Thermal Assg PRINT

7/23/2019 Thermal Assg PRINT

http://slidepdf.com/reader/full/thermal-assg-print 1/22

TABLE OF CONTENTS

Page

Table of Contents

i

1. Problem Statement

2

2. Objective 23. Introduction

2-34. ir Conditioning Process

4-!". #efrigeration C$cle

%-12&. Combustion C'amber for Internal Combustion (ngine

13-1%!. Conclusion

1)%. #eferences

1)). **endi+

1)

1



PROBLEM STATEMENT

A simple ideal vapor-compression heat-pump system is shown in Figure 1 to provide a

heating effect for a building. The heat from evaporator of the refrigeration system is

dissipated to the outdoor air flowing at 20,000 cfm from 0! to condition it to "0! and the

relative humidity is "0#, respectively. An unmi$ed cross flow heat e$changer is used to

increase the temperature to 1%!. The heat e$changer is made of tubes through which flows

hot fluid from the compressor of the refrigeration system. Then, the air is humidified by the

in&ection of hot steam in the humidifying section.

OBJECTIVE

The ob&ective of this assignment is for students to conduct preliminary design calculations of

an air conditioning system combining all the relevant topics in thermal engineering.

INTRODUCTION

Air conditioning (often referred to as A/C or AC) is the process of altering the

properties of air (primarily temperature and humidity) to more comfortable conditions,

typically with the aim of distributing the conditioned air to an occupied space such as

a building or a vehicle to improve thermal comfort and indoor air quality. In common

use, an air conditioner is a device that lowers the air temperature. he cooling is

typically achieved through a refrigeration cycle, but sometimes evaporation or free

cooling is used. Air conditioning systems can also be made based on desiccants. In

the most general sense, air conditioning can refer to any form of technology that

modifies the condition of air (heating, cooling, (de!)humidification, cleaning,

ventilation, or air movement). "owever, in construction, such a complete system of

heating, ventilation, and air conditioning is referred to as heating, ventilation, and air

conditioning ("#AC !as opposed to AC ).

Principles

2

https://en.wikipedia.org/wiki/Air

https://en.wikipedia.org/wiki/Temperature


https://en.wikipedia.org/wiki/Humidity

https://en.wikipedia.org/wiki/Thermal_comfort

https://en.wikipedia.org/wiki/Indoor_air_quality

https://en.wikipedia.org/wiki/Cooling_(disambiguation)

https://en.wikipedia.org/wiki/Refrigeration_cycle

https://en.wikipedia.org/wiki/Evaporative_cooler

https://en.wikipedia.org/wiki/Free_cooling


https://en.wikipedia.org/wiki/Construction

https://en.wikipedia.org/wiki/Ventilation_(architecture)

https://en.wikipedia.org/wiki/HVAC


https://en.wikipedia.org/wiki/Humidity

https://en.wikipedia.org/wiki/Thermal_comfort

https://en.wikipedia.org/wiki/Indoor_air_quality

https://en.wikipedia.org/wiki/Cooling_(disambiguation)

https://en.wikipedia.org/wiki/Refrigeration_cycle

https://en.wikipedia.org/wiki/Evaporative_cooler



https://en.wikipedia.org/wiki/Construction

https://en.wikipedia.org/wiki/Ventilation_(architecture)

https://en.wikipedia.org/wiki/HVAC

https://en.wikipedia.org/wiki/Air



$or an air conditioning system to operate with economy, the refrigerant must

be used repeatedly. $or this reason, all air conditioners use the same cycle of

compression, condensation, e%pansion, and evaporation in a closed circuit. he

same refrigerant is used to move the heat from one area, to cool this area, and to

e%pel this heat in another area.

&. he refrigerant comes into the compressor as a low!pressure gas, it is

compressed and then moves out of the compressor as a high!pressure

gas.

'. he gas then flows to the condenser. "ere the gas condenses to a

liquid, and gives off its heat to the outside air.. he liquid then moves to the e%pansion valve under high pressure.

his valve restricts the flow of the fluid, and lowers its pressure as it

leaves the e%pansion valve.. he low!pressure liquid then moves to the evaporator, where heat from

the inside air is absorbed and changes it from a liquid to a gas.*. As a hot low!pressure gas, the refrigerant moves to the compressor

where the entire cycle is repeated.

