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Chapter 11Hydrogen StorageProblem Solutions

Fund. of Renewable Energy Processes Prob. Sol. 11.1 Page 1 of 3 395

Prob 11.1 An alloy, Mg2Ni, interacts with hydrogen so thatthe following reversible reaction takes place:

Mg2Ni + 2H2 ↔ Mg2NiH4.

The thermodynamic data for the absorption reaction are (perkilomole of H2):

∆Hf0 = −64.4 MJ,

∆Sf0 = −122.3 kJ/K.

The relevant atomic masses (in daltons) are

H 1Mg 24.3Ni 58.7

In the questions below, all pressures are in the plateau region.

a. What is the dissociation pressure at 300 K?.....................................................................................................................

ln p = −∆S

R+

∆H

RT= −

−122.2× 103

8314+

−64.4 × 106

8314× 300= 14.7 − 25.82 = −11.12 (1)

p = 14.8 × 10−6 atmos. (2)

Owing to the great stability of the hydride (high energyof formation), the hydrogen equilibriumpressure at 300 K is surprisingly low:

15 millionths of an atmosphere.

b. At what temperature is the equilibrium pressure 1 MPa?.....................................................................................................................

Inverting Equation 2,

T =∆H

R ln p + ∆S=

−64.4 × 106

8314× 10 + 122.2× 103(3)

T =∆H

R ln p + ∆S=

−64.4× 106

−122.2 × 103 + 8314 ln 10= 625 K. (4)

The equilibrium pressure reaches 1 atmosphere at 625 K.

Solution of Problem 11.1

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396 Page 2 of 3 Prob. Sol. 11.1 Fund. of Renewable Energy Processes

c. What are the proportions, by mass, of the elements in Mg2Ni?.....................................................................................................................

The molecular mass of the alloy is 2 × 24.3 + 58.7 = 107.3 daltons.Hence, magnesium represents 2 × 24.3/107.3 = 0.453 or 45.3%, while Nirepresents 58.7/107.3 = 0.547 or 54.7%.

The composition of the alloy (by weight) is45.3% magnesium and 54.7% nickel.

d. A vessel with an internal volume of 1000 cm3 contains 1.56kg of Mg2NiH4. The density of this material is 2600 kg/m3.The saturated hydride is placed in the evacuated vessel. Thetemperature is raised to 400 C, but no hydrogen is allowed toescape. How many kg of hydrogen are desorbed? How manyremain in the hydride?

.....................................................................................................................1.56 kg of alloy occupy a volume of 1.56/2600 = 600× 10−6 m3 or 600

cm3. Thus, the dead space inside the vessel is 400 cm3.The pressure at 400 C (673 K) can be calculated the same way as in

Question 1:

ln p = −∆S

R+

∆H

RT= −

−122.2× 103

8314+

−64.4 × 106

8314× 673= 3.19 (5)

p = 24.2 atmos (6)or

p ≈ 2.4 × 106 Pa. (7)

From the perfect gas law, the amount of H2 in gas form is,

µ =pV

RT=

2.4 × 106 × 400 × 10−6

8314 × 673= 170 × 10−6 kmole. (8)

This corresponds to 340 mg of H2. The molecular mass of the hydride is107.3+4 = 111.3 The hydrogen represents 4/111.3=0.036 of the total mass.While cold, the amount, MH , of hydrogen in the hydride was 0.036×1.56 =0.056, or 56,000 mg. Thus, some 0.6% of the hydrogen was desorbed. Theamount remaining is 56,000 − 340 = 55,660 mg.

340 mg of hydrogen were desorbed,55,700 mg remain in the hydride.

e. How much heat must be added to desorb the rest of the hy-drogen? Assume that desorption stops when the empirical

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formula of the material reaches the Mg2NiH0.4 composition.The hydrogen desorbed is removed from the vessel so thatthe pressure remains constant.

.....................................................................................................................The amount of hydrogen in the material after desorption represents

10% of the amount in the saturated hydride, or 5600 mg. Thus, the amountdesorbed is 55500 − 5600 = 49900 mg or 0.0499 kg or 0.025 kmoles. Theenergy necessary to desorb this hydrogen is

W = 0.025× 64.4 = 1.6 MJ. (9)

The energy necessary for “total” desorption is 1.6 MJ.

f. If a hydrogen compressor were built using the vessel above,and if the hydrogen were fed into the system at 105 Pa and85 C, what would the theoretical efficiency of the compressorbe? The output gas is at 85 C and 5 MPa. Neglect all losses.

.....................................................................................................................The efficiency would be

η =RT ln r

∆H,

where r is the compression ratio, 50, in the present case.

η =8314 × (273 + 85) × ln 50

64.4 × 106= 0.18.

The efficiency is 18%.

Observe, however, that when all losses are included the efficiency maybe as low as some 5%.


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Prob 11.2 Figure 11.8 (text) shows the pressure-composi-tion isotherms for the reaction of hydrogen with LaNi5. Assumethat, in the plateau region, the isotherms are horizontal—that is,that the pressure does not depend on the composition. Assumefurther that the pressure at 40 C is 0.3 MPa and at 140 C is 5MPa.

One kg of LaNi5H5 is placed inside a pressure vessel andheated to 140 C. Hydrogen is slowly withdrawn until the hydrideis left with the empirical formula LaNi5H2.

Molecular masses: La, 138.9; Ni, 58.7.

The ∆H for absorption for this alloy is -32.6 MJ per kilomoleof H2, and the ∆S is -107.7 kilojoules per kelvin per kilomole.

a. To maintain the temperature at 140 C during the desorption,must heat be supplied to or withdrawn from the hydride?

.....................................................................................................................

Desorption is endothermic. If heat is not supplied, temperature falls.

To maintain the temperature at a constant 140 Cduring desorption, heat must be added to the hydride.

b. How many joules of heat must be exchanged with the hydrideto perform the desorption?

.....................................................................................................................

We must first calculate the density of LaNi5H5:

The molecular mass of LaNi5H5 is 1 × 138.9 + 5 × 58.7 + 5 × 1 =437.4 daltons. This means that 1 kilomole of the alloy masses 437.4 kg.Consequently, 1 kg corresponds to 2.29 × 10−3 kilomoles.

When the hydride changes from LaNi5H5 to LaNi5H2, 1.5 kilomoles ofH2 (3 kilomoles of H) are desorbed for each kilomole of hydride. Thus, thehydrogen desorbed is 2.29 × 10−3 × 1.5 = 3.43 × 10−3 kilomoles of H2.

The energy required for the desorption is 32.6 × 106 × 3.43 × 10−3 =112, 000 J.

112 kJ of heat are required to desorb the hydrogen.

c. What is the change in entropy of the system during desorp-tion? Assume that ∆S is temperature independent. Does theentropy increase or decrease during the desorption?

.....................................................................................................................

During desorption, the hydrogen goes from a highly ordered form inthe crystal to a highly disordered for in the gas—the entropy increases.

The entropy increases is 107.7 × 103 × 3.43 × 10−3 = 369 J K−1.

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Entropy change due to desorption is 369 J K−1.

d. How many kilograms of hydrogen were withdrawn?.....................................................................................................................

From Answer 2, we see that 3.43×10−3 kilomoles of H2 or 6.86×10−3

kg of H2 were desorbed.

6.86 grams of hydrogen were withdrawn.


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Prob 11.3 Hydrogen is to be transported. Two solutionsmust be considered:

1. Liquefy it.2. Convert it to ammonia. Later, when hydrogen is to be used,

the ammonia can be catalytically cracked with 85% recoveryof hydrogen. 46.2 MJ per kg of ammonia are required todissociate the gas.

Purely from the energy point of view, which is the more eco-nomical solution?

From the Textbook (Section 11.2), it can be seen that liquefying hy-drogen costs about 40 MJ/kg or 80 MJ/kmole. To obtain liquid hydrogenwe invest 286 MJ/kmole to obtain the gas and 80 MJ/kmole to liquefy it.We recover only 286 MJ/kmole when using the liquid. Thus, the processefficiency is

ηcryo =286

286 + 80= 0.78.

In the ammonia method, we produce 2/3 kilomole of NH3 for eachkilomole of H2. The reaction that produces ammonia is exothermic but theheat produced cannot be stored.

To recover the hydrogen, we must dissociate the ammonia into itselements and this takes 46.2 MJ per kilomole of ammonia or 2/3 × 46.2 =30.8 MJ per kilomole of hydrogen produced. Of this, only 85% of the gasis saved. The efficiency of the process is

ηammonia =286 × 0.85

286 + 30.8= 0.77.

From the energy point of view, the two processes are about equal.

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Prob 11.4 Calorimeter measurements show that when acompound, AB, reacts with H2 forming a hydride ABH, 18.7MJ of heat are released for each kilomole of H2 absorbed. Themanufacturer of this hydride consults you about the advisabil-ity of shipping it (saturated with hydrogen) inside a normal gaspressure cylinder (which is rated at 200 atmospheres). Duringshipment the cylinder may be exposed to the sun. Without hav-ing any additional data on the hydride, what would you advisethe manufacturer? Could you tell him the maximum temperaturethat the hydride can safely reach?.....................................................................................................................

You must be cautious in your advise to the manufacturer. Since you donot know the entropy change owing to hydration, you could make a guessat ∆S = −100 kJ per kelvin per kilomole of hydrogen. Then, the hydridewould reach the limiting pressure of 200 atmospheres at a temperature thatsatisfies

RT ln p = ∆H − T∆S

from which

T =∆H

R ln p + ∆S=

−18.7 × 106

8314 ln200 − 100, 000= 334 K.

This temperature (61 C) is uncomfortably low. Moreover, the estimateof the entropy change may be wrong: if the instead of -100 kJ K−1 kmole−1,the value is -110 K−1 kmole−1, then the maximum permissible temperatureis a cool 284 K (about 10 C). Recommendation: do not transport hydrogenthis way. You can use a more stable hydride, one with higher ∆H (inabsolute value).


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Prob 11.5a. Estimate the enthalpy change for the hydrogen absorption re-

action of an alloy that has a plateau pressure of 1 atmospherewhen T = 0 C.

.....................................................................................................................For all alloys, the temperature at which the plateau pressure is 1 at-

mosphere is given (roughly) by

T = 10∆H,

where ∆H is expressed in megajoules. Consequently,

∆H = T/10.

For T = 0 C or 273 K, ∆H ≈ −27.3 MJ/kmole of H2. The negativesign results from the absorption being exothermic.

By rough estimate, the enthalpy change owing to the absorptionof hydrogen in Alloy A is about -27.3 MJ/kmole of hydrogen.

b. Estimate the enthalpy change for the hydrogen absorption re-action of an alloy that has a plateau pressure of 1 atmospherewhen T = 30 C.

.....................................................................................................................For T = 30 C, i.e., T = 303 K, ∆H ≈ −30.3 MJ/kmole of H2.

By rough estimate, the enthalpy change owing to the absorptionof hydrogen in Alloy B is about -30.3 MJ/kmole of hydrogen.

Before proceeding with the rest of the solution, it is useful to tabulatethe different plateau pressures that will occur.

The plateau pressure of either alloy is

peq = exp

(

−∆S

R+

∆H

RT

)

atmos.

Using the two values of ∆H calculated above and our guess of −100 kJK−1kmoles−1 for the entropy change during absorption, we can constructthe table below:

Alloy TEMP. TEMP. PLATEAU.PRESS.

(C) (K) (atmos.)

A 0 273 1.0A 28 301 3.1B 28 301 0.9B 100 373 9.6

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Assume that Alloy A is in an environment that causes itstemperature to be 0 C (perhaps the outside of a house) and thatAlloy B is at 28 C (say, inside a room). The alloy containers areinterconnected by means of a pipe so that the hydrogen pressureis the same in both. The amount of hydrogen in the system issuch that, in any phase of the cycle, when one alloy is saturated,the other is depleted.

c. What is the system pressure?

.....................................................................................................................

1.0 atmos

3.1 atmos

A1

A2

Hydride A

10

5

2

1

0.5

0.2

0.1

Hydride B

9.6 atmos

0.9 atmos

28 C

100 C

B2

B1

28 C

0 C

At this phase of the cycle, the conditions are as follows:

TA = 273 K,

pA

= 1.0 atmos,

TB = 301 K,

pB

= 0.9 atmos.

Assume that initially the system contains no hydrogen and that athird container is used to load the gas. The pressure rises until it reaches0.9 atmos where it stabilizes as Alloy B starts absorbing the gas. Whensaturated, the pressure once again starts up. However, the problem specifiesthat when one alloy is saturated the other must remain depleted. Thus,loading must be stopped an the gas inside the system will be at a pressurelarger than 0.9 atmos (to saturate B) but less than 1.0 atmosphere (to keepA depleted). The alloys will be at points A1 and B1.

The system pressure will be between 0.9 and 1.0 atmos.


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d. Which alloy is depleted and which contains most of the hy-drogen?

.....................................................................................................................

Alloy A is depleted while Alloy B is saturated.

Raise the temperature of Alloy A to 28 C.

e. What is the plateau pressure? What is the actual hydrogenpressure?

.....................................................................................................................The plateau pressure of Alloy A is now raised to 3.1 atmos.However, the alloy was depleted, so, raising the plateau pressure only

accentuates this condition. Actually, the pressure might rise slightly be-cause the hydrogen is warmed up a bit.

The hydrogen pressure remains essentially unchanged.

Circulate hot water (from a hot source) through the heatexchanger of Alloy B and heat it up to 100 C (373 K). Maintainthis temperature.

f. What is the pressure of the system and what happens to thetwo alloys?

.....................................................................................................................At this new phase of the cycle, the conditions are now:

TA = 301 K,p

A= 3.1 atmos,

TB = 373 K,p

B= 9.6 atmos.

Hydrogen is desorbed from B and absorbed by A. Gas flows from Bto A. When the process is finished, the gas will be at a pressure somewhatabove 3.1 atmos and well below 9.6 atmos. The exact pressure depends onhow much gas is in the system.

The system pressure will be between 3.1 and 9.6 atmos.Alloy A will be saturated and Alloy B will be depleted.

Return the temperature of Alloy B to 28 C, and that of AlloyA to 0 C (by putting it in contact with the exterior).

g. What happens?.....................................................................................................................

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This re-establishes the initial conditions in which Alloy A was depletedand B was saturated. Thus desorption occurs in A and absorption, in B.Hydrogen flows from A to B.

Alloy A will desorb, becoming depleted.Alloy B will absorb becoming saturated.

h. Tabulate all the sorption heat inputs and outputs of the sys-tem and the corresponding temperatures. Ignore all otherheat inputs and outputs. For example, ignore the heat nec-essary to change the temperature of the system.

.....................................................................................................................

All heat quantities are per kilomole of hydrogen.

