q\Y1 t/ -/- t)'t< =' Expanslon Deu{cps @l Thus the bulb always cont:ols the valve operation even with a colder diaphragm or bellow. The charging of the refrigerant itself in tlre power element results in an increase in operating superheat as the evaporator temperature decreas6s. This limits its use for moderately high evaporator temperatures. For evaporator temperatures which are substantially below OoC,liquid cross-charged expansion valves may be used. They have superheat characteristics which remain fairly constant throughout the evaporator temperature range. CAPILII\RY TUBE AITD ITS SIZING The capillary tube is a fixed restriction-type device. It is a long and narrow tube connecting the condenser directly to the evaporator. Tlre pressure drop through the capillary tube is due tg the following two factors: (i) Friction, due to fluidJireosila, psu,ttjq.g infrictional pressure drop. (ii1 Acceleration, due to the flashing of the liquid refrigerant into vapour, resulting in momentum pressure drop. The cumulative piessure drop must be equal to the difference in pressure at the two ends of the rube. The mass flow through the capillai'y tube will, therefore, adjust so that the pressure drop through the tube just equals the difference in pressure between the condenser and the evaporator. For a given state of the refrigerant, the pressure drop is directly proportional to the length and inversely proportional to the bore diameter of the tube. A number of combinations cf length and bore are possible for a capillary tube to obtain the desired flow and pressure drop. However, once acapillary tube has been selected, it will be suitable only for the design pressure drop and flow. It cannot satisfy the flow requirements with changing condenser and evaporator pressures. Even then, the capillary tube is the most commonly used expansion device in small refri !gi, such as desrestic refrigerators, window-1ype alr iti6nE6, water coolers, etc. The advantages of a capillary tube are its simplicity, low cost and the absence of any moving pafts. Also, it is found most advantageous with on-off control because of its unloading characteristics. Thus, when the compressor stops, it allows high and low pressures to equalize, thereby enabling the compressor motor to re-start on no load. Accordingly, smaller lowstarting torque motors can be used. _,-->Ttre sizing of a capillary tube irnplies the selection of bore and length to provide the desired flow for the design condenser and evaporator pressures.The method employed by manufac$rers is usually that of cut and lry. The principle of design based on methods proposed by Stocker,6 and Hopkinsa and Cooper er al.z is presented here. A capillary of a particular bore dia D and area A is first selected. Step decrements in pressure are then assumed and the colresponding required increments of length calculated. These increments can be totalled to give the complete length of the tubing for a given pressure drop. @ Refiigeruflon ad. A{r tundttt'r'qg Consider that the state of the entering refrigerant rs saarrglg_d_Iqgg.The mass flow rate m is known. The condenser and evaporator ternperatures arc t*and lo, ana.o@-gp.''ssuresafePtcandporesiectively.Dividethistemieratus drop into a number of parts. Let the colresponding pressure drops be Ap1, Ap, and A p3, ..., etc., as shown in Fig. 8.1 1. Now there are {wo approaches to design. ti) Isenthalpic expansiono as shown by line &-a. {ii) Adiabatic or Fanno-line expansion, as strown by line &-b. h ----+ Fig. 8.11 Incremental Pressure Drops tn a Captllary Tube IjenthelpiSg[pg!$rg[-LslhggoJ1n[gq-tb"9J"g"r*o,-d-y.I.gl4ig 4q9umpren. [n actual practice, however, expansion takes place adiabatically, viz., according to Fanno:line flow. Thus enthalpy does not remain constant sinbe, with pressure drop, the volurne increases and"an increase in kinetic energy is obtained from a decrease in enthalpy. Nevertheless, it may be noted tiom Fig. 8.11 that in the first few steps of pressure drop, there is not much difference between isenthalpic and Fanno:line flow. The steps of calculations to be followed in both cases are the same and are as follows for the first element. (i) Determine the quality at the end of the decrement assuming isenthalpic flow. Then at point I at Pressurepl ho - hr, Y.=r.'