Complete Text (Part 2)

and we have from the mechanical theory of heat; —“ Q,=A(W, —Wnn) + The theoretical performance of the machine it is, calling it 1, or __ Q- T,—T, W, —Wn Any =n us —T,) w= > rT we get finally iA a a result already found in § 3 by suppos- 28 If T,>T, the useful effect -~ The efficiency of the machine will be all the greater as T, approaches in value to T,; that is to say as it is urged ata lower pressure into the reservoir. But as we lower the pressure of working, the quantity of negative heat produced dim- inishes also and becomes nothing when Ty=T,. The necessary driving power W;,—Wm which we proceed to calculate, should be augmented by the passive resistances. If we consider the refrigerating ma- chine as composed of two distinct ma- chines driven by the same shaft, we are led to consider that the work of the pass- ive resistances is proportional not to the final work W,—W» but rather to the sum of the work developed in the two cylinders W,+Wm. Considering the simplicity of the machine, the small amount of friction, and the absence of a stuffing box, we can admit that the work of the passive resistances should not ex- ceed eight per cent of the above total work. 29 The resistance of the machine is then 1.08W,—0.92Wm. The following table gives the amount of refrigeration obtained, and the work expended, by passing a cubic meter of dry air through the machine; the press- ures in the reservoir varying from 1} to 43 atmospheres. The temperature of the external air is taken at 15°; the temper- ature of the air leaving the condenser at 18°; temperature of the water about 13° V,=1, T,=288 and m=1'.266. § 12. An examination of the table shows the enormous influence that the passive resistances exert upon the efficiency of air machines. It is one of the conse- quences of the inherent cumbrousness which follows from the use of this body in a thermic machine. The useful effect produced is not in- creased in proportion to the increase of pressure. It is of no advantage to em- ploy pressures higher than about 44 at- mospheres. Aside from the diminution of efficiency of the air at high pressures, @ loss is occasioned by heat developed in 30 lhamod asioy x0gt ‘oneroam | B-| -moq sed B85 8 & leat asi0y Jog a ok = aS 58 3 |-- 3 3 |-papuadxe war | _SBE5RSS i PlomUTELTOT | 3Bses. ss 2 5 ‘Pwonasoey | 2 noy 49 z & 8 3 S A cojourueIZ0Ty, ha “aalpaya ‘pepuadxa 410.4 ~-gxajourures30q7 Toran | “papuadxa qo | “pautwigo a2ZsSZsRe , S| Soxeaana conmumg | SRBSERES s peaeeas ‘aye Zurmogino Segasos yoommendua, | STITT Tt 4 | $2582588 caopssordtuoo sarge | Eoaresteced ae joamunduas, | S6FZREEE (sardsomy) | eo oe a j Sada 31 the compressor, and which extends to other working parts of themachine. We have said above that, with a given ma chine we can vary at will the pressure P, by varying the length of time of the opening of the admission valve in the eylinder B. If the time be shortened the pressure and the cooling effect are both increased; and if the time be in- creased P, is diminished. It is necessary that we should vary at the same time the working of the emission valve, so that it opens at the moment when the piston shall have passed through a space equal , toV,qt-corresponding to the atmospheric pressure on the inside of the expansion cylinder. A machine whose dimensions and veloc- ity are such that it uses 1000 cubic meters of air per hour will produce from 8.548 to 29.375 negative. calories and upwards per hour, provided that the driving power varies from 4 to 34 horse power. Practically however the efficiency of air machines is not so great as is indica 32 ted by the above table as no account has yet been taken of watery vapor in the air, nor of lost spaces in the machine. We proceed to examine the influence of these two causes of loss. ’" INFLUENCE OF MOISTURE IN THE AIR. § 13. This influence is not to be neg- lected. The vapor contained in the air condenses on the sides of the expansion cylinder, and parts with its latent heat of vaporization so that the final temper- ature of the air is higher than it would have been if dry. Furthermore the snow produced from this moisture accumulates around the ori- fice of the cold air outlet and we cannot readily utilize the cold which is required to produce it. For these two reasons, but especially for the latter, the moisture of the air causes a notable loss. We proceed to calculate the volume and the temperature of the air at the end of the expansion under the supposition of a known hygrometric state of the at- mosphere, from which we can easily de- 33 duce by the tables the pressure of the vapor p, and its weight yu,. In the compression cylinder of watery vapour not being near the saturation point, and exerting a feeble pressure will behave nearly as a perfect gas; its vol- ume and its temperature are represented. by the relations pv'=a constant, in which k=141 and pu=R’mT; R 0.622 The total pressure of air and vapor being represented by P, the pressure of the vapor being p, that of the air alone will be P—p and we shall have preserving our former notation: R= =47.061. P,VE=P,VE, (21) PVi=pVi, (22) (P,—p,)V,=RmT,, (28) PV,=R'p,T, (24) @,—p,)V,=RmT,, (25) PV,=R'y,T,. (26) The work of the resistance to compress- 84 AG eY.