In Step 6, Alloy B is heated to 373 K. As soon as the rising temperatureexceeds 334 K (the temperature that causes its plateau pressure to exceedthe 3.1 atmospheres of Alloy A), Alloy B starts desorbing and draws 30.3MJ from the hot source. This energy drives the heat pump. The hydrogenis absorbed by Alloy A (at 301 K) and 27.3 MJ are rejected into the room.

In Step 7, Alloy B is returned to 301 K by stopping the hot watercirculation. The temperature falls toward 301 K while that of Alloy A fallstoward 273 K. As soon as the plateau pressure of Alloy A exceeds that ofAlloy B, the latter starts absorbing hydrogen and delivers 30.3 MJ of heatto the room. Alloy A, at 273 K, desorbs hydrogen and draws 27.3 MJ fromthe environment outside the house.

i. How much heat is delivered (per kmole of hydrogen) to theenvironment at 28 C in a complete cycle as described above.?

.....................................................................................................................

The room receives 30.3 MJ from Alloy B and 27.3 MJ from Alloy A,a total of 57.6 MJ (all per kilomole of H2).

The room receives 57.6 MJ of heat foreach kilomole of hydrogen handled.

j. Of this amount of heat, how much must come from the hotwater source (and must be paid for)?

.....................................................................................................................

The hot water source has to furnish 30.3 MJ foreach kilomole of hydrogen handled.


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k. What is the coefficient of performance of this heat pump—that is, what is the ratio of the heat delivered to the environ-ment at 28 C to the heat required from the hot water?

.....................................................................................................................

COP =57.6

30.3= 1.90.

The coefficient of performance is 1.90.

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Prob 11.6 A recently developed binary compound is to beused as a hydrogen storage medium because it forms a reversible(ternary) monohydride.

The data of the system include:

Molecular mass of the compound (no hydrogen): 88 daltons.

Density of the compound (hydrided or depleted): 8900 kgm−3.

Enthalpy of hydriding: −26.9 MJ per kilomole of H2.

Change of entropy owing to absorption of hydrogen: −100 kJK−1 per kilomole of H2.

Type of isotherm: single plateau.

Heat capacity of the compound: 200 J K−1 kg−1.

Heat capacity of the container and of the hydrogen gas: neg-ligible.

Heat capacity of water: 4180 J K −1 kg−1.

Density of hydrogen: 0.089 kg m−3.

2.5 kg of the compound (activated and in a fine powder ata temperature of 0 C) are placed inside a container measuringinternally 10 by 10 by 10 cm.

We want to adjust the system so that the alloy is saturatedand the gas pressure is exactly the plateau pressure. To findthis point we will have to observe the behavior of the pressurein the alloy canister as hydrogen is added. The pressure initiallyrises, then as the plateau is reached, it stabilizes until saturation isachieved. We will first overshoot this point, then, having removedthe hydrogen source, we will purge some of the gas, observingthe pressure and stopping the purge just when the pressure fallsto the previously observed plateau pressure. During this wholeoperation the temperature is carefully maintained at 0 C.

A 5-atmosphere hydrogen source is used to fill the container.The gas is applied for a time long enough to allow equilibrium tobe established. Any heat absorbed or released by this operationis removed or added to the system so that at the end of theoperation, the pressure is still 5 atmospheres and the temperatureis still 0 C. Next the hydrogen source is removed. The pressureinside the container remains at 5 atmospheres. A valve is crackedopen and hydrogen leaks out while the pressure is monitored. Assoon as the pressure stabilizes, the valve is closed.

a. How many grams of hydrogen were lost in the bleeding above?

.....................................................................................................................


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The volume occupied by the compound is

Vcompound =2.5

8900= 2.8 × 10−4 m3.

The volume of the container is 0.001 m3. Thus the “dead space” is720 × 10−6 m3. This is the volume occupied by the gaseous hydrogen.

The equilibrium pressure is

p = exp

(

−∆S

R+

∆H

RT

)

= exp

(

−−100× 103

8314+

−26.9 × 106

8314× 273.2

)

= 1.21 atmos.

Hydrogen will be bled until its pressure falls from 5 to 1.21 atmospheres(5.06 × 105 to 1.23 × 105 pascals). At the higher pressure the hydrogen ingaseous for corresponded to

µ =pV

RT=

5.06 × 105 × 720 × 10−6

8314× 273.2= 160 × 10−6 kilomoles

At the lower pressure, the hydrogen in the dead space is 39 × 10−6

kilomoles.Thus, 160×10−6−39×10−6 = 121×10−6 kilomoles of H2 (242×10−6

kg) were bled off.

242 milligrams of hydrogen were lost in the bleeding.

b. How much hydrogen remains in storage? Express the loss asa percentage of the stored gas.

.....................................................................................................................Assuming that the alloy is fully hydrided, there is one kilomole of H

for each kilomole of alloy.One kilomole of the alloy masses 88 kg. Hence, 2.5 kg of alloy cor-

responds to 2.5/88 = 0.0284 kilomoles of alloy and, consequently, 0.0284kilomoles of H (28.4 × 10−3 kg of H2). This is the amount of hydrogen inthe storage after the bleeding. The amount of hydrogen lost in bleeding is0.242 g.

28.4 grams of hydrogen remain in storage after the bleeding.The bleeding caused a loss of 0.85% of the stored gas.

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c. What is the pressure of the stored gas?.....................................................................................................................

The gas remaining in storage it at a pressure of 1.21 atmospheres.

The container is now immersed in a tank of water at 40 C.This tank contains 0.3 liters of water and is entirely adiabatic.Its walls have negligible heat storage capacity.

d. What is the temperature of the system (water tank plus alloycontainer) after equilibrium is reached?

.....................................................................................................................Let T be the equilibrium temperature.As the new temperature, T , is established, the water cooled down by

40 + 273.2 − T = 313.2 − T K. This means that an amount, Qwater =4180 × 0.3 × (313.2 − T ) = 392, 800 − 1, 254T joules, of heat energy wastransferred to the hydride.

The heat removed from the water heats up the hydride (Qhydride) andcauses some hydrogen to be desorbed (Qdesorb):

Qhydride = chydrideMhydride∆T J,

wherechydride = heat capacity of the hydride = 200 J/K per kilogram, andMhydride = 2.5 kg.Thus, Qhydride = 500(T − 273.2) = 500T − 136, 600 J.

Qdesorb = µdesorb × 26.9 × 106 J.

The heat balance requires

392, 800− 1, 254T = 500T − 136, 600) + µdesorb × 26.9 × 106,

1754T − 529, 400 + µdesorb × 26.9 × 106 = 0.

The amount of hydrogen in gaseous form is the sum of the amountthat was present at 0 C plus the amount desorbed:

µ = 39 × 10−6 + µdesorbed =pV

RT.

Hence,

µdesorb =pV

RT− 39 × 10−6,

1754T − 529, 400 + 26.9 × 106

(

pV

RT− 39 × 10−6

)

= 0,


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1754T − 530, 449 + 26.9 × 106

(

pV

RT

)

= 0,

1754T − 530, 449 + 2.33( p

T

)

= 0.

At T , the equilibrium pressure of the hydrogen is

p = exp

(

−∆S

R+

∆H

RT

)

= exp

(

−−100× 103

8314+

−26.9× 106

8314× T

)

= exp

(

12.03 −3236

T

)

= 167, 700 exp

(

−3236

T

)

atmos or 17 × 109 exp

(

−3236

T

)

Pa.

1754T 2 − 530, 450T + 39.6 × 109 exp

(

−3236

T

)

= 0.

A numerical solution leads to T = 300.8 K.

The equilibrium temperature is 300.8 K.

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HydrideA

HydrideB

Prob 11.7 An inventor proposes the following device to cooldrinks at a picnic. It consists of two sturdy containers (some-thing like small portable oxygen bottles) one of which (ContainerA) can be placed inside a styrofoam box in which there are 12beer cans. The other (Container B) is outside the box and canbe placed over a fire. A pipe interconnects the two containers.See the figure. The containers are filled with different alloys ca-pable of absorbing hydrogen. Alloy A is TiFe. Alloy B has to bedescribed.

When both containers are at the same temperature, 298 K,Alloy A is depleted and B is saturated.

a. For this to happen, what are the required characteristics ofAlloy B? Establish a relationship between the thermodynamicparameters of the two alloys.

.....................................................................................................................

If, when both alloys are at the same temperature, alloy A is depletedand B is saturated, then the plateau pressure of the former must be higherthan that of the latter.

ln pA > ln pB,

−∆SA

R+

∆HA

RT> −

∆SB

R+

∆HB

RT,

∆HB − T∆SB < ∆HA − T∆SA

∆GB < ∆GA

The free energy of desorption of Alloy B mustbe smaller than that of Alloy A.

The various ∆H and ∆S, above are for absorption. This means thatboth ∆H and ∆S are smaller than zero.

b. Which of the alloys from the table below can be used as AlloyB in the device under consideration?


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Hydride ∆ H ∆SMJ/kmole kJ/K

per kmoleAB -21.0 -96.5CD -26.1 -99.4EF -27.9 -106.8GH -32.1 -101.8IJ -32.6 -110.5KL -33.4 -98.3

.....................................................................................................................To answer this question, one must calculate the plateau pressure of the

different hydrides at T = 298. This is done by applying the formula

p = exp

(

−∆S

R+

∆H

RT

)

.

The result is

Hydride ∆ H ∆S pMJ/kmole kJ/K per kmole atmos.

AB -21 -96.5 22.9CD -26.1 -99.4 4.1EF -27.9 -106.8 4.9GH -32.1 -101.8 0.5IJ -32.6 -110.5 1.1KL -33.4 -98.3 0.2

Alloy EF is the TiFe of Container A. The only alloy, other than EF,that cannot be used is AB whose plateau pressure is higher than that ofEF......................................................................................................................

To operate the system, Container A is placed in a bucketof water and kept at 25 C. This requires refreshing the wateroccasionally. Container B is placed over the picnic fire so thathydrogen is transferred to Container A whose hydride becomessaturated.

Next, container A is placed inside the styrofoam box, in con-tact with the 355-ml beer cans which, for good thermal contact,are immersed in 4.5 liters of water.

Container B is now cooled to 298 K returning the system toits original state.

The heat required to desorb the hydrogen from A, will coolthe 12 cans of beer from 25 C to 10 C.

Assume that beer behaves as water, at least as far its ther-mal capacity is concerned. The styrofoam box is, essentially, adi-

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abatic. Assume that during the cycle the composition of thehydride in A varies from TiFeH0.95 to TiFeH0.4

The atomic mass of Ti is 47.9 daltons and that of Fe is 55.8daltons.

c. Estimate the minimum mass of TiFe required......................................................................................................................

Inside the styrofoam box, the volume of liquid to be cooled is 4.5 +12 × 0.355 = 8.76 liters or 0.00876 m3. The heat capacity of this liquid(essentially, all water) is 4.2 MJ/K per cubic meter. Thus the heat to beremoved is

Wcooling = 4.2 × 106 × 0.00876× (25 − 10) = 550, 000 J.

When 1 kmole of H2 is desorbed, it removes 27.9 MJ of heat. Toremove 550,000 J one needs 0.0197 kmoles of hydrogen.

Each kilomole of alloy will desorb (0.95 − 0.4)/2 = 0.275 kilomoles ofH2. For the amount of hydrogen to be handled, we need 0.0197/0.275 =0.0716 kilomoles of alloy.

The atomic mass of the alloy is 47.9 + 55.8 = 103.7 daltons. 0.0716kilomoles of alloy mass 0.0716× 103.7 = 7.42 kg.

7.4 kg of TiFe are needed.

Considering that the mass of the alloy in Canister B must be of thesame order as that in Canister A and considering also the mass of canistersand pipe, the whole system is a bit too heavy to become a popular picknick beer cooler.


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Prob 11.8 Two canisters are interconnected by a pipe.Canister “A” contains TiFe and is at 300 K, while Canister “B”contains CaNi5 and is at 350 K. The system is filled with hydrogenat a pressure of 4 atmospheres.

In which canister is the bulk of the hydrogen? No guesses,please! Use the thermodynamic data of Table 11.4 of the Text......................................................................................................................The plateau pressure of hydrogen in contact with a hydride is

ln p = −∆S

R+

∆H

RT.

For Canister “A”,

ln pA = −−106, 100

8314+

−28.0 × 106

8314× 300= 12.76− 11.23 = 1.534,

pA = 4.64 atmospheres.

Thus the hydrogen above the TiFe is a lower pressure than the plateaupressure. Consequently the alloy is depleted.

For Canister “B”,

ln pB = −−101, 200

8314+

−31.8× 106

8314 × 350= 12.17 − 10.93 = 1.240,

pB = 3.46 atmospheres.

Thus the hydrogen above the CaNi5 is at higher pressure than theequilibrium pressure. Consequently the hydride is saturated.

In all probability, most of the hydrogen is in Canister ˝B˝

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Prob 11.9 WARNING: This problem contains units, such asgrams, centimeters, and so on, that are not of the SI.

0 1 Stoichiometric index, x

ln px = 0.1 x = 0.9

Consider a vessel containing a hydrogen-absorbing alloy, AB.Vessel:

Volume: 200 cm3,Thermal insulation: adiabatic,Heat capacity: negligible.

Alloy:Formula mass: 120 daltons,Amount: 200 g,Heat capacity: 1700 J/K per kg of alloy (same for hydridedalloy).

Hydride:Heat of formation of ABH (absorption): −30.0 MJ per kmoleof H2,Entropy change owing to absorption: −110 kJ/K per kmoleof H2,Density (of ABH0.9): 1600 kg/m3.System:Initial temperature: 300 K,Initial hydrogen pressure: equilibrium.

Hydrogen is forced into the vessel until the average alloy com-position is ABH0.9 What is the minimum pressure required toforce all the needed hydrogen into the vessel? How much hydro-gen is forced in?.....................................................................................................................

Let us start by calculating the plateau pressure, p300.

p300 = exp

(

−∆S

R+

∆H

RT

)

= exp

(

−−110, 000

8314+

−30 × 106

8314× 300

)

= 3.33 atmos.


090723


Next, let us determine the number, N , of kmoles of alloy in the vessel.

N = 0.2 kg ×1

120

kmole

kg= 0.00167 kmole.

The amount of hydrogen (H) absorbed is 0.00167∆x, where ∆x is thechange in the stoichiometric index, In this case, the change is 0.8. Thus,1336 × 10−6 kilomoles of H or 668 × 10−6 kilomoles of H2 are absorbed.This causes a release of Q = 668 × 10−6 × 30 = 0.02 MJ of heat in thevessel.

Since the specific heat capacity of the alloy is 1700 J/K per kilogram,its total heat capacity is c = 0.2× 1700 = 340 J/K (for the amount of alloyin the vessel).

