Bai Tap Truyen Nhiet Tham Khao - Co Loi Giai

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Text of Bai Tap Truyen Nhiet Tham Khao - Co Loi Giai

  • BI 1: ( BAI 12.6 tr 276): Vach lo hi bang thep day 20 mm, = 58 W/mK; nhiet o kh lo tf1 = 1000 oC ; ap suat hi p = 34 bar. HSTN cua kh loti vach 1 = 116 W/m2K ; t vach lo en nc 2 = 2320 W/m2K.Xac nh q , nhiet o be mat trong va ngoai vach lo.

    Giai:Nhiet o nc soi: tf1 = 240 oC

    HSTN qua vach phang: k = 106,4 W/m2K

    MDN q = k(tf1 tf2) = 80864 W/m2K.

    Nhiet o be mat trong va ngoai vach lo:

    303 oC ; 275 oC

    21

    111k

    ++

    =

    11f1w

    1qtt =2

    221qtt fw +=

    NHAN XET:

    - Gi tr HSTN k so vi He so toa nhiet- Nhit o vach va chenh lech nhiet o vach

  • BI 2: Ong dan hi bang thep dtr/dng = 200 / 216 mm co 1 = 47 W/(mK)c boc mot lp cach nhiet day 120 mm, co 2 = 0,8 W/(mK). Nhiet o hi la t1= 360 oC ; he so TNL pha hi 1 = 120 W/(m2K) . Khong kh ben ngoai co t2 = 25 oC; 2 = 11 W/(m2K).- Hay tnh ton that nhiet tren 1 m ong qL

    - Xac nh nhiet o be mat trong va ngoai cua lp cach nhiet.

    Giai: He so truyen nhiet, tnh theo 1m chieu dai ong

    32

    1

    11

    1ln211

    1

    ddd

    d

    k

    i

    i

    i

    L

    ++= + mKWkL /412,1708,0

    1 ==

    qL = kL(tf1 tf2) = x 1,412(360 25) = 1485,28 kW/mTon that nhiet, tnh cho 1m ong:

    Nhiet o be mat lp cach nhiet:

    Cd

    qtt oLfw 3,1191

    3223 =+= Cd

    dqtt Lww0

    2

    3

    232 2,340ln2

    =+=

  • Giai: TN qua vach tru 3 lp

    BI 3: (BAI 12.9 tr 277): ong dan gio nong cho lo cao, toc o gio 1 = 35 m/s; tf1 = 800 oC. co 03 lp : gach chu la 1 = 250 mm , 1 = 1,17 W/m; thep 2 = 10 mm , 2 = 46,5 W/mK; cach nhiet ngoai 3 = 200 mm , 3 = 0,174 W/mK; ngknh trong d1 = 1000 mm, khong kh xung quanh tf2 = 10 oC , 2 = 4 m/s.

    Tnh ton that nhiet tren 1 m ng ong (bo qua bc xa).

    Ton that nhiet tren 1 m ng ong: ( )21 ffll ttkq = 121

    1

    11

    1ln2111

    +=+ ++==

    n

    n

    i i

    i

    ill dd

    ddk

    R Vi d2 = d1 + 21

    d3 = d2 + 22d4 = d2 + 23

    Can tnh HSTN pha trong va pha ngoai ong

  • Tnh 1: Khong kh chuyen ong cng bc trong ongKTXD la d1 = 1m; NX la tf1 = 800 oC

    Ref1 = 2,6*105 > 104 -> chay roiNuf1 = 0,018 Ref10,8 = 382 -> 1 = 27,4 W/m2K

    Tnh 2: Khong kh chuyen ong ngang ong nKTXD la d4 = 1,92m; NX la tf2 = 10 oC

    Ref2 = 5,45*105 > 104 -> chay roiNuf2 = 0,245 Ref20,6 = 680 -> 2 = 8,85 W/m2K

    KET QUA: kl = 0,94 W/m2K ql = 2330 W/m

  • tkFQ =

    C

    tt

    ttt o136

    100180ln

    100180

    lnmin

    max

    minmax ===

    tkQF =

    BI 4: Can gia nhiet dau G2 = 1000 kg/h t t2 = 20 oC en t2 = 180 oCbang khoi nong co t1 = 280 oC. Biet t1 = 200 oC; k = 35 W/(m2K); cp1 =1,1 kJ/(kgK); cp2 = 2,3 kJ/(kgK).Tnh dien tch TN (F) khi bo tr dong chuyen ong ngc chieu.