-P.Y,) Wee ga A (27) or Wr= | (me+pe')(T,—T,) ce’ is the specific heat under constant volume of the superheated vapor ’=0,8407. After cooling the volume becomes " V=v, 7 (28) and we have ‘ PV’ =R'y,T", From equations 21 and 22 we can de- duce the pressure in the reservoir. We can determine by examining a table of tensions of saturated steam whether the pressure p, is greater or less than the pressure which corresponds to the temperature T,’. If it be less the air will not be saturated with vapor when leaving the condenser, and the heat ab- sorbed by the latter will be: Q,=h(met y,c’)(T,—T,') If the pressure p, is greater than the 35, pressure p,’, corresponding to the tem- perature T,’ for saturated steam, there will be a condensation of some of the vapor in the condenser ; the amount con- densed will be ,(1—a,') and the pressure of the vapor entering into the cylinder B will be p,’, that of the air being P,—p,’. We shall have also: rai pi PB a/at =o oe ‘PPP, vi We see that the quantity of vapor not condensed by the cooling, and passing into the expansion cylinder, will continu- ally diminish in proportion as the work- ing pressure is raised. The influence of the humidity in the air will therefore be Jess as the pressure is made greater. The weight of the mixture of air and vapor, which is m+ y, if there is no con- densation in the cooler or m+ 4,2, if there is a condensation, is carried into the cylinder B where it encounters the surfaces cooled during the preceding 36 stroke. We can neglect the influence of these cold surfaces upon the air alone, but not upon the mixture of air and vapor. The latter is converted into frost which releases a certain amount of heat to be imparted to the metal, and which during the expansion is restored to the air. Suppose at first that there is no con- - densation in the cooler, there is conveyed to the cylinder a weight y, of saturated, or nearly saturated, vapor at the tem- perature T,’. We may assume, consider- ing the very low temperature of the sur- faces, that all the vapor is condensed here ; it will disengage a quantity of heat ©, which is approximately equal to #,(r,' +79). 7,’ being the latent heat of the vapor corresponding to the tempera- ture ¢,', 79 is the latent heat of water released on freezing. The heat C is gradually restored to the air during expansion. The pressure of the air becomes P,, and the volume introduced into the cyl- inder is 37 Rm’ _ Vi=—5. The differential equation of the work is S mat +% pat — __pay ann, _¢, being the specific heat of ice, =0,5 dT dc av or (ae aH i ART =—™ Vy" We do not know the law of relation between C and T,, that is, how to cém- municate to the air the heat released from the water and ice formed. We are forced to make a hypothesis which is not rigor- ously exact, but which is sufficiently ap- proximate. We will suppose that the transmission is proportioned to the fall of tempera- ture, and therefore that d0=—pydT r/ +79 “Ty=T, whence we have; in which 38 aL Gna aT__oa ER (mM HAY) HH integrating we get 22 (44 HatHY) 1 U7 Vs ae 2 PY) og =e e-1 we have furthermore oa (a+ 4 MAMIE ot Lye 9) , P, V, =RmT,, PV”, =RmT’, whence T,_P,v,” = Py Equation. 29 can then be written; 1 metmyyr\ , Ts _7Ps ear ( tt AY) rtp: 0) We can obtain the value of T, by suc- cessive approximations. An approximate value for T, is found to be T= {mke He) _&- Ayme™ BIT tole +19) E 1)me,P, ee mke + pe, + 39 Suppose now that condensation occurs in the cooler, we find by the tables the pressure of p,’ of saturated vapor of temperature T,’ and we can deduce the weight of the vapor condensed in the’ cooler. We shall have then; C= p,2,'(r' +79) rl +T9 and yan, 7 7+, } The equations 29 and 30 apply in this case as in the preceding. The quantity of disposable negative heat is; Q=mke(T,—T,) (31) since we suppose the negative heat of the snow formed to be lost. Finally the work produced by the expansion is; WnaPV,"4 (14 Mit My) (T’,-T)—P,V, (82) or Wn: att alte, "—T.), (33) It ‘there i is a condensation in the cooler, 40 we should replace «, in equations 32 and 33 by y, 2,’. $14. The following table gives the cooling and general effect obtained from a cubic meter of air supposing a hygro- metric state of 4 and a temperature of 15°. The weight of the air is then 1. 2157 instead of 1.4 226 which is the weight of dry air at this temperature, We have also p,=85.§8 and u,= 0. 00626. 