This causes a temperature rise of

∆T =Q

c=

0.02 × 106

340= 59 K.

The temperature reaches 300 + 59 = 359 K. The pressure is now

p418 = exp

(

−−110, 000

8314+

−30 × 106

8314× 359

)

= 27.1 atmos.

The minimum hydrogen feed pressure is 27.1 atmos.

The volume occupied by the alloy is

0.2 kg ×1

1600

m3

kg= 125 × 10−6 m3.

This is 125 cm3. Since the vessel has a capacity of 200 cm3, there isan “dead” space of 75 cm3.

At 27.1 atmospheres (2.75 MPa), the amount of gas in this space is

µ =pV

RT=

2.75 × 106 × 75 × 10−6

8314 × 359= 69 × 10−6 kmoles.

The amount of hydrogen pumped in must be 668 × 10−6 kmoles ab-sorbed plus 69 × 10−6 kilomoles of compressed gas, a total of 737 × 10−6

kilomoles.

The amount of hydrogen pumped into the vessel is 0.737mole.

090723



Prob 11.10 A perfectly adiabatic (heat insulated) vessel hasan internal volume of 100 cm3 and contains 240 g of an alloy pow-der, AB, that forms a monohydride, ABH. The thermodynamicdata for absorption are:

∆H = −28 MJ per kilomole of H2,∆S = −100 kJ/K per kilomole of H2,Heat capacity, cv = 400 J/K per kg of alloy.

Additional data includeFormula mass of the alloy = 150 daltons,Density of the alloy = 8000 kg/m3,Bulk density of the alloy powder = 4000 kg/m3.

The vessel has been charged with hydrogen so that the pres-sure is 10 atmospheres. The system is at 30 C.

a. Hydrogen is withdrawn. How many milligrams of the gas canbe removed without causing a change in the temperature ofthe hydride? Remember that the container is adiabatic. Theheat owing to the work that the withdrawn hydrogen mayexert is exchanged with the gas outside the hydride containerand does not influence the temperature of the latter.

.....................................................................................................................If the system is at 30 C (303 K), then the plateau pressure is

ln p = −∆S

R+

∆H

RT=

100 × 103

8314−

28 × 106

8314× 303= 0.913. (1)

This means that the plateau pressure is 2.49 atmospheres, much lowerthan the actual gas pressure. We cab release hydrogen until the pressurereaches the plateau value without causing any appreciable desorption.

The volume occupied by the alloy itself is 0.24 kg × 18000

m3

kg = 30×10−6

m3 or 30 cm3. Hence, the dead volume inside the vessel is 70 cm3.The amount of hydrogen in this dead volume is

µ =pV

RT=

106 × 70 × 10−6

8314 × 303= 27.8 × 10−6 kmole. (2)

When the pressure is down to 2.49 atmospheres (about 2.49 × 105

pascals), the amount of hydrogen left in the container is

µ =pV

RT=

2.49 × 105 × 70 × 10−6

8314 × 303= 6.92 × 10−6 kmole. (3)

Thus, 20.9 × 10−6 kilomoles of hydrogen or 41.7 mg can be removed.


090723


A total of 41.7 mg of hydrogen can be removedwithout changing the temperature of the hydride.

b. If more hydrogen is released, it will cause the cooling of thehydride. Assume that the vessel has no heat capacity. Howmuch hydrogen is released if the pressure falls to 1 atmo-sphere?

.....................................................................................................................The plateau pressure of 1 atmosphere corresponds to a temperature,

T1, of

ln 1 =100 × 103

8314−

28 × 106

8314T1= 0. (4)

From this, T1 = 280 K.Thus hydrogen will be released until the hydride reaches this temper-

ature. Since the only source of heat is that stored in the hydride itself, thelatter must loose an amount of heat

∆H ′ = 400J

K kg× 0.240 kg × (303 − 280) K = 2210 J. (5)

So, the amount of heat available for desorption is 2210 J. This is enoughto desorb 2210/28× 106 = 79× 10−6 kilomoles of H2 or 158 mg of the gas.

158 mg of hydrogen were released.

c. What is the value of x in the empirical formula ABHx afterthe above desorption?

.....................................................................................................................The amount of alloy in the container is

µ′ =0.240 kg

150 kg per kmole= 0.0016 kmole. (6)

For each kilomole of the alloy AB, there is half a kilomole of H2 atthe beginning of the desorption. Thus the amount of H2 in the hydridewas 800 × 10−6 kmole. At the end of the desorption, it was 800 × 10−6 −79× 10−6 = 721× 10−6 kilomoles. Thus x = 721/800 = 0.90. Only a smallfraction of the stored hydrogen has escaped.

After the release, x =0.90.

090723



Prob 11.11 The Pons and Fleishman cold fusion experimentemploys an electrolytic cell consisting of a palladium negativeelectrode and a platinum positive electrode. The electrolyte is aconcentrated solution of LiOH in D20. The palladium electrodeis a cylindrical rod 10 cm long and 1.2 cm in diameter. Just priorto the experiment, the rod is completely degassed by heating itup in a vacuum.

When a 0.5 A current is forced through the cell, nothingunusual happens for a long time. To be sure, normal electrolysisoccurs with D2 evolving at the palladium and 02 at the oxygenelectrodes. The D2O used up is continually replenished.

In some rare instances, it is claimed, after the electrolysis hasproceeded for a long time, heat suddenly begins to be producedin substantial amounts—73 W, in this case. This heat productionrate is sustained for 120 hours after which the cell is disconnected.

If you look up the enthalpies of formation of all palladiumcompounds, you will find that the largest value is associated withthe formation of palladium hydroxide: 706 MJ/kmole.

You will also find that the atomic mass of palladium is 106daltons and that the density of the metal is 12 g cm−3.

Can you prove that the energy generated is not chemical innature?

To explain the delay, assume that D2-D2 fusion occurs at arapid rate only if the deuterium is packed with sufficient den-sity and that this will happen only when the palladium is com-pletely saturated with deuterium and the formation of palladiumdi-deuteride begins. How long would you expect the cell to op-erate before it heats up?.....................................................................................................................

The volume of the palladium electrode is

V =πd2

4× L =

π × 1.22

4× 10 = 11.3 cm3,

where d is the diameter and L is the length of the electrode.The mass of the electrode is

M = V δ = 11.3 × 12 = 136 g.

Here, δ is the density in g/cm3.106 kg of palladium correspond to 1 kilomole, hence, the number of

kilomoles of palladium in the electrode is

N =0.136

106= 1.28 × 10−3 kmoles (Pd).


090723


If all the palladium in the electrode reacted chemically to producepalladium hydroxide (releasing 706 MJ/kmole), the released energy wouldne

Wchem = 706 × 106 × 1.28 × 10−3 = 906, 000 J.

This is about 1 MJ.However, the claim is that a total of 73 W were released over 120 hours.

This corresponds to an energy

Wexperiment = 73J

s× 120 hr × 3600

s

hr= 31.5 × 106 J.

Not only is the claimed energy release much larger than that you couldexpect from a chemical reaction, but, also, it was stated that there was noobvious corrosion of the electrode, in other words, the palladium did notreact chemically.

A 0.5 A current corresponds to a flow of

0.5coulombs

s×

electrons

1.60 × 10−19 coulombs= 3.1 × 1018 electrons

s,

Each electron corresponds to 1 deuterium atom, hence, the electrolysisliberated 3.1 × 1018 deuterium atoms.

The total number of palladium atoms in the electrode is 1.28× 10−3×6.02 × 10−26 = 770 × 1021

Assuming that all the liberated D2 was absorbed by the palladium, itwould take 770×1021/3.1×1018 = 247, 000 seconds (68 hours) to transformall the Pd into PdH.

If the fusion reaction would be initiated by the formation of PdH2,then you would expect a delay of some 68 hours between the beginning ofthe experiment and the beginning of massive heat production.

090723



Prob 11.12 A canister contains a mixture of two alloys (Al-loy 1 and Alloy 2). A hydrogen source equipped with a valve isconnected to this canister. A measured amount of the gas can bedelivered to it.

Describe the behavior of psystem (the hydrogen pressure, inpascals, read by a manometer connected to the canister) versusthe amount, µH2

of H2 (in moles) introduced into the system.Sketch a rough psystem vs µH2

graph.Estimate all the break points in the above sketch—that is, all

the values of µH2at which the curve changes abruptly its charac-

ter.Do the above for T=400 K, following the detailed instructions

in Items 1 through 5.Here are some data:Internal volume of the canister: 1 liter.The idealized p vs x characteristics of the alloys are as follows:Region 1 – The equilibrium pressure is proportional to the

stoichiometric coefficient, x, for 0 < p < pplateau. When x = xcritℓ, p

reaches pplateau.Region 2 – The plateau pressure is perfectly constant until x

reaches a critical value, xcritu.

Region 3 – For x > xcritu, the pressure rises linearly with x

with the same slope as that of Region 1.Alloy 1: 0.8 kg of alloy AB having a density of 2750 kg/m3,

∆S = −110 kJ K−1kmole−1, ∆H = −35 MJ kmole−1.At 400 K, xcritℓ1

(the minimum value of x in the plateau re-gion) is 0.3, and xcritu1

(the maximum value of x in the plateauregion) is 3.55.

Alloy 2: 1.0 kg of alloy CD having a density of 2750 kg/m3,∆S = −90 kJ K−1kmole−1, ∆H = −28 MJ kmole−1.

At 400 K, xcritℓ2(the minimum value of x in the plateau re-

gion) is 0.3, and xcritu2(the maximum value of x in the plateau

region) is 4.85.The formula masses areA – 48 daltons,B – 59 daltons,C – 139 daltons,D – 300 daltons.The density of the above alloys is (unrealistically) the same

whether hydrided or not.Define “gas-space” as the space inside the canister not occu-

pied by the alloys.


090723


a. Calculate the volume, Vgas−space of the gas space......................................................................................................................

Vcanister = 0.001 m3.

Valloy 1 =0.8 kg

2750 kg per m3 = 0.000291 m3.

Valloy 2 =1.0 kg

2750 kg per m3 = 0.000364 m3.

Vgas−space = 0.001 − 0.000291− 0.000364 = 0.000345 m3.

The volume of the gas-space inside the canister is 345 ml.

b. What is the number of kilomoles of each alloy contained inthe canister?

.....................................................................................................................The molecular mass of Alloy 1 is 48 + 59 = 107 daltons; that of Alloy

2 is 139 + 300 = 439 daltons.For Alloy 1, with a molecular mass of 107 daltons, the total number of

kilomoles is

N1 =0.8 kg

107 kg/kmole= 0.0075 kmoles.

For Alloy 2, with a molecular mass of 439 daltons, the total number ofkilomoles is

N2 =1.0 kg

439 kg/kmole= 0.0023 kmoles.

The canister contains 7.5 moles of Alloy 1 and2.3 moles of Alloy 2.

c. Tabulate the plateau pressures of the two alloys for 300 and400 K.

.....................................................................................................................

pplateau = exp

(

−∆S

R+

∆H

RT

)

090723



For T = 300 K:

pplateau1= exp

(

−−110× 103

8314+

−35 × 106

8314 × 300

)

= 0.448 atmos.

pplateau2= exp

(

−−90 × 103

8314+

−28 × 106

8314× 300

)

= 0.670 atmos.

For T = 400 K:

pplateau1= exp

(

−−110× 103

8314+

−35 × 106

8314 × 400

)

= 14.97 atmos.

pplateau2= exp

(

−−90 × 103

8314+

−28 × 106

8314× 400

)

= 11.08 atmos.

Alloy 300 K 400 K

1 0.448 14.972 0.670 11.08

The figure below shows the 400 K isotherms for the two alloys. Observethat in this figure, the ordinate is linear with pressure and not with thenatural log of pressure, as usual.

0 1 2 3 4 5 Stoichiometric index, x

BP3x = 0.3 14.97 atmos.

Alloy 1 (7.5 moles)

BP1x = 0.3 11.08 atmos.

Alloy 2 (2.3 moles)BP1BP2

BP3BP4x = 3.55

BP4

x = 4.85BP2

T = 400 K

Pre

ssur

e (

atm

os)

d. Describe, in words, the manner in which the system pressure,psystem, varies as hydrogen is gradually introduced in the sys-tem. Use µH2

as the measure of the number of moles of H2

introduced. Sketch a psystem vs µH2graph.

.....................................................................................................................


090723


At low values of µH2, both alloys are depleted and the psystem grows

linearly with µH2until it reaches 11.08 atmospheres (the plateau pressure

of Alloy 2). Increasing µH2will not alter psystem until Alloy 2 is saturated

which occurs when x2 = 4.85. From here on, psystem grows again linearlywith µH2

until it reaches 14.97 atmospheres (the plateau pressure of Alloy1). It then remains steady until Alloy 1 becomes saturated (x1 = 3.55).Finally, the pressure grows again with µH2

.

e. Calculate the values of µH2that mark the break-points (points

of abrupt change in the psystem behavior—that is, points wherethe pressure changes from growing to steady, or vice-versa).

.....................................................................................................................Under all circumstances the amount of H2 in the system is the sum of

1. The hydrogen, µgas−space in gaseous form stored in the gas-space.

2. The hydrogen, µ1 stored in Alloy 1.3. The hydrogen, µ2 stored in Alloy 2.

µH2= µ1 + µ2 + µgas−space.

.......................................... The 1st Break-Point

The first break-point occurs when the pressure (that was growing lin-early with µH2

) reaches the plateau value of 11.08 atmospheres (1.12 MPa)of Alloy 2. At this point, xcritℓ2 = 0.3. This means that for each kilomoleof alloy, there are 0.3 kilomoles of H or 0.15 kilomoles of H2.

Since there are 0.0023 kmoles of Alloy 2 in the canister, the amount ofH2 in this alloy is

µ2 = 0.15 × 0.0023 = 345 × 10−6 kmoles.When the hydrogen pressure is 11.08 atmospheres, Alloy 1 is depleted

(Region 1) because its plateau pressure is 14.97 atmospheres. In this region

p = αx,

where α is a proportionality constant that has to be determined. Region 1ends when x = xcritℓ1 = 0.3 and p=14.97 atmosphere, hence

α =14.97

0.3= 49.9.

At 11.08 atmospheres,

x =11.08

49.9= 0.222.

This means that for each kilomole of alloy, there are 0.222 kilomolesof H or 0.111 kilomoles of H2.

Since there are 0.0075 kmoles of Alloy 1 in the canister, the amount ofH2 in this alloy is

µ1 = 0.111 × 0.0075 = 832 × 10−6 kmoles.

090723



When the hydrogen pressure is 11.08 atmospheres (1.12 MPa), thegas-space contains

µgas−space =pV

RT=

1.12 × 106 × 0.000345

8314× 400= 116 × 10−6 kmoles.

