D. Keffer - ChE 240: Heat Transfer and Fluid Flow

Homework Assignment Number OneAssigned: Wednesday, January 13, 1999

Due: Wednesday, January 20, 1999 BEGINNING OF CLASS.

Problem 1. Unit Conversions between concentrations and densitiesWater has an approximate (mass) density of 1.0 grams/ml and a molecular weight of 18.0 grams/mole.(a) What is the molarity (molar density or molar concentration, or simply concentration) of water in mol/liter?(b) What is the mass density of water in kg/m3?(c) What is the molar volume of water in liter/mol?

Air has an approximate molecular weight of 28.84 grams/mol and roughly obeys the ideal gas law at ambientconditions, P = 101325 Pa and T = 298 K.(d) What is the molar density of air at these conditions in mol/liter?(e) What is the mass density of air at these conditions in kg/m3?

Solution:(a)

litermol556.55

literml1000

mlmol05556.0

molg0.18

mlg0.1

MWCn~ =⋅==ρ==

(b)

33

6

mkg0.1000

mml10

g1000kg

mlg0.1 =⋅⋅=ρ

(c)

molliter018.0

litermol556.55

1C1V~ ===

(d)

lmol0409.0

l0.1000m

mmol897.40

K298314.8Pa101325

RTP

Vnn~

3

3Kmol

J =⋅=⋅

===⋅

(e)

33 mkg736.0

l0.1000kg

molg18

mmol897.40MWn~ =⋅⋅=⋅=ρ

Problem 2. Dimensionally Consistent EquationsFor the following balance equations, (i) write down the units of each term in the equation, (ii) state whether theequation is dimensionally consistent, (iii) correct the equation to be dimensionally consistent.V is the volume, AC is the molar concentration of component A, t is the time, iF is the molar flowrate of the

ith stream, iF is the mass flowrate of the ith stream, A,ix is the mole fraction of component A in the ith stream,

D. Keffer - ChE 240: Heat Transfer and Fluid Flow

A,ix is the mass fraction of component A in the ith stream, k is a first order reaction rate constant with units of

inverse time, Aρ is the mass density of component A, AMW is the molecular weight of component A, ApC is

the molar heat capacity of component A, ApC is the mass heat capacity of component A, iT is the temperature of

the ith stream, n is the number of components in the system, rH∆ is the molar heat of reaction.

Solution:

(a) mole balance: AA,outoutA,ininA kVCxFxF

dtdC

V −−=mol/time = mol/time - mol/time - mol/time

The equation is dimensionally consistent as written.

(b) mass balance: AA,outoutAA,ininA

A kVxFMWxFdt

dCMWV ρ−−=⋅

mass/time = mass/time - mass2/mol/time - mass/timeThe equation is not dimensionally consistent as written.The equation would be correct if written as:

AA,outoutA,ininA

A kVxFxFdt

dCMWV ρ−−=⋅

(c) energy balance:

( )( )( )

( ) Ar

n

jirefoutjjjpout

n

jirefinj,injpin

n

ji

refjjp

kCHTTMWCF

TTxCFdt

TTCCdV

∆−−ρ−

−=−

∑

∑∑

=

==

energy/time = energy/time - energy*mass/volume/time - energy/time/volumeThe equation is not dimensionally consistent as written.The equation would be correct if written as:

( )( )( )

( ) Ar

n

jirefoutj,outjpout

n

jirefinj,injpin

n

ji

refjjp

kVCHTTxCF

TTxCFdt

TTCCdV

∆−−−

−=−

∑

∑∑

=

==

Problem 3. Pressure ConversionsConvert a pressure of 100 psia to absolute pressure in(a) Pa, (b) bar, (c) atmospheres, (d) feet of water, (e) feet of mercury (use specific gravity in text).

(a) Pa689300atm0.1

Pa101325psia7.14atm0.1psia0.100 =⋅⋅

D. Keffer - ChE 240: Heat Transfer and Fluid Flow

(b) bar893.6Pa10

bar1*Pa689300 5 =

(c) atm80.6psia7.14atm0.1psia0.100 =⋅

(d)g

pgh c

ρ= = ft8.230

mft2808.3m34.70

sm8.9

mkg1000

Pa689300

23

=⋅=⋅

(e)g

pgh c

ρ= = ft97.16

mft2808.3m17.5

sm8.95955.13

mkg1000

Pa689300

23

=⋅=⋅

⋅

Problem 4. U-Tube ManometersGeankoplis 2.2-4, pg. 104.

mmHg754pb = , 3amkg1000=ρ , 3b

mkg3.1=ρ , m0.0Z ≈ , m415.0R =

( ) 8.9*)3.11000(*415.0101325760754

ggRppc

BAba −+=ρ−ρ+=

kPa6.104Pa104600pa ==

psia17.15atm0.1psia7.14

Pa101325atm0.1Pa104600pa =⋅=

Problem 5. two-fluid U-tube Manometers with reservoirsGeankoplis gives a diagram of a two-fluid U-tube manometer with wide reservoirs in figure 2.2-4(b) on page 36.He gives the working equation for it in equation 2.2-15 on page 37. However, he does not provide a derivation forthe equation. Moreover, it is not a completely straightforward derivation. I would like you to derive equation 2.2-15. To do this we will take several steps.

(a) Consider:

h2

ρA

ρB

pa pb

h1 Ro

D. Keffer - ChE 240: Heat Transfer and Fluid Flow

oR is the initial difference between fluid levels at zero pressure difference. If the pressure at point a is the same as

the pressure at point b but the amount of B-fluid in the two legs of the manometer is different (so 21 hh ≠ ), find

oR in terms of BA21 ,,h,h ρρ .

Solution:The pressure at the bottom dotted line on each leg must be equal due to the principle of fluid statics.

21 pp =oA2B1B gRghgh ρ+ρ=ρ

( )21A

Bo hhR −

ρρ=

(b) In the above figure, if the actual reading is R . What is the pressure difference between points a and b interms of BAo ,,R,R ρρ ?

Solution:

( )( )c

BAoba ggRRpp ρ−ρ−=−

We must change our standard formula to include oR since the manometer does not read zero at zero pressuredifference.

(c) Consider some arbitrary change in pressure that cause a shift in a fluid from the figure on the left to the figureon the right.

A

aR

∆H

where a and A are cross-sectional areas. Express H∆ in terms of A,a,R . (The underlying principle here isconservation of volume. The volumes in both tubes are equal.)

Solution:

21 VolumeVolume =BOT,2TOP,2BOT,1TOP,1 VVVV +=+BOT,1BOT,2TOP,2TOP,1 VVVV −=−

aRHA =∆

D. Keffer - ChE 240: Heat Transfer and Fluid Flow

RAaH =∆

(d) Now, considering the correction factor from parts (a-b) for a non-zero default reading and considering thefactor for a change in tube cross-sectional area, derive equation 2.2-15 for the two-fluid U-Tube manometer withreservoirs.

h2

ρA

ρB

pa pb

h1 R

a

A

ρCρC

∆Η

Solution:

( )( ) ( )( )c

CBoc

BAoba ggRR

Aa

ggRRpp ρ−ρ−+ρ−ρ−=−

On the right-hand-side of the equation above, the first term accounts for the difference in height of the A-B fluidinterface in the two legs of the manometer and for the correction term due to a non-zero equilibrium reading. Thesecond term accounts for the difference in height of the B-C fluid interface in the two reservoirs of the manometerand for a correction term analogous to the one obtained in the legs. Rearranging yields:

( )c

CBBAoba gg

Aa

AaRRpp

ρ−ρ+ρ−ρ−=−

(e) Geankoplis says on page 38 that, for the two-fluid U-Tube manometer with reservoirs, “ oR is often adjusted tozero”. How is this done?

Solution:The amount of B-fluid in each of the reservoirs must be the same. (look at the result from part (a). What will givea zero oR ? When 21 hh = , which is equivalent to saying that the amount of B-fluid in each of the reservoirsis the same.