Solution manual for fundamentals of chemiscal engineering thermodynamics

Problem 2.7 Solution a Since the cooker contains both vapor and liquid, the state is saturated steam.Even though the vapor occupies 75% of the volume, it only represents 0.4% of the tota

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Solutions Manual for

Fundamentals of Chemical

Engineering Thermodynamics

Themis Matsoukas

Upper Saddle River, NJ • Boston • Indianapolis • San Francisco

New York • Toronto • Montreal • London • Munich • Paris • Madrid

Capetown • Sydney • Tokyo • Singapore • Mexico City

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Note to the Instructor

An effort was made to update all solutions requiring steam tables to conform with the tables in Appendix E of the book, which are based on IAPWS95) It is possible, however, that some problems may make use of older tables Be alert as such discrepancies could confuse students even though the final answers are not much different

Please report any mistakes or typos to Themis Matsoukas: matsoukas@psu.edu

The author and publisher have taken care in the preparation of this book, but make no expressed or implied warranty of any kind and assume no responsibility for errors or omissions No liability is assumed for incidental or consequential damages in connection with or arising out of the use of the information or programs contained herein

Visit us on the Web: InformIT.com/ph

This work is protected by United States copyright laws and is provided solely for the use

of instructors in teaching their courses and assessing student learning Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the

integrity of the work and is not permitted The work and materials from it should never

be made available to students except by instructors using the accompanying text in their classes All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials

ISBN-10: 0-13-269320-8

ISBN-13: 978-0-13-269320-2

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Contents

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1 Introduction

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Problem 1.2 Solution First we write the given equation as

14:5and substitute these values into the above equation After some manipulation the result is

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Problem 1.3 Solution a) The mean velocity is

b) The mean kinetic energy is

c) The above calculation shows that the mean kinetic energy depends only on temperature (it is independent

water at the same temperature:

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Problem 1.4 Solution

3 4 5 6 7 8 9 10 -4 ´ 10-22

-2 ´ 10-220

derivative of the potential equal to zero solve for the value of r To do this easily, we define a new variable

the volume occupied is

the following final formula:

AV.r/3

By numerical substitution we finally obtain the density:

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c) Specific volumes of saturated liquid methane are listed in Perry’s Handbook from which we can compute

corresponds to Perry’s tabulation at about 160 K Our calculation is approximate and does not incorporate

result will change quite a bit because if the third power to which this distance is raised But the importantconclusion is that the calculation placed the density right in the correct range between the lowest and highestvalues listed in the tables This says that our molecular picture of the liquid, however idealized, is fairly close

to reality

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Assuming molecules to be sitting at the center of a cubic lattice with spacing L, the volume occupied by N

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F

P0 P=0

2 R

The force needed to separate the two haves is equal to the force that is exerted by pressure on one hemisphere:

where P is the pressure difference between the atmospheric pressure and the contents of the sphere, and

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2 Phase Diagrams of Pure Fluids

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Problem 2.1 Solution a) 25ıC, 1 bar: liquid, because the temperature is below the saturation temperature

at 1 bar (99.63ıC)

lower the state is liquid.)

b) Liquid, because the pressure (1 atm = 760 mm Hg) is higher than the vapor pressure of bromobenzene(10 mg H) at the same temperature

c) Liquid, because the temperature is lower than the boiling point at the same pressure

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Problem 2.2 Solution a) From steam tables at 40 bar we collect the following data:

obtained by lever rule:

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Problem 2.3 Solution a) At 1 bar, 210ıC, the specific volume of steam is found by interpolation to be

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Problem 2.4 Solution Solution Initial state:

vol occupied by liq

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These results are quite different Which one should we pick?

For the molar volume we have

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Problem 2.7 Solution a) Since the cooker contains both vapor and liquid, the state is saturated steam.

Even though the vapor occupies 75% of the volume, it only represents 0.4% of the total mass

d) The quick solution is to take a look at the P V graph The initial state is at A and the final state, B, isreached by constant-volume cooling This state is obviously in the two-phase region because the originating

state A were in the superheated region, we would be able to tell if B is superheated or vapor/liquid and wewould have to do the solution in more detail as shown below.)

volume of the system is between these two values, therefore we still have a saturated system We conclude

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V initial state

const V process final state

25 °C 0.003175

bar

This solution is more general and will work regardless of where the initial state is

e) When the system has cooled, the outside pressure is 1 bar and the inside pressure is 0.03166 bar fore, the lid remains closed under the action of this pressure difference The force is

To put this force into perspective we calculate the mass whose weight is 12166 N:

If you can lift 2700–2800 lb then you could remove that lid! (Note: Whether you take the outside pressure

to be 1 bar or 1 atm or something similar, the conclusion remains that the required force is indeed very large

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C' D

E

F

G H

volume Assuming the isotherm to be vertical, the final state, B, is at the same temperature and on the

Notice that the temperature has not changed This is a consequence of the fact that we have approximatedthe isotherm with a vertical line In reality, the isotherm is not vertical and state B should be at a temperature

b) The initial state is a V/L mixure (state C) The process is under constant volume, therefore, the final state

is located at the intersection of the vertical line through C and the saturation line (state D) From steamtables:

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The specific volume at the initial state is (in cm3/g)

(saturated liquid)

Note 1: Even though the mass fraction of the vapor is very nearly 0, it would not be correct to set it equal to

saturated liquid which means that the final pressure is almost 1 bar Clearly, this approximation missesthe right pressure by more than 200 bar!!!

Note 2: In this case the specific volume of the vapor-liquid mixture was very close to the liquid side and for

this reason the final state was liquid In other words, under heating the vapor condenses and becomesliquid If, however, the initial volume was much closer the vapor side (state C’), then heating wouldproduce vapor In this case, heating would cause the liquid to evaporate That is, after heating thecontents of the vessel the final state might either saturated vapor or saturated liquid Can you establish

a criterion for the initial specific volume to determine whether the final state is vapor or liquid?c) The final state saturated vapor (state F) The process is cooling under constant volume, therefore, theinitial state must be somewhere on the vertical line through F and above point F (since cooling implies thatthe initial state is at higher T ) We conclude the initial state is superheated vapor

d) By similar arguments as above, we determine that the initial state is vapor/liquid mixture Notice thathere we are heating a vapor/liquid mixture and as a result the vapor condenses to produce saturated liquid!

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Problem 2.9 Solution The specific volume of water under these conditions in the tank is

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saturated vapor and it must be superheated

To locate this state, we interpolated in the steam tables between 1.0 bar and 1.5 bar at various temperaturesand construct the table below:

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Problem 2.10 Solution Solution

The graphs below show the P V in various combinations of linear and logarithmic coordinates and the ZPgraph

1.0 0.8 0.6 0.4 0.2 0.0

250 200 150 100 50 0

200 150 100 50 0

100 80 60 40 20 0

The volumes span a very wide range and in order to see the shape of the saturation line, we must plot

By doing the V axis in log coordinates we can now look at a very wide range of values withoutsqueezing the graph into nothingness Notice that in the log plot the volume goes from 1 to 100,000

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the truncated virial equation is valid.

b) The reduced conditions are

Tr D 244:15 K305:3 K D 0:80; Pr D 48:72 bar10 bar D 0:21

equation is not valid

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Problem 2.12 Solution a) Vapor.

b) The molar mass can be obtained from the relationship between molar volume and density:

We know density at two pressures, so we need the molar volume in one of them We choose the lowestpressure because at 0.01 we are justified (below) to assume ideal-gas state:

which is 64.3 bar That is, the reduced pressure is at most

0:01 bar

From generalized graphs is it clear that at such low reduced pressures the state is essentially ideal

The molar volume is

but is as good as we can do with the information we have

Solving the truncated virial for B:

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Solving the truncated virial for V and using P D 12 bar:

and the volume of the tank is

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Problem 2.13 Solution a) We collect the data for this problem:

second virial does not need to be recalculated):

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Problem 2.14 Solution a) The critical constants and acentric factor of nitrogen are:

