Cooling tower expt

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    SCHOOL OF CHEMICAL AND BIOMEDICAL

    ENGINEERING

    (Division of Chemical & Biomolecula En!ineein!"

    Nan#an! $echnolo!ical %nivesi#

    CH'()*

    E+,eimen CE-Coolin! $o.e

    Sho Re,o

    Gou, mem/es0 Chan 1ei Shan (%22*)3)34"

    Hu Sihui (%22**5)*L"

    Sim 6an! Ron! (%22*27354"

    8u 4im 9an! (%22**7*G"

    Gou, 0 G9

    Dae of E+,: 0 ;h

    Se,em/e *)2'< $hus=a#

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    1

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    Sho e,o > CE- Coolin! $o.e

    2" Ino=ucion an= O/?ecives

    Cooling tower is a heat removal device where heat exchanges happen between high

    temperature fluid and low temperature fluid. It consists of two main type, where heatis removed either by the evaporation of water or by low temperature air. An example

    of counter-current device is used in this experiment to demonstrate the operating

    theory of the cooling tower. The water stream enters the device from the top while air

    is introduced from the bottom due to density difference. Once in contact, heat

    exchange will happen at the interface, which will lead to water evaporation. Thus

    latent heat is transferred from hot water to bul air by water vapour. Also, throughout

    the cooling tower, pacing material is used to increase the contacting surface area

    between air and water. As a result of heat and mass transfer, the temperature of water

    outflow is lower than the temperature of water inflow, while air outflow is at higher

    temperature and humidity compared with air inflow.

    The frst objective o the experiment was to examine the mechanicso a bench-scale cooling tower under dierent cooling loads. Thesecond aim was to practice using a psychrometric chart and tocalculate mass and energy balances.

    *" Summa# of chan!es o ,oocol

    !. The fan inlet shutter was half open rather than fully open.

    ". The regular interval adopted was " minutes rather than !# minutes. Also the total

    experimental time is $# minutes for each run.

    %. The tan was refilled after each time interval rather than at the end of one run. Theaverage was taen to estimate the mae-up &uantity.

    $. The mae-up water &uantity was determined by taing the difference between the

    initial and final water level readings.

    2

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    '" Resuls

    ':2 Daa esuls fom e+,eimens

    'sing the raw data recorded in tables A! and A" in the appendix, the mean values of

    temperatures, water flow rate and orifice differential were calculated. These measured&uantities recorded in the appendix were taen at intervals of " minutes over a time

    period of $# minutes for cooling load !.#( and !.)(.

    Table !* Tabulation of the mean values of data of cooling load !.#( and !.)(

    Cooling load !.#( !.)(

    Time interval

    " minutes

    interval over a

    period of $#

    minutes

    " minutes

    interval over a

    period of $#

    minutes

    Air inlet dry bulb

    temperature, t!+C"%./ "$.%%

    Air inlet wet bulb

    temperature, t"+C!0.$! !0.$

    Air outlet dr bulb

    temperature, t%+C"$.)" "/.

    Air oulet wet bulbtemperature, t$+C

    "%./# ").0/

    (ater inlet

    temperature, t)+C%#.1/ %/."$

    (ater outlet

    temperature, t/+C"%. ").0#

    Orifice 2ifferential

    +mm 3"O!1./ !1.0#

    (ater flow rate +g4s %/."1 %/.##

    5ae-up &uantity

    +m6$1.)# //.")

    5ae-up &uantity

    +g#.#$1)# #.#//")

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    ':* Calculaions

    a. 'sing the psychometric chart provided +prepared for ! atm pressure, estimate the

    specific humidity, enthalpy and specific volume of inlet and outlet for each run.

    Table "* Tabulation of specific humidity, enthalpy and specific volume of inlet and outlet ofeach run

    b. 7erform a

    mass balance

    for each run to

    determine the

    extent to

    which the

    rate of mae-up addition agrees with the rate of evaporation.

    i: Mass /alance calculaions fo 2:)@1

    'sing the relation between pressure drop across the orifice and air flow rate given

    in the laboratory manual,

    ma=0.0137 x

    v top (1+utop )

    0.0137 17.86

    0.867 (1+0.0181 )

    ma=0 .0616 kg /s

    !

    Cooling load +( !.# !.)

    hbottom+84g )).") )/.)#

    ubottom+g 3"O 4 g

    dry air#.#!"" #.#!"/

    vbottom+m%4g #.) #.)0

    htop+84g 1#.1) #.##

    utop +g 3"O 4 g

    dry air#.#!! #.#"#0

    vtop+m%4g #./1 #.10

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    Sein! u, a s,ecies mass /alance on ai acoss he =oe= line /oun=a#