## Adiabatic Expansion Analysis

As the water escapes, the air volume decreases, causing a decrease in pressure and a corresponding decrease in thrust. We consider this process to be adiabatic (no transfer of heat during the split-second expansion process), which allows us to develop the equation for the variation of the pressure P to that of the volume V as presented in the following figure. In the analysis that follows we borrow heavily from the ME321 course "Introduction to Thermodynamics". For those of you who have done this course this should be a pleasant review. We assume that the compressed air behaves as an ideal gas throughout the process, and ignoring kinetic or potential energy effects we consider the differential form of the energy equation of a simple compressible system as follows:

dq - dw = du

where:
dq = 0 is the heat transfered (equals zero for an adiabatic process). [Note also the use of the d (delta) operator indicating that heat and work are path functions, and not properties.]
dw = P dv is the specific boundary work done by the expanding compressed air [J/kg], and v is the specific volume of the air [m3/kg]. Note that V = m v, where m is the mass of the compressed air.
du = Cv dT is the change of internal energy of the air, where Cv is the specific heat capacity at constant volume of the air [J/kg K], and T is the absolute temperature [K]

Thus P dv + Cv dT = 0

Differentiating the Ideal Gas Equation P v = R T (R is the gas constant for air) we obtain:

P dv + v dP = R dT

We substitue this relation into the energy equation above to obtain

P dv (1 + Cv/R) + v dP (Cv/R) = 0

We now substitute the various relations for an ideal gas, including Cv + Cp = R and Cp/Cv = k, where Cp is the specific heat capacity at constant pressure [J/kg K] and k is simply known as the ratio of specific heat capacities. Simplifying and replacing V = m v we obtain:

k dV/V + dP/P = 0

This equation can be easily integrated (recall MATH 263) leading to:

k ln(V/V0) = ln(P0/P)

where ln is the natural logarithm function, and the values of pressure and volume are those shown on the above figure.

Taking the inverse logarithm of this equation we obtain our required result: Note that for any diatomic gas (such as O2, N2 - the major components of air) the value of k is 1.4

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