense d ) lea ving th e condenser.

3.5 The Pump

In order to k e ep the sys tem flo wing and pressurized, w e need to add a pump. No w is a go o d time to set an initial system pressure. Let’s assume the system is at or near 1 MP a (10 bar), a mo dest p res sure for these systems. Dra w a b o x to r e p re sen t the pump, as s h o wn in the lo w er left of Figure 1. Connect the liquid w ater lea ving the condenser to one s id e of the pump, and dra w liquid w ater lea ving the pump from the other side. No w is a go o d time to fill in pressures for eac h no de. The liquid w ater lea ving the pump will b e at 1 MP a, while th e gas stream lea vin g the heat exc hanger will certainly undergo some pressure losses. Represen t these b y p enciling in a gas p re ssure of “ < 1 MP a.” W e’ll calculate this actual n um b er later, but w e’ll assume that

it is close to 1 MP a for en thalp y calculations.

3.6 Add an In terco oler

Remem b er the w ater v ap or en tering the CaC 2 reactor in Section 3.1? Notice that w e ha v e liquid w ater lea ving the pump and gaseous w ater en tering the CaC 2 reactor. Also notice that the gas stream that w ould ha v e en tered the condenser is probably coming in at a v ery high temp erature, whic h will b e w ell o v er 100C. This is a go o d place to add an inter c o oler , wh ic h remo v es heat from one flui d stream and transfers it to another. An in terco oler is esse n tially a heat exc hanger b et w ee n t w o parts of the same lo op.

Dra w a b o x with a d otte d lin e in the middle to represen t the in terco oler, as sho wn in the cen ter of Figure

1. Connect th e output of the gas/Na heat exc hanger and the input of the condenser to one side of the in terco oler. No w connect the output of the p ump and the steam input to the CaC 2 reactor to the other side. Notice that the w ater en ters as a liquid and lea v es as a solid. The h e at from the in terco oler will b e b oiling this w ater. Kee p this in mind when w e guess the temp eratures en tering and lea ving the in terco oler, whic h w e will do so on.

3.7 Accum ulate Useful P arameters

No w is the b est time to use lo okup tables and correlations to get whatev er th e r m o dynamic and ph ysical prop erties y ou thin k y ou will need. F or this example, this w ould include the follo wing:

• Saturation temp erature for 1 MP a steam (180C)

• Molar masses k g for all c hemical comp ounds in the system

• En thalpies of formation k J for all c hemical comp ounds in the system

• Sp ecific heat constan ts at c onstant pr essur e c p , c p

kJ &

kJ for all c hemical comp ounds in

• En thalp y of v ap orization h fg kJ for an y comp oun ds that ma y undergo a phase c hange in the system

• Key en th a l pies of comp ounds that wil l b e trac k ed as they flo w through the sys tem

– Here, “en thalp y” is a re l ativ e term. It re f e r s to the en thalp y ab ove the en thalp y of formation at 25C Δ 0 H f for the comp ound. This includes the terms for heating. Remem b e r that Δ 0 H f is th e en thalp y rele ased up on creating that comp ound from its constituen t elemen tal sp ecies, and that it is differ ent for liquid and gaseous phases. That also means that Δ 0 H f is zer o for elemen ts.

T ables of useful n um b ers for this problem ha v e b een compiled in T able 1.

3.8 Summary

Y ou should no w ha v e a blo c k di a gr am v ery similar to that in Figure 1. It seem s that w e ha v e a complete lo op, with ev erythin g accoun ted for. W e’ll see ho w true this is in the next section.

Before pro ceeding, fill in an y parameters that y ou kno w from the start without p erforming an y detailed calculations. These w ould include appro xima t e s ystem pressures (1 MP a in the lo op, 0.1 M P a for gases lea ving the condenser and en tering the atmosphere), all parame ters for so dium, and an y temp eratures that y ou w an t to sp ecify (25C acet ylene, for example).

Keep EVER YTHING in the same unit system, preferably SI.

4 Iteration #1

W e will no w b egin fi lling in the missing parameters f or the system, b y c ompu ting what w e can from prop erties b oth guess ed and giv en. Educated guess es based on the ph ysics of eac h sub-system will b e used to fill in missing parameters to pro ceed.

The metho dology will in v olv e p erforming the easiest calculati on that w e can do, and w orking either one step for w ard or bac kw ard in our thermo dynamic lo op to get more of the parameters w e are lo oking for. Changes made to the blo c k diagram from Figure 1 in this i te rati on will b e mark ed in blue .

