Document QkRaVQd3d3me99Bp8Yw1yJzx8

94 CHAPTER 6 1965 Guide And Data Book . " L * V M,/ V* . / J ; The kinematic viscosity of the vapor ia (36) , _ (V 2.48 x 10-1 *' " p, " 0.0402 - [7' 0 (7 - 0 r, * 6.16 X I0"* sq ft/sec. . Then, the associated Reynolds number ia (38) ' W*), VtD (14.7)(0.Q455)" ;6.16 X 107* 2--K(t-0]^ (39) }; : It is seen that the determination ofAp/r involves a numerical , ' Thus, the\vapor_is definitely flowing turbiilently since the iqu^acttuuaaUyyy- prescnt en.u.r.e_s__th__a_t_t_h_e_v_arporReynolds number evaluation of the integral in Equation 35 in which the into- ( ,, -- than tne.value calculated above.-*" grand factor 6i* is determined from Fig. 6. Furthermore, if the pressure drop is appreciable (say as'much as 10 percent of the absolute pressure) an iterative procedure is also required- to establish the pressure at each step of the numerical integration. 1 : In a similar wsy,_the'Reynolds number for the Squid'fraction st coil inlet, assuming'ho vapor present,-is found to be (N*,), ! + ,1*45.X.104 which assures that the liquid is flowing turbulently. -`-The dissipative flow model is, therefore, turbulentlicmid-turbu;lent-vapor'(W):J Although la rigorous'solution to the problem involves!the y The Tnckhart-Martinelli parameter to be used is therefore complicated procedure discussed above,:it should be remem-i, bered that extreme precision is unwarranted due to the many.' - (1- assumptions that have been made. In most cases, it is only ' an approximation of.the pressure drop which is required; This ; where can be obtained -very simply by approximating the integral - of Equation 35,by j " : - ~ 1 . : f %,'(1 - x)"dx (40) X, -- XiJXi . where is evaluated from Fig. 6 at the value Xw For turbulent-turbulent flow - . -. The average value to be used for the fluid properties is not known at this stage because the pressure drop is not yet known. For a first approximation, the fluid properties will be evaluated at mlet temperature (assumed to be 40 F). where > r/I.295W fl.236 \ / 1 y IA l) J' 86.4 / VQ.0119/ V 0.618 x, + Xi : , <?> A comparison of the more rigorous solution'arid the approxi mate solution will be given for a specific case in Example /. Example I. Calculate the two-phase pressure drop across a coil which is 11 tubes wide; 66 in. long 4 rows deep, fin. OD tubing, and has a 22-ton capacity using Refrigerant 12. Suction tern- peretur* is assumed as 40 F, with condensing temperature 105 F. The loading b 2 tons per circuit, and, since the standard vapor compression cycle b being assumed, the flow rate b 3.9 lb /(nun) (ton). Thus, the total flow rate, to, in the circuit b 7.8 lb/min, or 4.04 X 10"* slug/sec. - -- Solution: The quality entering the coil b found to be *< -- 0.235, so that the liquid fraction of the flow rate is w, -- 5.95 lb/min, and the .vapor fraction b to, -- 1.85.,lb/min.. It will, be assumed that the fluid leaving the coil'b saturated vapor, Le., x, 1X10. At 40 F, the viscosity of the liquid and vapor are: X n - 0.286 centipoise - S.95 X 10"* shig/(ft)(sec). Pt - 0.0119 oentipoise * 2.48 X lCT* dug/(ft) (sec). The densities are: Pt * 86.4 Ib/cu ft * 2.68 slug/cu ft. ' Pt TM 1.295 Ib/cu ft TM 0.0402 slug/cu ft. (X.-*).,, - 0.109' ' " From Fig. 6, it b found that ;-(*).., - is W)..,(l - x.,,)1* * (18)*(1 - 0.618)` - 57.4 ` ,, The liquid pressure drop Cp, b calculated next. Thb b given by Equation 31. The Velocity b given by: _ Ji. 7.8 ' V* " p/A " (86.4) (0.00162) (60) " 0.93 ft/sec ''TThe Reynolds number (7/a,), b: . _ VU> (0.93) (0.0455) f **'* " : -/ 2,22 X 10-* " (Nmm).,- 1.9 X 10* - c. From Fig. 5 (assuming a smooth tube) the friction factor /, b found to be: ' /. = 0.026 Neglecting the bends, the length of one circuit of the coil b: The inside diameter of f in. OD tubing b 0.0455 ft, and the inside area b 0.00162 eq ft. .The velocity of the.vapor at the coil inlet (assuming no liquid present) would be . , . -1.85 ': p,A. 1.295{.00162)(60) V, 14.7 ft/sec Substituting these various values into Equation 31: . ** - d o26(&) ("2) 13.4s)(0.93)* Ap. - 14.6 lb/sq ft Ap, ' 0.101 psi Fluid flow) Th, frictional component of the taro-phase pressure drop is tons found to be* -- np/r * Ap,(4i')(I " **#)** -:(0.101)(57.4) v Apfr`-5.8psi The deceleration component is given by Equation 28. Assumine ,, TfimtTpprorimatkin that ,,. - m "d s,, - sn, tt is found that: \ .. 