Chapter 08 (1)

CHAPTER 8

1

1. Calculate the specific volume of an air-vapor mixture in cubic meters per kilogram of dry air at the

following conditions Ct 32= , kgkgW 0160.= , kPapt 100= .

Solution:

st

s

pp

pW

−=

6220.

s

s

p

p

−=

100

62200160

..

kPaps 5082.=

kPappp sta 492975082100 .. =−=−=

( )( )kgm

p

RTv

a

aa

38990

49297

273322870.

.

.=

+==

2. Moist air at a dry bulb temperature of 25 C has a relative humidity of 50% when the barometric

pressure is 101.4 kPa. Determine (a) the partial pressures of water vapor and dry air, (b) the dew

point temperature, (c) the specific humidity, (d) the specific volume, and (e) the enthalpy.

Solution:

At 25 C, kPapd 1693.= , kgkJhg 22547.= ,

%50=φ

(a) ( )( ) kPapp ds 584511693500 ... === φ

kPappp sta 8299584514101 ... =−=−=

(b) Cpattt dsatdp 71358451 .. ===

(c) ( )

kgkgpp

pW

st

s009870

8299

5845162206220.

.

...==

−=

(d) ( )( )

kgmp

RTv

a

aa

38570

8299

273252870.

.

.=

+==

(e) ( )( ) ( )( ) kgkJWhtch gp 1450225470098702501 .... =+=+=

3. Air at a temperature of 33 C has a relative humidity of 50%. Determine (a) the wet bulb

temperature, (b) the dew point temperature, (c) the humidity ratio, (d) the enthalpy, and (e) the

specific volume.

CHAPTER 8

2

Solution:

From psychrometric chart, at 33 C, RH = 50%

(a) Ctwb 524.=

(b) Ctdp 221.=

(c) kgkgW 01580.=

(h) kgkJh 873.= ,

(e) kgmv3

8890.=

4. How much heat is required to raise the temperature of 0.50 m3/s of air from 19 C dry bulb and 15 C

wet bulb to 36 C? What is the final dew point temperature?

Solution:

From psychrometric chart,

At 19 C DB and 15 C WB

kgkJh 9411

.=

kgmv3

1840.=

skgv

Vm 5950

840

500

1

1 ..

.===

at 36 C, kgkgWW 0090012

.==

kgkJh 4592

.=

( ) ( ) kWhhmQ 410941459595021

.... =−=−=

CHAPTER 8

3

Ctt satdp 512.==

5. How much heat must be removed to cool 30 cu m per minute of air from 34 C dry bulb and 18 C dew

point to a wet bulb temperature of 19 C? What is the final relative humidity?

Solution:


At 34 C DB and 18 C Dew Point

kgkJh 3671

.=

kgmv3

18880.=

( )( )skg

v

Vm 5630

608880

30

1

1 ..

===&

&

at 19 C, kgkgWW 0129012

.==

kgkJh 0542

.=

%822

=RH

( ) ( ) kWhhmQ 5754367564021

... =−=−=

%822

=RH

6. How much heat and moisture must be added to 5 m3/minute of air at 21 C dry bulb and 30%

relative humidity to raise it to 37 C and 40% relative humidity?

Solution:

CHAPTER 8

4


At 21 C , 30 % RH

kgkJh 8321

.=

kgkgW 004601

.=

kgmv3

1840.=

( )( )skg

v

Vm 2980

60840

15

1

1 ..

===&

&

at 37 C, 40 % RH

kgkJh 8772

.=

kgkgW 015802

.=

heat added,

( ) ( ) kWhhmQ 4113832877298012

.... =−=−= &

moisture added ( ) ( ) skgWWm 0033400046001580298012

.... =−=−= &

7. How much heat must be removed to cool 50 m3/min of air at 29 C dry bulb and 21 C wet bulb

temperatures to 16 C dry bulb and 14 C wet bulb temperatures? How much moisture was removed?

Solution:

CHAPTER 8

5



kgkJh 6601

.=

kgkgW 012301

.=

kgmv3

18730.=


kgkgW 009102

.=

kgkJh 2392

.=

( )( )skg

v

Vm 9550

608730

50

1

1 ..