Useful thermo dynamic and c hemical prop erties
Comp ound	c p k J kg · K	c p k J mol · K	M . M . k g mol		Δ 0 H f k J mol
H 2 O (g )	2.44	0.044	0.018		-241.8
CO 2 (g )	0.843	0.037	0.044		-393.5
N 2 (g )	1.039	0.029	0.028
O 2 (g )			0.032		0
CaC 2 (s)			0.064		-59.8
C 2 H 2 (g )		0.0875	0.026		+227.4
Ca (OH) 2 (s)		0.044	0.074		-985.2
				Sat. H 2 O at 1 MP a (T sat = 180 C)
				Prop ert y		V alue
				c p k J l k g · K		4.405
				c p k J g k g · K		2.713
				h fg k J k g		2014.5

T able 1: Us efu l tables of prop erties f rom lo okup tables for the CaC 2 / acet ylene reactor example

4.1 Heat T ransfer Across the Gas/Na Heat Ex c hanger

W e k no w the temp erature c hange and mass fl o w rate across the so dium side of this heat exc hanger, so w e can calculate its heat transfer with the follo wing form ula:

Q ˙ = m ˙ c p ( T out − T in ) (3)

where Q ˙ is the h e at transfer rate in m ega w atts (MW), m ˙ is the mass flo w rate in kg , c p is the heat capacit y in k J , and T out & T in are the outlet and inlet te mp eratures, resp ectiv ely . W e kno w m ˙ , T out & T in . W e

can lo ok up that c p Na

= 1 . 228 k J . Plugging in all these v alues giv es us Q ˙ = 4 . 912 M W . W rite th is n um b er

inside the gas/Na heat exc hanger.

4.2 Molar Gas F ractions

W e can no w find the molar gas fractions of the gas streams en tering the gas/Na heat exc hanger. F rom Equation 2, w e kno w that there are t w o moles of CO 2 created for eac h mole of H 2 O created. W e also kno w that there are fiv e moles of O 2 consumed for ev ery four moles of CO 2 created. Finally , w e are mo deling air as an 80- 20 mix of N 2 − O 2 , so w e kno w that there are four moles of N 2 consumed for ev ery mole of O 2 consumed. That means that if fiv e moles of O 2 m ust en ter the reaction, then t w e n t y moles of N 2 get tak en along for the ride.

W e can use this information to mo dify Equati on 2 to ge t the new total equation:

2 C 2 H 2 ( g ) + 5 O 2 ( g ) + 20 N 2 ( g ) → 4 C O 2 ( g ) + 2 H 2 O ( g ) + 20 N 2 ( g ) Δ H R X N =??? (4)

It is imp ortan t to note that nitrogen is presen t in this reactor, ev en though it is not reac tin g. W e will see wh y v ery so on. Add the t w en t y moles of nitrogen to eac h side of the equation in th e ac et ylene burner on the blo c k diagram, and add N 2 ( g ) to eac h gas stream from the acet ylene reactor to the condenser, wh e re the nitrogen will escap e in to the atmosphere.

No w is a go o d time to get the mass fractions of the gases in this stream from their molar ratios and their molar masses. W e just found their molar ratios, and the molar masses w ere compiled in T able 1. Us in g that information, w e get the f ollo wing information:

(4 mol C O ) · 0 . 044 k g = 0 . 176 k g C O f or ev er y

(2 mol H O ) · 0 . 018 k g = 0 . 036 k g H O f or e v er y (5)

Image by MIT OpenCourseWare.

Figure 2: T ypical temp erature distribution in a coun terflo w heat exc hanger

(20 mol N ) · 0 . 028 k g = 0 . 560 k g H O

2 mol 2

If w e find the re l ativ e mass p ercen tages from the n u m b e rs in Equation 5, w e get a mix of 22.8% CO 2 , 4.7% H 2 O, and 72.5% N 2 . As a sanit y c hec k, mak e sure these fractions add up t o 100%. These mass fractions will b e useful when calculating the m ass flo w rate of the gas stream in Section 4.4 b elo w, esp e cially since the en thalpies in T ab le 1 are gi v en in kJ .

