ChE 201 Washington State University Chemical Process Principles and Calculations Voiland School of Chemical Engineering and Bioengineering Fall, 2009 Richard L. Zollars Final Exam You will have three hours (180 minutes) to complete this exam which consists of five (5) problems. You may use any books, notes, etc. that you may have brought with you. Remember that the majority of points on any problem will be given for clearly demonstrating that you know how to solve the problem. Thus, it is not necessary to complete all of the math needed to reach a final numerical answer in order to receive most of the points for any problem. Make sure that you have completed each problem to the point where only calculations remain before taking the time to complete the calculations on any problem. Each problem statement may also contain information that is not needed in the solution of the problem.

1) (20 pts) You are the first engineer assigned to work on Moonbase I, a permanent station on the moon’s surface (gmoon = 1/6 of the gravitational acceleration of the Earth). One of the vital pieces of equipment has an open-ended manometer attached to it that looks like the schematic below.

The fluid in the manometer is mercury (ρ = 13.5 g/cm3). The atmosphere inside Moonbase I is a pure oxygen atmosphere at a pressure of 155 mm Hg (3.00 psia). What is the pressure inside the process? (Think about how the gravitational pull will affect hydrostatic pressure.) SOLUTION Starting at the height of 10 cm we know that the pressure at that point is the same in both legs of the manometer. Thus

( )

HgmmHgmmHgmmP

HgmmPa

Hgmmmsm

mcm

gkg

cmgPhgP

process

ambientmoonprocess

1961554.41

155103.101

76025.06

8.91001000

15.13 3

23

3

=+=

+×

=+= ρ

Note that the pressure caused by 250 mmHg on the moon is not 250 mmHg but 41.4 mmHg since this pressure depends upon the acceleration of gravity.

Open to Interior To Process

25 cm

10 cm

45o

2) (25 pts) Coffee beans contain 1.00 % by weight caffeine. One process for decaffeinating coffee is to soak the beans in liquid CO2 at high pressure (supercritical CO2). Suppose you are treating 100 lbm of coffee beans with supercritical CO2. All of the caffeine in the beans is extracted by the CO2. The beans then are filtered from the CO2/caffeine solution. The beans carry with them 1 lbm of the CO2/caffeine solution for every 3 lbm of the decaffeinated beans. When the pressure on the beans is reduced all of the CO2 vaporizes leaving the caffeine from the solution that adhered to the beans on the treated beans. The amount of caffeine remaining on the beans must be only 1.00 % of the original amount in the beans in order for the beans to be considered decaffeinated. The liquid phase from the filtration process is sent to a vessel where the pressure is reduced. At the lower pressure all of the CO2 is converted to a gas phase, leaving all of the caffeine in the vessel. The now gaseous CO2 is taken from this vessel and repressurized to convert this back to supercritical CO2. This stream then is recycled to the start of the decaffeinating process where it is mixed with fresh supercritical CO2 and used to treat the next batch of coffee beans. What is the composition of the CO2/caffeine liquid leaving the filtration process? How much CO2 is recycled? How much fresh CO2 must be added to the recycled CO2 before being used to decaffeinating the beans? SOLUTION The process diagram for this process looks like the following

Since there are 0.99 (100) = 99 lbm of solids in the beans the amount of the beans out in the stream from the filter must also be 99 lbm. This stream carries with it 1 lbm of the solution per 3 lbm of beans. Therefore there must be 99/3 = 33 lbm of the CO2/caffeine solution. In addition the amount of caffeine in this stream cannot be greater than 1% of the original amount of caffeine since this caffeine will show up as decaffeinated coffee. Thus the amount

Recycled CO2

CO2 + Beans 100 lbm Coffee Beans

0.99 solids 0.01 caffeine

Extraction

Filter

Beans CO2 caffeine

Fresh CO2

Reduced pressure

caffeine

of caffeine is (0.01) (100) (0.01) = 0.01 lbm. If 0.01 lbm of the solution is caffeine the remainder must be CO2 so the amount of CO2 is 33 – 0.01 = 32.99 lbm. Thus the process diagram is

The composition of the liquid out of the filter must be the same as than adhering to the beans. Thus the composition is 0.01/33 = 0.000303 caffeine and 0.9997 CO2 (mass fractions). The stream carrying the caffeine into the reduced pressure chamber must have the remainder of caffeine 1.00 – 0.01 = 0.99 lbm. Since it is the same composition as the liquid adhering to the beans the amount of CO2 must be (0.99/0.000303)0.9997 = 3266 lbm. This is the amount of CO2 recycled. The amount of fresh CO2 required is equal to the amount of CO2 lost in the stream that contains the decaffeinated beans. Thus the fresh CO2 feed must be 32.99 lbm.

Recycled CO2

CO2 + Beans 100 lbm Coffee Beans

99 lbm solids 1 lbm caffeine

Extraction

Filter

99 lbm Beans 32.99 lbm CO2 0.01 lbm caffeine

Fresh CO2

Reduced pressure

caffeine

3) (35 pts) The Columbia generating station in the Tri-Cities is a nuclear powered, electrical generating facility. In nuclear generating facilities water is brought into thermal contact with the nuclear core, converting it to steam at high temperature and pressure. This steam is sent to a turbine where the pressure is reduced and work is extracted from the steam. The work obtained in the turbine is sent to a generator where the work is converted to electricity. Columbia has a rated output of 1190 MW (electrical). Assume that the generator is 35% efficient in converting thermal energy into electricity. Assume that the turbine is adiabatic and is 100% efficient. If the steam coming from the nuclear reactor is at 700ºC and 40 bar (abs) and if the steam leaving the turbine has a quality of 99% (i.e., 99 wt% of the steam is in the vapor state, 1 wt % is in the liquid state) and is at 1 bar (abs) how much steam must pass through the turbine (kg/s)? If the water entering the core is at 20ºC and 1 bar (abs) and it leaves at 700ºC and 40 bar (abs) how much heat is transferred to the water (kW)? SOLUTION The process diagram for this system would be

An energy balance around the generator would give the shaft work input as

kWMWMWWs31034003400

35.01190

×===

An energy balance around the turbine gives

( )s

kJHHm

WH

inout

s

3103400ˆˆ ×=−

−=∆

The enthalpies for the steam can be found in Tables B.6 and B.7. The steam coming out is 99% quality. Since it is a vapor liquid mixture it must be saturated at 1 bar. Looking up the enthalpy values for both saturated water and saturated steam in Table B.6 gives

( ) ( )kgkJHHH liquidsatvaporsatout 8.26525.41701.04.267599.0ˆ01.0ˆ99.0ˆ =+=+=

Steam 99% quality 1 bar

Steam 700°C 40 bar

Water 20°C 1 bar

Core

Turbine

Generator

Ws

1190 MW

Q

From Table B.7 you get inH = 3904 kJ/kg. Thus

( ) ( )

skgm

skJmHHm inout

2717

10340039048.2652ˆˆ 3

=

×=−=−

An energy balance around the core gives

( ) ( )

MWs

kJQ

HHmQH inout

36 1038.101038.10

9.8339042717ˆˆ

×=×=

−=−==∆

where the enthalpy of the water ( inH ) was obtained from Table B.5.

4) (55 pts) Ethyl acetate is commonly used as a solvent in various industrial painting operations. Because of toxicity concerns the paint booths used for painting, when the paints contain ethyl acetate, must be well ventilated. Suppose the gas phase coming from a paint booth contains 5.00 mole% ethyl acetate vapors with air forming the remainder of the gas phase. This gas stream is at a temperature of 25°C and a pressure of 1.00 atm. The OSHA requirement is that the ethyl acetate content of this gas stream must be no greater than 400 ppm before it can be discharged to the atmosphere. To reduce the ethyl acetate content the plan is to remove the ethyl acetate vapors by reducing the temperature, condensing enough of the ethyl acetate to meet the OSHA standard, before discharging the paint booth gases. What is the highest temperature of the gas stream if the OSHA requirements are to be met? If 1.00 mole of gas is to be treated how much heat must be transferred from the gas stream in this process? Additional Information:

Cp (ethyl acetate, vapor) = 0.126/(mol•ºC) Cp (ethyl acetate, liquid) = 0.170 kJ/(mol•ºC) Normal Boiling Point (ethyl acetate) = 77.0ºC Heat of Vaporization (ethyl acetate, @ 77.0ºC) = 31.94 kJ/mol Cp (air) = 0.0304 kJ/(mol•ºC) Vapor Pressure of ethyl acetate:

( )0.217

71.123809808.7log *10 +

−=T

p acetateethyl (T in °C, p* in mm Hg)

SOLUTION The process diagram for this system is

Where the composition of the gas stream leaving the condenser is given as the OSHA limit (400 ppm = 400x106 = 0.0004 mole fraction). This must be saturated with ethyl acetate. Thus the partial pressure and vapor pressure of ethyl acetate must be

( ) HgmmHgmmpEA 304.07600004.0* ==

EA (l) xEA = 1.0 Tout 1 atm

Air + EA yEA = 0.0004 yair = 0.9996 Tout 1 atm

Air + EA yEA = 0.05 yair = 0.95 25°C 1 atm

Q

Using the Antoine equation for ethyl acetate gives

( ) ( )

CTT

p

out

outacetateethyl

3.54

304.0log0.217

71.123809808.7log 10*

10

−=

=+

−=

If there is one mole of gas coming in there must be 0.95 mole air in both the input and output streams. Thus the total amount of the output stream must be 0.95/0.9996 = 0.95038 mole and the amount of ethyl acetate in the vapor state would be 0.95038(0.0004) = 0.000380 mole. The ethyl acetate that liquefies is 0.05 – 0.00038 = 0.04962 mole. The input/output enthalpy data then gives

nin inH nout

outH Air 0.95 0.95 EA (l) 0 - 0.04962 EA (v) 0.05 0.00038

Pick air and ethyl acetate (v) at 25°C as the reference states. The enthalpy table then is