16(4.04'X 10"*)' rl.00 - 0.235 _ 1.00 -- 0.235?] Ap*-TM' T*(,0455)4 rL 0.0402 *" 2.68' Ap. --100 lb/sq ft Ap,*0.7Pf; The total two-phase pressure drop is thus found to be: Aprr " Apt + Apm . 5.8 +.0.7 ,. - Aprr - 6:5 psi .. Thb value could now be improved by reworking the problem using properties evalifated for the average evaporator tempern-. tore" which b about 36 F. If tins m done, it a found that a more^accurate numerical integration of Equation 35 b Carried out rather than using the average value of Equation 40, it b found that Aprs ** 5.2 psi. ,. - Tbb gives an idea of the order of magnitude of dinerencea with different computational methods. It also may be of interest to note that the manufacturer of thb particular cod states that the pressure drop is 5.6 psi. . * : t-' DYNAMIC PRESSURE LOSSES As was discussed previously; total pressure,"P,'is equal to the sum of the static pressure and kinetic energy head (velocity pressure). If there are no losses in a system, the total pressure remains constant, but if losses exist, the total pressure decreases in the direction of flow. The loss of total pressure due to wall friction in ,uniform pipes baa been discussed previously. In addition to the fric tional losses, losses can be imposed on the flow, due to;separ ation of the boundary layer and the subsequent dissipation into turbulence. ------ 95 Dynamic Losses ~ " ." Dynamic now losses represent a coayei^on of kinetic energy of a vector g*>n<*> parallel to. the direction of flow to kinetic energy of a random-vector apnss or turbulence. Whenever-any factor in a flow system changes the magnitude or direction of the fluid kinetic energy,'turbulence occurs with an accompany ing in total pressure. Flow' losses, in general, are propor tional to the kinetic energy of.the fluid for any given.system configuration. Because flow losses are proportional to the kinetic energy of the fluid it-is convenient to express them non-dimensionally in terins.of toe velocity pressure or dynamic head existing at some'point, in toe system; usually immediately upstream of the section under consideration. Thb non-dimensional ex pression, commonly referred to as.a loss coefficient, b-defined as: % ' , ' -1 . 7 : "cl ' ' - P* (> trAerc Cl loss coefficient.' ' -AP -- pressure loss. V* velocity pressure. The units in Equation 43 should be consistent for all terms;-wi *. - - If the losses are expressed as loss coefficients based on.local velocities; they are nearly independent of scale, .velocity, and fluid properties so long as the Reynolds number b well into the turbulent region, no changein toe mode of flow occurs, and, for compressible fluids, the Mach number b less than 0.3. Loss coefficients for duct elbowsiand duct area changes are given in Tables 3 and 4 of Chapter 31. Additional loss coeffi cient data for elbows, bends, area changes,-diffusers, turning vanes, branch cnnnwt.inni,- madifolds7 valves, and screens are' given in Section 1, of Reference 12. .... . , Pressure loss data for ducts and duct fittings and various types of pipe and pipe fittings, expressed as equivalent lengths rather as loss coefficients, are included in other chapters of the 1964 and 1965 volumes of the Guide.Ano Data Book as shown in Table 5. Table 5 .... Location of Pressure Loss Data Jn the 1964 . and 1965-GUIDE AND DATA BOOK. , '3 Location Chapter Pages; Ducts and duct fittings Duct fittinss and intakes Registers Pipe fittings 1965 1964 1964 - 1964 1964 1964 31 562-569 7 78r79 7 85-86 -, 8 97 , 10 - -139-140 - '76 ' 849 - - Pipe friction losses, copper and iron 1964 10 136-137 Pipe friction losses, oommer- 1964 Pipe and tubing friction losses 1964 valves and fittings friction losses - - ' ' 1964 : ' 10 138 82 - 921 82' i922-: