Pj PzT3T2 p2r V V2

(c) Isothermal expansion process (3-4).

In this process, heat is supplied to the cycle at a high temperature 7"roiw, during expansion of the working fluid. Work is done, by the working lluid. equal in magnitude to the heat supplied. There is no change in the internal energy, but an increase in the entropy of the working fluid.

Heat transfer (O) ~ work done (W) = p3V3 In r = RT, In r. Change in entropy (S.,-S;,) = R In r.

(d) Constant-volume regenerative transfer process (*l- I).

In this process, heat is transferred from the working lluid to the regenerative matrix, decreasing the temperature of the working fluid from 7"niax to T„,,„. No work is done, and there is a decrease in the internal energy and entropy of the working fluid.

Heat transfer (O) = CV(T, - 7'.,). Chaniie in entroov — = C In -r

In the regenerative processes, the heat transferred from the matrix it) the working fluid in process 2-3 is restored from the working fluid to the matrix in process 4-1. There is no net gain or loss of heat by the working fluid or the matrix. Therefore the total heat supplied (at Tm;u)= K7'3 In r. the total heat rejected (at 7mln) - KT, In (l/r).

... . . heat supplied-heat rejected and thermal cfhcieucy =-71---

heat supplied work done heat supplied — {KT3 In r-RT, In r)/RT3 In r

This value corresponds to the Carnot efficiency between the same temperature limits.

rußscHMim cycle

The classical analysis of the operation of Stirling engines is due to Schmidt (1871). The theory provides for harmonic motion of the reciprocating elements, but retains the major assumptions of isothermal compression and expansion and of perfect regeneration. It, thus, remains highly idealized, but is certainly more realistic than the ideal Stirling cycle. Provided a reasonable level of caution is exercised in interpretation, the predictions of the Schmidt theory can be a useful tool for engine design. _____

Principal assumptions of the Schmidt cycle

1. I he regenerative process is perfect.

2. The instantaneous pressure is the same throughout the system.

3. Hie working fluid obeys Hie characteristic gas equation, PV- RT.

4. There is no leakage, and the mass of working lluid remains constant.

5. The volume variations in the working space occur sinusoidally.

0. There are 110 temperature gradients in the heat exchangers.

7. The cylinder-wall and piston temperatures are constant.

8. There is perfect mixing of lite cylinder contents.

9. The temperature of the working lluid in the ancillary spaces is constant.

10. The speed of the machine is constant.

Nomenclature used in the following analysis!

A = a factor (rJ + 2tk cos at + k~)'. H = a factor (t + k+2S). Kr = constant.

M = total mass of working lluid.

N = machine speed.

p = instantaneous cycle-pressure.

Pmnx = maximum cycle-pressure.

Pmin = minimum cycle-pressure.

11 = engine output per cycle.

Pmw P/MRT€ dimensionless power parameter based on the unit mass of working fluid.

= Pf(pn,nx V-r)» dimensioiiless power parameter, based on the maximum cycle-pressure and combined swept volume per cycle.

O =hcat transferred to the working fluid in the expansion space, the heat lifted.

Q . - Q/MRTC the dimensionless cooling parameter, based on the unit mass of working fluid.

Omax = 0/(pmnxV-t), the dimensionless heat lifted, based on the maximum cycle pressure.

R = characteristic gas constant of the working fluid.

I'c = temperature of the working fluid in the compression space, generally assumed to be 300 K.

i0 temperature of the working fluid in the dead space.

7'r = temperature of the working fluid in the expansion space.

V<; = swept volume in the compression space.

Vp = swept volume in the expansion space.

VD total internal volume of heat exchangers, volume of regenerator. and associated ducts and ports.

Vw = J V,,( I + cos <\>) f \ \\ [ I + cos{«/> - «) j » V„, volume of total working space.

Vwnuu = maximum volume of total working space.

• N'oJc: Lower case suffixes indicate instantaneous values of temperature, pressure volume, and mass. Upper case suffices Indicate maximum (or constant) values Thus: L or c refers to expansion space.

a- = angle by which volume variations in the expansion space lead those in the compression space (in fractions of tt radians, or degrees).

/> =(t2 + k2 + 2tk cos u)»/(t + k + 2s). 0 = tan '((k sin a)/(r+rc cos ex)). k ~ VJ VP, swept-volume ratio, r = Tc/Te, temperature ratio. ({/ = crank angle.

Haste equations

Volume of expansion space Vc = I VF( 1 + cos </»). (4.1)

Volume of compression space Vc = JVC[1+ cos(<i>-a:)] (4.2)

Volume of dead space, being that constant volume of the total working space not included in the volumes of the expansion or compression space.

Mass of working fluid in expansion space, A/c - (pe Ve)/(R'l\). Mass of working fluid in compression space Mc= (pc Ve)/(RTC). Mass of working fluid in dead space Mt, = (p„Vv)/(RTd). Since the total mass of the working fluid remains constant,

M, =<Pr Ve)/( RTe)+(peVc)/(RTe)+(pa Vd)/(KTa) - (K Ve)/(2RTe).

If the instantaneous pressure is the same throughout the system, and equal to p. say. and if Te and T, are constant at T,- and Tc then, substituting for the volumes, eliminating R, and re-arranging,

K/p = (Tc/Ty)( 1 + cos »/>)+*[! + cos(c/» <*)] + (2 VnTc)/( VeTD).

If the temperature variation in the dead space is linear in the axial direction, then the mean temperature

7 „ = rc + Ta - Tc) = (H- TJTC){Tc/2) and, since TcfTr.= r then, from eepi (4.6)

KJp = t( 1 + cos iff) + k[ 1 + cos(<£ - «)j + 25 (4.7)

where S (the reduced dead volume) = 2Xt/(tH 1). To simplify eqn (4.7) consider first v = v rrte J» » cti*

then whore and

SitJCC

y = v r7 cos(<f» 0), tan ß = z/x, z = r sin ß, x = r cos ß,

-Jr cos(</> ß) = Jr2(cos <i> cos ß + sin </> sin ß) - r cos </> cos ß -r r sin c/> sin ß = x cos ('» I z sin </>.

Eqn (4.8) is similar in form to eqn (4.7) therefore, by analogy,

K/p = T(r t k cos ct)2 + (k sin a)2]' cos((/> íJ) + t + k I- 25

= (t2 + 2tk cos a i k ')^ COs(</> - 0) -t- r + k i- 25. (4.9)

Let /\ - <t7-» 2tk cos a - k2)>, B = t+k-» 25, and fi = /\/R then

The instantaneous pressure p is

(a) a minimum, when <f> = (I, i.e. (</> 0) = ().

(b) a maximum, when </> = (0 + 7r), i.e. (<,6- í)) = tt. therefore, /7mln = Kl\ li( I + 5)], and pmn% = Kf| IH 1 Ä)]. Thus,

= P„.in( I + «)/l 1+ S cos(</> - tf)J, (4.1 Ob)

and the pressure ratio

Me«" cycic pressure The mean cycle pressure is given by

/1 ími i which can be resolved to

I [cut transferred and work done Since the processes of expansion and compression take place isothermally the heat transferred Q is equal to the work done P. therefore

If V=¿VK(1 +cos </>), d V= -.i VEsin </> d</> (4.14)

and. if

P = Pmcaitl I - A cos(</> - 0)], approximately, (4.15)

where

= ~ 2Pmean^K ^ tS'n ~ A(«JS <¿> COS 0 Sill <f) + 51*11 O SÍn'</')] d</> = -Ornean VA f/> - Af - COS ft • \ COS 2c/» + SÍH 0tj«/» SÍiríMjJf,"

Solar Stirling Engine Basics Explained

Solar Stirling Engine Basics Explained

The solar Stirling engine is progressively becoming a viable alternative to solar panels for its higher efficiency. Stirling engines might be the best way to harvest the power provided by the sun. This is an easy-to-understand explanation of how Stirling engines work, the different types, and why they are more efficient than steam engines.

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