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Energy stored in a pressurized Fluid – The Expansion process

How much mechanical energy, i.e. exergy could one extract from a gaseous, pressurized  fluid at a pressure P1 and temperature T1 when it expands in an adequate engine?

We assume the equilibrium the surrounding condition to be P0, T0.

The first question we have to answer is how an ideal gas behaves when it is expanded. The thermodynamic theory of an ideal process without heat input (adiabatic) and without friction (isentropic) yields to a set of the following equations:

P2/P1 = (V1/V2) ^ n
<=>
V1/V2 = (P2/P1) ^ (1/n)

V1/V2 = (T2/T1) ^ 1/(n-1)
<=>
T2/T1  = V1/V2^ (n-1)

T1/T2 = (P1/P2) ^ ((n-1)/n)
<=>
P1/P2 = (T1/T2) ^ (n/(n-1))

The equations relate the initial state 1 characterized by temperature T1, pressure P1 and volume V1 to the final state 2 (T2, P2, V2).  n is characterising the gas and depending related on the heat capacities at constant pressure Cp and at constant volume Cv:

n = Cp/Cv

This ratio is called the isentropic exponent. Here some characteristic values:

• Mono atomic gas -  Argon: n= 1.67
• Diatomic gas – nitrogen, Hydrogen , “normal air”: n = 1.4
• “Multi-atomic” gas – organic fluid – refrigerant R11: n =1.13
• “Multi-atomic” gas – organic fluid – Butane: n =1.09

The above equations only hold for a situation, where energy is reversible transferred from the gas to a macroscopic body, i.e. a piston.  So the gas is cooled and decreases its temperature. In a situation where an ideal gas freely expands without energy transfer, the temperature will remain constant. In contrast to the isentropic expansion this is a irreversible process.

Using the above equations with  c_exp = V/V1, the expansion ratio, yields to

P(V) = P1  (V1/V) ^ n = P1  c_exp^(-n) = P(c_exp)

The following plot shows the pressure decrease depending on the expansion ration during an expansion process for the four typical fluids Argon,  Hydrogen, refrigerant R11 and Butane. Initial pressure is 10 bar. Note: In technical applications, e.g. characterizing air motors, people normally talk about pressure differences to atmospheric pressure. So a  pressure of 1 bar in a data sheet of an air motor means actually an absolute pressure of 2 bar. Analogous there is a temperature decrease. The following diagram shows to typical expansion processes: Air is expanded via an air motor starting at low temperature T1 = 25°C , a temperature you  have in a compressed air tank. The second process starts at 100°C a situation that may arise in an Organic Rankine cycle.

T(V) = T1  (V1/V) ^(n-1) = T1  c_exp^(1-n) = T(c_exp) To calculate the maximal energy we can extract from the pressurized fluid we have to use a total isentropic  process. Isentropic means, that every step of the process has to be reversible and finally the fluid ends with atmospheric conditions (T0,P0). Let’s further assume we are using  an ideal  positive displacement engine, for example a “massive” piston sliding frictionless in a cylinder. The whole process can be divided in 4 steps:

1.) Input process, filling the piston to volume V1

2.) Expansion of  the fluid from V1 -> V2, choose V2 so that the pressure becomes equal to the surrounding atmospheric pressure: P2 = P0

3.) In most of the cases after the expansion the gas temperature T2 is not equal to the surrounding air temperature. So there is  a potential source to extract further exergy via a heat engine: T2 ->  T0

4.) Output exhaust process, emptying the piston in pushing the fluid to the surrounding

The total pressure P-P0 acting on the piston 1.) During the filling process of the empty piston to a volume V1 a total constant pressure of P1-P0 acts on the piston. So we get

L_input = (P1-P0)  V1

2.) The generated mechanical energy is calculated via integrating this pressure over the expansion volume

L_expansion
= Integral(P(V)-P0, V1, V2)
= Integral(P(V), V1, V2) – P0 (V2 – V1)
= 1/(1-n)  P1 V1  ((V2/V1)^(1-n) – 1) – P0 (V2 – V1)

Import point here is, that for an ideal expansion the final volume has to be chosen in a way, that the end pressure after expansion equals the surrounding’s pressure. So:

(V2/V1)^(1-n) = P2 V2/(P1*V1)

and finally

L_expansion = 1/(1-n) (P2 V2 -  P1 V1) – P0 (V2 – V1)
= Cv (T2 – T1) -
P0 (V2 – V1)

3.)  We neglect for the moment

L_heatengine = ?

4.) During the output process after opening the exhaust no force is acting on the piston. So

L_output = 0

The following diagram shows input cycle and expansion work as corresponding areas in the pressure volume chart: A more intuitive formulation is to transfer to the analogous equations for power instead of energy and normalize by dividing by the input volume flow Vflow1:

Lp_spec_input =  Lp_input/Vflow1 = (P1-P0)

Lp_spec_expansion =  Lp_expansion/Vflow1
= 1/(1-n) (P2 Vflow2/Vflow1 -  P1) – P0 (Vflow2/Vflow1 – 1)

Using the relations for  optimal expansion

c_exp_opti = Vflow2/Vflow1 = (P1/P2)^(1/n)

and

P0 = P2

yields to

Lp_spec_expansion = 1/(1-n) (P2  c_exp_opti -  P1) – P2 (c_exp_opti – 1)

The following plot shows the two parts of the specific expansion power of air resulting from the input cycle (red)  and isentropic expansion (blue). The blue line shows the sum of both. Let’s see an example: To generate 1000 Watt of mechanical power with an ideal engine you need an air input stream of 1 l/sec at approx. 7 bar. In this case it splits into 600 Watt resulting from the input cycle and  400 Watt from the expansion. For real machines the inlet air consumption may be 3 to 15 times higher. Join In 