B GIO DC V O TOTRNG I HC NNG LM TP. H CH MINHB MN CNG NGH HA HC

K THUT LNH V NG DNG

Thit k kho lnh bo qun100 tn tm ng lnh

iu kin bo qun:Nhit , m mi trng thit k kho lnh: Kho lnh t ti TP.HCM, ta c: Nhit ma h: 37.3(oC) m: 74(%) Nhit bu t: 33 (oC) Nhit ng sng: 32 (oC)

I. XC NH KCH THC BUNG LNH:

1.Dung tch kho lnh:

E = V.gv E : dung tch kho lnh (t) V : th tch kho lnh (m3)gv : nh mc cht ti th tch (t/m3). gv= 0.45 (t/m3)Dung tch tht s cc bung : Sn phm l to Espv thng g EbbChn Ebb =10% Esp Esp = 100 (t) (u ) dung tch tht s ca bung lnh :E = Esp + Ebb = 100 + 10 = 110 (t)th tch ca bung lnh :V = E/gv = 110/0.45 = 244.44 (m3)

2.Din tch bung lnh :

Din tch cht ti F = V/h h: l chiu cao cht ti (m) , chn h= 5 ( m) F=V/h =244.444/3 =81.48 (m2 )Chn chiu cao cht ti lnh 3m v chiu cao bung lnh l 5m

3.Ti trng ca nn v trn: gF> gv * h => gF> 0.45*3 =1.35 t/m24.Din tch lnh cn xy dng: F1= F/F = 81.48/0.725 = 112.386 (m2)Vi F=0.725 (tra bng 2-5 trang 34)Din tch bung lnh quy chun (bi ca 36 m2) nn chn F1= 108 m2 (912)Chn kch thc kho nh sau: chiu di 12m, rng 9m, cao 5mII. Tnh chn vch, tnh ng sng , ng m: Tnh chiu dy cch nhit: cn= cn( Trong :Chn vt liu cch nhit l polystitol cn=0,047 W/mK.(bng 3.1). K: h s truyn nhit.Vi nhit bung lnh = -25oC ,tra bng 3-3 trang 84 ta c k = 0.21W/m2Ka1: h s to nhit ca mi trng ngoi ti tng cch nhit,W/m2K.a1= 23,3 W/m2K.(bng 3.7).a2: h s to nhit ca vch bung lnh vo bung lnh,a2= 10.5 W/m2K.(bng 3.7). Chn vch:

Lp vt liu xy dng,m ,W/m2K (g/mhMPa)

Lp va xi mng 0.020.890

Lp gch 0.380.82105

Lp cch m bitum0.00260.180.86

Cch nhit polystirol ???0.0477.5

Chiu dy cch nhit l:

cn= cn(

= 0,047(= 0,191 (m).Ta chn chiu dy l : 0.2 m H s truyn nhit thc t :

K== = 0.202 W/m2K Kim tra ng sng :Nhit mi trng tmtat = 37.3oC, m 74% =>ts = 32oC v t =33oC (ma h ti tp.HCM tra bng 1-1)Nhit ca bung lnh tb=-250C.Ta c Ks= 0,95* 1(tmt-ts)/(tmt-tb) = 0,95 x 23,3 x (37.3-32)/(37.3-(-25)) = 1.88 W/m2KKs=1.88 > K=0.202 W/m2K cho nn vch ngoi khng b ng sng. Kim tra ng m.H s dn nhit v m ca polystirol l : =0,15m ;=0,047W/mK;=7,5g/mhMPa

Mt dng nhit qua kt cu cch nhit :

q= k .t= 0.202.(37.3-(-25)) = 12.615 W/m2.Xc nh nhit b mt ti cc lp vch:

q = i .tf =tw t1 = tf1 - q/1 =37.3 - 12.615/23.3 =36.76

t2= t1 -q= 36.76 oC- (12.615* )=36.47oC t3= 30.63 oC t4= 30.33 oC t5= 30.1 oC t6= -23.53oC t7= -23.81oC +tbl= t7-= -25 oCTra bng c cc p sut nh bng sau:VchNhit ( oC)p sut P''x (Pa)

136.758687976174

236.472038656077

330.627189034397

430.34053974327

530.158358134174

6-23.5121535192

7-23.7988028389

Dng hi thm thu qua kt cu:

Ph1: phn p sut thc ca hi nc bn ngoiPh2: phn p sut thc ca hi nc bn trongTa c: T1=37.3oC ; 1 = 74% ; Px (T1)= 6358 Pa Ph1= Px (T1). 1 = 6358*0,74 = 4704.92PaT2 = -25oC ; 2 = 100% ; Px (T2) = 80 Pa Ph2= Px(T2) . 2 = 80*1=80 PaH s tr khng thm hi ca kt cu:

Phn p sut thc ca hi nc trn b mt cc lp:

p sut thc ca hi nc u nh hn phn p sut hi nc bo ha nn khng c hin tng ng m trong c cu cch nhit. Chn nn: chn nn c cc lp nh sau: Nn nhn bng cc tm b tng lt 1 = 40 mm ; 1 =1.4 w/mK Lp b tng 2 =100 mm ; 2=1.4 w/mK Lp cch nhit bng t st xp , si: 3 =??? ; 3=0.2 w/mK Lp b tng c si in : mm Lp cch m Lp b tng dm lm kn nn t

Tra bng 3-6: k = 0.21 w /m2KTra bng 3-7: 1 =23.3 w /m2K2 = 10.5 w /m2KChiu dy lp cch nhit l:

cn = 0,2(=0.904 (m)Chn 4 = 1 (m).H s truyn nhit thc t:

K==

kt= 0.2 w/m2k Chn trn:

cn= lcn(Tra bng dng lp cch nhit t st,si :+ cn = 0,2 w/mK : (bng 3.1 P 61 [1]);+ K : h s truyn nhit ( bng 3.3 p 63 ,[1]) K = 0,2 w /m2K+ 1 h s ta nhit ca mi trng bn ngoi ti trng cch nhit 1 =23,3 w/m2K + 2 h s ta nhit ca vch bung lnh vo bung lnh 2 = 10,5 w/m2K + i : chiu dy cc lp xy dng th i (m) + i : h s dn nhit ca lp vt liu xy dng th i w/m2K Lp ph ng thi l lp cch m bng vt liu xy dng v borulin 1= 12 mm ; 1 =0,3 w/mK Lp b tng ging c ct 2 =40 mm ;2=1,4 w/mK Lp cch nhit chiu dy 3=? ; 3=0,2 w/mK Tm cch nhit bng xp polystrirol 4 =50 mm ;4=0,047 w/mK Lp b tng ct thp chu lc 5 =220 mm ;5=1,5 w/mK Chiu dy cch nhit ca trn:

cn= cn(

= 0,2( =0.72 (m)Chn dcn =0.75 (m).H s truyn nhit thc t:

K== K=0.194 w/m2kIII. Tnh nhit kho lnh:1. Tnh nhit tht thot qua vch:Bao cheK (w/m2k)F (m2)T (k)Q1 =K*F*T (w)

Tng ngoi0.2026062.3755.076

Tng ngoi0.2024562.3566.307

Tng ngoi0.2026062.3755.076

Tng ngoi0.2024562.3566.307

Nn0.210862.31345.68

Trn0.19410862.31305.3096

Tng Q15293.7556

2. Tnh dng nhit do sn phm to ra:

a. Dng nhit do tm ta ra:Q21 = M* (h1 h2 )* 1000/(24*3600) (kw)M: nng sut bung bo qun lnh ng (t/24h)Q21 :Dng nhit do tm ta rah1 ,h2 : enthapi ca sn phm trc v sau khi x l lnh :Theo bng 4-2 p.81 ,[1] Chn nhit hng nhp thng vo kho bo qun lnh ng l:t1 = -8 oC h1 = 43.5 kJ/kgChn nhit hng sau khi c lm lnh l:t2 =-25 oC h2 = 0 kJ/kgM = 8%E = 0.8*100 = 8 ( t/24h)Vi: M: khi lng hng nhp vo vo qun lnh ngE: dung tch phng bo qun lnh ngVy : Q21 = M*(h1 h2)*1000/(24*3600) = 8*(43.5 0 )*1000/(24*3600) = 4.0278 (kW)b. Dng nhit do bao b ta ra:

Q21 = Vi: Mb : khi lng bao b a vo cng sn phm (t/24h)Cb : Nhit dung ring ca bao b (kJ/kgK)t1, t2 : nhit bao b trc v sau khi bo qun lnh ng1000/(24*3600) : h s chuyn i t/24h ra kg/sTa c :Khi lng bao b cac tng : Mb = 30%M = 30% *8 =2.4 (t/24h)Nhit dung ring ca bao b cc tng : Cb = 1.46 (kJ/kgK)Nhit bao b trc khi bo qun: t1= -8 oCNhit bao b sau khi bo qun : t2 = -25oCQ22 = 2.4*1.46*(-8 + 25 )*1000/ (24*3600) = 0.689 (kW)TngNhit do sn phm ta ra l :Q2 = Q21 + Q22 = 4.0278 + 0.689 = 4.7168 (kW)3. Dng nhit do vn hnh kho:a. Dng nhit do n chiu sng:Q31 = A.F A: nh mc chiu sng trn mt m2 phng, A = 1.2 W/m2F: din tch phng lnh, F= 108 m2Q31= 1.2*108= 129.6 Wb. Dng nhit do ngi ta raQ32= 350.n350: nhit lng do ngi ta ra khi lm vic nngn: s ngi lm vic trong phng, chn n=3Q42= 350*3 = 1050 Wc. Dng nhit do ng c inQ33= 1000.NN: cng sut ng c in , N = 816 Q33= 1000*16= 16000 Wd. Dng nhit khi m caQ34 = B.FB: dng nhit do tn tht kho lnh m ca cho 1 m2 phng lnh, B= 12 F: din tch phng lnh, F = 108 m2Q34= 12*108 = 1296 W Vy dng nhit vn hnhQ3 = 129.6 +1050 +16000 +1296 =18475.6 (W) =18.4756 (kW)4. Nhit ti ca thit b:Qtb = Q = 5.2937556 + 4.7168 + 18.4756 = 28.48 (kW)Xc nh ti nhit cho my nn:

Ta c QMN l tng nhit ti ca my nn i vi 1 nhit d bay hi QMN= Q1+ Q2*60%+Q3*75%= 5.2937556+ 4.7168*0.6+ 18.4756*0.75 = 21.98 (kW)Chn k=1.07, chn b=0.7 i vi thit b lnh nh nng sut lnh ca my nnQ0 = = = 33598(W)

IV. TNH CHU TRNH LNHChn mi cht lm lm lnh l: NH3Tnh ton thng s lm vic ca my lnh1. Nhit si ca mi cht lnh:

t0 = tb - t0 = - 25 - 10 = - 35 0C 2. Nhit ngng t tk :tk =tw2 + Tk tk : nhit ngng t (oC)tw2 : nhit nc ra khi bnh ngngTk = 5 ktw2 = tw1 +tw tw1 : nhit nc vo bnh ngngtw = 5 k tw1 = t+ 5(oC) t : nhit ban u vi t1 =38 oC =74% t =33oC tw2 = tw1 + 5 oC = 38oC + 5oC =43oCtk =tw2 + 5 oC = 43oC + 5oC = 48oC3. Nhit qu lnhtql = tw1 + tqltql = (3 5 oC)tql = 38+ 5 =43 oC4. Nhit qu nhit ( hi ht)tqn = t0 + (5 15 oC) = -35 + 10 = - 25 oC p sut ngng t pk ca NH3 nhit tk = 48oC l pk = 1,93 (Mpa)p sut bay hi ca NH3 nhit t0 = -35oC l p0 = 0.093 (Mpa)T s nn : = = = 20.75My nn lnh 1 cp :

+ Nhit si ca mi cht lnh (to): - 350C+ Nhit ngng t (tk): 480C+ Nhit qu nhit (th):- 250C+ Nhit qu lnh (tql): 430C

Theo gin R717 ta c:Nhit bay hi l: -35 oC H1 = 1435 kJ/kgv1 = 1.2 (m3/kg)Nhit ngng t l: 48 oC H2 = 1940 kJ/kgH3 = H4 =400 kJ/kgLu lng dng mi cht lnh (kg/s):m= Qv : cng sut lnh = Nhit ti lnh Qe = 28.48( kW) = 28.48 (kJ/s)

m= = 0,0275 (kg/s)Th tch ht thc t .Vtt =m *v1 = 0,0275* 1.2 = 0,033 ( m3/s)Th tch ht l thuyt Vlt = Vtt / : h s cp ca my nn ph thuc vo t s nnVi =6,48 tra th (H 7.4 p 215[1]) = 0.65Vlt =Vtt / =0,033 /0.65 =0,0508 (m3/s)

Chn my nn AYY90 theoOCT 6492-76 do Nga sn xut c th tch ht l thuyt l: Vlt = 0.0716 (m3/s)S lng my nn : Z = = = 0,71 myVy cng sut my nn l thuyt :Qcomp=Ns = m * (H2 - H1 ) = 0,0275* (1940 1435) = 13,888 kW

Hiu sut ch th;

- Cng sut ch th (cng sut nn thc t) Ni = kW - Cng sut ma st Nmas=VttPms chn Pms=0.06 Mpa; = 0.033*0.06*1000=1,98 kW; - Cng sut hu ch Ne= Ni+Nmas=19,67+1,98 = 21,65 kW;

- Cng sut tip in Nel =chn

Nel = = 25,32 kW; - Cng sut ng c lp t NdcChn h s an ton l 1,7

Ndc= 1,7*Nel = 1,7 *25,32= 43,1 kW

V. THIT B NGNG TTnh v chn thit b ngng tChn thit b ngng t nm ngang c v, bnh ngng gm mt v hnh tr bn trong c b tr mt chm ng, hai u c hai mt sng. Hai pha c hai np.Hi amoniac trong khng gian gia cc ng ngng t trn b cc ng chm. Nc vo theo ng ng b tr trn mt np, i pha trong cc ng chm theo cc li b tr sn ri ra theo ng pha trn

Tnh thit b ngng tNhit thi ngng t = Cng sut qu trnh ngng t: QkQk = m*(H2 H3) = 0,033*( 1940 - 400 ) = 50,82 kWDin tch b mt trao i nhit:

F = ( m2)Trong :k: h s truyn nhit i vi thit b. i vi bnh ngng nm ngang NH3 k= 7001000Chn k = 900(W/m2K)ttb: hiu nhit trung bnh ca qu trnh ngng t

ttb =

tmax: hiu nhit ln nht ( pha nc vo )tmin : hiu nhit b nht ( pha nc ra )tmax = tk tw1 = 48 38 = 10 Ktmin = tk tw2 = 48 43 = 5 Kttb = 7,21K

F= = 7,83(m2)Tnh lu lng nc qua thit b ngng t:

Vn = (m3/s)C nhit dung ring ca nc C= 4,19 kJ/kgK - khi lng ring ca nc =1000 kg/m3tw tng nhit trong thit b ngng ttw = 48 38 = 5 oC

Vn = = 0,0024 (m3/s)= 8,64 (m3/h)Chn thit b ngng t:Cc thng s ca thit b ngng t:K hiuF(m2)K D (mm)Di L (mm)Rng B (mm)Cao H (mm)S ngKhi lng (kg)S lng

KT-1094081880535760995551

VI. THIT B BAY HIChn thit b bay hi.Chn bnh bay hi lm lnh cht ti lnh lng (nc, nc mui ) cn gi l h thng lnh gin tip. Tnh thit b bay hi:Nhit ti ca thit b bay hi Q0= Cng sut lnh Qv = 24,48 kWDin tch b mt trao i nhit

F = ttb: hiu nhit trung bnh logarit gia cht ti lnh (nc mui) v mi cht lnh si

ttb = =

tn1 v tn2 nhit nc mui vo v ra khi bnh bay hiNhit si ca mi cht lnh NH3 la -35oCttb = 40CChn bnh bay hi ng v amoniacTheo bng 8-7, chn k= 500W/m2KDin tch b mt trao i nhit:

F == 14,24 m2Theo bng 8-8 p283[1] chn bnh bay hi ng v nm ngang KT-40 vi din tch b mt trao i nhit 40,7 m2Cc thng s ca thit b bay hiK hiuF(m2)K D (mm)Di L (mm)Rng B (mm)Cao H (mm)S ngTh tch khng gian gia cc ng, m3

KT-4040,7600 x 83580107515902160,52

Lu lng nc mui tun hon:

Vn = Cn : nhit dung ring nc mui, kJ/kgKn : khi lng ring nc mui, kg/m3tn hiu nhit nc mui vo v ra khi TBBH, Ktn = tn1 - tn2 = 2 K

Vn =

= =0, 004m3/s = 14,4 m3/h