The total amount of H2 at the first break-point is µ1+µ2+µgas−space =832 × 10−6 + 345 × 10−6 + 116 × 10−6 = 1.293× 10−3 kilomoles of H2.

.......................................... The 2nd Break-Point

The next break-point occurs when Alloy 2 saturates at x = 4.85 andthe gas pressure is still 11.08 atmospheres.

The amount of hydrogen stored in Alloy 2 is

µ2 =4.85

2× 0.0023 = 5, 580 × 10−6 kmoles.

Since the pressure is still 10.08 atmosphere, the amounts of hydrogenin Alloy 1 and in the gas-space are the same as before. Thus, the totalamount of H2 at the second break-point is µ1 + µ2 + µgas−space = 832 ×10−6 + 5, 580 × 10−6 + 116 × 10−6 = 6.53 × 10−3 kilomoles of H2.

.......................................... The 3rd Break-Point

The third break-point occurs when the pressure, having risen from11.08 atmospheres, reaches 14.97 atmospheres (1.52 MPa), the plateau pres-sure of Alloy 1.

µ1 =0.3

2× 0.0075 = 1, 120 × 10−6 kmoles.

The amount of hydrogen in Alloy 2 is the same as in the previousbreak-point (µ2 = 5, 580 × 10−6 kmoles.), plus the small amount of gasdissolved in the saturated alloy.

The slope of the p-x line is

α =11.08

0.3= 36.9,

hence, the amount dissolved is

xdissolved =14.97 − 11.08

36.9= 0.10.

Thus the value of x at this break-point is x = 4.85 + 0.10 = 4.95, andthe amount of hydrogen in Alloy 2 is

µ2 =4, 95

2× 0.0023 = 5, 690 × 10−6 kmoles.


090723


µgas−space =pV

RT=

1.52 × 106 × 0.000345

8314× 400= 158 × 10−6 kmoles.

Thus, the total amount of H2 at the third break-point is µ1 + µ2 +µgas−space = 1, 120 × 10−6 + 5, 690 × 10−6 + 158 × 10−6 = 6, 970 × 10−6

kmoles of H2.

.......................................... The 4th Break-Point

The amounts of gas in the gas-space and in Alloy 2 are the same as inthe previous step because the gas pressure is the same.

µgas−space = 158 × 10−6 kmoles.

µ1 =3.55

2× 0.0075 = 13, 300× 10−6 kmoles.

µ2 = 11, 400× 10−6 kmoles.

Break- Pres. µ1 µ2 µgs µpoint (atmos.) (moles)

1 11.08 0.83 0.34 0.12 1.292 11.08 0.83 5.58 0.12 6.533 14.97 1.12 5.69 0.16 6.974 14.97 13.30 5.69 0.16 19.15

0 5 10 15 20 25 Hydrogen in the system (moles)

Pre

ssur

e (

atm

os.)

BP3

BP2

BP1

BP4

11.08 atmos

14.97 atmos

T = 400 K

090723



Prob 11.13 A container able to withstand high pressures hasan internal capacity of 0.1 m3. It contains 490 kg of an alloy, AB,used to store hydrogen.

The properties of this alloy are:Atomic mass of A 60 daltons

Atomic mass of B 70 daltons

Density of AB 8900 kg/m3

Heat capacity 1 kJ K−1kg−1

∆H -25 MJ

/kmole of H2

∆S -105 kJ K−1

/kmole of H2

Depletion end

of plateau x=0.01

Saturation end

of plateau x=1

x is the stoichiometric coefficient of hydrogen in ABHx. Thevalues correspond to 300 K. The plateau pressure is essentially in-dependent of x in the above interval. Above x = 1, the equilibriumpressure rises very rapidly with x so that x does not appreciablydepend on the hydrogen pressure.

a. What volume inside the container can be occupied by gas?

.....................................................................................................................

490 kg of AB occupy 490/8900 = 0.055 m3 so that the space for thegas inside the container is 0.100 − 0.055 = 0.045 m3.

In the container, the space that can be filled by gas is 45 liters.

b. How many kilomoles of alloy are inside the container?

.....................................................................................................................

The molecular mass of AB is 130 daltons or 130 kg/kilomole. Thus,490 kg of alloy correspond to 490/130 = 3.77 kilomoles.

In the container, there are 3.77 kilomoles of the Alloy AB.

c. What is the plateau pressure of the hydrogen in the alloy at300 k?

.....................................................................................................................

ln p = −∆S

R+

∆H

RT= −

−105× 103

8314+

−25 × 106

8314 × 300= 2.606, (1)


090723


p = 13.5 atmos. (2)

The plateau pressure is 13.5 atmospheres.

d. Introduce 10 g of H2. Give an upper bound for the pressureof the gas in the container.

.....................................................................................................................10 g of H2 is the same as 0.01 kilomoles of H. To reach the plateau

(which starts when x=0.01), the amount of hydrogen required is 0.01 ×3.77 = 0.0377 kilomoles of H. The amount introduced is insufficient toreach the plateau. A small amount of gas will dissolve in the alloy, the restwill remain in gaseous form. An upper bound for the pressure can be foundby assuming that no gas is in the alloy, all of it being in gaseous form.

p =µRT

V=

0.005× 8314 × 300

0.045= 277 kPa. (3)

Notice that in the above equation, the number of kilomoles of hydrogenwas taken as 0.005, not 0.01, because the hydrogen gas is in the H2, not H,form.

The gas pressure will be 2.77 atmospheres or less.

e. Now introduce additional hydrogen so that the total amountintroduced (Steps 4 and 5) is 100 g. The temperature is keptat 300 K. What is the pressure of the gas?

.....................................................................................................................100 g of hydrogen correspond to 0.10 kilomoles of H or 0.05 kilomoles

of H2.In the plateau region, the alloy must have more than 0.01 × 3.77 =

0.0377 and less than 3.77 kmoles of H. The amount introduced seems to besufficient to cause the alloy to operate in the plateau, hence the gas pressureis probably the plateau pressure (13.5 atmospheres). We must, however,check how much hydrogen is gaseous form.

At 13.5 atmospheres (≈ 1.35 MPa), the amount of gas in the containeris

µ =pV

RT=

1.35 × 106 × 0.045

8314 × 300= 0.024 kmoles of H2 (4)

The rest of the hydrogen (0.05−0.024 = 0.026 kilomoles of H2 or 0.052kilomoles of H) is in the hydride. Thus, there is sufficient hydrogen to causethe alloy to reach its plateau. The gas pressure is 13.5 atmospheres.

The gas pressure is 13.5 atmospheres.

090723



f. What is the stoichiometric index, x, of the hydride, ABHx?

.....................................................................................................................From the previous question, the amount of hydrogen absorbed is 0.052kilomoles of H. Since there are 3.77 kilomoles of alloy, the stoichiometricindex is 0.052/3.77 = 0.014.

The empirical formula for the hydride is ABH0.014.

g. Finally, introduce sufficient hydrogen so that the totalamounts to 4 kg. The temperature remains at 300 K. Whatis the pressure of the gas?

.....................................................................................................................

4 kg of hydrogen correspond to 4 kilomole of H or 2 kilomoles of H2.Since the alloy, at the saturated end of the plateau has an empirical formula,ABH1, the 3.77 kilomoles of alloy can hold, at this point, 3.77 kilomoles ofH. If the amount of H is increased further, no more H is absorbed becauseof the very steep p vs x characteristic. All the excess (4 − 3.77 = 0.23kilomoles of H or 0.115 kilomoles of H2) hydrogen must be in gaseous form.Its pressure is

p =µRT

V=

0.115 × 8314× 300

0.045= 6.37 × 106 Pa. (5)

The gas pressure is 63.7 atmospheres.

h. Assume that the container has negligible heat capacity. With-draw adiabatically (that is, without adding any heat to thesystem) 1 kg of hydrogen. What will be the pressure of thegas inside the container after the 1 kg of hydrogen have beenremoved?

.....................................................................................................................

A certain amount of hydrogen, µdes, will be desorbed owing to theremoval of 1 kg of gas from the container. Qdes joules of heat are requiredfor this desorption:

Qdes = µdes∆H = 25 × 106µdes joules.

Notice that µdes is the number of kilomoles of H2, not of H.


090723


Since the processes is adiabatic and assuming that the hydrogen gashas negligible heat capacity, the heat necessary for desorption must comefrom the alloy itself.

Qdes = 25 × 106µdes = cM∆T = 1000× 490(300− T2)

Where T2 is the temperature after desorption, c is the heat capacity of thealloy and M is the mass of the alloy.

µdes = 5.88 − 19.6 × 10−3T2 kmoles of H2.

Before the removal of the hydrogen, the amount of gas in the containerwas 0.115 kmoles of H2. After the removal, the temperature falls to T2 andthe plateau pressure is

p2 = 105 exp

(

12.63 −3007

T2

)

.

The 105 factor, above, converted the pressure from atmospheres topascals.

At that pressure, the amount of gas in the container is

µ2 =p2V

RT2=

105 exp(

12.63 − 3007T2

)

× 0.0.45

8314T2= 0.541

exp(

12.63− 3007T2

)

T2

Hence, the amount of hydrogen in gaseous form was reduced by

∆µ = 0.115 − 0.541exp

(

12.63 − 3007T2

)

T2kmoles of H2.

3 kg of hydrogen removed corresponds to 1.5 kilomoles of H2. Conse-quently, the amount desorbed is

µdes = 1.5 − ∆µ = 1.385 + 0.541exp

(

12.63− 3007T2

)

T2kmoles of H2.

But we had already determined that

µdes = 5.88 − 19.6 × 10−3T2 kmoles of H2

hence

5.88 − 19.6 × 10−3T2 = 1.385 + 0.541exp

(

12.63− 3007T2

)

T2

090723



A numerical solution yields, T2 = 229 K.The pressure of the gas is the plateau pressure at 229 K:

p2 = 105 exp

(

12.63 −3007

229

)

= 61.4 kPa.

The gas pressure is 61.4 kPa or 0.606 atmos.


090723


Prob 11.14 Two hydrogen storing alloys have the followingproperties:

Hydride ∆H ∆SMJ kmole−1 kJ K−1 kmole−1

A -28 -100B -20 ?

a – (10 pts) –What must the value of ∆SB be to make theplateau pressure of the two hydrides be the same when T = 400K?

∆SB is, of course, the ∆S of Alloy B......................................................................................................................

The plateau pressure is given by

exp p = −∆S

R+

∆H

RT,

hence, the two hydrides have the same plateau pressure when

−∆SA

R+

∆HA

RT= −

∆SB

R+

∆HB

RT,

−∆SB =1

T(∆HA − ∆HB) − ∆SA =

1

400(−28 + 20) × 106 + 100 × 103

= −80 × 103 kJ K−1 kmole−1

The ∆S of Hydride B must be 80 kJ−1 kmole−1.

b – (5 pts) – How do you rate your chances of finding an alloywith the properties of Alloy B? Explain......................................................................................................................

The ∆S of alloy B seems to be substantially lower than that of anyalloy listed.

Chances of finding an alloy with such a low ∆S are poor.

090723



Prob 11.15 TiFe is sold by Energics, Inc under the labelHYSTORE 101. Pertinent data are found in Table 11.4 through11.6 in the Text. Treat this alloy as “ideal” (no hysteresis).

The atomic mass of iron is 55.8 daltons and that of titaniumis 47.9 daltons. The saturated alloy (TiFeH0.95) is at 350 K andis in a perfectly adiabatic container with negligible heat capacity.

A valve is opened and hydrogen is allowed to leak out untilthe pressure reaches 2 atmospheres. What is the composition ofthe hydride at the end of the experiment—that is, what is thevalue of x in TiFeHx?

To simplify the problem, assume that there is no“gas space”in the container (patently impossible). Also, neglect the heat ca-pacity of the hydrogen gas and any Joule-Thomson heating owingto the escaping gas......................................................................................................................The data for HYSTORE 101 are

∆S = −106.1 kJ K-1 kmole−1.∆H = −28 MJ kmole−1.Heat capacity of the alloy, c = 540 J K−1 kg−1.Since the formula mass of the alloy is 103.7 daltons, c = 540× 103.7 =

56 kJ K−1 kmole−1.The plateau pressure is

p = exp

(

106.1 × 103

8314−

28.0 × 106

8314×

1

T

)

= exp

(

12.762− 3367.81

T

)

.

For T = 350 K, p = 23.1 atmos.A 2 atmos pressure corresponds to a temperature of 279.0 K.Hydrogen will be desorbed cooling the alloy until it reaches 279.0 K,

i.e, until the temperature falls by 350−279 = 71 K. Since the container hasnegligible heat capacity, all the heat must come from the alloy. The heatremoved per kilomole of alloy is

Q = c∆T = 56, 000× 71 = 3.98 × 106 joules/kmole of alloy.

To remove this much heat the amount, N, of hydrogen that must bedesorbed is

N =3.98 × 106

28 × 106= 0.14 kmoles (H2)/kmole alloy.

or

0.28 kmoles (H)/kmole alloy

Since initially the hydride had the empirical formula, TiFeH0.95 it willhave a formula TiFeH0.67 after the release of the gas.

The stoichiometric index of H in the alloy is 0.67.


090723


Prob 11.16 A canister contains an alloy, AB, that forms amonohydride, ABH. The canister is perfectly heat-insulated—that is, adiabatic and contains 0.01 kmoles of the alloy and a freespace of 600 ml which is, initially, totally empty (a vacuum).

The molecular mass of the alloy is 100 daltons and its ther-modynamic characteristics for absorption are ∆H = −25 MJ and∆S = −100 kJ/K all per kilomole of H2. For simplicity, make theunrealistic assumption that the plateau extents from x = 1 all theway to x = 0. x is the stoichiometric coefficient in ABHx. As-sume also that the plateau is perfectly horizontal and that thereis no hysteresis. The heat capacity of the alloy is 500 J kg−1K−1.Again, for simplicity, assume that neither the canister itself northe hydrogen gas has significant heat capacity. Finally, assumethat the volume of the alloy is independent of x.

Introduce 0.002763 kmoles of H2 into the canister. Both can-ister and hydrogen are at 300 K. What will the gas pressure beinside the canister?.....................................................................................................................

Inside the canister, part of the gas will occupy the free space (designatethis quantity as µFree) and part, µHydride, will be absorbed causing theaverage composition of the alloy to become ABHx. These two parts are

µFree =pVFree

RT,

where VFree is the volume of the free space, and T is the final temperatureof the system, and

µHydride = 12xµAlloy ,

where µAlloy is the number of kilomoles of the alloy inside the canister.When the hydrogen is introduced and µHydride kilomoles of hydrogen

is absorbed, an amount of heat

Q = 12xµAlloy |∆H |

is released. This causes a temperature rise of

∆T =Q

cM,

where c is the heat capacity of the alloy and M is its mass in kg. Sinceeach kilomole of AB masses 100 kg,

M = 100µAlloy = 100 × 0.01 = 1 kg.

090723



The heat capacity of the alloy is c = 500 J kg−1K−1 so that

∆T =Q

cM=

12xµAlloy|∆H |

500,

and the final temperature of the alloy will be

T = 300 +Q

cM=

12xµAlloy |∆H |

500= 300 +

12.5 × 104

500x = 300 + 250x.

The plateau pressure will be

p = exp

(

∆H

RT−

∆S

R

)

= exp

(

−25 × 106

8314× (300 + 250x)−

100 × 103

8314

)

= exp

(

−3007

(300 + 250x)+ 12.03

)

The amount of free hydrogen will be

µFree =pVFree

RT=

105 × exp(

−3007(300+250x) + 12.03

)

× 6 × 10−4

8314 × (300 + 250x)

=exp

(

−3007(300+250x) + 12.03

)

× 7.22 × 10−3

(300 + 250x)

The amount of H2 in the hydride is

µHydride = 12µAlloyx = 0.005x.

Finally, the total amount of hydrogen introduced into the canister is

µTotal = 0.002763 = µFree + µHydride

=exp

(

−3007(300+250x) + 12.03

)

× 7.22 × 10−3

(300 + 250x)+ 0.005x.

A numerical solution of the above equation leads to x = 0.32. Thisallows the calculation of the pressure,

p = exp

(

−3007

(300 + 250x)+ 12.03

)

= 61.4 atmos.

The temperature is

T = 300 + 250x = 380 K.

The final hydrogen pressure is 61.4 atmospheres.


090723


Prob 11.17 HELIOS is an electric airplane developed byAeroVironment to serve as a radio relay platform. It is supposedto climb to fairly high altitudes (some 30 km) and orbit for aprolonged time (months) over a given population center fulfillingthe role usually performed by satellites. Although its geographiccoverage is much smaller, HELIOS promises to be substantiallymore economical.

The plane is propelled by 14 electric motors of 1.5 kW each.Power is derived from photovoltaic cells that cover much of thewing surface. In order to stay aloft for many days, the planemust store energy obtained during the day to provide power fornighttime operation. The solution to this problem is to use a wa-ter electrolyzer that converts the excess energy, provided duringdaytime hours by the photovoltaics, into hydrogen and oxygen.The gases are then stored and, during darkness feed a fuel cellthat provides the power required by the airplane.

Although the specifications of the HELIOS are not know, letus take a stab into providing the outline of a possible energystorage system. To that end, we will have to make a number ofassumptions that may depart substantially from the real solutionbeing created by AeroViroment.

a. Calculate the amount of hydrogen and of oxygen that mustbe stored. Assume that during take off and climbing to cruisealtitude, the full 1.5 kW per motor is required, but, for orbit-ing at altitude only half the above power is required. Assumealso that the power needed for the operation of the plane(other than propulsion, but including the energy for the ra-dio equipment) is 3 kW. Assume also that the longest periodof darkness last 12 hours. The fuel cell has an efficiency of80%.

.....................................................................................................................14 motors operating at half power require 14× 1

21.5 = 10.5 kW. Addingto this the 3 kW for house keeping, leads to a power of 13.5 kW requiredfor orbiting. This calls for a 15 kW fuel cell that, in an emergency cansupply much more power for a short time. The total energy spent duringa 12 hour night would be 13.5 × 12 × 3600 = 583, 000 kJ or 583 MJ.

Since the fuel cell has an 80% efficiency, the required fuel input is583/.8 = 729 MJ. The free energy of 1 kilomole of hydrogen combining withhalf a kilomole of oxygen is 228 MJ. Thus, 729/228 = 3.2 kmol of hydrogen(6.4 kg). Correspondingly, 1.6 kmol of oxygen (51.2 kg) are needed.

090723



Fuel needed:6.4 kg of hydrogen,51.2 kg of oxygen.

b. The amount of fuel calculated in Item 1 must be stored. As-suming STP conditions, what is the volume required?

.....................................................................................................................The density of hydrogen at STP is

ρH2=

2 kg/kmol

22.4 km3/kmol= 0.0893 kg/m3,

and that of oxygen is

ρO2=

32 kg/kmol

22.4 km3/kmol= 1.43 kg/m

3.

The volume of hydrogen is

VH2=

6.4

0.0893= 71.7 m3,

and that of oxygen is

VO2=

51.2

1.43= 35.8 m3.

Of course, the volume of hydrogen is twice that of oxygen.

The volumes areHydrogen: 71.7 m3

Oxygen: 35.8 m3.

c. Clearly, the volumes calculated in Item 2 are too large tofit into HELIOS. The fuel cell busses being operated exper-imentally in Chicago, have hydrogen tanks that operate at500 atmospheres and that allow a gravimetric concentrationof 6.7% when storing hydrogen. If such tanks were adoptedfor the HELIOS, what would be the mass of the fully chargedfuel storage system?

.....................................................................................................................At 500 atmosphere the volume of hydrogen is reduced to 71.7/500 =

0.143 m3.. With a gravimetric concentration of 6.7%, the tank would mass6.4/0.067 = 95.5 kg and the total (tank plus gas) would be 102 kg. Themass-to-volume ratio of the tank is 95.5/0.143 = 668 kg/m3.


090723


The 51.2 kg of oxygen would occupy 35.8/500 = 0.07 m3. The mass ofthe tank is 668 × 0.07 = 46.8 kg.

The total mass of the fully charged fuel storage system (not countingcompressors) is 95.5 + 6.4 + 46.8 + 51.2 = 149 kg.

The mass of the fully charge fuel storage system is 149 kg.

d. Assume that the efficiency of a mechanical hydrogen compres-sor is 60% and that of an oxygen compressor is 80%. Howmuch energy do these compressors require to compress thegases isothermally to their 500 atmosphere operating tem-perature. Assume, for simplicity that the electrolyzer thatproduces these gases is pressurized to 5 atmospheres.

.....................................................................................................................The energy to compress a gas isothermally is

W = µRT lnp1

p0.

The compression ratio is, for both hydrogen and oxygen, 500:5, thus,ln 100 = 4.61. The RT product is 8314 × 273 = 2.27 × 106.

For hydrogen,

WH2=

1

0.6× 3.2 × 2.27 × 106 × 4.61 = 55.8 × 106 J.

For oxygen,

WO2=

1

0.8× 1.6 × 2.27 × 106 × 4.61 = 20.9 × 106 J.

The total compression energy is 76.7 MJ

e. While orbiting, during daylight, what is the total energy thatthe photovoltaic collectors have to deliver? The electrolyzeris 80% efficient.

.....................................................................................................................

090723



The total amount of hydrogen that has to be produced is 3.2 kilomoles.An ideal electrolyzer produces 1 kilomole of hydrogen for each 237.2 MJ ofelectric energy used. An electrolyzer with 80% efficiency needs 237.2/0.8 =297 MJ per kilomole or a total of 950 MJ.

The photovoltaic collectors have to supply:

950 MJ for propulsion and housekeeping.76.7 MJ for the gas compressors,or 1.03 GJ total. which are expended during the 12 daylight hours,and correspond to some 1.8 kW.

The photovoltaic system has to generate a total of 1.03 GJ daily.


090723


Prob 11.18 Two 100-liter canisters are interconnected by apipe (with negligible internal volume). Canister “A” contains37.2 kg of FeTi and Canister “B”, 37.8 kg of Fe0.8Ni0.2Ti.

Although the gas can freely move from one canister to theother, there is negligible heat transfer between them. Thus thegas can be at different temperatures in the two canisters. Thegas always assumes the temperature of the alloy it is in contactwith.

The pertinent data are summarized in the two boxes below:

Element Atomic DensityMass

(daltons) (kg/m−3)

Ti 47.90 4540Fe 55.85 7870Ni 58.71 8900

Alloy ∆H ∆S(MJ (kJ

kmol−1) K−1kmol−1)

FeTi -28.0 -106.1Fe0.8Ni0.2Ti -41.0 -118.8

Plateau

ln p

x0 1

To simplify the solution, assume that the ln p vs x characteris-tics of the alloys consist of a perfectly horizontal plateau followedby a vertical line in the saturated region. In other words, thecharacteristics look as sketched in the figure. The alloys are de-pleted when x = 0 and are saturated when x = 1. Also, neglect allthe hydrogen dissolved in the saturated alloy.

Initially, Canisters A is at 300 K and Canister B, at 400 K.They have been carefully evacuated (the gas pressure is zero).

Assume that the density of the alloys is the average of thatof the component elements.

Enough hydrogen is introduced into the system so that oneof the alloys becomes saturated. This will cause the temperatureof the alloys to change.

To simplify things, assume that the alloys and the canistersthemselves have negligible heat capacity.

a. Does the temperature increase or decrease?.....................................................................................................................

090723



Hydrogen will be absorbed by the alloy. This is an exothermic process.The temperature will rise.

The temperature will rise.

b. The temperature is now adjusted to the values of 300 K and400 K, as before. How many kg of hydrogen had to be intro-duced into the system to make the gas pressure, at this stage,10% higher than the plateau pressure of the saturated alloywhile leaving the other alloy depleted? Please be accurate tothe gram.

.....................................................................................................................

The density of FeTi is

δFeTi =7870 + 4540

2= 6200 kg/m

3.

The density of Fe0.8Ni0.2Ti is

δFe0.8Ni0.2Ti =0.8 × 7870 + 0.2 × 8900 + 4540

2= 6310 kg/m3.

The volume of 37.2 kg of FeTi is

VFeTi =37.2

6200= 6 × 10−3 m3 or 6 liters.

The volume of 37.8 kg of Fe0.8Ni0.2Ti is

VFe0.8Ni0.2Ti =37.8

6200= 6 × 10−3 m3 or 6 liters.

Thus, the “dead space” in each canister is 94 liters.

The plateau pressure is

pP lateau

= exp

(

−∆S

R+

∆H

RT

)

.

For the two alloys,

pP lateauA

= exp

(

12.76 −3368

T

)

,

pP lateauB

= exp

(

14.29 −4931

T

)

,


090723


This leads to a plateau pressure for Alloy A of 4.63 atmos at 300 K and,for Alloy B, at 400 K, 7.11 atmos. We have to introduce sufficient hydrogenso that the pressure of the system reaches 1.1 × 4.63 = 5.09 atmospheres.

Since each canister has a 94 liter dead space, when the gas pressureis 5.09 atmospheres (509 kPa), the amount of hydrogen in gas form in theCanisters A is

µ =pV

RT=

509, 000× 0.094

8314 × 300= 0.0192 kmoles of H2.

and in Canister B is

µ =pV

RT=

509, 000× 0.094

8314 × 400= 0.0144 kmoles of H2.

Alloy A is completely saturated. Its stoichiometric index is x = 1. Themass of one kilomole of FeTi is

mFeTi = 55.85 + 47.90 = 103.8 kg/kmole,

and of the Fe0.8Ni0.2Ti alloy is

mFe0.8Ni0.2Ti = 0.8 × 55.85 + 0.2 × 58.71 + 47.90 = 104.3 kg/kmole.

37.2 kg of FeTi correspond to 37.2/103.8 = 0.358 kmoles, while 37.8kg of Fe0.8Ni0.2Ti correspond to 37.8/104.3 = 0.362 kmoles of H.

Thus in the just saturated FeTi, we have 0.358 kmoles of H absorbedor, 0.179 kmoles of H2.

Alloy B is depleted and holds no hydrogen.The total amount of hydrogen in the system is 0.0192+0.0144+0.179 =

0.213 kmoles of H2.

The system must be filled with 0.425 kg of H2.

c. Now raise the temperature of Alloy A to 400 K. Describewhat happens:3.1 what is the final gas pressure,3.2 what is the stoichiometric value of H in each alloy,3.3 how many joules of heat had to be added,3.4 how many joules of heat had to be removed.To simplify things, assume that the alloys and the canistersthemselves have negligible heat capacity.

.....................................................................................................................When the temperature of Alloy A is raised to 400 K, its plateau pres-

sure becomes 76.9 atmos. Hydrogen desorbs and is absorbed by Alloy B

090723



which clamps the pressure at 7.1 atmospheres while it absorbs the gas. Isthere enough hydrogen to saturate this alloy?

At a gas pressure of 7.1 atmosphere and 400 K (in both canisters),the 94 liters of dead space in each canister will hold 0.020 kmoles of H2,a total of 0.04 kmoles of hydrogen. Since the total hydrogen introducedwas 0.213 kmoles, that leaves 0.213 − 0.040 = 0.173 kmoles of H2 (0.346kmoles of H) to be absorbed by Alloy B. There are 0.362 kmoles of thisalloy and, thus, when saturated (x = 1) it can hold 0.362 kmoles of H.Consequently, there is not enough hydrogen in the system to saturate thisalloy. Its stoichiometric index will be

x =0.346

0.362= 0.956.

The stoichiometric index in Alloy A is zero (totally depleted).

The stoichiometric index in Alloy B is 0.956 (not quite saturated).

Since there is not enough hydrogen to saturate Alloy B, the pressureof the gas will be equal to the plateau pressure of this alloy.

The gas pressure will be 7.1 atmospheres.

Alloy A was saturated before its temperature was raised. It then be-came completely depleted. All hydrogen (0.358 kmol of H) was desorbed.This corresponds to 0.178 kmoles of H2. It take 28 MJ to desorb 1 kmoleof H2, thus the amount of heat that had to be introduced was

QA = 0.178× 28 = 4.98 MJ.

The amount of heat to desorb the gas from Alloy A is about 5 MJ.

Alloy B has to absorb 0.346 kmoles of H (0.173 kmoles of H2). Theheat that has to be removed is

Q = 0.173× 41 = 7.09 MJ.

The amount of heat that must be removed from Alloy B is 7.1 MJ.


090723


Prob 11.19 A canister with 1 m3 capacity contains 3000 kgLaNi

5. Although initially the system was evacuated, an amount,

µ1, of hydrogen is introduced so that the pressure (when thecanister and the alloy are at 298 K) is exactly 2 atmospheres.Assume that once the alloy is saturated, it cannot dissolve anymore hydrogen, i.e., assume that the p-x characteristic just afterthe beta phase is vertical.

The density of lanthanum (atomic mass 138.90 daltons) is6145 kg/m3 and that of nickel (atomic mass 58.71 daltons) is8902 kg/m3. Assume that the density of the alloy is equal to thatof a mixture of 1 kilomole of lanthanum with 5 kilomoles of nickel.

a. What is the value of µ1?.....................................................................................................................

One kilomole of LaNi5

masses 138.90 + 5 × 58.71 = 432.5 kg.