4.3 Gas/Na Heat Exc hanger T emp eratures

W e don’t y et kno w the temp eratures of the gases en tering and lea ving the gas/Na heat exc hanger. No w is the time to mak e an educated gues s. Recall that for a coun terflo w heat e xc hanger, the graph of the temp eratures lo oks lik e that in Figure 2. The gas stream’s outlet temp erature T out gas cannot lie b elo w the

inlet temp erature of the so dium (T in Na ), and the gas stream’s inlet temp erature

T in gas

m ust lie ab o v e the

so dium’s outlet temp erature (T out Na ). Since the so dium en ters at 100C and lea v es at 500C, and T out gas will probably b e b et w een them, let’s c ho ose the midp oin t of 300C for T out gas for no w. Also, since T in gas m ust exceed T out Na , and the gases will b e c oming from a hot acet ylene flame, let’s just assume t hat T in gas = 900 C . Note that the actual guess for these parameters are not imp ortan t no w, as they will b e iterated up on and refined later.

4.4 Mass Flo w Rate Lea ving the Acet ylene Burner

No w w e ha v e enough information to find the mass flo w rate of the three-gas mixture lea ving the acet ylene burner, and consequen tly the mass flo w rate of eac h of the three consituen t gases. W e can mo dify Equation 3 to use sp ecific en thalpies as follo ws:

Q ˙ = m ˙ ( h out − h in ) (6)

where h out and h in are the a v erage sp ecific en thalpies of the gas mixture outlet and inlet, resp ectiv ely . Since w e ha v e three gases, their mass fractions and the en thalpies at the outlet & inlet temp eratu re s, w e can mo dify

Equation 6 to accoun t for these three gase s:

3

Q ˙ = m ˙ total [ x n ( h o n − h i n )] (7)

n =1

where m t ˙ otal is th e total m ass flo w rate of the gas stream in kg , x n is the mass fraction of one gas, and h o

sec n

& h i n are the outlet and inlet en thalpies, resp ectiv ely , of that gas. Substituting in the gas mass fractions and

en thalpies, alon g with the heat exc hanger’s heat transfer rate from Section 4.1, w e get:

4 , 912 k W = m ˙ total · [(0 . 228) (1485 − 771 . 6) + (0 . 047) (4340 − 3055 . 4) + (0 . 725) (1287 − 600 . 3)] (8)

⇓

4912 k W = m ˙ total · [162 . 66 + 60 . 38 + 497 . 86]

⇓

4912 k W k g

m ˙ total = 720 . 92 k J

= 6 . 814

sec

W rite this m ass fl o w rate as that lea ving the ac et ylene reactor and en tering the gas/Na heat e xc hanger. This is also the mass flo w rate en tering & lea ving the in terco oler. No w, since w e ha v e the mass fraction of w ater (4.7%), w e kn o w that

k g m ˙ H 2 O = m ˙ total × 0 . 047 = 0 . 320 sec

(9)

W rite this as the mass flo w rate of w ater lea ving the condenser, en tering and lea ving the pump, en te ri ng and lea ving th e in terco oler, and en tering the CaC 2 reactor.

4.5 Mass Flo w Rates of Acet ylene and Remaining Sp ecies

The only ma jor mass flo w rate m i s sing inside the lo op is that of acet ylene gas. T o find it, w e note that the mass flo w rate of w ater (0.320 kg ) corresp onds to 17.78 mol (see T able 1 for w ater’s molar mass). W e also kno w that the molar ratios of H 2 O and C 2 H 2 in the acet ylene burner are 1:1, so 17.78 mol of C 2 H 2 flo w through th e lo op. Using the molar mass of C 2 H 2 from T able 1, w e find a mass flo w rate of 0.462 kg for C 2 H 2 .

W rite this as the mass flo w rate of C 2 H 2 on the blo c k diagram.

W e can easily find the mass flo w rates of CaC 2 in and C a (OH) 2 out of the CaC 2 reactor, since they are all in a 1:1:1 molar ratio with C 2 H 2 . Us in g the molar masses in T able 1, w e find mass flo w rates of 1.138 kg

for CaC 2 and 1.316 kg for Ca (OH) . W rite these do wn on the blo c k diagram, as they are k ey in puts and

sec 2

outputs for th e system.

W e can also use these molar ratios to find the mass flo w rate of air required b y the acet ylene burner, and therefore the individual mass flo w rates of nitrogen and o xygen in to the acet ylene burner. Note that

acet ylene has a 2:5 molar ratio with o xygen, so th e r e are 5 · 17 . 78 = 44 . 45 mol of o xygen en tering, or 1.42 kg .

