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Department of Chemical and Biochemical Engineering CBE2224: Chemical Engineering Thermodynamics

Assignment 1: Review Problems (Due Friday January 30, 2009: Locker # 420)

Problem 1: Two large gas storage spheres (0.1 m3) each contain air at 2 bar (Figure below). They are connected across a small reversible compressor. The tanks, connecting lines, and the compressor are immersed in a constant temperature bath at 280 K. The compressor will take suction from one sphere, compress the gas, and discharge to the other sphere. Heat transfer between the bath and the tanks, lines and compressor is excellent. Assume that air is an ideal gas with Cp = 29.30 J/mol K.

(a) What is the work requirement to compress the gas in one sphere to 3 bar? (b) What is the heat interaction with the constant-temperature bath? Solution (Method 1):To begin with, let us consider tank A is our system. Then according to the problem, the compressor is taking gas from tank B and discharges it to tank A, isothermally (the problem states that the heat transfer is excellent). The work of the compressor from the analysis of energy balance for open system will give:

A

B

P

Rev PW VdP

Because the gas is ideal, nRT

VP

. Since we can not assume 1 mole, we want to know how the number

of moles change from one Tank to the other. If we look at the entire system, whatever leaves Tank A gets into Tank B.

But we are told that the volume of the two tanks is the same.A B

A B A B

dn dn dn

V Vdn dP dP dP dP

RT RT

2

ln

ln

ln ln

ln ln ; noting that

ln ln

ln ln

ARev

B

ARev A

B

Rev A B A

Rev A A B A A B

Rev A A B B

Rev A A B B

RT PW dP dP RT dP

P P

P VW RT dP

P RT

VW RT P RT P dP

RTW V P dP V P dP dP dP

W V P dP V P dP

W V P dP P dP

Since the volume of both tanks is the same and the temperature of the gas is also the same, when the pressure in one of the Tanks is increased the pressure on the other Tank decreases proportionally. This means, if the pressure inside one of the tanks is increased by 1 bar, the other will reduce by 1 bar. Thus,

3 1 3 1

2 22 2

3 5

3

3

ln ln ln ln

0.1m 10 Pa 3ln 3 3 2ln 2 2 1ln1 1 2ln 2 2

5.23 10 J

; 0 Isothetmal Process

5.23 10 J

bar bar

Rev A A B B A A A B B Bbar bar

Rev

Rev

W V P dP P dP V P P P P P P

W x

W x

dU dQ dW dU CvdT

Q W x

Method 2:

3

4

5

Problem 2: While relaxing near a large tank of nitrogen gas (A) at 687kPa and 298K, I began reviewing some of my knowledge about thermodynamics. A rather interesting experiment suggested itself and I thought I would compare theory with real field data (see figure below).

I obtained a small high-pressure vessel (B) and two valves (C) and (D). I first filled B with nitrogen gas at 101kPa and 298K and connected it as shown. Then working quickly, I opened valve C (with D closed) and allowed the pressures in B and A to equalize. Then, I quickly closed C and opened D to blow down vessel B to its original pressure. I repeated this sequence a number of times. Assume tank A was so large that I did not cause any significant drop in pressure in it by my experiments. Also suppose

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that, I pressurized and blew down B so rapidly that little heat transfer probably occurred this time.

Nitrogen is an ideal gas with J

29.33mol. K

Cp

(a) Guess the temperature of the gas in B after the second pressurization and after the second

blow down. (b) What do you think these temperatures were after a large number of cycles?

Solution:

We denote the large tank properties as ', ', 'H T P

We also denote the small tank properties as 1 1, ,U T P

We also know 2P of the small tank after pressurization

Take your system as the small tank and write the total energy balance equation.