nin inH nout

outH Air 0.95 0 0.95 EA (l) 0 - 0.04962 EA (v) 0.05 0 0.00038

The enthalpy of liquid EA at -54.3°C on this basis then is

( ) ( )kgkJH

dTdTdTCHdTCH

lEA

lEApvvEAplEA

71.47773.54170.094.312577126.0ˆ

170.094.31126.0ˆ

3.54),(

3.54

77

77

25

3.54

77)(,

77

25)(,3.54),(

−=−−+−−=

∫+−∫=∫+∆−∫=

−

−−

−

The enthalpy of EA vapor at -54.3°C on this basis then is

( )kgkJH

dTdTCH

vEA

vEApvEA

99.9253.54126.0ˆ

126.0

3.54),(

3.54

25

3.54

25)(,3.54),(

−=−−=

==

−

−−

− ∫∫

The enthalpy of air at -54.3°C on this basis then is

( )kgkJH

dTdTCH

air

airpair

41.2253.540304.0ˆ

0304.0

3.54,

3.54

25

3.54

25,3.54,

−=−−=

==

−

−−

− ∫∫

The enthalpy table finally gives

nin inH nout

outH Air 0.95 0 0.95 -2.41 EA (l) 0 - 0.04962 -47.71 EA (v) 0.05 0 0.00038 -9.99

The energy balance on the condenser gives

( ) ( ) ( )kJQ

HHQH inout

66.4099.900038.071.4704962.041.295.0

−=−−+−+−=−==∆

5) (65 pts) Air (21.0 mole% O2, 79.0 mole% N2) and ethane (C2H6) vapor, both at 1 atm (abs) and 25ºC, are mixed and fed into a furnace. The ethane is burned completely. The gases come out of the furnace at 200ºC and 1 atm (abs) and contain 8.51 mole% CO2, 7.45 mole% O2 and 84.0 mole % N2 on a dry basis. The relative humidity of the gases leaving the furnace is 0.782%. The total flow rate (wet basis) of the gases leaving the furnace is 118 standard liters per minute. What is the flow rate of ethane (mole/min) into the furnace? What is the percentage of excess air fed to the furnace? What is the heat transfer rate (kW) for the furnace? Additional Information: Vapor Pressure (water, 200ºC) = 11659 mm Hg MW (ethane) = 30.07 ΔHc (ethane, 25ºC) = -1559.9 kJ/mol Boiling Point (ethane) = -88.6 ºC Heat Capacity (ethane, vapor) = 0.050 kJ/(mol•ºC) Heat Capacity (ethane, liquid) = 0.069 kJ/(mol•ºC) ΔHf (ethane, 25ºC) = -84.67 kJ/mol ΔHf (CO2(g), 25ºC) = -393.5 kJ/mol ΔHc (H2O (l), 25ºC) = -285.84 kJ/mol SOLUTION The process diagram for this system would look like

If the flow rate for the exhaust gases (wet basis) is 118 standard liters/minute the molar flow rate (wet) is 118/22.4 = 5.27 mole/min. Since the vapor pressure of water at 200°C is 11659 mmHg and the relative humidity is 0.782% the partial pressure of water in the exhaust gases must be

( ) Hgmmp OH 2.911165900782.02 ==

Q

Ethane 25°C 1 atm

Furnace

Air 25°C 1 atm

Gases yCO2 = 0.0851 yO2 = 0.0745 yN2 = 0.840 200°C 1 atm

Water (v) 200°C 1 atm

Therefore, the mole fraction of water vapor (wet basis) in the exhaust gases is

120.0760

2.912 ==OHy

Therefore, the number of moles of water leaving the furnace is

( )minmolenwater 631.027.5120.0 ==

So the number of moles of gas (dry basis) is 5.27 – 0.631 = 4.64 mole/min. Therefore the number of moles of each species in the exhaust gas is

( )

( )

( )minmolen

minmolen

minmolen

N

O

CO

89.364.4840.0

345.064.40745.0

395.064.40851.0

2

2

2

==

==

==

A carbon balance gives

minmolen

nn

inHC

outCOinHC

197.0

395.02

,62

,2,62

=

==

A nitrogen balance gives

( )

minmolen

nn

inN

outNinN

89.3

89.3222

,2

,2,2

=

==

From the composition of the air you get

minmolen

n inNinO 04.1

79.089.321.0

79.021.0 ,2

,2 ===

The stoichiometry of the reaction that is taking place is

OHCOOHC 22262 3227

+→+

So if there are 0.197 moles of C2H6 entering 7/2(0.197) = 0.690 mole/min of O2 are required for complete reaction. Thus the %excess air is

%8.50%100690.0

690.004.1% =−

=excess

All of the molar quantities are now known. The enthalpy input/output table would now be

nin inH nout

outH C2H6 0.197 0 CO2 0 0.395 H2O 0 0.631 O2 1.04 0.345 N2 3.89 3.89

We need to pick a reference state for the calculations. Let’s pick C2H6 (v), CO2 (g), H2O (l), O2 (g), and N2 (g) at 25°C for the reference states. Thus the heat of combustion data given in the problem statement can be used. Using these reference states all of the input enthalpies are zero. The output enthalpies then can be determined from Table B.8. They are:

molkJH

molkJH

molkJ

molkJ

molkg

kgkJH

molkJH

outN

outO

outOH

outCO

13.5ˆ

31.5ˆ

01.5001.6018016.05.2442ˆ

08.7ˆ

,2

,2

,2

,2

=

=

=+=

=

Where, for water, the enthalpy of evaporation was obtained from Table B.5 at 25°C. The input/output table thus becomes

nin inH nout

outH C2H6 0.197 0 0 - CO2 0 - 0.395 7.08 H2O 0 - 0.631 50.01 O2 1.04 0 0.345 5.31 N2 3.89 0 3.89 5.13

The energy balance on the furnace gives

( ) ( ) ( ) ( ) ( )

kWs

kJminkJQ

Q

HnHnHQH iin

iiout

ioc

19.419.42.251

13.589.3`3.5345.001.50631.008.7395.09.1559197.0

ˆˆˆ

−=−=−=

++++−=

−+∆==∆ ∑∑ξ