    GIAI: S dung PT TRUYEN NHIETChenh lech nhiet o tai hai au thiet b:

    t1 - t2 = 280 180 = 100 oC = tmint1 - t2 = 200 20 = 180 oC = tmax

    o chenh nh/o trung bnh:

    Nhiet lng cap cho dau: Q = G2 cp2 (t2 - t2 ) = 1000 . 2,3 (180 20)/3600 = 102,2 kW

    Vay = 102,2 / (0,035 . 136,1) = 21,46 m2

  • BI 5: Thiet b un nc nong bang khoi thai t CT, loai lu ong ngc chieu, co cac thong so sau:Pha khoi nong: G1 = 0,8 kg/s; cp1 = 1,12 kJ/(kgK); t1 = 450 oC.Pha nc: G2 = 3,2 kg/s; cp2 = 4,18 kJ/(kgK); t2 = 50 oCDien tch truyen nhiet F = 15 m2; k = 85 W/(m2K). a) Hay tnh Q; nhiet o ra cua khoi va nc.b) Neu ong c van hanh non tai vi G1* = 0,5G1, cac ieu kien ban au khac khong thay oi th nhiet lng trao oi va nhiet o nc ra se la bao nhieu?

    GIAI: a) Tnh Q va nhiet o cuoi cac chat, dung pp NTUChenh lech nhiet o tai hai au thiet b:

    C1 = G1 cp1 = 0,8 . 1,12 = 0,896 kW/K = CminC2 = G2 cp2 = 3,2 . 4,18 = 13,376 kW/K = Cmax

    Nhiet lng truyen cc ai: Qmax = Cmin (t1 - t2 ) = 0,896 (450 50) = 358,4 kW

  • maxmin

    *CCC = = 0,896 / 13,376 = 0,067

    NTU = kF/Cmin = 85 . 15 / 896 = 1,42Hieu suat thiet b (tra o th): = 0,72Nhiet lng trao oi: Q = Qmax = 0,72 . 358,4 = 258 kWNhiet o khoi thoat: t1 = t1 Q/C1 = 450 258 / 0,896 = 162 oCNhiet o nc ra: t2 = t2 + Q/C2 = 50 + 258 / 13,376 = 69,3 oC

    b) Neu ong c van hanh non tai vi G1* = 0,5G1Cmin giam 2 lan -> C* giam 2 lan = 0,033

    NTU tang 2 lan = 2,84Qmax giam 2 lan

    HS truyen nhiet luc o = 0,92

    Do vay: Q = 0,92. 358,4 / 2 = 164,8 kWt2 = 50 + 164,8 / 13,376 = 62,3 oC

  • BI 6 : Thiet b un nc nong bang hi nc, kieu chum ong.- Tong lu lng nc Gn = 10 kg/s; toc o nc trong ong 19/17 la w = 1m/s; t2 = 30 oC va t2 =70 oC.- Hi nc co p = 2 bar, x = 0,9. Nc ngng ra khoi bnh co tK = 90 oC.Biet HS toa nhiet pha hi ngng tu 1 = 8000 W/(m2K); nhiet tr dan nhiet cua vach ong

    Tnh dien tch F cua thiet b va lu lng hi can cung cap. (Cho phep bo qua anh hng phng hng dong nhiet va coi he so t =1

    WKm200017,0=

    tkQF =NHAN XET:S dung PT TRUYEN NHIET e tnh FONG MONG nen co the tnh HSTN k theo CT vach phang. (Nhiet tr dan nhiet a cho la cua vach phang)Can tnh 2 pha nc chay trong ong e xac nh HSTN kTm nhiet o hi vao e tnh ttb. Ap suat hi la a/s tuyet oiTnh Q: nhiet lng nc nhan c

  • cp = 4,174 kJ/kgK ; = 64,8.10-2 W/mK Prf = 3,54

    TC Reynolds: 46 103060010.556,0017,01Re >===

    trf

    wd-> chay roi

    Ong thang: R = 1; Gia thiet L/d > 50: l = 1. PTTC:Nuf = 0,021Ref0,8Prf0,43 = 0,021 x 306000,8 X 3,540,43= 137,5

    TSVL cua nc

    = 0,556.10-6 m2/s ;

    NX: t2 = 0,5(30 + 70) = 50 oC

    ( )KmWd

    Nutr

    f2

    2

    2 /5241017,010.8,645,137 ===

    2. He so truyen nhiet

    21

    111k

    ++

    = =2062 W/(m2K)

    Giai:1. Tnh he so toa nhiet 2 pha nc chay trong ong

  • Ctt

    ttt o5,55

    23,5060ln

    23,5060

    lnmin

    max

    minmax ===

    tkQF =

    5. Cong suat thiet b = Nhiet lng can e gia nhiet ncQ = Gn cpn (t2 - t2 ) = 10 . 4,174 (70 30) = 1672 kW

    6. Dien tch BMTN:= 1672 / (2,062. 55,5) = 14,6 m2

    7. Lu lng hi can: G = Q/(ih ik) = 1672 / (2486,6 377) = 0,79 kg/s= 2853 kg/h

    3. Thong so hi nc: p = 2 bar, co ts = 120,23 oC; i= 504,8 kJ/kgr = 2202 kJ/kg. Hi bao hoa am, x = 0,9: ih = i+ rx = 2486,6 kJ/kg

    4. o chenh nhiet o TB: tmax = 60 oC ; tmin = 50,23 oC

  • BI 7: Trong mot TBTN, nc chay trong ong co dtr/dng = 28/32 mm viG2 = 1 kg/s, nhiet o nc vao: tf2 = 25 oC ; ra tf2 = 95 oC ; Hi: tf1 = 120 oC. He so TNL pha hi 1 = 4000 W/m2K ; vach co v = 45 W/mK.