Rate of energy Energy per unit mol Inlet mol Energy per unit mol outlet mol

accumlation of fluid at inlet flow rate of fluid at outlet flow rate

Rate of heat flow Rate that work is

into the system done on the system

1) During the pressurization cycle, valve D is closed so there is no outflow from our system which is the small tank. There is also no heat transfer because he conducted the experiment rapidly (thus q is zero). There is no shaft or any mechanical device for work, so the work term is also zero. With these assumptions,

Rate of energy Inlet mol Enthalpy

accumlation flow rate

The mole balance is:

in in

in out

d nU H dn

dt dt

dn dn dn

But we know that during the pressurization nothing leaves the small tank. Thus indn dn

inind nU H dn

d nU H dndt dt

If we denote 'inH H , then we will have: 'd nU H dn

Note that the enthalpy of the large tank stays the same. This is evident from the problem that says “Assume tank A was so large that I did not cause any significant drop in pressure in it by my experiments”. By integrating,

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2 2 1 1 2 1 2 2 1 1

' '

2 2 1 1 2

2 2 1 1

2 12

2 1

' or ' ' 0

' ' ' ' 0 ; noticing that '

' ' ' ' 0 ;noticing that

' '

H H

n U nU H n n n U H n U H

n U U RT n U U RT U U Cv T

PVn Cv T T RT n Cv T T RT n

RTPV PV

Cv T T RTRT RT

1

2 12 1

2 1

' ' 0

' ' ' ' 0

Cv T T RT

PV PVCvT CvT RT CvT CvT RT

RT RT

Now we also notice that Cv R Cp which is given.

2 12 1

2 1

' ' 0PV PV

Cp R T T Cv R Cp R T T Cv RRT RT

(A)

1 1 2' 298K, 298K, - , 101kPa and 687kPaT T Cp Cv R P P , the only known is 2 393.03KT

Note that when he depressurize (blow down) the small tank with valve C closed, the initial conditions are P2 and T2. Do not forget this. 2) Blow down Process Energy and mass balance:

2 2

1 1

2 2

1 1

( ) ; mass balance:

( )

ln ln

out out in out out

T n

T n

d nU H dn dn dn dn dn

d nU Hdn ndU Udn Hdn

dndU H U

ndn

CvdT RTn

T nCv dT dn Cv

R T n R T n

2

22 1 2 2 12 1

2 1 1 1 21

1

2 2 1

1 1 2

2 2 2

1 1 1

2

But we know that and ln ln ln

ln ln ln

ln ln ln

ln

PVRTPV PV T PTCv

n nRT RT R T PTPV

RT

T P TCv

R T P T

T T PCv

R T T P

T

2 2 2

1 1 1 1

1 ln ln 1 lnP T PCv Cp R

T R P T R P

8

22 1

1

R

CpPT T

P

(B)

Note that T1 for the depressurization step is actually the value of T2 we obtained during the pressurization step, which is 393.03K.

8.314

29.33

2

101kPa393.03K 228K

687kPaT

Thus the temperature would be 228K when he blows down the tank to 101kPa. For the second cycle (he is going to close valve D and open C to pressurize the tank), the small tank initial conditions will be:

228K and 101kPaT P . Thus 228K will be the value of T1 in equation (A). With this value we calculate another T2 which will be used to find yet, another T for the depressurization step using (B). The repeated cycle is conducted between equations (A) and (B) with the only change being temperature at each cycle. The final temperature in the small tank during the pressurization step will converge to 364K whereas it converges to 211K during the depressurization step.

Problem 3: Elevator Design Competition for the Green Engineering Building at Western:

A clever Chemical Engineer has devised a thermally operated elevator shown in the Figure. The elevator compartment is made to rise by electrically heating the air contained in the piston-cylinder drive mechanism and the elevator is lowered by opening a valve at the side of the cylinder, allowing the air in the cylinder to slowly escape. Once the elevator compartment is back to the lower level, a small pump forces out the air remaining in the cylinder and replaces it with air at 20oC and a pressure just sufficient to support the elevator compartment. The cycle can then be repeated. There is no heat transfer between the piston, cylinder and the gas; the weight of the piston, elevator and its contents is 4000kg; the piston has a surface area of 2.5m2 and the volume contained in the cylinder when the elevator is at the lowest level is 25m3. There is no friction between the piston and cylinder and the air in

the cylinder is assumed to be an ideal gas with J

30mol.KpC

.

Fig. A1.1

(a) What is the pressure in the cylinder throughout the process?

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(b) How much heat must be added to the air during the process of raising the elevator 3m and what is the final temperature of the gas?