One kg of La occupies 1/6145 m3, hence 1 kilomole (138.90 kg) occupies0.0226 m3. One kg of Ni occupies 1/8902 m3, hence 5 kilomoles (293.6 kg)occupie 0.0330 m3. The density of LaNi

5is 432.5/(0.0226+0.0330)=7780

kg/m3.3000 kg of LaNi

5correspond to 3000/432.5 = 6.94 kilomoles and oc-

cupy 3000/7780 = 0.386 m3. This means that in the canister there is anempty space of 0.614 m3.

From a table in the Text, one obtains for the alloy used, ∆Hf = −31.0MJ/kilomole (H2) and ∆Sf = −107.7 kJ/K−1kmole−1 (H2). This allowsus to calculate the plateau pressure at 298 K.

p = exp

(

∆H

RT−

∆S

R

)

=exp

(

−31.0× 106

8314× 298−

107.7 × 103

8314

)

=1.556 atmos.

The hydride formed when LaNi5 fully combines with hydrogen isLaNi5H5. Since the gas pressure inside the canister exceeds the plateaupressure, the alloy must be saturated an must hold 5 kilomoles of H foreach kilomole of alloy. Thus the total amount of monoatomic hydrogenabsorbed by the alloy is 6.94× 5 = 34.68 kilomoles of H or 17.34 kilomolesof H2.

The rest of the gas fills the dead space in the canister which has avolume. Vdead = 0.614 m3. At a pressure of 2 atmospheres, the amount ofhydrogen in this space is

µdead =pV

RT=

2 × 105 × 0.614

8314× 298= 49.56 × 10−3 kmoles. (1)

The total amount of hydrogen introduced into the canister was 17.34+0.05 = 17.39 kmoles.

090723



The amount of gas introduced was 17.39 kmoles of H2

b. 100 MJ of heat are introduced into the canister whose wallsare adiabatic and have no heat capacity. Ignore also the heatcapacity of the free hydrogen gas.Calculate the pressure of the free hydrogen gas.

.....................................................................................................................After the heat injection and after a new equilibrium has been estab-

lished, an amount, µ, of hydrogen has been desorbed. The pressure of thefree gas must equal the plateau pressure of the alloy:

1.0133× 105 exp

(

−31 × 106

8314×

1

T+

107700

8314

)

=µ + 0.05

0.614× 8314T. (2)

This simplifies to

exp

(

−3728.7

T+ 12.954

)

= 0.13363(µ + 0.05)T (3)

The 100 MJ of heat injected into the alloy will be used up in providingthe desorption energy and in heating up the alloy:

108 = 31 × 106µ + 420 × 3000(T − 298). (4)

Solving this equation for µ,

µ = 15.338− 0.040645T. (5)

And introducing this value of µ into Equation 3,

exp

(

−3728.7

T+ 12.954

)

= 2.0563T − 0.0054314T 2, (4)

whose numerical solution leads toT = 369.8 K and µ = 0.307 kilomoles.Using this temperature, the plateau pressure equation yields

p = exp

(

−3728.7

369.8+ 12.954

)

= 17.654 atmos. (5)

The gas pressure in the canister is 17.7 atmospheres.


090723


Prob 11.20

Please use R = 8314.

Here is the setup:

A hydrogen source is connected to a canister containing analloy, A, that can be fully hydrided to AH.

The hydrogen source is a high pressure container with an in-ternal capacity of Vs = 1 liter. It is charged with enough hydrogento have the gas at p0 = 500 atmospheres when the temperature isT0 = 300 K. The container is in thermal contact with the hydridecanister. In steady state, the container, the hydrogen and thecanister are all at the same temperature. The heat capacity ofthe container is 300 J/K.

The whole system— container and canister—is completelyadiabatic: no heat is exchanged with the environment.

There is a pipe connecting the hydrogen source to the hydridecanister. A valve controls the hydrogen flow. According to theJoule-Thomson law, the gas escaping from the source and flowinginto the canister will warm up. However, in this problem, assumethat there is no Joule-Thomson effect.

Initially, while the hydrogen delivery valve is still shut off,there is no gas pressure inside the canister and canister and con-tents are at 300 K.

Canister:

Volume, V = 1 liter.

Heat capacity, ccan = 700 J/K.

The alloy has the following characteristics:

Amount of hydride, m = 5.4 kg.

Density of hydride, δ = 9000 kg/m3.

Molecular mass of alloy, A, 100 daltons.

Heat of absorption: -28 MJ/kmoles.

Entropy change of absorption: -110 kJ K−1kmole−1.

Heat capacity, chyd = 500 J kg−1 K−1,

The ln p versus x characteristics of the alloy are a perfectlyhorizontal plateau bound by vertical lines corresponding tothe depleted and the saturated regions.

The valve is opened and hydrogen flows into the canister. Ifyou wait long enough for the transients to settle down, what isthe pressure of the gas in the hydrogen source? Is the hydride inthe plateau region?

.....................................................................................................................

090723



The alloy occupies a volume of

Valloy =m

δ=

5.4

9000= 0.0006 m3, (1)

hence the volume of the dead space is

Vds = 0.001 − 0.0006 = 0.0004 m3. (2)

The total volume available for has is Vgas = 0.0014 m3 (the sum of thedead space in the alloy canister plus the space in the hydrogen container).

The initial amount of hydrogen in the source is

µ0 =p0Vs

RT0=

500 × 1.0133× 105 × 0.001

8314 × 300= 20.3 × 10−3 kmoles (H2).

(3)Hydrogen from the source will flow into the canister filling the dead

space and hydriding the alloy.There are 5.4 kg of alloy (molecular mass, 100 daltons) in the canister.

This amounts to 5.4/100 = 0.054 kilomoles of alloy. Since when fullyhydrided, the hydride has an empirical formula, AH, the alloy can, at amaximum, absorb 0.054 kilomoles of H or 0.027 kmoles of H2. This is morethan the available hydrogen (0.0203 kilomoles of H2). Thus the systemcannot move into the saturated region—it must be in the plateau region.This means that, after all settles down, the gas pressure will be equal to theplateau pressure, p. Owing to the hydrogen absorption, the temperatureof the system (hydrogen container, alloy canister, and alloy) will rise fromthe original 300 K to a final temperature, T .

p = 1.013 × 105 exp

[

−∆S

R+

∆H

RT

]

= 1.013 × 105 exp

[

−−1.1× 105

8314+

−28 × 106

8314T

]

= 1.013 × 105 exp

[

13.2307−3367.8

T

]

Pa. (4)

The amount of hydrogen in gas form is

µgas =pVgas

RT=

0.0014p

8314T= 168.4× 10−9 p

T. (5)

Since no gas was lost,

µhy = µ0 − µgas = 20.3 × 10−3 − 168.4 × 10−9 p

T(6)


090723


Here, µhy is the amount of hydrogen (H2) stored in the hydride.We need more independent information relating µhy to T , so as to

eliminate µhy from the above formula.The heat source is the absorption of hydrogen by the alloy. This

amounts toQabs = ∆Hµhy = 28 × 106µhy. (7)

This amount of heat must go to raising the temperature of the sourcecontainer, the canister, the alloy and the gas in the dead space,

Qin = (T − 300) (Csource + Ccan + mChy + µdsCH)

= (T − 300)(

300 + 700 + 500 × 5.4 + 29 × 103µds

)

= (T − 300)(

3700 + 29 × 103 × 48.1 × 10−9 p

T

)

= (T − 300)(

3700 + 1.395 × 10−3 p

T

)

. (8)

Since, in steady state, Qabs = Qin,

28 × 106µhy = (T − 300)(

3700 + 1.395× 10−3 p

T

)

(9)

µhy =T − 300

28 × 106

(

3700 + 1.395× 10−3 p

T

)

. (10)

Equation 6 becomes,

20.3 × 10−3 − 168.4× 10−9 p

T

−T − 300

28 × 106

(

3700 + 1.395 × 10−3 p

T

)

= 0 (11)

1 − 8.296× 10−6 p

T− (T − 300)

(

0.006509 + 2.454 × 10−9 p

T

)

= 0 (12)

By combining Equations 4 and 12, we can solve numerically for T (bytrial and error). The result is (to an unwarranted precision): T = 407.1 K.A this temperature, the plateau pressure is 142.2 atmosphers.

The final hydrogen pressure is 142 atmospheresand the alloy is in the plateau region.

090723



Prob 11.21 A hydrogen source consists of a 5-liter con-tainer with hydrogen at an initial pressure of 200 atmospheres.Throughout the experiment the hydrogen in this container re-mains at a constant temperature of 300 K.

a - How many kilomoles, µ0, of hydrogen are initially in the con-tainer?.....................................................................................................................

µ0 =pV

RT=

200 × 1.013× 105 × 0.005

8314× 300= 0.0406 kilomoles. (1)

The source contains 0.0406 kilomoles of hydrogen.

A separate 2-liter canister (“alloy canister”) contains 3.5 kgof a metallic alloy, AB, with the following properties:

Density δ 3.5 kg/literHeat of absorption ∆H -30 MJ/kmole (H2)Entropy change (absorp.) ∆S -100 kJ/(kmole K)Formula mass 100 daltonsComposition (fully hydrided) ABH

Assume an extremely simplified ln(p) vs x characteristic:- Horizontal plateau.- No depletion region.- Vertical saturation region.

Plateau

ln p

x0 1

The following heat capacities are relevant:

Canister 100 J/KAlloy 440 J/K per kgHydrogen (H2) 20,800 J/K per kmole

b - If the alloy were fully hydrided, how many kilomoles of hydro-gen would be absorbed by the 3.5 kg of the alloy in the canister?.....................................................................................................................


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When fully hydrided, there is one kilomole of hydrogen per kilomole ofAB. Since there are 3.5 kg of the latter in the canister and since its formulamass is 100 daltons, then the canister contains 0.035 kmoles of AB or 0.035kmoles of H or 0.0175 kilomoles of H2.

If fully hydrided, the alloy would hold 0.0175 kmoles of hydrogen.

c - The alloy does not completely fill the canister. A “deadspace” is left which is initially a vacuum but will later containsome hydrogen. What is the volume of this dead space?.....................................................................................................................

The density of the alloy is 3,500 kg/m3, hence 3.5 kg occupy 0.001 m3.The canister has a volume of 0.002 m3, hence the dead space has a volumeof 0.001 m3.

The volume of the dead space is 0.001 m3.

Before the beginning of the experiment, the alloy is com-pletely degased—there is hydrogen neither in the dead space norabsorbed by the alloy. Both canister and the alloy are at 300 K.

The experiment consists of opening the valve and waitinguntil the system settles into steady state.

As hydrogen flows from source to alloy canister, the pressurein the former drops (but the temperature stays put) while thepressure in the dead space rises (and so does the temperature).

d - When steady state is reached, there is an unique pressurein the dead space. The calculation of this pressure is laborious.Assume that it is 10 atmospheres. Calculate the number of kilo-moles, µds of H2 in the dead space and the number of kilomoles,µabs, of H2 absorbed by the alloy. Prove that the assumed pres-sure cannot correspond to a steady state situation......................................................................................................................

[ Assume p = 10 atmos.]

Note that, unless the alloy is saturated, the gas pressure must equalthe plateau pressure, which is

p = 1.013× 105 exp

(

−∆S

R+

∆H

RT

)

= 1.013× 105 exp

(

100,000

8314+

−30 × 106

8314T

)

= 1.013× 105 exp

(

12.03 −3608

T

)

Pa. (1)

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From this,

T =3608

12.03 − lnp

1.013× 105

. (2)

[T = 370.9 K.]

The amount of gas in the dead space must be:

µds =pVds

RT=

10 × 1.013 × 105 × 0.001

8314 × 370, 9= 3.285 × 10−4 (3)

[ µds = 0.0003285 kmoles of H2.]

The introduction of hydrogen into the alloy canister resulted in a tem-perature rise of

∆T = T − 300 = 370.9− 300 = 70.9 K. (4)

[ ∆T = 70.9 K.]

Since the alloy canister is adiabatic, this temperature rise must be theresult of the release of an amount Q of absorption heat, assuming that Cis the total heat capacity of the system.

∆T =Q

C. (5)

Here, C = Calloy +Chyd+Ccan is the total head capacity of the system.The 3.5 kg of alloy contribute

Calloy = 440 × 3.5 = 1540 J/K. (6)

The µds kmoles of hydrogen in the dead space contribute

Chyd = 20,800µds, J/K. (7)

and the rest of the canister contributes Ccan = 100 J/K. Consequently,

C = 1640 + 20,800µds. (8)

Since we have a values for µds, we can calculate C.

[ C=1646.8. J/K]


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Q = C∆T = (1640 + 20,800 × 0.0003285)× 70.9 = 1.1676× 105 J. (9)

[ Q = 116,760 J.]

Each kmole of H2 absorbed will release ∆H joules of heat, thus

µabs =Q

0.0175∆H=

116,760

×30 × 106= 0.003892 kmoles of H2. (10)

[ µabs = 0.003892 kmoles of H2.]

The total amount of hydrogen removed from the hydrogen source is

µtrans = µds + µabs = 0.0003285 + 0.003892

= 4.22 × 10−3 kmoles of H2. (11)

[ µtrans = 0.00422 kmoles of H2.]

The original amount of gas in the source as calculated in part a of thisproblem is µ0 = 0.0406 kmoles of H2.

The amount, µ1, of hydrogen left in the source is

µ1 = µ0 − µ = 0.0406− 0.00422 = 0.03638 kmoles of H2. (12)

[ µ1 = 0.03638 kmoles of H2.]

Consequently, the pressure becomes

p =µ1RTsource

Vsource

=0.03638× 8314× 300

0.005= 1.815× 107 Pa or 179.1 atmos.

(13)

Here we have a contradiction: the pressure of the gas in the dead spaceis 10 atmos while that of the hydrogen source is 179 atmos, yet the twocontainers are interconnected. A stream of hydrogen must be flowing fromsource to alloy canister—it cannot be a steady state situation.

e - Calculate the correct steady state pressure in the dead space.Make sure you check the correctness of any assumption you mayhave made.....................................................................................................................

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Initially, the source canister contains µ0 kmoles of hydrogen, under apressure, pinit, and a temperature of 300 K.

µ0 =pinitV

RT=

0.005

8314 × 300pinit = 2.00 × 10−9pinit. (14)

When Valve “a” is opened, an amount, µtrans, of hydrogen flows intothe alloy canister. Of this, an amount, µds will fill the dead space in thatcanister and an amount, µabs, will be absorbed by the alloy.

µtrans = µds + µabs. (15)

After the transfer, a total of µ0 − µtrans is left in the source.At the end of the experiment, the pressure in the source is

pf =8314 × 300

0.005(µ0 − µtrans) = 4.988× 108(µ0 − µtrans). (16)

This pressure is the same as that in the alloy canister because thesource canister and the alloy canisters are interconnected. Since the deadspace has a volume, Vds = 0.001 m3, it now contains

µds =pfVds

RT=

4.988 × 105(µ0 − µtrans)

8314T=

60(µ0 − µtrans)

Tkmoles (H2).