(8.1) hrr, -'| Determine the specific volume Erpansio n Deutres |ffi| (8.2) t)r=o!r+ xr(arr-aI,) lnlt, (iii) Calculate the velocities frorn the continuity equation at both the ends of the : -. element . u*- irc1 A and n, = ,A ---{ (8.3t or wher. c tt = Y- = G =Constant (iv) ir'*urr1rto.i,y ,-':,YF*'I$ 'i q. For Fanno-line flow, an iteration procedure is necessary. This is done by applying the correction to enthalpy since hr+ h*. Thus '.^(8.+; The calculations for quality, spe cific volume, vclocity and enthalpy may be repeated until the final value of /r, is equal to its value in the preceding iteration. /Uote Ap: fost. The procedure oppeors time'consuming bul, fortunately, il convergds veA (v) Determine the pressure drop due to the acceleration, Apo. frorn the momentum equation Adp = whencc - fudu ,orri1) = S1tt1 Apt= 'A 4 - ur) (8 5) (vi.r Determine the pressure drop due to the friction'Lpp, fiorn a'Pr= LP - .LPe (8.6t lvii) Equate the required frictional pressure drop to Lpr where p = f LLu2 (8.7) 2D I 7' p AL = length of the element Substituting .it it =.puA, pu = G, we'have ; = ^Gf Apr (8.8) (8.11 =#frLL=YfuLL(say) where Y= * (8.10) ,:4- @ Refitgeratton at'd, Air cffiaffionng j ia # the section. The friction factor is a turn ls expressecl as nction ofReynolds number which in il ii I Dup_DG_Z Re=ff=;--;(sav) { (8.1 l) Z = DG (8.12) Niaz and Daviss haue proposed the following correlation for evaluating the friction factbr: where - -,' -t=;;m 0.324 They mention that the length ohained by using this correlation is about l0Zo greater than the experimental length. Based on available analytical and experimental data, lbe expreg$eq for-higtion f I terms_of a straight capillarv 9 siven bv Gorasi a et aI' is: . t _ -__t7.24 and /r = tr= where ^ t7.24 IEtr'' Fy I i.i I l. ll Eo=l Fr : r Rep= GD 'and Re.= g i I I " where subscripts/and g refer to liquid and gas phases respectively. The friction faefir for the liquid-vapour mixture flowing in the capillary is found by taking into account the percentage weightage of each phase. Thus f=ft(l -x) +fr, The examples that follow illustrate ttre method of calculation as it can be applied for the design of capillari.es for air conditionels and retrigerators. !r I I etcample 8.2 Design of I TR R 22 Air Conditioner Capillary A capillary tube in a one-ton R 22 aft conditioner has a bore of 2.3 mm. Saturated liquid from the condenser enters at a temperature of 48'C and flows adiabatically through the tube until its temperature is 5oC. Determine its length. The friction factor is given by /^= 1;or 0.32 Assume intermediate sections at 40, 30, 20 and l0oC Solution Frsrn the sirnple saturat;on cycle, the mass flow rate is tit = A.02417 kgls The cross-sectional area of the capillary tube is A = !t- (o.oo23)? = 4.15 x lOa 4 7t m2 Expottston Deutces @l I*t (J=-= ^m A u 0,u2417 4.15 x 10{ = 5.83 x 103 kgs-t 1r-z a y 2(0.0023) _ Z= DG = (0.0023) (5.83 = *= 2D t='l: Itg: = t.z674x 106 kgs-r p-r x 103) = 13.41 kgs-r 6-r t rt In actual practice, the enthalpy does not remain constant in flow through a capillary.Assuming isenthalpic flow, however, the properties and velocities at vari" o'us sections are found and are given in Table 8.1. The last column in the table gives the actual decrease in enthalpy due to increase in kinetic energy, e.g., at the first point A,t = ht -ht = - A(KE) = [0.492 - 5322 = 4l I/kB = 0.041 kJ/kg Table 8.1 calculauons for Example 8.2-Isenthalpic Flow but Ah = _ A(I(E) ion 3ng Section l03t 48 40 30 20 18.548 15.335 u=Ga 10.49 19.63 33.18 53.33 67.A3 Ah k I 2 3 4 q l0 5 1r.819 9.099 6.807 5.84 0 0.07 0.144 0.209 0.267 o.292 0.9137 l.32 3.375 11.572 1.804 0 0.04r 9.167 5.7A4 0.t79 0.536 r.408 2.232 rs adi h. Calculations can nglv.be done for actual Fanno-line flow, starting from .h*= zffi.314 kJlkg. Point I At 40oC, from the table of properties for R2Z hr=2497 kJ/kg 0.0884 x 10-3 m3lkg "/= ht, = 166'9 kYkg ug = 15.1 x 10-3 m3/kg hr = h*- Lh =26A314 260.273 - 'tl - - 166.9 0.041 =26A.273 kJ/kg 24s.i = 0.0633 "r = 0.884 x 10-3 + 0.0633 (15.1 = l.Z8 x l0-3 m3lkg - 0.884)10-3 @ Refiigeration and. Att &ndtfioning ut = Got= 5.83 x 103 x l.?8 x 10-3 = 10.38 m/s Recheck, 8\ lAbfu h: 10.3g2 = - 5322 = 40 Jlkg = 0.04 kJlkg Furlher iterotion con be dona for grealer eccuracf. Proceeding in tlris monne4 ileroled