41 ‘xopurt£o worsuudxe O7U payleo sodea Jo 10310 “remod 08104 |_ser0ay3 Jog g | eanoogo sg | 3 3 | eg | estooue sod | S55 aa | amoqgaad | owe ex 2 | somodonoy ageeeeee 3 s a ‘onovoomy | ae aa popuadxa OM OATHOOT ssxnowuresZ0ny “q10m onasoaq, ‘gayioywo oaryesony “aye poredxe jo aanjeioduia, “sossardw0o wy amerodura, ‘saroydsomye amnseaiy 42 In comparing this table with the table of § 11 we see that the influence of the humidity of the air upon the results obtained is the greater when the press- ure is low. We have made a similar remark in reference to the passive resist- ances. The theoretical advantage there- fore of low pressures is practically much diminished by these causes of loss. “ It is possible to neutralize almost com- pletely the influence of moisture in the air. To accomplish this it would suffice to employ the air after it had produced its cooling effect and had parted with its moisture. It would be necessary to make the refrigerating machine a closed machine, making the same quantity of air serve indefinitely. The cooling would be produced by causing the cooled air to pass through an apparatus surrounded by some liquid not easily frozen, such as a solution of calcium or magnesium chloride. A part of the neg- ative calories would thus be used, as well as by direct contact, and so many as are not used would not be lost, as the air 43 passes directly to the compressor A, not at 15° as before, but a—8° or—10° of temperature. We think that it is only in this way that we can improve the air machine so that it can compare favorably with the machines using a liquefrable gas. INFLUENCE OF WASTE SPACES. § 15. We will suppose the air to be dry in order to avoid complexity in our calculations. Preserving our previous notation and calling v the amount of useless space in the compression cylinder, and v’ that of the expansion cylinder; y the weight of air enclosed in the space v at the end of the compression, we have; PCV, +o)F=P(V,+0)F (84) P(V,+2)=Rom+y)T, (85) PV,=RmT, (36) Po,=RpT, (37) m being the weight of dry air driven out of the compressor. Equations (84), (35), (36) and (37) give by elimination of 44 aang ow» and T=T, @)* (39) The work of resistance to compression, taking account of the work restored to the piston as it begins to ascend, by the air in the waste space, expanding from P, to P,, is; k We = 2-7 (V,-P.V,) ZL v.-V,, | Py Vito? (40) For the cooling a yy =V\n (41) and PV’, neh ’ (42) The heat Q, absorbed by the water of the condenser is; Q,=mke(T,—T,’) (48) Pertop or Expanston.—The air coming from the reservoir R’ under pressure P, and the temperature P,’, should at the moment of opening of the inlet valve 45 cause the air in the waste space and whose volume is v', to change its press- ure from P,toP,. This influences the temperature T,"’ of the mixture, also the weight m’ of the air which passes from the reservoir into the waste space. 7 The dimensions of the reservoir being very large in comparison to the waste spaces, we may assume that ho change occurs either in temperature or pressure of the reservoir, while the waste spaces are filled with air at the pressure P,. ~~ Calling y’ the weight of the air en- closed in the waste space at the moment that the inlet valve opens. We have; Pw'=RwT; (44) T, being the final temperature of the expanded air. . The stored up work of this air is; La TH T,. The weight m’ of air filling the waste space, and having a temperature T,’ and @ pressure P, has a stored energy of Loca qr. 46 After the waste space is filled, the stored up energy of the total quantity of air m’+ y’ contained there is © nt yh" qt ve), and we have furthermore; Piv'=R(m' + wT," (45) As we suppose there is neither loss nor gain of heat from the exterior, the differ- ence between the stored energy of the mixture after the mass m’ is introduced, and the sum of the stored energies of the masses m’ and y’ before mixing is equal to the external work performed. This exterior work is evidently P, v,’, and calling the volume of m’ before its introduction into the cylinder under pressure P, and temperature T,’ equal to »,', then; P,o,’=RmT,’. We have also c Com RAE —gm'T +5 (m' + wT =RmT,’, AT Replacing re by fea and combining with equations 44 and 45 m= PSP (46) n_kem'T,' + pT, and T, mth a oe te PDT, a : ~ PYT,+P@T/-T,) When the inlet valve closes, the piston has described a volume V,’’, which has been filled by the weight m” of air at pressure P, and temperature T,’. We have then; m! +m!" =m There is no external work performed upon the total mass of air, since the negative work of the piston P,V,” is exactly equal to the positive work ex- erted by the air of the reservoir. The weights and temperatures of the air at the beginning and the end of the intro- duction possess the following relations: 48 e 2», Een T+ gD HS (ml tm 4 Ty". T,’” being the temperature at the end of the introduction. This equation gives; pie FHT, ¢m"'T! . m+ yl or PLY +0) E@,-P,)o! TS —syrm pet NT. iV’T,+Pv'T, ae, (48) we also have PV, +0)=Rm+ y’)T,” or PV," +0)=P(V,'+0') . 1 {PHP's 9) V,’ is given equations 38 and 41. Equa- tion 49 gives the value of V,”. The inlet valve being closed, the mass of air m+ yu which is at pressure P, and temperature T,’”’ expands without gain or loss of heat since we neglect the 49 influence of the sides of the cylinder. At the end of the stroke, this volume * becomes V,+v’, its temperature T, and its pressure P,, We have then PAV, +0/)F=P,(V," +0 or PAV, +0')F=P, (50) : {q+} (@- z) v 7 P(V,+0')=Rim+ pT, (61) Equations 50 and 51 give V, and T, if P, be known, or P, and T, if V, is known ; this latter being the volume de- scribed by the piston of cylinder B. We have k-1 When there is no waste space we have AsT,’” is greater than T,’, it results that for a given weight of air passed through the machine, at a given working