tutorial sheet 01 answers 2014

8/12/2019 Tutorial Sheet 01 Answers 2014

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1

Tutorial Sheet No. 1 2014: - Answers

Ideal and Non-Ideal Process Systems

1) I have some ethylene gas at a pressure 10 MPa and a temperature of 47C. Calculate;

a) The specific volume, (V/m), in m3/kg, using the ideal gas model: PV = mRT .

[4 marks]

b) The specific volume via the non-ideal gas function: PV = ZmRT .

[5 marks]

c) Using the non-ideal gas function, what pressure is required in order that the ethylene

have a specific volume of 0.0062 m3/kg at a temperature of 47C.

[8 marks]

d) Using the non-ideal gas function, what will be the temperature of ethylene when it has

a specific volume of 0.01 m3/kg and a pressure of 10 MPa

[8 marks]

Supplied Data: R = 0.29637 kJ / kg K for ethylene

T (K) = T (C) + 273

Data Sheet No. 2.1 Table of Critical Constants of Gases.

Data Sheet No. 2.2 Generalized Compressibility Factor Plot.

Answer:

a) Ideal gas:

PV = mRT

P

RT

m

V ν

Substituting in the data (note the units):

kg N

mJ 0.0095

)(N/m1010

(K)47)(273K)(J/kg)10(0.29637

2

26

3

n

V

1 J = 1 N m, so:

/kgm0.0095kg N

mm)(N 0.0095

32

b) For ethylene, from Data Sheet 2.1:

T C = 282.4 K

P C = 5.12 MPa = 51.2 × 105 Pa

1.13282.4320

C

r T T T



2

1.951051.2

1010

5

6

C

r P

P P

From Data Sheet 2.2:

Z = 0.45

kg/m0.0043)(N/m1010

(K)47)(273K)(J/kg)10(0.296370.45

3

26

3

P

ZRT

m

V ν

c) First calculate T r

1.13282.4

320

C

r T

T T

Next calculate P r , however, in order to calculate P r we need to know P . Substituting for P

from the non-ideal gas equation, gives:

ν

ZRT P

Therefore, P r becomes:

C C

r νP

ZRT

P

P P

Substituting in the known data gives:

Z Z

P r 2.9881051.20.0062

320100.29637

5

3

We now need to use a trial and error procedure to find P r . First guess a value of P r , calculate

the value of Z from the above equation. Then read off the value of Z from the chart for the

given P r and T r . Compare the two values, when they are equal we have the answer.

P r guess Z Calc Z Chart

1.25 0.42 0.64

1.50 0.50 0.53

1.75 0.59 0.48

From the graph (see next page), P r = 1.55, giving

P = P r P C = 1.55 × 51.2 × 105 = 7.936 × 106 Pa = 7.94 MPa



3

Pr

1.20 1.25 1.30 1.35 1.40 1.45 1.50 1.55 1.60 1.65 1.70 1.75 1.80

Z

0.40

0.45

0.50

0.55

0.60

0.65

0.70

ZCalc

ZChart

d) First calculate P r

1.9531051.2

1010

5

6

C

r P

P P

Next calculate T r , however, in order to calculate T r we need to know T . Substituting for T

from the non-ideal gas equation, gives:

ZR

P νT

Therefore, T r becomes:

C C

r ZRT

P ν

T

T T

Substituting in the known data gives:

Z Z T r

1.195

282.4100.29637

0.011010

3

6



4

We now need to use a trial and error procedure to find T r . First guess a value of T r , calculate

the value of Z from the above equation. Then read off the value of Z from the chart for the

given P r and T r . Compare the two values, when they are equal we have the answer.

T r guess Z Calc Z Chart

1.3 0.92 0.721.4 0.85 0.81

1.5 0.80 0.86

Tr

1.25 1.30 1.35 1.40 1.45 1.50 1.55

Z

0.70

0.75

0.80

0.85

0.90

0.95

ZCalc

ZChart

From the graph , T r = 1.44, giving

T = T r T C = 1.44 × 282.4 = 406.66 K



5

2) Calculate the molar volume, v = V/n, (m3/kmol) of a mixture of gases containing 59.39

mol% CO2 and 40.61 mol% methane (CH4) at 310.94 K and 86.19 bar ab pressure using:

a) The ideal gas law equation: PV = nRT

[3 marks]

b) Van der Waals equation:2

ν

abν

RT P m

m

where 2 j jiim a ya ya , j jiim b yb yb and

i

i

C

C

i P

T Ra

64

2722

,

i

i

C

C

i

P

RT b

8

[10 marks]

c) Psuedo-crictical constants (Kays Rule) and the compressibility factor chart.

[10 marks]

d) The measured specific volume is 0.2205 m3/kmol. What is the percentage deviation of

each model from the real value?

[2 marks]

Supplied Data: Data Sheet No. 2.1 Table of Critical Constants of Gases.

Data Sheet No. 2.2 Generalized Compressibility Factor Plot.

Universal Gas Constant: R = 8.314 kJ kmol-1 K -1

Answer:

a) Using the ideal gas law we can assume the mixture is ideal so:

PV = nRT

P RT

nV ν

T = 310.94 K, P = 86.19 bar ab, R = 8.314 kJ kmol-1 K -1

/kmolm0.29991086.19

310.94108.314 3

5

3

ν

b) Van der Waals equation:

CO2: T C = 304.2 K P C = 7.39 MPaCH4: T C = 191.1 K P C = 4.64 MPa

2-6

6

22322

kmolmPa365155.5)10(7.3964

(304.2))10(8.31427

64

27

2

2

2

CO

CO

C

C

CO P

T Ra

/kmolm0.04278)10(7.398

304.2108.314

8

3

6

3

2

2

2

CO

CO

C

C

CO P

RT b



6

2-6

6

22322

kmolmPa229513)10(4.6464

(191.1))10(8.31427

64

27

4

4

4

CH

CH

C

C

CH P

T Ra

/kmolm0.04280)10(4.648

191.1108.314

8

3

6

3

4

4

4

CH

CH

C

C

CH P

RT

b

2-6

22

kmolmPa306290.1

2295130.4061365155.50.59394422

CH CH COCOm a ya ya

/kmolm0.042790.04280)(0.40610.04278)(0.59393

4422 CH CH COCOm b yb yb

2

ν

a

ν - b

RT P m

m

2

35 306290.1

-0.04279-

310.94108.314 1086.19

νν

Solve by trial and error:

Guess v RHS 105 Pa

0.2 87.87

0.3 66.47

0.21 85.150.205 86.49

0.206 86.21

Therefore:

v = 0.206 m3/kmol

c) Pseudo critical constant or Kays Law

K 258.27191.1)(0.4061304.2)(0.59394422

CH COi

C CH C COC iCM T yT yT yT

MPa6.274.64)(0.40617.39)(0.59394422

CH COi C CH C COC iCM P y P y P y P

1.203258.27

310.94

CM

rM T

T T

1.375106.27

1086.19

6

5

CM

rM P

P P

From Data Sheet 2.2a: Z M = 0.71



7

kmol/m0.2130 N/m1086.19

K)(310.94K)(J/kmol)10(8.3140.71

3

25

3

P

RT Z ν

M

d) The percentage deviation is given by:

100 valueactual

valuecalculated-valueactual Deviation%

Model % Deviation

Ideal Gas - 36.0

Van Der Waals + 6.6

Pseudocritical constant + 3.4



8

3) In the soft drinks industry, there is considerable demand for food grade carbon dioxide for

product carbonation. The carbon dioxide is supplied as a gas at low pressure, and the first

step in the production process of the food grade carbon dioxide is to pressurize the carbon

dioxide gas.

Accordingly, estimate the work (W in kJ per kmol) required to compress 1.0 kmol CO2 from P 1 = 14.7 psia to P 2 = 4042.5 psia in a compressor unit operating reversibly with inter-stage

cooling so that the gas temperature is constant at T = 61.6 C, using the followingmathematical models:

a) An ideal gas, such that:

1

2ln P

P nRT W

[4 marks]

b) A non-ideal gas, such that the compressibility factor, Z , has to be used to

evaluate the actual CO2 volumes of V 1 and V 2 for:

2

1lnV

V nRT W

[10 marks]

Existing compressors indicate that the actual compressibility factor for the carbon dioxide

over this range of system pressures is Z = 0.72.

c) Hence, comment on how the predicted Z values obtained in part (b) relate to

this actual Z value.

[4 marks]

d) When the carbon dioxide gas flow to the suction inlet of the compressor is

1020 m3 h-1 (at T = 61.6 C and P 1 = 14.7 psia) and the compressor has an

operational efficiency of 35%, specify the total power rating (in kW) of the

compressor motor.

[7 marks]

Supplied Data: Data Sheet No. 2.1 Table of Critical Constants of Gases.

Data Sheet No. 2.2 Generalized Compressibility Factor Plot

Universal Gas Constant: R = 8.314 kJ kmol-1 K -1

1 bar = 1 × 105 Pa = 14.5 psi

T (K) = T (C) + 273

Answer:

The objective is the engineering design evaluation of a dry carbon dioxide compressor duty in

terms of work energy.

ASSUMPTIONS:Steady state / isothermal / reversible / ideal and non-ideal operation / no chemical reaction.



9

BASIS OF CALCULATION:

1.0 kmol CO2 at initial conditions (suction point) State [1].

First draw a diagram of the system.

DIAGRAM:

Effective isothermal compression change.

a) For an ideal gas, the mathematical model is:

nRT PV

and

1

2ln

P

P nRTW

Now n = 1.0 kmol CO2

Then the work done on the gas by the compressor is:

psia14.7

psia4042.5lnK)(334.6

K kmol

kJ 8.314)COkmol(1.0 2W

kJ15625W

b) For a non-ideal gas with the non-ideality being described by the compressibility factor, Z ,

then the mathematical model is:

ZnRT PV

We need to apply this model at the inlet state (State [1]) and the outlet state (State [2])

conditions to determine the value of Z at each state and hence the magnitude of the actual real

volume of CO2 gas, V 1, and V 2 for the basis of calculation of n = 1.0 kmol CO2.

From Data Sheet No. 2.1, critical constants of carbon dioxide gas are:

P 2 = 4042.5 psia

T 2 = 61.6 C (334.6 K)

Discharge

Q

Suction

n = 1 kmol CO2

P 1 = 14.7 psia

T 1 = 61.6 C (334.6 K)



10

MPa7.39C P K 304.2C T

AT STATE [1]

1.10K 304.2

K 334.6 1

T

T T

C

r

0.0137107.39

10114.5)/(14.7

6

5

1

C

r P

P P

Then from Data Sheet No 2.2b,

0.991 Z

(Note: Close to ideal behaviour with gas at 14.7psia and 61.6 C)

The volume occupied by the non-ideal CO2 gas at State [1] is then:

1

111

P

RT nZ V

Now watch the units, pressure must be in N m -2 and the universal gas constant in J / kg mole,

so:

2-5-25

1

m N101.013 bar 1.0

m N101

psia/bar 14.5

psia14.7 P

2

3

2-5

21 COm27.19

)m N10(1.013

K)(334.6K)kmol/J314(8(0.99))COkmol(1.0

V

AT STATE [2]

1.10K 304.2

K 334.6 2

C

r T

T T

3.77107.39

101.5)(4042.5/14

6

5

1

C

r P

P P

Then from Data Sheet No 2.2a,

0.542 Z

(Note: Now at high pressure, non-ideal as more intermolecular force effects)

The volume occupied by the non-ideal CO2 gas at State [2] is then:



11

2

3

2-5

22 COm0.0539

)m N101.5)(4042.5/14

K)(334.6K)kmol/J314(8(0.54))COkmol(1.0

V

The work done by the on the gas by the compressor is:

2

1lnV

V nRTW

the non-ideality is incorporated in the volume calculations.

3

3

22m0.0539

m27.19lnK)(334.6K)COkmol/kJ(8.314)COkmol(1.0W

kJ31317W

Note the extra energy needed to overcome the non-ideality of the system and to push the CO2

molecules together.

c) Comparison of the compressibility factors, Z , is:

Ideal Gas Z = 1.00

Non-ideal P 1 = 14.7 psia Z = 0.99

P 2 = 4042.5 psia Z = 0.54

Measured value Z = 0.72

The reported value is at an unknown pressure on the existing compressors but supports the

predicted values in this design problem solution for Z , illustrating how non-ideality increases

with increasing pressure as intermolecular forces increase.

d) The carbon dioxide flow at the suction inlet is 1020 m3 / h, at:

T 1 = 343 K and P 1 = 14.7 psia = 1.013 × 105 N m-2

The molar mass flow into the compressor of the nearly ideal gas is:

K)(334.6K)kmol/J(8314(0.99)

)hm(1020)m N10(1.013

-13-25

11

11

RT Z

V P n

-1-2-1 skmol101.042hkmol37.52 n

Then the energy demand rate (power) of the compressor for handling a real (non-ideal) CO2

gas is:

)sCOkmol10(1.042)COkmol/kJ(17313Power-1

2

-2

2 NON-IDEALn W

kW180.44skJ44.180Power -1

The compressor has an operational efficiency of 35%, hence, the power rating of compressormotor is:



12

kW515.50.35

kW180.2 PowerMotor

This is a big power demand.



13

4) Hydrogen is under consideration as an alternative fuel to petrol in vehicles. High pressure

hydrogen tanks make it possible for the hybrid vehicle to travel further distances on a single

tank of fuel. A typical tank will contain 4.5 kg of hydrogen stored at a pressure of 35 MPa. If

the average temperature for storage is 20 C calculate the volume of the tank required (in

litres) using the following mathematical models, namely:

(a) The Ideal Gas Law:

nRT PV

[3 marks]

(b) The Van der Waals Equation, namely:

T Rbvv

a P w2w -

where

C

2

C

2

w P 64

T R27 a

andC

C w P 8

T Rb

with T C being the critical temperature of hydrogen, P C the critical pressure of

hydrogen and n / V v , the molar volume of hydrogen (V is the volume and n

the number of kmols).