2 sec sec

Nitrogen has a 4:1 molar ratio with o xygen, so it requires (4) · 44 . 45 = 177 . 8 mol of flo w, whic h corresp onds

to 4.98 kg of nitrogen. W rite these do wn on the blo c k diagram, as they will determine the size an d p o w er of the air in tak e unit to b e sp ecified.

4.6 W ater/Steam T emp eratures En tering & Lea ving the In terco oler

W e ha v e n ’t y et sp ecified the temp eratures of t he w ater e n tering and the steam lea ving the in terco oler. W e kno w from Section 3.5 that the w ater en tering the in terco oler is a liquid, and w e ha v e as sumed in Section

3.1 that the w ater lea ving the in terco oler m ust b e a gas . The saturation temp erature of w ater at 1 MP a w as found to b e 180C. Le t us ass u m e th at the liquid w ater is co oled as little as p ossible with some small margin, sa y to 170C. Let us also as sume that the in terco oler will pro vide just enough heat to v ap orize this w ater, plus ten degrees of sensible heat. This giv es an exit temp erature of 190C. W rite these t w o temp eratures do wn on the blo c k diagram.

4.7 In terco oler P o w er

W e no w ha v e e n ough information to find the p o w er transferred in the in terco oler. W e kno w that enough heat has to b e transferr e d to transform 170C l iquid w ater to 190C w ater v ap or at 1 MP a. The equation for heat transfer across the in terco oler is as follo ws:

Q ˙ I nter c oo l er = m H O · c p (180 C − 170 C ) ˙ + h f g + c p (190 C − 180 C ) (10)

W e kn o w ev ery term on the righ t side of Equation 10, from a com bi nation of v alues from Section 4.4 and T able

1. Plu gging these v alues in, w e arriv e at Q ˙ I nter cool er = 667.4 kW. W rite this n u m b e r inside the in terco oler on

the blo c k diagram. Y ou can no w use this v alue to dete r m in e the temp erature of the gas stream lea ving the hot side of the in terc o oler, as w ell as the temp erature of the exhaust gases lea ving the condenser.

4.8 Something Is Missing!

Lo ok carefully at the inputs and ou tputs for the CaC 2 reactor. Do es conserv ation of mass apply? Th e answ er is NO! Note that w e only ha v e half the required w ater en tering the CaC 2 reactor! Chec k th e n um b er of moles of acet ylene lea ving the CaC 2 reactor, compared with the n um b er of moles of w ater en terin g it. Thi s w ater m ust come from somew h e r e .

Add an extra 0.320 kg of w ater e n tering the CaC 2 reactor. Let’s assume that w e’v e heated it to 190C, and pressurized it to 1 MP a for simplicit y’s sak e. Dra w thi s as a new input on y our blo c k diagram.

4.9 Summary of Iteration #1

No w y our blo c k diagram should resem ble th at in Figure 3. A t this p oin t, w e ha v e b alance d the lo op from a thermofluid p oin t of view. Ho w ev er, note th at there are still a few k ey missing features to the diagram. The en thalpies of reaction ha v e not b een filled in, so w e don’t kno w th e outlet temp eratures from e i ther the CaC 2 reactor or the acet ylene burner. W e will ha v e to find these missing n um b ers, figure out ho w hot the outlet gas streams are, and c hec k our assumptions accordingly .

5 Iteration #2

As w e men tioned in Section 4.9, w e will b e adv ancing our diagram from a purely thermofluid analysis to a com bined thermofluid an d thermo chemic al balance. This will in v olv e finding the e n thalpies of reaction for eac h ma jor c h e mical reaction o ccurring, and propagating the consequences along our lo op. Changes made to the blo c k diagram from Figure 3 in this iteration will b e mark ed in red .

5.1 En thalp y of Reaction for Acet ylene Com bustion

Using the en th alpies of formation from T able 1, w e can determine the en thalp y of reaction p er mole of acet ylene. This en thalp y is the energy a v ailable for heating the p ro ducts ( b oth c hemical pro ducts and those “coming along for the ride”) lea ving the acet ylene burner. Recall Equation 2, and sub s tit ute the en thalpies of formation for eac h c hemical sp ec ies, lea ving the stoic hiometric constan ts i n place . T o sa v e a s tep, w e will divide eac h stoic hiometric constan t b y t w o, since t w o moles of acet ylene are consumed, and this equation giv es the en thalp y p er mole of acet ylene:

