Question 5.10: Bernoulli Equation for Compressible Flow Derive the Bernoull...
Bernoulli Equation for Compressible Flow
Derive the Bernoulli equation when the compressibility effects are not negligible for an ideal gas undergoing (a) an isothermal process and (b) an isentropic process.
The Bernoulli equation for compressible flow is to be obtained for an ideal gas for isothermal and isentropic processes.
Assumptions 1 The flow is steady and frictional effects are negligible. 2 The fluid is an ideal gas, so the relation P = 𝜌RT is applicable. 3 The specific heats are constant so that P/\rho ^k = constant during an isentropic process.
Analysis (a) When the compressibility effects are significant and the flow cannot be assumed to be incompressible, the Bernoulli equation is given by Eq. 5–40 as
\int{\frac{dP}{\rho } } + \frac{V^2}{2} + gz = constant (along a streamline) (1)
The compressibility effects can be properly accounted for by performing the integration ∫dP/𝜌 in Eq. 1. But this requires a relation between P and 𝜌 for the process. For the isothermal expansion or compression of an ideal gas, the integral in Eq. 1 is performed easily by noting that T = constant and substituting 𝜌 = P/RT,
\int{\frac{dP}{\rho } }= \int{\frac{dP}{P/RT} } = RT \ln P
Substituting into Eq. 1 gives the desired relation,
Isothermal process: RT \ln P + \frac{V^2}{2} +gz = constant (2)
(b) A more practical case of compressible flow is the isentropic flow of ideal gases through equipment that involves high-speed fluid flow such as nozzles, diffusers, and the passages between turbine blades (Fig. 5–45). Isentropic (i.e., reversible and adiabatic) flow is closely approximated by these devices, and it is characterized by the relation P/\rho ^k = C = constant, where k is the specific heat ratio of the gas. Solving for 𝜌 from P/\rho ^k = C gives \rho = C^{−1/k}P^{1/k}. Performing the integration,
\int{\frac{dP}{\rho } } = \int{C^{1/k} P^{-1/k}} dP = C^{1/k} \frac{P^{-1/k+1}}{-1/k + 1} = \frac{P^{1/k}}{\rho } \frac{P^{-1/k + 1}}{-1/k + 1} =\left(\frac{k}{k – 1} \right)\frac{P}{\rho } (3)
Substituting, the Bernoulli equation for steady, isentropic, compressible flow of an ideal gas becomes
Isentropic flow: \left(\frac{k}{k – 1} \right)\frac{P}{\rho } + \frac{V^2}{2} + gz = constant (4a)
or
\left(\frac{k}{k – 1} \right)\frac{P_1}{\rho_1 } + \frac{V^2_1}{2} + gz_1 = \left(\frac{k}{k – 1} \right)\frac{P_2}{\rho _2}+\frac{V^2_2}{2} + gz_2 (4b)
A common practical situation involves the acceleration of a gas from rest ( stagnation conditions at state 1) with negligible change in elevation. In that case we have z_1 = z_2 and V_1 = 0. Noting that 𝜌 = P/RT for ideal gases, P/\rho ^k = constant for isentropic flow, and the Mach number is defined as Ma = V/c where c = \sqrt{kRT} is the local speed of sound for ideal gases, Eq. 4b simplifies to
\frac{P_1}{P_2} = \left[1 + \left(\frac{k – 1}{2} \right)Ma_2^2 \right]^{k/(k – 1)} (4c)
where state 1 is the stagnation state and state 2 is any state along the flow.
Discussion It can be shown that the results obtained using the compressible and incompressible equations deviate no more than 2 percent when the Mach number is less than 0.3. Therefore, the flow of an ideal gas can be considered to be incompressible when Ma ≲ 0.3. For atmospheric air at normal conditions, this corresponds to a flow speed of about 100 m/s or 360 km/h.
