Throttling in Closed Systems

6. Throttling in Closed Systems

Consider a large insulated tank that is divided into two sections A and B. Section A consists of high pressure gases at the conditions (P A,1 ,T A,1 ) and section B consists of low pres- sure gases at the state (P B,1 ,T B,1 ). If the partition is ruptured, the tank will assume a new equi- librium state. The state change occurs irreversibly and the entropy reaches a maximum value.

The new equilibrium state can be obtained either by differentiating S = S A +S B with respect to T A subject to the constraints that U,V and m are fixed, or by applying the energy balance equations.

nn. Example 40 Eight kmole of molecular nitrogen is stored in sections A and B of a rigid tank. Sec- tion A corresponds to a pressure P A,1 = 119 bar and temperature T A,1 = 177 K. In sec- tion B, P B,1 = 51 bar and T B,1 = 151 K. The partition is suddenly ruptured. Determine

the final equilibrium temperature T . Assume that c =c = 12.5 kJ kmole –1 K –1 2 v vo (c vo does not depend upon the temperature), and that the gas behavior can be described by 1/2 6 the RK equation of state P = RT/(v–b) – a/(T

v(v+b)), where a = 15.59 bar m kmole –2 , b = 0.02681 m 3 kmole –1 ,T c = 126.2 K , and P c = 33.9 bar. Assume that the

volume V B = 3V A .

Solution Applying the RK equation to section A

119 = 0.08314 ×177÷( v A,1 –0.02681) – 15.59

A,1 ( v A,1 +0.02681) bar. (A) Therefore, , v A,1 = 0.0915 m 3 kmole –1 . (Alternatively, we can use the values T R = 1.4

÷(177 1/2 v

and P R = 3.5 to obtain Z = 0.74 from the appropriate charts. Thereafter, using the re- lation P v = ZRT, that value of v can be obtained.)

Similarly, v B,1 = 0.167 m 3 kmole –1 (for which, P R,B,1 = 1.5 T R,B,1 = 1.2, and Z = 0.68). Using the relations, V A / v A,1 +V B / v B,1 = 8 and V B = 3V A , we obtain the expression

V A (1/v A,1 + 3/v B,1 ) = 8, i.e.,

A = 0.277 m ,V B = 0.831 m so that V = 1.108 m .

Thereafter, v

2 = 0.139 m kmole ,N A = 3.024 kmole, and N B = 4.976 kmole. Recall that the internal energy u – u = (3/2 a/bT 1/2

) ln(v/(v+b)). In section A, u–u o = –1684.7 kJ kmole , i.e., U A,1 –U A,1,o = –5094.8 kJ. Similarly, in section B,

u–u –1 o = –1056.8 kJ kmole , i.e., U B,1 –U B,1,o = –5258.6 kJ. Applying the First law to the tank, Q – W = ∆U = 0, so that

U A,1 +U B,1 =U 2 , i.e., U 2 =U A,1,o – 5094.8 + U B,1,o – 5258.6 = U A,1,o +U B,1,o –10353.4 kJ. Now,

U 1/2

2 –U 2,o =U A,1,o +U B,1,o – 10353.4 – U 2o = N(3/2 a/bT ) ln (v/(v+b))

2 kJ. If c vo is a constant, then N A c

= 697799/T 1/2 2 ln(0.139

÷(0.139+0.02681) = –123,070/T 1/2

vo T A1 +N B c vo T B1 – Nc vo T 2 – 10353.4 = –123,070/T 2 kJ.

Therefore,

2 ) = 10353.4 – 123,070/T 2 kJ, or T 2 = 156 K.

12.5 ×(3.024 × 177 + 4.976×151 – 8× T

Using the RK equation, P 1/2

= 61.45 bars Remarks

Likewise, for isenthalpic throttling in sssf devices we can use the relation for (h – h o ).

The entropy generated during adiabatic throttling can be determined using a similar procedure. Such calculations are useful in determining the final pressures and temperatures for shock tube experiments. These experiments involve a pressurized gas in a section A that is separated by a diaphragm from section B. During the experiment, this dia- phragm is ruptured.