Δ H R X N = Σ H P r odcuts − Σ H R eactants (11) 2 ( − 393 . 5) 1 ( − 241 . 8) 1 (227 . 4) 2 . 5 (0)

Δ H = `

2 C ˛ ¸ O x + `

H ˛ ¸ O

x − ` C ˛ H ¸ x − ` 2 . 5 ˛ ¸ O x = − 1256 . 4 k J

W e can ignore nitrogen for no w, s in c e in either case it w ould cancel out. W rite this en thalp y of reaction on the blo c k diagram. This en th alp y is no w a v ailable to heat up the CO 2 , H 2 O, and N 2 lea ving the acet ylene

F i g u r e 3: B l o ck diagram for i t e r a t i o n #1 of t h i s system . Changes since i n i t i a l setup h av e b een mark ed i n

b l u e .

burner. Using a v arian t of E quation 7, but using mole fr actions instead of mass fractions (b ecause our v alues for c p in T able 1 are giv en in k J ), w e arriv e at the follo wing equation :

Q ˙ = Δ T ·

3

n =1

# mol es n # mol es C 2 H 2

· c p n

(12)

1256 . 4 k J = ( T out − 25 C ) · ⎢ ⎣ `

2 C ˛ ¸ O 2

x + `

H ˛ 2 ¸ O

x + `

1 0 ˛ N ¸ 2

x ⎥ ⎦

T out = 3054 C

That’s quite hot, F AAAAR hotter than our assumed outlet temp erature of 900C! As a sanit y c hec k, w e can lo ok up that the flame temp erature of an air-acet ylene mixture is indeed in the range of 2400C, since there are some losses (radiativ e and other) in the real w or ld.

5.2 What Do W e Do No w?

Either w a y , w e will ha v e to c hange something to a v oid gases en tering our heat exc hanger at 3054C, wh ic h w ould s u re l y melt it. There are a n um b er of p ossibilities, some more elegan t that others. A few include:

• Changing fuel s ou rc es

• Dumping the excess heat

• Limiting the com bustion efficiency (incomplete reaction)

• Storing excess heat

• Adding thermal m ass to absorb energy

The first option is n ot an option, since the free calcium carbide is the reason w e’re here in the first place ! Dumping the e xcess heat through a heat sink is p ossible, though w e w ould b e w asting a large amoun t of our p o w er. Limiting the com bustion efficiency do es n ’t dump an y heat, but it DOES pu m p explosiv e acet ylene through our system, w h ic h w e m ust a v oid. Storing e xcess heat only w orks as long as there is storage capacit y a v ailable. The on ly option that seem s elegan t enough is to add extra thermal mass to ab s orb energy , k eeping all of our energy at a lo w er temp e rat ure.

5.3 Adding Thermal Mass to the Acet ylene Burner

T ak e a close lo ok at the reaction in Equation 4. As w e men tioned in Section 5.1, nitrogen is “going along for the ride” in this system. It d o esn’t react, but it DOES absorb heat and lo w er the o v erall temp erature of the pro ducts, as the same enthalpy is a v ailable to heat mor e pr o ducts . T h is explai ns th e d iff erence in flame temp eratures b et w ee n air-acet ylene (˜2400C) and o xy-acet ylene (˜3050C).

If w e w an t to k eep our flame outlet temp erature at 900C, where w e w ould lik e it to b e, w e need to figure out ho w m uc h extra thermal mass to add to l o w er the temp erature. Let’s assume that w e ha v e more w ater (w e m ust b e near a lak e or a riv er!), en tering as 25C co ol ing w ater. This w ould require a complex burner/b oiler design, but w e’ll ignore the sp ecifics for no w.

W e can mo dify Equation 12 to include this new amoun t of w ater ( x mor e mol es ), b y fixing the outlet temp erature of the acet ylene burner at 900C:

1256 . 4 k J = (900 C − 25 C ) · ⎢ ⎣ `

2 C ˛ ¸ O 2

x + `

H ˛ 2 ¸ O

x + `

1 0 ˛ N ¸ 2

x ⎥ ⎦

(13)

W e find that x m ust equal 24.88 mol of excess w ater p er mole of ac etylene . Remem b e r that Equation 13 is normalized to one m ol e of acet ylene. If w e m ultiply this n um b er b y the actual moles of acet ylene flo wing through the system (17.88 mol), w e arriv e at a required 442.37 mol , or 7.97 kg , of w ater en tering the acet ylene burner. W rite this as a new input to the acet ylene burner on the blo c k diagram.