class examples

8/13/2019 Class Examples

1/32

Class Examples

Energy Balances

Q1 A copper bar of mass 2 kg at an initial temperature of 180C is quenched in oilof a mass of 10 kg and a temperature of 25C. i!en the follo"ing data#

C p $copper% &'85 ()kg* & 0.'85 k()kg* C p $oil% & 2.' k()kg*

1180C T = C and+ 1 25OT = C

Calculate the final mean temperature of the oil and the bar and determine theheat transfer from the bar to the oil. ,tate an- assumptions.

Assuming no heat loss to surroundings.

0 E =

0C O E E + =

et f T =final mean temperature

0c oc p c o p o

m C T m C T + =

1 1$ % $ % 0

c oc p f c o p f om C T T m C T T + =

1 10

c c o oc p f c p c o p f o p om C T m C T m C T m C T + =

1 1c o c oc p f o p f c p c o p om C T m C T m C T m C T + = +

1 1c o

c o

c p c o p o f

c p o p

m C T m C T T

m C m C

+= +

$2 0.'85 180% $10 2.' 25%'0$2 0.'85% $10 2.'% f T

+ = = + C

/eat transfer from the bar & 1$ % 2 0.'85 $180 '0% 115.5cc p c f m C T T = = k(

Q2 etermine the design po"er rating for an electric kettle element gi!en thefollo"ing data.

ass of "ater to be heated & 1.5 kg,pecific heat capacit- of "ater & .18 k()kg* 3nitial "ater temperature & 10C4inal "ater temperature & 100C

1


2/32

ime to reach boiling point & ' minutes

,tate an- assumptions.

etermine the time of heating if a 2.5 k6 kettle is a!ailable for the same dut-.

Assumptions# 7o heat transfer to surroundings. 7o internal energ- change ofthe element. All po"er input is transferred to heating the "ater.

1.5 .18 $100 10% 5 .' p E mC T = = = k(

5 .''.1'5

180w E work Power

time time= = = = k6

4or a 2.5 k6 kettle noting E w is unchanged.

ime5 .'

225.2.5

= = s

Alternati!el- using ratios#

ime &'.1'5

180 225.2.5

= s

Conduction ExamplesQ1 9ne "all of a house measures 5m 'm and is constructed of a single brick

skin+ e:cept for a 2m 1m "indo" more or less in the middle.

etermine the heat loss through the "all if the brick is 0.25m thick and theglass is 5mm thick+ the inside and outside temperatures are 25C and 5Crespecti!el-+ and the thermal conducti!it- !alues are as gi!en in the coursenotes#

Area 1' m 2

Area 2 m 2

5 m

'm

2

T through "all & 20C


3/32

0. ; 1' 20


4/32

ra"ing a diagram of the "all as follo"s#

kdT Q

dx= $unit area%

dT dx

=constant

4or the plastic ''$50 85%

1.


5/32

Convection Examples

Q1 An electricall- heated plate dissipates heat b- con!ection at a rate of 5006)m 2 to air "hich is at an ambient temperature of 22C. 3f the surface of thehot plate is maintained at 110C calculate the heat transfer coefficient for

con!ection bet"een the plate and the air.

500Q = 6)m 2

110 22 88T = = C

Q h T =

500


6/32

58

08.' 10

0.'5 5. < 10 $2 0% A = = m

2

Q2 A domestic hot "ater c-linder 0.5 m in diameter and 1 m high is situated in alarge enclosed cupboard. he c-linder surface temperature is 80C and thesurrounding surfaces in the cupboard area are at 25C. etermine the heattransfer due to radiation in each of the follo"ing cases+ and the reduction inthe radiation heat transfer from the c-lindrical surface of the c-linder if theinitial o:idised copper surface is then coated "ith aluminium paint. heemissi!it- of o:idised copper and the aluminium paint are 0.8 and 0.'respecti!el-.

( )urr Q A T T =

emperatures must be in K

( )80.8 5. < 10 0.5 1 '5' 2;8 5Q = =6

After coating "ith aluminium paint

0.' 520

0.8Q

= = 6

Beduction in radiation transfer 5 20 ' 0= = 6

Beduction is 2.5=.

7ote+ heat transfer "ill also occur b- natural con!ection from the c-lindersurface to the air in the cupboard.

Q' A sheet of glass at a temperature of 00C is cooled b- passing a flo" of airo!er its surface. he temperature of the air is the same as the temperature ofthe surroundings. o pre!ent cracking it has been found that the temperaturegradient must not e:ceed 15C)mm an-"here in the glass during the cooling

process. 3f the thermal conducti!it- for glass is 1.5 6)m* and its surfaceemissi!it- is 0.8+ "hat is the appro:imate lo"est temperature of the air thatcan be initiall- used for cooling> ake the con!ection heat transfer coefficientas 5 6)m 2*.

urr T T =

T = emperature of surface

0Co!d Co!" #ad

Q Q Q =


7/32

sing unit area#

( ) ( ) 0urr dT

k h T T T T dx

=

15 15000 15000

max

dT C C K dx mm m m

= = =

o o

T must be in *

00 2


8/32

Heat Exchanger calculation using Log Mean Temperature Difference

Q1. A multi tube single pass heat e:changer uses hot "ater on the shell side toheat a flo" of cold "ater on the tube side in a co current mode of operation.

i!en the follo"ing data determine the appro:imate cold "ater flo" outlettemperature and using this !alue+ the cold "ater flo" rate per hour.