[6 marks]

(c) The Non-ideal Gas function, namely:

ZnRT PV

where Z is the generalised compressibility factor.

[9 marks]

(d) The Department of Energy are aiming for the storage tank size to be 62 litres

by 2015. Using the non-ideal gas function calculate the temperature at which

the hydrogen needs to be stored assuming the pressure remains the same. What

implication does this have for hydrogen powered vehicles?

[7 marks]

Data Supplied:

1 Pa = 1 N m-2 = 1 10-5 bar

R = 8.314 kJ / kmol K

T (K) = T (C) + 273

Data Sheet No. 2.1 Table of Critical Constants of Gases.

Data Sheet No. 3 Generalized Compressibility Factor Plot for the high pressure range.



14

Answer:

(a) BASIS OF CALCULATION: Hydrogen mass in tank = 4.5 kg

kmol/Hkg2.016

Hkg4.5

Hof RMM

Hof Mass

2

2

2

2n

2Hkmol2.232n

For the ideal gas law, the mathematical model is:

nRT PV

then

P

nRT

V

Watch out for the pressure units and units of R.

litres155.3m0.1553(Pa)1035

K 273][20K)kmol/(J8314)H(kmol2.232

3

6

2

V

(b) For the van der Waals equation;

RT bvv

a

P ww

2

2v

a

P R

bv

T ww

First calculate the values of the van der Waals constants;

C

C

w P

T Ra

64

2722

C

C w

P

RT b

8

By using Data Sheet No. 2.1 to obtain the critical data for hydrogen of:

Pa101.3MPa1.3

6C P K 33.3

C T

Then

24

6

22

kmol/m N24874.19)10(1.364

(33.3)(8314)27

wa

kmol/m0.02662)10(1.38

33.38314

3

6

wb

Hence,



15

)20273(831402662.019.24874

)1035(2

6

v

v

243600202662.019.24874

)1035(2

6

v

v

Best to solve by trial and error, as a first guess for v use the ideal gas value, i.e.

kmol/m0696.0232.2

1554.0 3n

V v

Substituting into the above equation gives:

24360022.1724997

243600202662.00696.0

0696.0

19.24874)1035(

2

6

Thus v = 0.0696 m3 / kmol is too low, so increase the value say to 0.1 m3 / kmol and try

again:

24360028.2750826

243600202662.01.01.0

19.24874)1035(

2

6

This time the value is too high so the answer is between 0.0696 and 0.1 m3

/ kmol. Nextguess by linear interpolation:

kmol/m0907.0)2.17249978.2750826(

)2.17249972436002()0696.01.0(0696.0

3

newv

24360027.2436556

243600202662.00907.00907.0

19.24874)1035(

2

6

Close enough, so:

v = 0.0907 m3 / kmol

and

V = n v = 2.232 × 0.0907 = 0.2024 m3 = 202.4 litres

(c) For the non-ideal gas system;

ZnRT PV



16

1553.01035

)20273(8314232.26

Z

Z

P

ZnRT V

In order to calculate the compressibility factor, Z , we need T r and P r :

8.83.33

20273r T

26.9Pa101.3

Pa1035

6

6

C

r P

P P

Using Data Sheet No. 3,

Z = 1.3

So,

V = 1.3 × 0.1553 = 0.2019 m3 = 201.9 litres

Note the non-ideal gas equation and the van-der Waals equation give virtually the same

answer.

(d) For the non-ideal gas system P r is the same as before;

26.9

Pa101.3

Pa1035

6

6

C

r

P

P P

C

r T

T T and

C r T T T

Therefore:

RT nT

PV

nTR

PV Z

C r

Putting in the available data into this equation;

r r

calcT T

Z 3.512

(8314)(33.3)(2.232)

(0.062))10(35

6

A Trial and error scheme is required:

Select a value for T r , get Z CHART from Data Sheet No 3, calculate Z calc from equation above

and compare with value from chart, when they are equal the value of T r is correct.

Therefore with P r = 26.9



17

Test T r Zcalc = 3.512 / T r Z CHART

2.00 1.756 2.00

1.80 1.950 2.10

1.60 2.195 2.225

1.55 2.266 2.25

This is close enough, therefore: T r = 1.55 (can check convergence graphically if you want)

C)221.4-(K51.633.31.55 oT

This result means that there needs to be a considerable cooling system in the car as well in

order to store 4.5 kg of hydrogen in 62 litres at 35 MPa. This is obviously impractical and

currently the technology is not available to do this (Note: the smallest tank volume so far at

35 MPa is 148 litres).

tutorial sheet 01 answers 2014

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