(17)Assuming that the alloy is not saturated, the pressure of the gas, both

in the plateau and in the dead space, is

pf = 1.013 × 105 exp

(

−∆S

R+

∆H

RT

)

= 1.013 × 105 exp

(

100,000

8314+

−30 × 106

8314T

)

= 1.013 × 105 exp

(

12.03−3608

T

)

Pa. (18)

Observe that the pressure is in pascals and that the pressure in thealloy canister is the same as that in the source, that is, Equations 16 and18, represent the same pressure:

4.988 × 108(µ0 − µtrans) = 1.013 × 105 × exp

(

12.03−3608

T

)

(19)

µ0 − µtrans = 2.031× 10−4 × exp

(

12.03−3608

T

)

(20)


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T =3608

12.03− ln(

µ0−µtrans

2.031×10−4

) (21)

From the above equation, one can find the temperature of the alloycanister for any chosen value of µtrans. Of course, only a single temperature(and a single value of µtrans) is correct, because the temperature has alsoto satisfy

T = 300 +Q

C= 300 +

∆Hµabs

Calloy + Chyd + Ccan

, (22)

where Q is the heat of absorption of µabs kilomoles of H2, and C = Calloy +Chyd + Ccan, is the total head capacity of the system.

The 3.5 kg of alloy contribute

Calloy = 440 × 3.5 = 1540 J/K. (23)

The µds kmoles of hydrogen in the dead space contribute

Chyd = 20,800µds, J/K. (24)

and the rest of the canister contributes Ccan = 100 J/K. Consequently,

C = 1640 + 20,800µds = 1640 + 20,80060(µ0 − µtrans)

T, (25)

and

T = 300 +30 × 106µabs

1640 + 20,800 60(µ0−µtrans)T

= 300 +30 × 106

(

µtrans −60(µ0−µtrans)

T

)

1640 + 20,800 60(µ0−µtrans)T

(26)

The temperature we are seeking must satisfy simultaneously Equations21 and 26. One easy way to find T is to use a numerical method:

Use µtrans as a parameter. Start by assigning to it an arbitrary value(which must be smaller than µ0). Using Equation 21, calculate T . Intro-duce this value of T into the right hand side of Equation 21.This will yilda value of T that, in all probability, is different from the one found fromEquation 21. Change the value of µtransuntil the T from the two equationsis the same.

This leads to the following results (when the initial pressure in thesource is 200 atmos):

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H2 transferred to alloy canister: µtrans = 0.01460 kmoles of H2.H2 in dead space: µds = 0.003097 kmoles of H2.

H2 absorbed by the alloy: µabs = 0.01120 kmoles of H2.Temperature in the alloy canister: T = 502.5 kelvins.

System pressure: pf = 127.6 atmos.

Note that the amount of H2 in the alloy is less that necessary to satu-rate it (0.0175 kmoles).


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Prob 11.22 The (very idealized) data of the TiFe alloy are:∆H= -28 MJ kmoles−1, for absorption.∆S= -106.1 KJ K−1kmoles−1, for absorption.Specific heat capacity, c = 540 J kg−1K−1

Density = 6200 kg/m3.The plateau pressure extends all the way from x = 0 to x = 1.

A this latter point, the pressure rises independently of x.100 kg of the above alloy are placed inside a 160-liter canister

which is then carefully pumped out so that the internal pressureis, essentially, zero. The container is perfectly heat insulated fromthe environment, and is (as well as the alloy inside) at 298.0 K.

A large, 1000-liter, separate container filled with hydrogen(pressure = 500.00 atmospheres) is kept at a constant 298.0 Kthroughout the whole experiment.

A pipe, equipped with a shut-off valve interconnects the twocontainers above. The valve is, initially, closed.

The experiment begins with the momentary opening of thevalve, allowing some hydrogen to enter the alloy-containing canis-ter. This process lasts long enough for the hydrogen source pres-sure to fall to 475.54 atmospheres, at which moment the valveis shut. Disregard any Joule-Thomson effect on the temperatureof the gas entering the alloy container, i. e., assume that thehydrogen enters this container at 298.0 K.

How much hydrogen was allowed into the alloy-containingcanister?.....................................................................................................................

The volume of the hydrogen source is Vsource = 1 m3.The temperature of the gas in the hydrogen source is Tsource = 298 K

(always).The initial pressure of the gas in the hydrogen source is psource0

=500 × 1.013× 105 = 50.65 MPa.

As a consequence, the initial amount of hydrogen in the source is

µsource0=

50.65 × 106 × 1

8314 × 298= 20.443 kmole (H2). (1)

When the pressure falls to psource1= 475.54 × 1.013 × 105 = 48.172

MPa, the amount of gas is

µsource1=

48.172× 106 × 1

8314 × 298= 19.443 kmole (H2). (2)

Exactly 1 kmole of H2 has been withdrawn and introduced in the alloy-containing canister.

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1.00 kmoles of H2 was allowed into the alloy-containing canister.

What is the final pressure of the hydrogen in this alloy-contain-ing canister after steady state has been reached?

Ignore the heat capacity of the H2. The heat capacity of thecanister is 20,000 J K−1......................................................................................................................

Since the alloy has no α-phase, the hydride will either be in the plateauor in the saturated region. Let us assume the former (if it proves wrong,we will assume saturation).

The amount of hydrogen, µintro, introduced will be divided into twoparts:

– one part, µalloy, will be absorbed by the alloy, forming a hydride,and

– another part, µds, will fill the dead space whose volume is

Vds = 0.160 − Valloy . (3)

The volume of the alloy, Valloy is

Valloy =Malloy

δalloy

=100

6200= 0.01613 m3. (4)

henceVds = 0.160− 0.01613 = 0.14387 m3. (5)

The pressure of the gas in he dead space must be equal to the plateaupressure (if the hydride is in the β-phase):

p = exp

(

−∆S

R+

∆H

(298 + ∆T )R

)

= exp

(

12.762−3367.8

298 + ∆T

)

. (6)

The amount of hydrogen in the dead space is

µds =pVds

8314(298 + ∆T )=

exp(

12.762− 3367.8298+∆T

)

× 1.013 × 105 × 0.14387

8314(298 + ∆T ).

(7)In the last two equations, we represented the temperature of the con-

tents of the canister as the sum of the initial temperature, 298 K, and thetemperature elevation, ∆T , owing to the absorption process.

The amount of gas absorbed by he alloy is

µalloy = µintro − µds = 1.0 −exp

(

12.762− 3367.8298+∆T

)

1.753

298 + ∆T. (8)


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The absorption of µalloy kmoles of H2 causes the release of an amount,Qrelsd, of heat

Qrelsd = 28 × 106

1.0 −exp

(

12.762− 3367.8298+∆T

)

1.753

298 + ∆T

. (9)

This heat causes temperature of both the alloy and the canister itselfto rise by an amount, ∆T .

Ignoring the heat capacity of the gas, the total heat capacity, Csys, ofthe system is the sum of the heat capacity, Ccan, of the canister itself andthe heat capacity, Calloy, of the alloy which is

Malloy × Calloy = 100 × 540 = 54, 000 J/K. (10)

Thus,

Csys = 20, 000 + 54, 000 = 74, 000 J/K, (11)

Intermediate Results

∆T 139.23 K

µds 0.6316 kmole

µalloy 0.3683 kmole

Qrelsd 10.31 MJ

and,Qreleased = 74, 000∆T, (12)

74, 000∆T = 28 × 106

1.0 −exp

(

12.762− 3367.8298+∆T

)

1.753

298 + ∆T

, (13)

∆T = 378.38

1.0 −exp

(

12.762− 3367.8298+∆T

)

1.753

298 + ∆T

. (14)

The solution of the above equation is ∆T = 139.25 which leads toT = 437.27 K.

Introducing the correct value for ∆T into Equation 6, one finds thatthe gas pressure is 157.57 atmospheres,

The pressure of the hydrogen in the dead space is 158 atmospheres.

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Prob 11.23 Solving this problem “exactly” is somewhat laborious.What we want is simply an estimate of the final temperature. You mustmake some simplifying assumptions. If you do this the hard way, you willlose some points (even if you get the right answer) because you will bewasting time. Reasonable simplifying assumptions will lead to estimateswith less than 2% error which is good enough for government work.

A small metal canister with a volume of 2 liters contains 3kg of TiFeH0.95 (HY-STOR alloy 101) at 300 K. Alloy density is6200 kg/m3. The atomic mass of titanium is 47.90 daltons andthat of iron is 55.85 daltons.The canister has no heat capacityand is completely heat-insolated from the environment. Estimatethe temperature of the gas in the cylinder after 2 g of H2 arewithdrawn......................................................................................................................

The atomic mass of TiFeH0.95 is 104.3 daltons. The number of kilo-moles of alloy in the canister is 3/104.3 = 28.8 × 10−3 kilomoles and theamount of hydrogen in the alloy is 1

228.8 × 10−3 × 0.95 = 13.66 × 10−3

kmoles of H2.Volume of the “dead space” is

Vds = 0.002−3

6200= 0.001516 m3. (1)

Clearly, if there are 0.95 kmoles of H in each kilomole (formula) ofalloy, the latter must be in the plateau region, i.e., the gas must be at apressure of

p = exp

(

−−106,100

8314+

−28 × 106

8314T

)

= exp

(

12.76 −3368

T

)

. (2)

For T = 300 K, p=4.64 atmos.Assume that the 2 g (0.001 kilomoles) of hydrogen withdrawn all come

from the hydrogen in the hydride. This ignores the fact that there is achange in the amount of hydrogen in gas form, which, in this problem willbe considered negligible. Under this assumption, the energy absorbed bythe alloy to permit the desorption is Qdes = 28 × 106µrem, where µrem =0.001 kmoles (H2), is the number of kilomoles desorbed. Consequently,Qdes = 28,000 J. This amount of energy must come from the cooling of thealloy whose heat capacity is 3 × 540 = 1620 J/K,

1620× (300 − Tf ) = 28,000. (3)

where Tf is the final temperature of the alloy and is, in this case, 282.7 Kor, since we are only estimating this temperature, Tf = 283 K.

The gas temperature in the cylinder will be 283 K.


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Prob 11.24 An adiabatic canister having a volume of 0.05m3, contains 30 kg of alloy, HY-STOR 205 at 300 K. Howeverthere is no gas of any kind in this canister.

The heat capacity of the canister, by itself, is negligible.Admit µin = 0.1 kilomoles of hydrogen into the canister. After

a while, things will settle to a new equilibrium. What is thepressure of the gas?

The density of HY-STOR 205(totallydepleted) is 8400 kg/m3......................................................................................................................

The volume of the depleted alloy is

Valloy =30

8400= 0.00357 m3. (1)

The dead space has a volume, Vds, of 0.05 − 0.00357 = 0.0464 m3.A fraction, α, of the hydrogen will be absorbed by the alloy, and a

fraction, 1 − α, will occupy the dead space.The absorption will cause the release of heat from he alloy:

Q = µinα|∆H | = 0.1α× 31 × 106 = 3.1 × 106α J. (2)

This heat will raise the temperature of the alloy and of the free gas.They have a heat capacity of 30 × 420 = 12,600 J/K for the alloy and29,000× 0.1× (1−α) = 2,900(1−α) J/K for the gas. This totals 12,600+2,900(1 − α) J/K.

Consequently, a temperature raise of

∆T =3.1 × 106α

12,600 + 2,900(1 − α)(3)

results leading to a final temperature of,

T = 300 + ∆T = 300 +3.1 × 106α

12,600 + 2,900(1 − α)(4)

At this temperature, the plateau pressure will be,

p = 1.013×105×exp

(

−∆S

R+

∆H

RT

)

= 1.013×105×exp

(

12.95−3729

T

)

(5)The gas in the dead space will be at the same temperature as the alloy.

From the perfect gas law,

µds = µin(1 − α) =Vdsp

RT. (6)

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0.1(1−α) =

0.0464×1.013×105×exp

12.95 −3729

300 +3.1 × 106α

12,600 + 2,900(1 − α)

8314×

(

300 +3.1 × 106α

12,600 + 2,900(1 − α)

)

(7)

(1 − α) =

5.654× exp

12.95− 3729

300 +3.1 × 106α

12,600 + 2,900(1 − α)

300 +3.1 × 106α

12,600 + 2,900(1 − α)

(8)

Although Equation 8 looks complicated, it can be seen that the onlyunknown is α, and that it can be solved numerically. The result is α =0.423.

We can now go to Equation 4 and find out T which turns out to beT = 403.9 K. This value can, in turn, be applied to Equation 5. . Thisyields p = 4,166,000 pascals or 41.1 atmospheres.

The hydrogen pressure is 41.1 atmospheres.


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Prob 11.25 The hydrogen distribution system of a given citydelivers the gas at a pressure of 20 atmos. An automobile refuel-ing station must increase the pressure to 400 atmospheres to fillthe pressurized gas containers in the cars which, typically, has astorage capacity of 6 kg of the gas.

Design a hydride hydrogen compressor for this application.The compressor should have the ability of delivering 6 kg of com-pressed hydrogen (or somewhat less) in one single “stroke”, i.e.,one single compression cycle. To make sure that there is an ade-quate gas flow from the city pipeline to the input of the compres-sor, the plateau pressure of the hydride used, at 25 C, should bea bit lower than the 20 atmosphere pipe line pressure. Assume,however that the intake plateau pressure is 20 atmos.

Normally, you should have complete freedom to select thealloy to be used, however, to avoid disparate solutions to thisproblem, we will impose some constraint in this choice. The alloymust have:

∆Sabsorption = −106.8 kJ K−1kmole−1,

A beginning of the plateau (transition between depletion andplateau) when the stoichiometric coefficient, xbeg = 0.15.

An end of the plateau (beyond which the alloy saturates)when xend = 1.05−0.00033T where T is the temperature in kelvins.

Use idealized characteristics of an alloy to build a hydro-gen compressor capable of raising the pressure from 0.5 atmo-spheres to 50 atmospheres. Idealize characteristics have horizon-tal plateaus. The alloy has a heat capacity of 540 J kg−1K−1.

The hydrogen compressor has the following phases:

– Intake. Start from point “A” of the characteristics and moveto Point “B”, at constant pressure.

– Compression. Begins at point “B and goes to Point “C. Tem-perature and pressure increase.

– Exhaust. Begins at Point “C” and goes to Point “D”. Gas isdelivered at constant pressure.

– Reset. Goes from Point “D” to Point “A” completing thecycle. Both pressure and temperature are reduced.