3



A. Air Conditioning Process

Question 2: 'sing (ry mass balance ) m

1 *m

2+, ater mass balance ) m

1 ω

1 *m

2 ω

2+, nergy balance , Q m ∆ h

State 1

Pv1 ∅ 1Pg1/ P,sat0 T ℃ ,refer table -4

,.)",.&11

."%112Pa

Pa1 P1-Pv1

4



)).41)Pa

1 RaT 1

Pa1 (0.287 k )(273)

99.419 k .!%%m

3

kgdry air

ω1

0.662 Pv1

P1− Pv1 0.662(0.58112k )100k −0.58112k 0.00364

kg h20

kgdry air / ,

ω1=ω2

State 2

ω1

=ω2

.3&4

kJ h20

kgdry

'1 ,C*T1 , ω 1 'g1 ,'g refer table -4 for 'g sat

,1.", ,.3&4,2".) 9.0925 kJ

kg

'2 ,C*T2 , ω 2 'g2 , 'g refer table -4 for 'g sat

,1.",1" ,.3&4,2"2%.3 24.278kJ

kg

5etermine t'e ṁ 1 /

61 V 1

V 1 566.337

0.788 /, V 1 from converting 2cfm to m37min

!1%.) kg

min / convert tokg

s !1%.)×1min

60 s 11.978 kg

s

T'erefore/

Q 61 ∆ ' ,11.)!%,24.2!% 8 ).)2" 18.1892 9 ,:eat

transfer

"



Question 3: Analye rate of water vapour at t * 20 ℃

until 2/ ℃

)with increament 1 ℃

+

elative humidity * "0#-/# )assume 0#+

Water vapour at t20 ℃

6v 6a , ω 3- ω 2 / 62 11.)!%kg

s / ω 2 .3&4kg h20

kgdry air

5etermine ω 3/

ω 3

0.662∅ Pg

P−∅ Pg

0.662(0.6)2.3392 k

100k −(0.6)(2.3392 k ) %.%"42 ×10

−3

kg h20

kgdry air

m v ,11.)!%,%.%"42 ×10−3

8 .3&4 0.06246 kg

s

Water vapour at t21 ℃

6v 6a , ω 3- ω 2 / 62 11.)!%kgs / ω 2 .3&4

kg h20kgdry air

5etermine ω 3/

ω 3 0.662∅ Pg

P−∅ Pg !"# $ro% &'terpo(at&o' ta)(e A*4+/ Pg 2.""32Pa

0.662(0.6)(2.50532k )

100k −(0.6 )(2.50532k ) .)4)2"

kg h20

kgdry air

m v ,11.)!%,.)4)2" 8 .3&4 0.07010 kg

s

Water vapour at t 22 ℃

&



6v 6a , ω 3- ω 2 / 62 11.)!%kg

s / ω 2 .3&4kg h20

kgdry air

5etermine ω 3/

ω 3 0.662∅ Pg

P−∅ Pg !"# $ro% &'terpo(at&o' ta)(e A*4+/ Pg 2.&!144Pa

0.662(0.6)(2.67144 k )100k −(0.6 )(2.67144k ) .113

kg h20

kgdry air

m v ,11.)!%,.113 8 .3&4 0.07774 kg

s


6v 6a , ω 3- ω 2 / 62 11.)!%kg

s / ω 2 .3&4kg h20

kgdry air

5etermine ω 3/

ω 3 0.662∅ Pg

P−∅ Pg !"# $ro% &'terpo(at&o' ta)(e A*4+/ Pg 2.%3!&Pa

0.662(0.6)(2.3876 k )

100k −(0.6 )(2.3876 k ) .1!!kg h20

kgdry air

m v ,11.)!%,.1!! 8 .3&4 0.08540 kg

s


6v 6a , ω 3- ω 2 / 62 11.)!%kg

s / ω 2 .3&4kg h20

kgdry air

5etermine ω 3/

!



ω 3 0.662∅ Pg

P−∅ Pg !"# $ro% &'terpo(at&o' ta)(e A*4+/ Pg 3.3!Pa

0.622(0.6)(3.0037 k )

100k −(0.6)(3.0037 k ) .1142

kg h20

kgdry air

m v ,11.)!%,.1142 8 .3&4 0.09319 kg

s


6v 6a , ω 3- ω 2 / 62 11.)!% kgs / ω 2 .3&4 kg h

20

kgdry air

5etermine ω 3/

ω 3 0.662∅ Pg

P−∅ Pg !"# $ro% &'terpo(at&o' ta)(e A*4+/ Pg 3.1&)%Pa

0.622(0.6)(3.1698 k )

100k −(0.6)(3.1698 k ) 0.01206

kg h20

kgdry air

m v ,11.)!%,.12& 8 .3&4 0.10085 kg

s

%



B. Refrigeration cycle

Select 2 working refrigerants for the system and explain the reasons of selection based onsafety and thermal properties.

