Boyle Temperature and Boyle Curves

6. Boyle Temperature and Boyle Curves

a. Boyle Temperature Using the RK equation, the Boyle temperature can be determined as follows. First, multiply Eq. (37) by v to obtain

Pv = RTv/(v–b) – a/(T 1/2 (v+b)). (48a) Dividing by RT, Z = Pv/RT = v/(v–b) – a/(RT 3/2 (v+b)).

(48b) Since Z = Pv/RT, ∂Z/∂P = (1/RT) ∂(Pv)/∂P. Therefore, if ∂Z/∂P = 0, it implies that ∂(Pv)/∂P =

0, and ∂(Pv)/∂P = (RT/(v–b) – RTv/(v–b) 2 + a/(T 1/2 (v+b) 2 )) ( ∂v/∂P)

(49) Since

1/2 = (a/(T (v+b) 2 – RTb/(v–b) 2 )( ∂v/∂P).

(50) As P → 0, the volume becomes large, and Eq (50) assumes the form ∂(Pv)/∂P → (b – a/RT 3/2 ), i.e., ∂(Z)/∂P → (b/RT – a/R 2 T 5/2 ).

∂v/∂P = 1/((a(2v+b)/T 1/2 v 2 (v+b) 2 ) – RT/(v–b) 2 )

(51) Expressed in terms of reduced variables.

∂Z/∂P = 0.08664/T R – 0.4275/T 5/2 R R as P R → 0. (52) Recall from Abbott’s correlation, (Eq. 16) that ∂Z/∂P R = B 1 (T R ) = 0.083/T R –

0.4275/T R . Hence, using the RK equation, when T R = 1, the slope ∂Z/∂P R = – 0.3409. This slope tends to zero when T 3/2 R → 2.8983. The corresponding temperature is called the Boyle temperature. (This result compares well with the value of 2.76 that is obtained using the em-

pirical virial expressions.) The slope is negative when T R < 2.8983 and is positive when T R > 2.8983. It can be shown that the slope has a maximum value of 0.009737 at T R = 5.33873 and decreases slowly to 0.004093 as T R → 20, indicating that the value of Z ≈ 1 even at high pres-

sures and temperatures. For instance, in diesel engines pressures as high as 80 bars are attained but the gaseous mixtures behave like an ideal gas, since for many combustion–related species T R > 20.

b. Boyle Curve If T R < 2.8983, Z approaches a minimum value at a particular reduced pressure. The loci of all the minima of Z = Z minimum constitute the Boyle curve along which the gas behavior is similar over a wide pressure range. The Boyle curve may be characterized as follows. The Z minimum condition occurs where the product (Pv) has a minimum value at a specified tempera- ture. Applying Eq. (49a),

’ ,Z 1.5

Figure 9: The conditions at Z minimum for a RK gas.

(53) at the minima ∂(Pv)/∂P = 0. Therefore, at this condition, either ∂v/∂P = 0, which implies that

( ∂(Pv)/∂P) = (a/(T 1/2 (v+b) 2 – R T b/(v – b) 2 )( ∂v/∂P),

there is no volumetric change during compression (an unrealistic supposition), or

(54) Using Eq. (55) to solve for T,

(R T/(v – b) – R T v/(v – b) 2 + a/(T 1/2 (v + b)) = 0.

T 3/2 = a (v – b) 2 /(R b(v + b) 2 ).

In terms of reduced variables T 3/2 = 4.9342 ( v – 0.08664) 2 /( v + 0.08664) R 2 ′ R ′ R .

(55) Further, solving for v ′ R ,

= 0.08664 (1 + 0.4502 T R )/(1 – 0.4502 T R ) (56) Therefore, at specified values of T R , v ′ R can be determined using Eq. (57) and the result sub-

v 3/4 ′

stituted into Eq. (45) to determine P R at Z minimum . There is no solution for v ′ R and Z minimum for values of T R > 2.898 (i.e., above the Boyle temperature), since Z always increases. These re- sults are illustrated in Figure 9 .

c. The Z = 1 Island For specified values of T R , the corresponding values of Z first decrease below unity as the pressure is increased, pass through a minima, and then increase to cross over Z = 1. On a pressure–temperature graph there is restricted regime or an island on which Z = 1. Here the real gas behaves as an ideal gas. Using Eq. (49b) at this condition, we obtain the relation

Pv/RT = Z = 1/(1 – b/v) – a/(R T 3/2 v (1 + b/v)) = 1, i.e., (57)

Figure 10: The variation in T R and v ′ R with respect to P R at the condition Z = 1.

(v R T 3/2 /a) = (1 – b/v)/(1 + b/v), (58) which yields a relation between v and T at Z = 1. Solving for the product (bv –1 ), we obtain b/v = (A – 1)/(A + 1),

(59) where A = (a/vRT 3/2 ). Substituting Eq. (60) into the RK equation of state, P = (RT/(2 A b)) (A – 1) – (a/(2 b 2 T 1/2 A(1 + A))) (1 – A) 2 .

(60) In the terms of reduced parameters, A = 4.9342/T 3/2 R , and v ′ R = 0.08664 (A + 1)/(A – 1), so

that

A (1 + A))). (61) These results are illustrated in Figure 10 .

P R = 5.77101 T R (A – 1) – 28.47536(A – 1) 2 /(T 1/2 R