===&

&

heat removed ( ) ( ) kWhhm 4420239660955021

.... =−=−= &

moisture removed ( ) ( ) skgWWm 0030600091001230955021

.... =−=−= &

8. Air at 32 C and 20 percent relative humidity is cooled and humidified by means of an air washer until

the relative humidity becomes 90%. How much moisture was added per kg of dry air. What was the

air washer efficiency and the dew point temperature of the leaving air?

Solution:

CHAPTER 8

6


At 1, 32 C and 20% RH

kgkgW 005901

.=

At 2, 90% RH

kgkgW 011602

.=

moisture added kgkgWW 00570005900116012

... =−=−=

air washer efficiency = wbdb

dbdb

tt

tt

−

−

1

21

Ctwb 0172

.= , Ctdb 0182

.=

air washer efficiency = ( ) %.% 3931001732

1832=

−

−

Dew point of leaving air = Ctdp 4162

.=

9. A stream of outdoor air is mixed with a stream of return air in an air conditioning system that

operates at 101 kPa pressure. The flow rate of outdoor air is 2 kg/s, and its condition is 35 C dry bulb

temperature and 25 C wet bulb temperature. The flow rate of return air is 3 kg/s, and its condition is

24 C and 50 percent relative humidity. Determine (a) the enthalpy of the mixture, (b) the humidity

ratio of the mixture, and (c) the dry bulb temperature of the mixture.

Solution:

CHAPTER 8

7


At 35 C DB, 25 C WB

kgkJho 975.=

kgkgWo 01590.=

at 24 C, 50% RH

kgkJhr 847.=

kgkgWr 00930.=

(a) ( )( ) ( )( )

kgkJm

hmhmh

m

rroom 059

5

84739752.

..=

+=

+=

(b) ( )( ) ( )( )

kgkgm

WmWmW

m

rroom 01190

5

009303015902.

..=

+=

+=

(c) ( )( ) ( )( )

Cm

hmhmt

m

ro

m

db

dbrdbo

db 4285

243352.=

+=

+=

10. An auditorium is to be maintained at 25 C dry bulb temperature and 50% relative humidity. The

supply air enters the auditorium at 17 C. The sensible and latent heat loads are 150 kW and 61 kW,

respectively. Determine the wet bulb temperature, relative humidity, and volume flow rate of the

supply air.

Solution:

CHAPTER 8

8

kWQS 150= , kWQL 61=

At 25 C, 50% RH

kgkJh 3502

.=

kgkgW 009902

.=

( )12

dbdbpS ttcmQ −= &

( )( )172500621150 −= .m&

skgm 6318.=&

kgkJm

QQhh LS

0396318

61150350

21.

.. =

+−=

+−=

&

( )12

2500 WWmQL −= &

( )( )1

009906318250061 W−= ..

kgkgW 008601

.=


At kgkJh 0391

.= , kgkgW 008601

.=

Then,

Wet bulb at 1, Ctwb 9131

.=

Relative humidity, %.5711

=RH

CHAPTER 8

9

kgmv3

183340.=

( )( ) smvmV3

115315833406318 ... === &&

11. In a certain space to be air conditioned the sensible and latent heat loads are 20.60 kW and 6.78

kW, respectively. Outside air is at 33 C dry bulb and 24 C wet bulb temperatures. The space is to be

maintained at 25 C with a relative humidity not exceeding 50%. All outside air is supplied with

reheater. The conditioned air enters at 18 C. Determine (a) the refrigeration load required, (b) the

capacity of the supply fan, and (c) the heat supplied in the reheater.

Solution:

From pyschrometric chart.

At 1, Ctdb 331

= , Ctwb 241

=

kgkJh 9711

.=

At 4, Ctdb 254

= , %504

=RH

kgkJh 3504

.=

kgkgW 009904

.=

CHAPTER 8

10

At 3, Ctdb 183

=

( )34

dbdbpS ttcmQ −= &

( )( )1825006216020 −= .. m&

skgm 9252.=&

( )34hhmQQ LS −=+ &

( )3

35092527866020 h−=+ ....

kgkJh 9403

.=

at Ctdb 183

= , kgkJh 9403

.=

kgkgW 0090.03

=

kgmv3

38370.=

At 2, with kgkgWW 0090032

.==

kgkJh 1352

.=

(a) Refrigeration load required ( ) ( ) kWhhm 6107135971925221

.... =−=−= &

(b) Capacity ( )( ) smvm3

345283709252 ... ==&

(c) Heat supplied in the reheater ( ) ( ) kWhhm 9716135940925223

.... =−=−= &

12. An air conditioned auditorium with a capacity of 1000 persons is to be maintained at 24 C dry bulb

temperature and 55% relative humidity. The sensible and latent heat loads are 115 kW and 42 kW,

respectively. The conditioned air enters the auditorium at 17 C. For proper ventilation, 40% of the

supply air is fresh air and the rest is recirculated air. Outside air is at 34 C and 50% relative humidity.