/eat D:changer length 1mCold inlet temperature 10C/ot inlet temperature 0CQh & 8k6/ot "ater mass flo" rate 00kg)hour C p & .18 k()kg* tubes ')8E 9 .% & 11 5 6)m 2*

hh h p hQ m C T =

cc c p cQ m C T =

L& Q %A =

1 2

1

2

ln L&

=

"here 1 is the temperature difference of the streams at one end of the heate:changer and 2 is the temperature difference of the streams at the oppositeend of the heat e:changer.

,olution.

ra"ing a diagram as follo"s#

00 kg)hour

/ot 6ater at 0C

10C

C9 C BBD7

tubes ')8E 9 .

C

ength 1mC

Cold 6ater

8


9/32

3n this e:ample#

1 & hot in temperature F cold in temperature.

2 & hot outlet temperature F cold outlet temperature.

Cold out Cold in $10C%/ot out /ot in $ 0C% 00kg)hour

hh h p hQ mass C T =

00 1808000

' 00T =

11.5T C = o

0C F hot outlet temperature & 11.5C

/ot outlet temperature & 8.5C

h c L& Q Q Q %A = = =

''1 25. 10 0.1


10/32

se trial and error to determine appro:imate !alue of 2 and hence this !alueto determine cold "ater outlet temperature.

o guess the !alue of 2 take the !alue of L& as the !alue of an arithmetic

mean. Calculated L& is '8.' C+ therefore as an arithmetic mean1 2 2 '8.' < .


11/32

Class Test

?art 1. A multi tube single pass heat e:changer uses hot "ater on the shell sideto heat a flo" of cold "ater on the tube side in a co current mode of operation.

i!en the follo"ing data determine the appro:imate cold "ater flo" outlettemperature and using this !alue the cold "ater flo" rate per hour.

/eat D:changer length 110 cmCold inlet temperature 10C/ot outlet temperature ;CQh & 8k6/ot "ater mass flo" rate 10kg)minuteC p & 180 ()kg* tubes ')8E 9 .% & 11 5 6)m 2*

hh h pQ m C T =

cc c pQ m C T =

L& Q %A =

1 2

1

2

ln L&

=

"here 1 is the temperature difference of the streams at one end of the heate:changer and 2 is the temperature difference of the streams at the oppositeend of the heat e:changer.

3n this e:ample#

1 & hot in temperature F cold in temperature.

2 & hot outlet temperature F cold outlet temperature.

,olution.

Cold out Cold in $10C%/ot out $ ;C% /ot in $10kg)min%

hh h p hQ mass C T =

00 1808000

' 00T =

11. 8h

T C =

o

11


12/32

/ot inlet temperature & ;C @11. 8C

$7ote the temperature difference is added in this case%

/ot inlet temperature & 0.5C

h c L& Q Q Q %A = = =

''1.1 25. 10 0.1; ;8

A = = m2

800011 5 0.1; ;

L&

L&

Q%A

=

=

' .88 L& = C

1 2

1

2

ln L&

=

1 0.5 10 50.5C C C = =o o o

2 u!k!ow! =

2

2

50.5' .8850.5

ln

=

se trial and error to determine appro:imate !alue of 2 and hence this !alueto determine cold "ater outlet temperature.

o guess the !alue of 2 take the !alue of L& as the !alue of an arithmeticmean. Calculated L& is ' .88C+ therefore as an arithmetic mean.

1 2 2 ' .88 ;.< C + = = o hence guess


13/32

cc c p cQ m C T =

'8000 .18 10 $2 10%cm=

0.11;cm = kg)s

' 00 2 11; 10''cm = = kg)da-

?art 2

3f the number of tubes "as reduced to 5 and the siGe of each tube reduced to1) inches 9 + "hat "ould be the length of the heat e:changer required tomeet the same dut- as the original+ if the o!erall heat transfer coefficient is11 5 6)m 2* as before.

,ame dut- so L& +% + and Q must be the same. herefore the area must be thesame as that of the original.

3t follo"s#

'0.1; ; 5 0.25 25. 10 A L = =

1.;< L = m

?art '

9utline one factor that should be considered "hen changing the length+ andtube siGe of a shell and tube heat e:changer.