During the operation of the compressor, you must alwaysremain in the plateau region. You are not allowed to go intoeither the depletion or the saturation region. It makes sense todesign the system so that the end of the exhaust phase (Point D)is exactly at the beginning of the plateau (this allows the deliveryof the largest amount of hydrogen). Since Point A must have astoichiometric index larger than that of Point D, Point A cannotbe at the beginning of the the plateau. Use a stoichiometric index

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at A, xa = 0.36.The alloy you are going to employ has the formula AB and

forms an hydride ABH. “A” has an atomic mass of 48 and “B”, of56 daltons. The alloy has a density of 7000 kg/m3. The internal,empty, volume of the compressor is Vempty. Owing to the granularnature of the alloy, only 60% of this volume is actually occupiedby the alloy, the rest is “dead space”, usually occupied by H2 gas.The granules exactly fill Vempty, hence the “dead space” is onlythe intergranular space.

a – What is the ∆H (absorption) that the alloy must have......................................................................................................................The requirement that at 25 C (298 K) the plateau pressure must be 20atmos translates into

pin = exp

(

−∆S

R+

∆H

RT

)

= exp

(

−−106,800

8314+

∆H

8314 × 298

)

= exp

(

12.84 +∆H

2.478 × 106

)

= 20. (1)

12.84 +∆H

2.478 × 106= ln 20 = 3.0, (2)

from which, ∆H = −24.4 MJ/kmole.

The alloy must have an absorption enthalpy of -24.4 MJ/kmole

b – Determine to what temperature must the alloy be heatedto achieve the desired compression......................................................................................................................

The requirement that the output pressure be 400 atmos, leads to

ln pout = 12.84 +−24.4× 106

8314Tout

= ln 400 = 6.0 (3)

from which, Tout = 429 K or 156 C.

To effect the desired compression, the alloy must be heated to 156 C.

c – Assume that in the intake phase (A→B) the compressortakes in 6 kg of hydrogen. Calculate how many kilograms of thealloy are required......................................................................................................................To maximize the amount of hydrogen handled in the intake phase, the endof that phase, Point B, must be at the edge of the plateau, i.e., xB = 1.05−


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0.00033 × 298 = 0.952. Between Point A and Point B, the stoichiometricindex changed by ∆x = 0.952 − 0.36 = 0.591 and the amount of H-gastaken in is 0.591 × µAlloy = 6 kmoles of H (6 kg of hydrogen as specifiedby the problem statement. Hence, µAlloy = 6/0.591 = 10.1 kilomoles ofAB. Each kmole of alloy masses 48 + 56 = 104 kg. The total mass of alloyneeded is 10.1 × 104 = 1050 kg.

1050 kg of alloy are needed.

d – The above mass of alloy granules, exactly fills the internalvolume of the compressor. This leaves, as dead space, the inter-granular volume. Calculate the internal volume of the compressorand the volume of the dead space......................................................................................................................The density of the alloy is 7000 kg/m3, but owing to the granularity ofthe product, the alloy has an effective density of 0.6× 7000 = 4200 kg/m3.Hence the granules will occupy a volume of 1050/4200= 0.250 m3. Thedead space volume is 0.250 × 0.4 = 0.100 m3.

The compressor has a volume of 0.250 m3.The dead space has a volume of 0.100 m3.

e – How much H2 is desorbed while going from Point B toPoint C?.....................................................................................................................The amount of gas in the dead space at Point B is

µdsB=

pBVds

RTB

=20 × 105 × 0.1

8314 × 298= 0.081 kmole (H2). (4)

Repeating for Point C, we get µdsC= 1.125 kmole (H2).

The amount desorbed is 1.125− 0.081 = 1.04 kmole (H2).

1.04 kmoles of hydrogen (H2) gas (2.08 kg) are desorbed.

f – Considering only the energy for desorption form PointB to Point C and that from Point C to Point D plus the energynecessary to heat up the alloy from the intake temperature to theexhaust temperature, estimate the efficiency of the compressor.The compressor itself has zero heat capacity; ignore it......................................................................................................................

We have, from the previous item, knowledge of the amount of gasdesorbed in Step B→C: 1.04 kmoles of H2. The amount desorbed in StepC→D ought to be equal to the 6 kg (6 kmoles of H or 3 kmoles of H2),

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absorbed in Step A→B†. Thus the total desorbed from B→D is 4.04 kmolesof H2. This takes an energy of

Wdesor = 24.4 × 106 × 4.04 = 98.6 × 106 . (5)

In addition, we need energy to raise the temperature of 1050 kg of alloyfrom 298 K to 429 K (Step B→C), a ∆T of 131 K:

Wtemp = 1050 × 540 × 131 = 74.3 × 106. (6)

The total energy spent is 98.6 × 106 + 74.3 × 106 = 173 × 106 J.The useful output is the energy, Wout, to isothermally compress the

gas from 20 to 400 atmos:

Wout = 3 × 8314× 298 × ln(400/20) = 22.3 × 106 J. (7)

The efficiency is

η =22.3 × 106

172 × 106= 0.129. (8)

The efficiency of the compressor is 12.9%.

† If you calculate exactly the amount of hydrogen pumped out by the com-pressor, you will find that it is slightly larger than the amount taken in. In otherwords, the compressor is not in exact steady state. The reason for this minordiscrepancy is that the initial value for xA = 0.36 that I suggested you use is notthe precise value for steady operation—the correct value is 0.3563.


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*************************** **************************** Prob11.26 A system consists of two perfectly adiabatic 1-liter canis-ters interconnected by a pipe. A closed valve impedes the trans-fer of gas from one canister to another. There is no heat transferfrom one canister to the other even when the valve is opened.Both canisters are at 300 K.

Canister A contains 5 kg of TiFe alloy and a total of 0.04 kg ofhydrogen (part in the form of gas and part chemically combinedwith the alloy.) Canister B contains 5 kg of MgNi5 alloy and atotal of 0.02 kg of hydrogen.

Canisters and hydrogen have negligible heat capacity.

If the valve is opened, what will be the temperature of thealloy in A and that of the alloy in B after equilibrium has beenestablished?

Data

xmin

xmax

ln p

x

Material ∆Sf ∆Hf xmin xmax Heat Density MolecularkJ K−1 MJ J K−1 kg masskmole−1 kmole−1 kg−1 m−3 daltons

FeTi -106.1 -28.0 0.1 1.0 540 7,800 103.7MgNi5 -107.7 -31.0 0.6 5.0 420 6,600 317.8

.....................................................................................................................

In solving this problem it is a good idea to make an educated guess ofwhat is going to happen. A plausible guess is that both alloys are in theplateau region. To make sure, determine the initial conditions (before thevalve interconnecting the two canisters is opened). This is ease to do.

Next, consider that the plateau pressure of the two alloys—in this case,the hydrogen pressure in the dead space of the two canisters—are going tobe different. When the valve is opened, hydrogen will be desorbed from onealloy and absorbed by the other. Consequently, one alloy will cool down (itstemperature will be below 300 K) and the other will warm up (its tempera-ture will be above 300 K). You can’t expect that the temperature will changevery much because there is only a modest amount of hydrogen in the system.A preliminary wild guess would suggest something like 250 K for the colder

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alloy and 350 for the warmer.To proceed with the calculations we assume that the amount of hydrogen

in the free space is small compared with that in the alloy (check later to seeif this is true). Write down the obvious condition that when the valve isopened, the hydrogen pressure must be the same in the two canisters. Thisestablishes a relation between the two final temperatures, TA and TB.

Now we need another piece of information. The amount of hydrogendesorbed by one alloy must equal (to a first approximation) the amountabsorbed by the other. This will give you the second relationship betweenTA and TB.

Initial conditionsBoth canisters are at 300 K. The plateau pressure is given by

p = exp

(

−∆S

R+

∆H

RT

)

(1)

For Alloy A this leads to pplateauA0= 4.64 atmos. and for Canister B,

pplateauB0= 1.69 atmos.

The volume of the 5 kg TiFe (5/103.7=0.0482 kmoles) in Canister Ais 5/7800 = 6.41× 10−4 m3 or 0.641 liters. That leaves 0.359 liters of deadspace (3.59 × 10−4 m3).

The 0.04 kg of hydrogen are partially in the dead space as normalhydrogen gas at a pressure of 4.64 atmos or 4.70 × 105 Pa. From theperfect gas law:

µdead spaceA0=

4.7 × 105 × 3.59 × 10−4

8314 × 300= 6.76 × 10−5 kilomoles of H2

(2)This leaves 0.02-0.00007 or 0.0199 kmoles of H2 or 0.0399 kilomoles of

H as part of the hydride. The stoichiometry is

xA0=

0.0399

0.0482= 0.83. (3)

So, the alloy is not quite saturated.The volume of the 5 kg MgNi5 (5/317.8=0.0157 kmoles) in Canister A

is 5/6600 = 7.58× 10−4 m3 or 0.758 liters. That leaves 0.242 liters of deadspace (2.42 × 10−4 m3).

The 0.02 kg of hydrogen are partially in the dead space as normalhydrogen gas at a pressure of 1.69 atmos or 1.71 × 105 Pa. From theperfect gas law:

µdead spaceB0=

1.71 × 105 × 2.42 × 10−4

8314 × 300= 1.66 × 10−5 kilomoles of H2

(4)


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This leaves 0.01-0.00002 or 0.01 kmoles of H2 or 0.02 kilomoles of H aspart of the hydride. The stoichiometry is

xA0=

0.02

0.0157= 1.27. (5)

Since the alloy allows x to be as larger as 5, the alloy is quite farfrom saturation and can absorb a large amount of hydrogen. It can hold5 × 0.0157 = 0.0785 kmoles of H. It has capacity to hold an additional0.0785− 0.02 = 0.0585 kmoles of H.

The initial conditions are

Canister Plateau Alloy Dead Alloy H2 H2 H xPress. volume space Total Total D.S. Alloy

(atmos) (liters) (liters) (moles) (moles) (moles) (moles)

A 4.64 0.641 0.359 48.22 20 0.0677 39.9 0.83B 1.68 0.758 0.242 15.73 10 0.0166 20.0 1.27

We see that the hydride in Canister A, which is at 4.64 atmos, has theempirical formula TiFeH0.83 and is nearly saturated but not quite, while inCanister B, which is at 1.69 atmos, the hydride is MgNi5H1.27 and, althoughon the plateau, it is very far from saturation.

This suggests that when interconnected, hydrogen will flow from A toB, and, plausibly, both alloys may wind up in the plateau region. Thissimplifies calculations, but requires confirmation.

When the valve is opened, hydrogen at higher pressure in CanisterA will flow to Canister B. Temperature of A falls, reducing the plateaupressure of A, while the temperature of B rises (owing to hydrogen absorp-tion) and so will the plateau pressure. Equilibrium is re-established whenpplateauA

= pplateauB. This occurs at temperature TE .

exp

(

−∆SA

R+

∆HA

RTA

)

= exp

(

−∆SB

R+

∆HB

RTB

)

(6)

−∆SA +∆HA

TA

= −∆SB +∆HB

TB

(7)

106,100−28 × 106

TA

= 107,700 −31 × 106

TB

(8)

From this,

TB = 31TA

28 + 0.0016TA

. (9)

We need another independent piece of information relating TA to TB:

TA = 300 −µA × ∆HA

2700= 300 − 10,370 µA. (10)

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In the above, 2700 is the heat capacity of the 5 kg of TiFe, and µA isthe number of kilomoles of H2 desorbed.

Similarly,

TB = 300 +µB × ∆HB

2100= 300 + 14762 µB. (11)

The amount of hydrogen in the dead spaces is small compared withthat in the hydride, therefore on can take

µA ≈ µB (12)

This allows eliminating µ from Equations 10 and 11,thus establishinganother relationship between TB and TA:

TB = 727.06− 1.4235TA. (13)

Equating Equation 9 to Equation 13, and thus eliminating TB, we getthe quadratic,

T 2A + 30,599.9TA − 8,938, 042 = 0, (14)

whose solution isTA = 289.4 K. (15)

Using Equation 13,TB = 315.1 K. (15)

Alloy A cools down to 289.4 K, while Alloy B warms up to 315.1 K.

These are the solutions sought, however, it remains to prove that thepremise that both alloys are still in the plateau region is correct.

Using either Equation 10 or Equation 11, we can find the amount, µ,of hydrogen desorbed from A or absorbed by B. It works out to

µ = 0.00102 kmole of H2, (12)

which corresponds to 0.00204 kmoles of H.Hydride A was TiFeH0.83. It contained 0.0399 kmoles of H. After the

valve was open this amount fell to 0.0379 kilomoles of H and the hydridecomposition became TiFeH0.78. Alloy A is still in the plateau region.

Hydride B was MgNi5H1.27. It contained 0.020 kmoles of H. After thevalve was open this amount grew to 0.022 kilomoles of H and the hydridecomposition became MgNi5H1.40. Alloy B is still in the plateau region.

Conclusion: Both alloys are still in the plateau region, as assumed.Actually, the relative amount of hydrogen transfered from Canister A toCanister B, is quite small (see figure below).


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0 2 4 6 Stoichiometric index, x

x = 0.83

x = 1.27

x = 0.37

x = 1.40

TiFe300 K

TiFe (289.4 K) and

MgNi5 (315.1 K)

MgNi5 (300 K)

ln (p

)

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Prob 11.27 1 kg of the alloy AB is used to store hydrogen.Its characteristics for adsorption of hydrogen are:

∆S = −110 kJ per K per kilomole.

∆H = −30MJ per kilomole.

Heat capacity, c = 500 J per kg.

Both A and B have atomic masses of 50 daltons.

The alloy is charged with hydrogen so that its average com-position is ABH0.9. It is at 400 K and is inside an adiabaticcontainer with negligible heat capacity. Neglect the heat capac-ity of the gas and assume that the alloy fills the container leavingno dead space.

How much hydrogen is released if a valve is cracked open untilthe pressure falls to 1.5 atmosphere?

.....................................................................................................................

The plateau pressure is

p = exp

(

110 × 103

8314−

30 × 106

8314T

)

= exp

(

13.231− 36081

T

)

(0)

For T = 400 K, p = 67.4 atmos. 2 atmos correspond to T = 288 K.

The desorption of hydrogen will proceed until the alloy cools to 288 K,a temperature drop of 400 − 288 = 112 K.

All the heat must come from the alloy:

Q = m¸∆T = 1 × 500 × 112 = 56 × 1036 J (0)

This amount of heat must come from the desorption of

N =56103

30= 0.00187 kmoles or 3.7 grams of hydrogen. (1)

3.7 grams of hydrogen are released.


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472 Prob. Sol. Chapter 11 Fund. of Renewable Energy Processes

fep11 conssol elsevier - stanford universityweb.stanford.edu/.../fep11_conssol_elsevier.pdf ·...

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