Pressure-volume relations

The gas being compressed in a refrigeration machine can sometimes be regarded as an ideal one where pressures and volumes are related as follows:

PeVen=PiVJ (9.12)

Three cases can then be distinguished:

(i) The case of reversible adiabatic compression, in which the process is isentropic and

We have for the value of n.

N = cp/cv = y (9.13)

Here cp and cv are the specific heats of the refrigerant gas at constant pressure and constant volume respectively.

(ii) Reversible but non-adiabatic compression in which there is heat transfer between the refrigerant and its surroundings. Reciprocating compressors with cylinder cooling may approach this case and in such circumstances n will be less than y.

(iii) Irreversible but adiabatic compression, in which there are thermal effects due to gas friction and turbulence but no actual heat transfer between the refrigerant and its enclosing surfaces. Centrifugal machines approach this type of compression, in which case n will be greater than y.

When n & y the process is called polytropic compression. With polytropic compression the value of n depends upon the process — it is not a property of the refrigerant. During polytropic compression the entropy of the gas increases in case (iii) and diminishes in case

(ii) . Thus, the process departs from the simple saturation cycle which we have hitherto considered.

If the compression is reversible, as in cases (i) and (ii), the work done on 1 kg mass of refrigerant in Nm during steady flow is given by the following expression, where pressures are in Pa absolute and volumes are in m3.

Pi

(9.14)

подпись: (9.14)Work done = J Vdp

Pt

Using equation (9.12) we have:

PVn = c (where c is a constant), or

V = cUnp-Un

Thus,

_ Ain

P-l, ndp

JPe (n—)ln

Pd_

Pe

Pc

_ n r

(n — 1) ^

Work done = c

(n — 1)

1

Pe Ve

Pressure-volume relations

(9.15)

 

Pressure-volume relations

It should be noted that the work done on 1 kg mass of refrigerant in Nm is here equal to the head in metres against which the compressor is working. Developing the expression by using equation (9.5) we get the rate of work done in compression, in w, on m kg of refrigerant

Where hve is the enthalpy of saturated vapour at the evaporating pressure and hlc is the enthalpy of saturated liquid at the condensing pressure.

Pressure-volume relationsIf the compression process is the isentropic one described as case (i) above, the equation becomes:

(9.17)

In case (iii), where the process is irreversible but adiabatic, equation (9.14) still gives the

Fluid output in Nm kg-1 of refrigerant, or the head in metres against which the machine is working, but it no longer represents the required input energy to the fluid.

This exceeds the quantity

Pressure-volume relations

Pc

Pressure-volume relationsAnd is calculated from the following expression:

(9.18)

Comparing this with equation (9.17) we can define what is termed the isentropic efficiency

Ols) as

Theoretical work for case (i)

J“| = V /..

Theoretical work for case (iii)

(9.19)

подпись: (9.19)(Pd/Pe)^ ~ 1 (pd/pc)(n-l)ln — 1

The quantity

N

(n — 1)

подпись: n
(n- 1)
(J- 1) Y

Is called the poly tropic efficiency and that represented by equation (9.15) is called the polytropic head. Polytropic head has to be divided by the polytropic efficiency to give the power required by case (iii) in Nm kg-1 of gas pumped.

Other forms of equations (9.15) to (9.18) are possible. Thus, for an ideal gas we can write RTe in place of peVe, R being the particular gas constant for the refrigerant.