    - Hay tnh HSTN ve pha nc 2. Xem (Prf/Prw)0.25 = 1.- Xac nh HSTN k va mat o dong nhiet trung bnh tren 1 m ong qL.

    - Chieu dai can thiet cua ong la bao nhieu?

    Giai:

    TSVL cua nc = 983,2 kg/m3 ;cp = 4,179 kJ/kgK ; = 65,9.10-2 W/mK

    Prf = 2,98

    Van toc nc : s/m,,,d

    Gwtr

    65102802983

    14422 =

    ==

    TC Reynolds: 46 10966521047800280651 >=== .,,,wdRe trf

    1. He so toa nhiet 2Nhiet o TB cua nc trong ong: tf2 = 0,5(25 + 95) = 60 oC

    = 0,478.10-6 m2/s ;

  • Do che o chay roi nen ta co PTTC:

    Nuf = 0,021Ref0,8Prf0,43 = 0,021 x 966520,8 x 2,980,43 = 326,76

    Km/W,,.,,

    dNu

    trf

    22

    2 5769002801096576326 ===

    2. He so truyen nhiet va nhiet lng truyen

    trtr

    ng

    vng

    L

    ddd

    lnd

    k

    21

    1211

    1

    ++=

    mK/W,

    ,,ln

    ,

    kL 2473

    0280576901

    2832

    4521

    032040001

    1 =++

    =

    qL = kL(tf1 tf2) = x 73,24(120 60) = 13,798 kW/m3. Chieu dai ong can thiet

    Nhiet can cap cho nc:

    kW,)(,)tt(cGQ 'f''fp 532922595179412222 ===

    Chieu dai ong: (Kiem tra L/d !) m,,

    ,qQL

    L

    22179813

    53292 ===

  • Bai 1 (nh 10.2) Tng lo nung: h = 2,5 m; F = 39 m2; tw = 90 oC ; khong kh xung quanh tf = 30 oC.

    Tnh va ton that nhiet Q.

    KTX: l = h = 2,5 m

    NX:2

    ttt wfm

    += C602

    3090 o=+= Tm = 333 oK

    Chenh lech N:fw ttt = C603090 o==

    TSVL tai tm = 60 oC ( phu luc Bang 22):

    KmW 109,2 2m = sm1097,18 26m =696,0Prm =

    ( ) = mPrGr 1010 1034,5696,01067,7 =

    Giai

    2

    3

    mm

    thgGr

    = = 7,67 x 1010

  • Chay roi 135,0C =31

    n =

    PTTC:

    ( ) 31mm PrGr135,0Nu = ( ) 508101034,5135,0 31 ==HSTN:

    h

    Nu mm= KmW 9,55,2

    109,2508 22

    ==

    Nhiet lng ton that:

    ( )fw ttFQ = kW 13,8 W1380606399,5 ===

  • Bai 2 (nh 10.7)Ong dan kh nong nam ngang : d = 0,5 m; tw = 470 oC ; khong kh

    xung quanh tf = 30 oC. Tnh va ton that nhiet ql.Neu d* = d/2 th * = ?

    KTX:

    NX:2

    ttt wfm

    += C2502

    30470 o=+= Tm = 523 oK

    Chenh lech N: fw ttt = Co44030470 ==TSVL tai tm = 250 oC ( phu luc Bang 22):

    KmW 1027,4 2m = sm1061,4026

    m=

    68,0Prm = 2m

    3

    mtdg

    Gr = ( )

    8122

    3

    10255,61061,40 523.

    4405,081,9 ==

    ( )mPrGr 88 10253,468,010255,6 ==

    Giai l = d = 0,5 m

  • Chay roi, C = 0,135 31

    n =

    PTTC:

    ( ) 311350 mm PrGr,Nu = ( ) 510181025341350 31 ,,, ==HSTN:

    d

    Nu mm= KmW67,85,0

    1027,45,1012

    2

    ==

    Nhiet lng ton that tren 1m ong:

    ( )fw ttFq === 0445,014,367,8 kW/m 6 W/m5989,2 =

    Khi d* = d/2 Gr giam 23 = 8 lan, tc Ra* = (Gr*.Pr) = 5,3 x 107 . Van chay roi. Che o automodel : khong oi

  • Bai 3: (10.12)Tm t va q qua lp khong kh trong mot khe hep = 20 mm, tw1 = 200 oC,tw2 = 80 oC.

    Neu be day giam 2 lan th t se the nao?

    2

    21 wwf

    ttt

    += C1402

    80200 o=+=

    Theo phu luc thong so vat ly cua khong kh ta c:

    Km.W 0349,0f = sm108,27 26f =684,0Prf =

    Tieu chuan Gr:

    21 ww ttt = Co12080200 ==

    2f

    3

    f

    tlgGr

    = 4122

    3

    109521082741