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 m 3) 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 C p = 29.30 J/mol K.
(a) (b)
What is the work requirement to compress the gas in one sphere to 3 bar? What is the heat interaction with the constant-temperature constant-temperat ure bath?
Solution (Method 1):To 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: W Rev
Because the gas is ideal, V
P A
P B
VdP
nRT
. Since we can not assume 1 mole, we want to know how the number P of moles change from one Tank to the other. If we look at the entire system, whatever leaves Tank A gets into Tank B. dn dn A dnB But we are told that the volume of the two tanks is the same. dn
V
dP A RT
V
dP B dP A dP dP B RT
1
W Rev
A P RT ln dP P B
RT dP P dP
P V RTRev ln A dP P RT B
W
A
RT ln P A RT ln P B
W
V
dP A RT W Rev V ln P dP A A V ln P dP B A ; noting that dP A dP B Rev
dP B V ln P dP A A V ln P B
W
Rev
W
Rev
V ln P AdP A ln P BdP B
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, W Rev V
3bar
2bar
W Rev
ln P AdP A
3 1 ln P BdP B V P Aln P A P A PBln PB PB
1bar
2 0.1m3 x105 Pa 3ln 3 3 2 ln 2 2 1ln1 1 2 ln 2 2 2 bar
W Rev 5.23x10 J 3
dU dQ dW ; dU CvdT 0 Isothetmal Process Q W 5.23 x10 J 3
Method 2:
2
2
3
4
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 5
that, I pressurized and blew down B so rapidly that little heat transfer probably occurred this time. J Nitrogen is an ideal gas with Cp 29.33 mol. K (a)
Guess the temperature of the gas in B after the second pressurization and after the second blow down. What do you think these temperatures were after a large number of cycles?
(b) Solution:
We denote the large tank properties as H ', T ', P'
We also denote the small tank properties as U , T1 , P1
We also know P2 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 done on the system into 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 d nU H in dnin accumlation Enthalpy flow rate dt dt The mole balance is: dn dnin dn out But we know that during the pressurization nothing leaves the small tank . Thus dn dnin d nU H in dn d nU H in dn dt dt If we denote H in 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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n2U 2 n1U 1 H ' n2 n1 or n2 U 2 H ' n1 U1 H ' 0 H ' H ' n2 U 2 U ' RT ' n1 U1 U ' RT ' 0 ; noticing that U 2 U ' CvT
n2 Cv T2 T ' RT ' n1 Cv T1 T ' RT ' 0 ; noticing that n P2V
T T ' RT ' Cv RT
PV 1
P2V
CvT2 CvT ' RT ' RT
2
RT
2
2
PV
T T ' RT ' 0 Cv RT 1
1
PV 1
CvT1 CvT ' RT ' 0
1
RT
Now we also notice that Cv R Cp which is given. P2V PV R T2 T ' Cv R 1 Cp R T1 T ' Cv R 0 Cp RT RT 2 1
(A)
T ' 298K, T1 298K, Cp - Cv R, P1 101kPa and P2 687kPa , the only known is T 2 393.03K
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: out out in out out d (nU ) H dn ; mass balance: dn dn dn dn
d (nU ) Hdn ndU Udn Hdn dU H U
dn n
dn CvdT RT n Cv R
T2
T1
dT T
n2
n1
dn n
Cv R
T2 n2 ln T1 n1
ln
P2V RT 2 PT PV PV Cv T2 2 1 ln 2 1 But we know that n2 and n1 ln ln PV RT RT R 1 T 2 1 2 1 PT 1 RT 1 T2 P2 T 1 ln ln R T1 1 P T2 T P Cv T2 ln ln 2 ln 2 R T1 1 T1 P T Cv P T 1 ln 2 ln 2 ln 2 T1 R P1 T1 Cv
ln
Cp R P 1 ln 2 R P1
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P2 P1
T2 T 1
R Cp
(B)
Note that T1 for the depressurization step is actually the value of T 2 we obtained during the pressurization step, which is 393.03K. 8.314
101kPa 29.33 228K 687kPa
T 2 393.03K
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: T 228K and P 101kPa . Thus 228K will be the value of T 1 in equation (A). With this value we calculate another T 2 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 o small pump forces out the air remaining in the cylinder and replaces it with air at 20 C 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 2 4000kg; the piston has a surface area of 2.5m and the volume contained in the cylinder when the 3. elevator is at the lowest level is 25m There is no friction between the piston and cylinder and the air in J the cylinder is assumed to be an ideal gas with C p 30 . mol.K
Fig. A1.1 (a) What is the pressure in the cylinder throughout the process? 8
(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 2 Total mass = 4000kg; Piston area = 2.5m 3 Volume of the cylinder when the elevator is at the lower floor = 25m 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): F mg Total P Patm Pa 1.01x105 Pa A A P
4000kg*9.8m/s 2
2
1.01x105 Pa 1.17 x10 5 Pa
2.5m When the elevator is on the top floor, the gas has expanded such that the volume change is: V Axh 2.5m2 x3m 7.5 m3 . At this point the total volume occupied by the gas is: Total (7.5 25) m3 V Then the work done by the gas will be:
5 3 W PdV P(V2 V1 ) PV 1.17 x10 Pa 7.5m 875.1 kJ
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: V1 V 2 T1
T 2
T 2
V2T 1
380.9K 3 25m The number of moles in the system using ideal gas law is PV1 1.17 x105 Pa 25m3 1200 mol n Pa.m3 RT 1 8.314 x 293K K .mol For ideal gases, we know that internal energy is only a function of temperature and the change can be expressed as follows. U Cv(T2 T1 ) V 1
32.5m 3 293K
But CV=CP-R CV=30-8.314=21.686
U 21.686(380.9 293) 1.906
kJ
mol The total internal energy change will be then U U n 1.906 1200 2287.2kJ Using first Law of thermodynamics for closed system:
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U Q W
Q U W Q 2287.2kJ ( 875.1kJ) Q 3162.5kJ c) The % of Q used to increase the temperature of the system is U 2287.2kJ *100 72.3% Q 3162.5kJ % 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 3 initial position V=7.5 m amount of gas should be released while the gas temperature is at 380.9K. So using Ideal gas law. 3 5 V=7.5 m ; P=1.17 x10 Pa; T=380.9K PV nRT
n
PV RT
1.17 x10 Pa 7.5m 5
8.314
Pa.m 3 K.mol
3
277.1mol
380.9K
(Option 2) Mathematical argument: Here we note that the system is open. The mass and energy balance is: dn d dn out dV H Q P n and nU 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. dU dn dn d nRT dn dT d n H P H U nCv R U nT dt dt dt dt P dt dt dt
H U
dn dt
Noting that
dn dT R n T dt dt dt
dT
nCv
H Ufor an ideal gas is
RT :
dn dT dT dn dT dT dU 0 0 RT nCv nR RT n Cv R dt dt dt dt dt dt dt Thus T3=T2=380.1K
10
H U dn
or,
dt
dn dt
dn
RT
dt P dV
dV
P
dt
RT dt
with P and T constant n3 n2
V3 V2
25m3 32.5m3
0.7692 0.7692 x1200mol 923mol
n 277mol Problem 4: Consider 2 moles of an ideal gas in a piston-cylinder assembly at 25 oC, 0.5MPa and volume V. The gas is expanded isothermally to twice its original volume, then cooled isobarically to its o original V. It was then heated at constant volume to 25 C. 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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