,iGe+ space required. P ' pressure drop% on the tube side.Affect on %

P $pressure drop% on the shell side.Cost

1'


14/32

!orced Convection

Q1 9il flo"s through a pipe of 5mm 9 and a length of 1m+ "ith the "alltemperature of the pipe maintained at a constant temperature of 100C. he oil

enters at 20C and e:its at 0C. he a!erage !alues of the ph-sical propertiesof the oil are as follo"s#

ensit- 880 kg)m '

,pecific heat capacit- 1;00 ()kg* hermal conducti!it- 0.25 6)m*Hiscosit- 10 m7s)m 2

sing the follo"ing (usselt correlation#

0.8 0.''0.02' Be ?r (u =

6here the dimensionless groups are as follo"s#

Be

?r p

hd (u

k ud

C

k

=

=

=

Calculate#

the mass flo" rate of oil if the heat transferred to the oil is 1000 6+ theo!erall heat transfer coefficient % + and the !alue of h i and hence ho if theo!erall heat transfer coefficient+ is made up from contributions from the insideand outside heat transfer coefficients+ and the resistance of the pipe "all isneglected. Comment on the !alue of ho+ and "h- this has resulted.

,olution

9il in 20C 9il out 0C/eating fluid in 100C /eating fluid out 100C

pQ mC T =

ass of oil1000

1;00 $ 0 20% p

Qm

C T = =

& 0.02 kg)s

/eat gained b- the oil is equal to the heat transferred to the oil across the pipe"all.

1


15/32

L& Q %A =

$100 20% $100 0%80

ln0

L& =

& ;.52C

'

10005 10 ;.52 L&

Q%

A = =

% & ;15.


16/32

0.02' 180.58 .1< 1


17/32

"team Ta#les

1. 6hat is the enthalp- of steam at 50 bar>

& 2

& 85; k()kg

'. 6hat is the boiling point of "ater at 0.5 bar>

& 81.'C

. 6ater boils at

& 0.'2 bar.

5. 6hat is the boiling point of "ater at 1 bar>

& ;;. C

. 6hat is the specific !olume of saturated !apour at 0 bar>

& 0.0 ;

& 1. ' k()kg

10. 6hat is the internal energ- of saturated "ater !apour at 10 bar>

& 258 k()kg

11. Calculate the heat e!ol!ed "hen dr- saturated steam at ' bar is con!erted tosaturated liquid at 0.8 bar.

3nitial h g & 2


18/32

12. Calculate the proportion of liquid that "ould flash to steam if saturated liquidat


19/32

"team Ta#les $"team "parging%

Q1 A stainless steel !essel contains 000 litres $ 000kg% of "ater at an initialtemperature of 2;C+ "hich must be heated to a final temperature of 8 C in a

time period of 0 minutes. ,team is a!ailable at bar+ and direct steminKection is to be used.

$a% Calculate the final "ater !olume at the end of the heating time.$b% Calculate the steam flo" !elocit- if the steam suppl- pipe is 0 mm 3 .

,olution.

$a% Assuming no heat is lost to the en!ironment.

6ater# 3nitiall- J 2;C h l & 121 k()kg4inall- J 8 C hl & ' 0 k()kg

h & ' 0 F 121 & 2'; k()kg

,team# 3nitiall- J bar h g & 2


20/32

'

0.0511.0

he coefficient for condensing steam hi together "ith an- scale is taken as8500 6)m 2*+ and the thermal conducti!ities of the pipe "all and the lagging+k w and k l as 5 6)m* and 0.0


26/32

2 8 1 821

2 8ln

1 8

md = =

mm & 0.21 m

hermal resistances can no" be calculated $unit length of pipe%

1 10.00025

8000 0.150i ih d = =

0.00;0.000 0

5 0.15;w

w w

xk d

= =

0.0501.01;

0.0


27/32

7ote in the abo!e it is assumed the steam temperature and the outside of thesteam pipe are the same at * as resistance of the pipe "all is !er- small.

Checking the !alues of hr and hc based on the outside surface temperature ofthe lagging+ at sa- '10 * is required. $basis unit area%

80.; 5. < 10 $'10 2; %$'10 2; %r

h = &


28/32

Condensation Example

Q1 A condenser is used to condense "ater !apour at a feed pressure of 0. bar g+and a flo"rate of 500 m ' )hour+ for an operating time of < hours. hecondensate flo"s b- gra!it- into a storage !essel+ "hich is emptied at the end

of the operating period. he condenser tubes are positioned in a bank!erticall- at normal to the direction of the !apour flo".

i!en the follo"ing data#

g & ;.81 m)sCondenser tubes outside diameter d o & 1; mmCondensate 0 & 1 c? $1 : 10 ' 7s)m 2%T coola!t & 1 C,uppl- pipe a!ailable 1E+ 2E+ 'E+ E and E 7I $nominal bore%Condenser tube length & 1 m

etermine#

$a% he suppl- pipe required for the "ater !apour feed to the condenser ifthe site standard ma:imum !elocit- is '0m)sL

$b% he heat released per hourL

$c% he o!erall !alue of hmL

$d% he number of tubes requiredL$e% he !olumetric siGe of the condensate storage !esselL

$f% he number of tubes if positioned horiGontall-.

,olution.

$a% !olumetric flo" & !elocit- : cross sectional area

2

500 '0d

= d & 0.0< 8 m

< .


29/32

T & 10;.'C+ g & 1.2' m ' )kg

2 & 22'2 k()kg

densit- of condensate & 1000 kg)m '500

0.112'

class examples

Documents