Determine (a) the volume flow rate of recirculated air, (b) the apparatus dew point, and (c) the

refrigeration load.

Solution:

CHAPTER 8

11

mmo&& 400.=

mmr&& 600.=

At 4, Ctdb 244

= , %553

=φ

kgkJh 2504

.=

kgkgW 010204

.=

kgmv3

48560.=

At 1, Ctdb 341

= , %501

=φ

kgkJh 2771

.=

CHAPTER 8

12

At 5, ( ) ( ) Ctdb 2824600344005

=+= ..

( ) ( ) kgkJh 612506002774005

=+= ....

( )34

dbdbpS ttcmQ −= &

( )( )172400621115 −= .m&

skgm 3316.=&

( )34

2500 WWmQL −= &

( )( )3

010203316250042 W−= ..

kgkgW 009203

.=

At 2, Ctdb 8122

.= , kgkgWW 0092032

.==

kgkJh 1362

.=

(a) Volume flow rate of recirculated air = ( )( )( ) smvmV rr

3

439885603316600 .... === &&

(b) Apparatus dew point = Ctdb 6122

.=

(c) Refrigeration Load = ( ) ( ) kWhhm 640613661331625

... =−=−= &

13. A store to be maintained at 25 C and 50% relative humidity has a sensible heat load of 18.90 kW and

a latent heat load of 6.30 kW. Outside air is at 32 C dry bulb and 23 C wet bulb temperatures. The

conditioned air enters at 17 C. If 30% of the supply air is fresh air and the bypass system is used,

determine (a) the refrigeration required, and (b) the volume of the bypass air at supply condition.

Solution:

Ctdb 25

4

= , %50=φ

CHAPTER 8

13

kWQS 9018.=

kWQL 306.=

Ctdb 173

=

Ctdb 321

=

Ctwb 231

=

mmo 300.=

mmm br 700.=+

From psychrometric chart

At 4, Ctdb 254

= , %50=φ

kgkJh 3504

.=

kgkgW 009904

.=

kgmv3

48580.=

( )34

dbdbpS ttcmQ −= &

( )( )1725006219018 −= .. m&

skgm 3482.=&

( )34hhmQQ LS −=+ &

( )3

35034823069018 h−=+ ....

kgkJh 6393

.=

At 3, Ctdb 173

= , kgkJh 6393

.=

kgmv3

38340.=

CHAPTER 8

14

at 1, Ctdb 321

= , Ctwb 231

=

kgkJh 0681

.=

To solve for 2

dbt :

Let m

mb b= ,

m

mc c=

(1) 1=+ cb

( ) ( )24

3dbdbdb tctbt +=

(2) ( )2

2517 dbtcb +=

( ) ( )24

3wbwbwb tctbt +=

Ctwb 143

= , Ctwb 8174

.=

22

dbwb tt =

(3) ( )2

81714 dbtcb+= .

(3) – (2) ( )b8.17251417 −=−

417.0=b

5830417011 .. =−=−= bc

Substitute in (2)

( )2

2517 dbtcb +=

Ctt dbwb 31122

.==

kgkJh 4321

.=

∴ ( ) skgmcmc 367134825830 ... === &&

( ) skgmbmb 979034824170 ... === &&

( ) skgmmo 70403482300300 .... === &&

( )3482700700 ... ==+ mmm br&&&

( )34827009790 ... =+rm&

skgmr 6650.=&

∴ at 5,

( ) ( )kgkJ

mm

hmhmh

or

or459

70406650

0687040350665014

5.

..

....=

+

+=

+

+=

&&

&&

(a) Refrigeration load ( ) ( ) kWhhmc 836432359367125

.... =−=−= &

(b) Volume of the bypass air at supply condition

( ) smvmb

3

3816083409790 ... === &

CHAPTER 8

15

Chapter 08 (1)

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