Jamil Ghojel - Fundamentals of Heat Engines

(1.59)

Since specific enthalpy h = u + pv , Eq. (1.59) can be rewritten as

(1.60)

For a fluid flowing steadily at the rate of kg/s , the energy equation becomes

(1.61)

where

All terms in Eq. (1.61) have units of power ( ).

For a control volume with multiple flows into and out of the system, the general steady‐flow energy equation can be written as

(1.62)

Stagnation properties are those thermodynamic properties a flowing compressible fluid would possess if it were brought to rest adiabatically and reversibly, i.e. isentropically, and without heat and work transfer. The stagnation state is a convenient hypothetical state that simplifies many of the equations involving flow by taking account of the kinetic energy terms in the steady flow energy equation implicitly.

The stagnation enthalpy h tis the enthalpy that a gas stream of enthalpy h and velocity C would possess when brought to rest adiabatically and without work transfer. The energy equation thus becomes

(1.63)

For a perfect gas, h = c p T and the corresponding stagnation temperature T tis

(1.64)

Applying the concept of stagnation properties to an adiabatic compression, the energy Eq. (1.60) becomes

Rearranging, we get

(1.65)

Temperature‐measuring devices such as thermometers and thermocouples in reality measure the stagnation temperature of the flow and not the static temperature. Thus, introduction of stagnation temperatures simplifies solving the energy equation by eliminating the kinetic energy term and the need to measure flow velocity.

The stagnation pressure p tis defined as the pressure the gas stream would possess if the gas were brought to rest adiabatically and reversibly. Using Eqs. (1.48) and (1.64), p tcan be written as

(1.66)

Examples include flow in ducts, nozzles, and diffusers without heat transfer and work being done. Knowing , where a is the speed of sound and M ais the Mach number at the inlet, and rewriting Eq. (1.64) as

we obtain

Also,

Combining the last two equations, we get

(1.67)

From Eqs. (1.48) and (1.67)

(1.68)

The pressure ratio for M a= 1 at γ = 1.4 is equal to 1.893. This is the critical pressure ratio for air. To achieve supersonic flow, the stagnation pressure needs to be such that p t> 1.893 p .

It was shown in Section 1.2that dimensional analysis and similitude can be used to derive functional representations of complex flow systems, such as compressors, using a reduced number of nondimensional groups of properties. Among the groups discussed were the velocity parameter and mass flow parameter ( Eq. 1.39a). To find a physical interpretation of these seemingly arbitrary combinations of physical properties, consider first the ratio :

Combining this equation with Eq. (1.67) we obtain

(1.69)

Now, for a given compressor blade design and impeller tip speed U , C = f ( U ) and U = f ( ND ); hence, from Eq. (1.69) for the compressor inlet conditions

The compressor speed parameter is a function of the flow Mach number at the inlet (flight Mach number for a turbojet engine) and thermodynamic properties of the fluid. For a given gas with known thermodynamic properties and a compressor of fixed size,