volues ore obloined ol various poinls which ore given in'foble 8.2. Table 8.2 Calculations (h)iterated for Example 8.2-Fanno-line Flow Id xfu) itersted (u)iterated 0.91 1.8 (Ah) iterarcd kJ/kg 0.0 0.04 0.18 0.52 1.34 U k I 2 3 0.0 0.064 zffi.5 260.46 260.324 259.977 5.317 10..474 0.r33 0.191 4 5 0.244 0.261 259.t63 258.424 3.3s6 5.634 8.937 I l9:525 32.773 51.99 64.6i55 l.l .j 14 2.08 Friction Fac to r Calculatiorts : Point t (40"C) rlt '! Viscosities, l*= 0.221 cP /s = 0'0134 cP = I r=(l-rl)ft+xr1tr (l 0.064)0.221 + (0.064) (0.0134) = 0.2071 cP 13.4t Re,= *-- 0.2071x l0-3 0.32 = 64,?80 f' = Tab[e - *r*pr It1 =0'02 The calculations for various points are given in Table 8.3. 8.3 Friction Factor Calculations for Example 8.2 ys cP k I 2 3 48 40 30 20 4 5 t0 5 0 0.064 0. r33 0.191 0.240 0.261 0.215. 0.01368 0.22r o.229 0.239 o.25 o.257 0.0r34 0.0131 0.0127 0.0124 0.0123 0.215 0.2071 9,780 62,400 o.tg7g 0.1917 0.1934 0.1973 66,800 68,400 68,400 69.300 0.020 0.02 0.02 0.0197 0.0197 0.0198 I*ngth Calculatiorts: Consider section ft-1. Total pressure drop Erpansion Deuices @l Am= l8.if3 Acceleration Pressure droP 15.331 =i.212 bar 103 N/rn2 Lpe= G au= (5.83 x 101 (00.474- 5.317) = 32.8 x Friction Pressure droP APr = A'P - APe= 3.212 x ro5-0.328xr05 = 2.884 x lOs N/m2 Mean friction factor -Mean velocitY o.o20z+0.02 \=_ -, 5.317 = 0.0201 u= +t4.474 = 7.82 m/s -= 1.389 m '" ' -r_ Illtt/tgvuwr lvrrlour ' The calculations for various secdond ari given in Table 8.4. {"-1r.; i;;, 'Table Sections 8.4 Caplllary Tube kngth CalculaUons for Exampie S.2 Lpr AL Lh Ap bar 3.2t2: 3.416 2.818 2.2y2 0.969 bar 0.328 a.574 0.857 1.283 0.844 bar 2.884 2.8A2 1.961 k-1 t-2 2-3 34 4-5 1.009 o.t?5 1.389 0.706 0.276 0.087 0.008 Total leogth rcquired = EAL =2.466 m Andersen, S A, Automatic Refrigeration, Maclaren and Sons Ltd. for Danfoss, Nordborg, Denmark, 1959. 2. Cooper L, C K Chu and W R Brisken, "Simple selection method for capillaries derived from physical flow conditions" Refrigerating Engineering, Vol.65. 3. Gorasia, J N, N Dubey and K K, ,Jain, "Computer-aided design of capillaries of different configurations, ASHP,I^E Transaction 1991, pp. 132-138. l. -:-:#t' ReJrigeratton i +. and Arr &ndttirrtirtg Iiopkins N E, 'Rating tha resuictor iubeo, Refrigerating Engineering, Vo!. 58, No. 11, Nov. 1950, p. 1087. 5. Niaz, R H, and G, Davis, 'Adiabatic two-phase flow in a capillary tube', Synzp. Ser. of Cen. Soc. for Chent. Eng., Vol. i, 1969, pp.2>9-269. 6. Stoecker, W F, Refrigeration and Air Canditioiiirzg, McGraw-Hili, New York 1958, pp. lll-t?g. 8.1 (a) Calculate the pressure of R 12 as it expands through a capillary tube having a bore of'l .05 mm if 4.028 x i 0-2 kg/s of saturated liquid enters the capillary tube at a temperature of 45'C. Neglect heat transfer. (b) Determine the temperature at which the condition of choked flow 8.2 occurs as the expansion proceeds. An R 12 thermostatic expansion valve uses R l2 itself as the power fluid set to fade out at a pressure of 0-362 MN/m2. The internal volume of the power element 1, 15 cm3. Cal*late the mass of Rl2 contained in the element. ::rion valve is factory set for a superheat of 7oC when (a) An R22 supplying refrigc:ant to an evaporator at 5oC. If the evaporator is operating at - 15oC, what will be the effective superheat of the suction vapour? (b) If the valve is cross-charged with R 12 as the power fiuid, what will be the effective superheat? An R 22 thermostatic expansion valve with R 22 itself as the power fluid is not equipped with an external qualizer. It supplies a coil in which there is pressure drop due to friction, The superheat setting made on the valve at the factory is 5'C with OoC evaporator. (i) What is the difference in pressure on opposite sides of the valve required to open the valve? (ii) When the pressure at evaporatorinletis 4.22 bar, how many degrees is rhe suction gas superheatecl at I 1.5'C evaporator exit temperature if the pressure drop through th" coil is 0.53 bar? How much would have been the snperheat in Prob. 8.4 (ii) if the valve had been equipped with an external equalizer? 8.3 e, 8.4 8.5 8.6 Design a capillary for a 165L (Qo = 89 W) refrigerator working on (a) R12 tb) R134a The Rl2 compressor is 4.33 CC. The Rl34a eompressor i-< 5.48 CC. Use l;g. 3.21 for cYcle 'r'':r:\',iis.