(c) What fraction of the heat added is used in doing work and what fraction is used in raising the temperature of the gas?

(d) How many moles of air must be allowed to escape in order for the elevator to return to the lowest level?

System: Air is heated to raise the elevator; Air is released to lower the elevator T of the air at the inlet is 293K; Q=0 (no heat transfer); Cp=30J/K.mol Total mass = 4000kg; Piston area = 2.5m2

Volume of the cylinder when the elevator is at the lower floor = 25m3 Solution:

a) The system is the gas contained in the cylinder. The pressure of the elevator on the piston is the sum of the pressure due to the weight of the elevator and atmospheric pressure (Note that atmospheric pressure is exerted against the cylinder too):

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25 5

2

Pa 1.01 10 Pa

4000kg *9.8m/s1.01 10 Pa 1.17 10 Pa

2.5m

Totalatm

F mgP P x

A A

P x x

When the elevator is on the top floor, the gas has expanded such that the volume change is: 2 32.5 3 7.5V Axh m x m m . At this point the total volume occupied by the gas is:

3(7.5 25)TotalV m Then the work done by the gas will be:

5 32 1( ) 1.17 10 Pa 7.5 875.1 kJW PdV P V V P V x m

b) Since the number of moles of the gas and the pressure are constant, the final temperature of the system using the ideal gas law is:

1 2

1 2

32 1

2 31

32.5m 293K380.9K

25m

V V

T T

V TT

V

The number of moles in the system using ideal gas law is

5 3

13

1

1.17 10 251200 mol

Pa.m8.314 293

.

PV x Pa mn

RTx K

K mol

For ideal gases, we know that internal energy is only a function of temperature and the change can be expressed as follows.

2 1( )U Cv T T

But CV=CP-R CV=30-8.314=21.686

kJ21.686(380.9 293) 1.906

molU

The total internal energy change will be then 1.906 1200 2287.2kJU U n

Using first Law of thermodynamics for closed system:

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2287.2kJ ( 875.1kJ)

3162.5kJ

U Q W

Q U W

Q

Q

c) The % of Q used to increase the temperature of the system is

2287.2 kJ*100 72.3%

3162.5 kJ

U

Q

% of Q converted to Work 100% - 72.3% = 27.7% d) (Option 1) Logical arguments followed by ideal gas law: The system is adiabatic so there is

no heat transfer; Q=0. Since the piston-cylinder arrangement still exists, the pressure is constant. We don’t need any heat to bring the elevator to its initial position so there is no heat supply. Due to these reasons, the temperature will be constant. So to bring the elevator to its initial position V=7.5 m3 amount of gas should be released while the gas temperature is at 380.9K. So using Ideal gas law.

V=7.5 m3; P=1.17 x105Pa; T=380.9K

5 3

3

1.17 10 Pa 7.5m277.1 mol

Pa.m8.314 380.9K

K.mol

PV nRT

PV xn

RT

(Option 2) Mathematical argument: Here we note that the system is open. The mass and energy balance is:

and outdn d dn dVn nU H Q P

dt dt dt dt

(i) Note that Q=0 (ii) P=constant; (iii) the gas is ideal (iv) The Pressure and temperature of the gas leaving is the same as the gas inside.

Noting that for an ideal gas is :

0 0

dU dn dn d nRT dn dT dn U H P H U nCv R nT

dt dt dt dt P dt dt dt

dn dT dT dnH U nCv R n T

dt dt dt dt

H U RT

dn dT dT dn dT dT dURT nCv nR RT n Cv R

dt dt dt dt dt dt dt

Thus T3=T2=380.1K

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3

3 33

2 2

or,

with P and T constant

250.7692 0.7692 1200mol 923mol

32.5

277mol

dn dn dVH U RT P

dt dt dtdn P dV

dt RT dt

n V mx

n V m

n

Problem 4: Consider 2 moles of an ideal gas in a piston-cylinder assembly at 25oC, 0.5MPa and volume V. The gas is expanded isothermally to twice its original volume, then cooled isobarically to its original V. It was then heated at constant volume to 25oC. Sketch the process; calculate the work done at each step and the total work done on the system. You may assume that each process is reversible.

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