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The volume solved in part a) is 1.16810 3m3/mol Therefore the tempeture is 96.48 K, and the pressure

therefore the truncated virial is acceptable

therefore, the truncated virial equation is valid

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Problem 2.15 Solution a) The second virial coefficient is directly related to the slope of an isotherm onthe ZP graph Specifically,

P D0

This suggests the following graphical solution: calculate Z from the steam tables at constant T , plot them

Pressure (kPa)

200 C

this line is to fit a straight line through the points closest to the origin, say below 200 kPa A smarter way is

to use all the points and a quadratic equation:

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The second virial coefficient is

b) Using the above value of B we can calculate the molar volume of water at 14 bar as follows:

virial equation is valid at these conditions

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Problem 2.16 Solution We need the density of methane under these conditions We will calculate it viathe compressibility factor using the Lee-Kesler tables The critical parameters of methane are:

Here, numbers in regular font are form the Lee-Kesler tables and those in bold are interpolations First we

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Problem 2.17 Solution a) At the given conditions, Tr D 0:963675, Pr D 0:949281 The state is veryclose to the critical, therefore, far removed from the ideal-gas state.

b) Using the Lee Kesler method we find

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Outline:

b) With the compressibility factor known, calculate the specific volume and then the total volume of the tank(since the total mass is known)

c) Calculate the new molar volume after 90% (45 kg) is removed Calculate Z Since we know Z and T weshould be able to obtain P If we use the Lee-Kesler graph for Z we must do a trial-and-error procedure:choose P , calculate Z, if it doesn’t match the known Z try another pressure and continue

Alternatively, use the truncated virial equation: estimate B using the Pitzer correlation and use

and the total volume is

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and the new molar volume is

book looks “linear enough.” But we want to be more precise, so we will calculate the compressibilityfactor form the virial equation and from the Lee-Kesler graphs and will compare:

Using the virial equation:

is more than 5% way from the ideal-gas state, therefore, we reject the calculation

Lee-Kesler tables we find

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We compare this to the known value V0D 0:00101128 m3=mol by calculating the ratio

Vguess

ratio is sufficiently close to 1 The table below summarizes the results of these iterations

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Problem 2.19 Solution We need to calculate densities, i.e we need V or Z The ideal-gas law is out ofthe question because the pressure is too high Same for the truncated virial equation We could use either theLee-Kesler tables or an equation of state Both methods would be appropriate since krypton is a non-polarcompound.

a) We need the density of krypton, so we will first find the compressibility factor at the indicated conditions.From tables we find

therefore, safe to store 2500 kg

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Problem 2.20 We collect the following information for n-butane:

a) We need the molar volume of the liquid Our options are: Lee-Kesler, and Rackett We choose the Rackettequation because it is known to be fairly accurate while the accuracy of the Lee-Kesler is not very good inthe liquid side Still, if you did the problem using L-K I will consider the solution correct

the final result is very close to the above This calculation is given at the end of this solution b) For thevolume of the vapor we use Lee-Kesler The required interpolation is shown below

B

B' T

T + d

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Both sides of the tank undergo constant-volume processes as indicated by the dashed lines The graph

It is obvious that the pressure in the liquid side will always be higher than the pressure of the vapor side.Therefore, the pressure of 40 bar will be reached first in the liquid side, causing that alarm to go off.d) To calculate the temperature at the state we recall that for liquids with constant ˇ and , we have

while that of the vapor will be not much higher than 2 bar!

Calculation of liquid V using Lee-Kesler:

If you opted to do the calculation using the Lee-Kesler tables, the correct solution is shown below First we calculate the reduced temperature and pressure.

T r D 0:69 0:7; P r D 0:545 Note: because the phase is liquid, one must extrapolate to P r D 0:0545 from the listed values for the liquid (shown in the tables in italics):

T r D 0:7

P r D 0:0545 P r D 0:2 P r D 0:4

Z0 9:45 1 3 0:0344 0:0687

Z1 4:1785 10 3 0:0148 0:0294 With these values we obtain the following:

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