It should be noted that although equation (9.7) is applicable to cases (i) and (iii), it cannot be applied to case (ii), since the quantity hwi — hve does not, in that case, represent all the work of compression. Instead, we must write for case (ii):

(9.20)

подпись: (9.20)Wr — ^[(Avd Ave) + ^j]

Here, q] is the heat lost to the jacket in J kg 1 of refrigerant in circulation. Its value can be calculated from

The temperature of the hot gas leaving the compressor can be calculated from the formula:

F ^(n-l)/n

TA = T„

подпись: ta = t„

(9.22)

подпись: (9.22)El

Pt

EXAMPLE 9.7

A refrigerant which behaves as an ideal gas has a molecular mass of 64.06 and a specific heat ratio of 1.26 (equal to cp/cv). If the compression ratio (pjpc) is 3.119 and the refrigerating

Effect is 322 kJ kg-1 with evaporation at 4.5°C, find the theoretical COP, the enthalpy gain

During compression and the temperature of the discharge gas for the following cases:

(i) isentropic compression,

(ii) polytropic compression with n = 1.22,

(iii) polytropic compression with n = 1.30.

Answer

(i) Write RTe in place of peVe and set R = 8314.41/64.06 (see section 2.6).

Equation (9.17) expresses the work done in compression for a refrigeration duty of Qr kW. Hence, re-arranging the equation:

Wr _ 8314.41 x (273 + 4.5) x (1.26) iq(o.26/i.26) _ it _ n 14.^4.

QT 64.06 x 1000 x 322 x (1.26 — 1) 1 J ‘

COP = %-= 1/0.1434 = 6.97 Wr

By means of equation (9.10), the enthalpy gain during compression is determined as ^vd -/*ve = f§ = 46.20 kJ kg-1

Where hvd is the enthalpy of superheated vapour at the discharge condition.

From equation (9.22) the temperature of the discharged gas is

Td = (273 + 4.5) x 3.ii9(i-26-i)/i.26 _ 350 9 K

Which is equivalent to 77.9°C.

(ii) Here we proceed as in case (i) but re-arrange equation (9.16) instead of (9.17):

Wr _ 8314.41 x (273 + 4.5) x 1.22 ,0(0.22/1.22) _ 11 = 0 141?

Qt ~ 64.06 x 1000 x 322 x (1.22 — 1) 1 J ‘

COP = %- = 1/0.1412 = 7.08 Wr

Equation (9.5) yields the mass flow rate of refrigerant:

Qr_ 322

By equation (2.20)

Whence

подпись: whenceW,

(/ivd — hve) + qj = 322 = 322 x 0.1412 = 45.47 kJkg“1

N, i q(1.22—1 )/i.22 n x (8314.41 x (273 + 4.5))

-Ux 6406

подпись: n , i q(1.22—1 )/i.22 n x (8314.41 x (273 + 4.5))
-ux 6406
Equation (9.21) gives the jacket loss: 1.22 1.26

=

(1.22 — 1) (1.26 — 1)

= 5734.48 J kg"1 = 5.73 kJ kg“1 The enthalpy gain during compression is then (hvd — hve) = 45.47 — 5.73 = 39.74 kJ kg“1 By equation (9.22) the temperature of the discharged gas is Td = (273 + 4.5) x 3.H9(1-22-1)/i.22 = 34Q 7 K

Which is equivalent to 67.7°C.

(iii) Proceeding as for case (i) but re-arranging equation (9.18) instead of (9.17) we have

Wf _ 8314.41 x (273 + 4.5) x 1.26 „ | q(i.3—ivi-3 _ n _ r> 1^97

Gr 64.06 x 322 x 1000 x (1.26- 1) 1 J

Rnp = Q±- — —i— = 6 15 WT 0.1627

By equation (9.5)

By equation (9.7)

Wr — hve)

Whence the enthalpy gain during compression is

(.hvd — hye) = 322 -^ = 322 x 0.1627 = 52.39 kJ kg“1

By equation (9.22) the temperature of the discharged gas is

TA = (273 + 4.5) x 3.119(1,3_1)/1’3 = 360.8 K

Which is equivalent to 87.7°C.

Summarising the results:

Case

N

COP

Enthalpy gain during compression kJ kg’[2]

Discharge gas temperature

°C

(i)

1.26

6.97

46.20

77.9

(ii)

1.22

7.08

39.74

67.7

(iii)

1.3

6.15

52.39

87.8

Posted in Air Conditioning Engineering


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