First and foremost, -13"a do not contain chlorine atom so that it afford to undermine

the role of atmospheric oone4 besides, -13"a has a good safety performance )non-

flammable, non-e$plosive, non-to$ic, non-irritating no rot resistance+, in addition, -13"a is

easier to retrofit refrigeration system so that the heat transfer performance is closer. 5ast but

not least, -13"a heat transfer performance better than the -12 which can help the amount

of refrigerant greatly reduced.

As a refrigerant, ammonia offers three distinct advantages over other commonly usedindustrial refrigerants. First, ammonia is environmentally compatible. 6t does not deplete the

oone layer and does not contribute to global warming. 7econd, ammonia has superior

thermodynamic 8ualities, as result ammonia refrigeration systems use less electricity. Third,

ammonia9s recogniable odor is its greatest safety asset. 'nli:e most other industrial

refrigerants that have no odor, ammonia refrigeration has a proven safety record in part

because lea:s are not li:ely to escape detection.

The 2 important parameters that need to be considered in the selection of refrigerant

are the temperatures of 2 media )refrigerated space and the environment+ with which the

refrigerant e$change heat. Furthermore, the characteristics of the refrigerant include be non-

to$ic, non-corrosive, non-flammable and chemically stable.

6n the case of heat pump, the minimum temperature for the refrigerant maybe

considerably higher since heat is usually e$tracted from media that are well above the

temperature encountered in refrigeration systems.

Choose operating conditions for the refrigeration cycle such as the evaporator and

condenser pressure if the surrounding temperature is 0°C.

)



The most common energy source for heat pumps is atmospheric air )air-to-air

system+, although water and soil also used. The ma&or problems with air source system isfrosting, which occurs in humid climates when the temperature falls below 2-/;!. the !<= of

heat pump usually range between 1./ and ", depending on the particular system used and the

temperature of the source. <ne system can be used as a heat pump in winter and an air

conditional in summer. This is accomplished by adding a reversing valve to the cycle, so it

can be in heating mode and cooling mode.

efrigerant Analytical Analysis>

The T-s diagram of the refrigeration cycle is drawn below.

This an ideal vapor compression refrigerant cycle, and thus the compressor are isentropic and

the refrigerants leave the condenser as a saturated li8uid and enter the compressor at saturated

vapor.

A refrigerator used refrigerant-13"a as the wor:ing fluid and the assuming design pressures

for the condenser are %%.3:=a while the evaporator is at 1.2?=a. =ressure of evaporator is

assumed below from pressure of condenser to allow heat transfer, Q L from surrounding

into the refrigerant andQ H from refrigerant into surrounding.

Assumptions>

1 7teady operating conditions e$ist

2 @inetic and potential energy changes are negligible.

1



From the refrigerant -13"a tables, the enthalpies of the refrigerant at all four states are

determined as below>

State 1: Saturated vapor refrigerant-13a

Assumptions> =ressure * %%.3:=a

Temperature * 3/ !

eat transfer efficiently )100#+

By referring to the efrigerant-13"a tables>

To findhg and

sg @T 1 we have done the interpolation from table A-11 at

T 1=35° C

(35−34)(36−34)

= (hg – 268.57 )

(269.49 – 268.57)

h1=hg=269.03

kJ

kg

(35−34)(36−34)

= (s – 0.91743)

(0.91675 – 0.91743)

s=sg @ T 1

=0.91709 kJ

kg . K

State 2: Super!eated vapor refrigerant-13a

The chosen pressure is 1.2 ?=a due to the pressure on the condenser. e pic:ed this pressure

to allow heat transfer from refrigerant to surrounding. At state 2, it is isentropic process

S1=S

2 .

P2=1.20 Pa(ass!m"#i$n)

s2=s

1=sg@P

1

=0.91709kJ

kg. K

To find h, we have done the interpolation from table A-13 at P=1.2 Pa

(0.91709−0.9130)(0.9267 – 0.9130)

= (h2 – 273.87)(278.27 – 273.87)

h2 * 2/.21 :CD:g

11



State 2S: Super!eated vapor refrigerant-13a !2s

The compressor efficiency is assumed at /#.

n%=h2s−h1

h2−h1

0.75=h2 s−269.03

kJ

kg

275.21 kJ /kg−296.03kJ

kg

h2s * 2%0."2 :CD:g

State : Saturated li"uid refrigerant-13a

)efer Table A-12+

P3=1.2 Pa

h3=h

4=h& @"3=117.77

kJ

kg

State !: #!rottling valve

'−(=h4−h

3

h4=h

3

h4=117.77

kJ

kg

Calculate the re"uired refrigerant mass flow rate to obtain the desired cooling effect.

´Q L=

´m(h1−h

4)

12



Q L is obtained from 8uestion 1.Q L 6ndicate that the heat transfer rate from refrigerant

to surrounding.

´18.1892= ´m(269.03−117.77)

´m=0.12

kg

s

Calculate the maximum C#$ and actual C#$ of the cycle if the compressor efficiency is

assumed at %&'.

!<= defines the performance of the refrigeration cycle. To calculate !<=, we use this

formula

C)P= Q L

* ¿

?a$imum !<=

(¿= m(h2−h1)

( ¿=0.12 (275.18−269.03 )

(¿=780*

C)Pma+=18.1892

0.738=24.65

Actual !<=

( ¿= m(h2 s−h1)

(¿=0.12 (280.42−269.03 )

(¿=44.36k(

C)Pa%# =18.1892

1.366=13.31

13



Suggest an innovative system that can improve the current C#$ i.e multistages or cascade

refrigeration cycle. $rove your suggestion using analytical analysis.

6t is obvious that the lower-temperature unit of the cascade system absorbs less power thanthe single stage system. This originates from the fact that the pressure ratio across the

compressor in the lower unit of the cascade system is less than that in the single-stage system

for the same refrigeration capacity. !<=s for the lower unit of the cascade system are higher

than those for the single-stage system.

(stimate the cost of running the system )single cycle and multistage or cascade* for a 12

hour operation )based only on the compressor work input* under steady conditions and

actual +alaysian daylight electrical tariff by ,- ),enaga -asional erhad*.

Tar&, - * Lo /o(ta#e 'utr&a( Tar&,

For Overa(( o't( Co'u%pt&o' Betee' 0*200 W%o't

;or all <' 3%. sen7<'

%$s# =!sag, ×38

%$s# =18.1892k ×38= R 691189.6 0

C. Co$%ustion C!a$%er for &nternal Co$%ustion 'ngine

14

Q H

To e+'aust

-C )2+. H 2)+/ 0 2

C 12

H 26

25℃

a ) +3.76 0



Figure 2 > Eatural as !ombustor

Question 1: iven > G * 1./?att

Q= mair

m&!,1

C 12

H 26+1.2a#h ()2

+3.76 0 2) 2 - C )

2+ H

2)+/ 0

2

C :12= - (1 ) 3 - =12

H :26=. (2 ) 3. =13

) :1.2a (2 )= - (2 )+ (1)

1.2a (2 )=12 (2 )+13 (1 )

a=15.4167

0 :1.2a (3.76 )=/ (1 )

1.2(15.4167 ) (3.76)=/ 3 / =69.56

C 12

H 26+1.2 (15.4167) ()2

+3.76 0 2) 212C )

2+13 H

2)+69.56 0

2

Hrate of heat transfer

4¿= 4$!#

1"

h%=h R−h "

h R= 5 0 1

h&

h & =r,&,r Ta61, 7−26

h RC 12

H 26=¿

- 2)11

=97=mol

h R)2=0



8ind h R¿ #a61,a# 273 K

1&

Standard

#eferences

C )2

2! %33"2!3 '2% %&)!273−270

280−270=

h−8335

8697−8335

)2

2! !%"%2!3 '2% %1"273−270

280−270=

h−7858

8150−7858

H 2

)

2! %)&12!3 '2% )2)&273−270

280−270=

h−8961

9296−8961

0 2

2! !%4)2!3 '2% %141273−270

280−270=

h−7849

8141−7849



8ind h R¿ #a61,a# 273 K

4 R= 0 (h& 0+h−h

$)

¿ (1 ) (−291010+0−0)+18.5 (0+7945.6−8682 )+69.56 (0+7936.6−8669 )

¿−291010−13623.4−50945.744

¿−355579.144 *a##

1!

C )2

12& "!24412!3 '12% "%3%11273−1260

1280−1260=

h−57244

58381−57244

)2

12& 4")412!3 '12% 413121273−1260

1280−1260=

h−40594

41312−40594

H 2

)

12& 4!2212!3 :12% 4!)121273−1260

1280−1260=

h−47022

47912−47022

0 2

12& 3%"!12!3 '12% 3)4%%1273−1260

1280−1260=

h−38807

39488−38807

h=39249.65 97gC 12

H 26≫ 0 =1

)2≫ 0 =18.5

0 2≫ 0 =69.56

C )2≫ 0 =12



4 "= 0 (h& 0+h−h1)

¿12 (−393520+57983.05−9364 )+13 (−2858.30+47600.5−9904 )+69.56 (0+39249.65−8669)

¿−4138811.4−3225735.5+2127190.014

¿−5237356.586 *a##

Q$!# = 4 R− 4 "

¿−3555790144+5237356.586

¿4881777.442 kJ

km$1

C 12

H 26=12 (12 )+26 (1 )

¿170 kg

km$1

'=4881777.442

170

¿28716.34 kJ /kg

Question 2: Assumptions> 1+ Tout of the saturated vapor is 100!

2+ 'se ! p average of water

3+ ṁv varied because of the temperature

Tout * 100!

1%

28716.34 × m&!,1=1.5 *a##

m =0.052k /s



Tin * /

Average temperature * )Tout I Tin+D2

* )100 I /+D2

* /2./!

6nterpolation>

Temperature )!+ 7pecific heat, ! p )@&D:g. @+

/0 ".1%0

/2./ ".1%1

/ ".1J0

Q * ṁv ! p )Tout K Tin+

Temperature )!+ ?ass flow rate water vapor, ):gDs+ṁ

20 0.02"

21 0.0010

22 0.0"

23 0.0%/"0

2" 0.0J31J

2/ 0.100%/

At 20!,

Q * )0.02"+)".1%1 $ 10L3+)33 K 2%+ * 2"%0%.%0 :C

At 21!,

Q * )0.0010+)".1%1 $ 10L3+)33 K 2%+ * 2%"3.3 :C

At 22!,

Q * )0.0"+)".1%1 $ 10L3+)33 K 2%+ * 30%.J" :C

At 23!,

Q * )0.0%/"0+)".1%1 $ 10L3+)33 K 2%+ * 33J20."/ :C

1)



At 2"!,

Q * )0.0J31J+)".1%1 $ 10L3+)33 K 2%+ * 301".0 :C

At 2/!,

Q * )0. 100%/+)".1%1 $ 10L3+)33 K 2%+ * "00/.12 :C

?ass flow rate water vapor, ):gDs+ṁ eat Transfer ate, Q

):C+

0.02" 24%%.%0.0010 2!%43.4

0.0" 3%!!.)

0.0%/"0 33)2."

0.0J31J 3!14.&

0.100%/ 4"!.1

2



CONCLUSION

The ob&ective of this assignment is for students to conduct preliminary design calculations of

an air conditioning system combining all the relevant topics in thermal engineering has been

achieved.

REFERENCES

1. Jan F. Kreider. Handbook of heating, ventilation, and air conditioning.

2. Winnick, J (1996). Chemical engineering thermodnamic!. John Wile and "on!

3. #ao$, K% Wang, &ia ('). *#e!iccant cooling air conditioning+ arevie* . -eneable and "$!tainable nerg -evie!.

". /c#oall, -obert ('6). F$ndamental! of H0C "!tem!. l!evier. 2. 3.

/. 4a, C.K. ('6). Comb$!tion 5h!ic!. Cambridge, K+ Cambridge niver!it

5re!!.

. Arora, amesh !handra. */echanical va2o$r com2re!!ion

refrigeration*. -efrigeration and ir Conditioning. 7e #elhi, 8ndia+ 5H8 4earning.

2. 3.

APPENDIX

+or -oad istribution

o

.

ame o I 0ercentage

&. Ahmad Fai Fi:ri Abdul 5atif )5eader+ '1&2'&34' 15

'. ?ohd Fahmi Aiman bin Abdul ashid '1&4'44 '*5

. ?ohammad 7yahan bin EorMAman '1&'314' '*5. Abdul adi Bin . AB. ahim '1&'*'3'6 '15

21

http://www.sciencedirect.com/science/article/pii/S1364032104001200







otal &115

22

Thermal Assg PRINT

Documents

Transcript of Thermal Assg PRINT