Question 3.11: Bernoulli’s Equation Revisited Show that for adiabatic frict......
Bernoulli’s Equation Revisited
Show that for adiabatic frictionless (i.e., isentropic) flow the Bernoulli equation, derived in Sect. 3.2 from the momentum equation, is identical to the energy equation.
Equation (3.27) with \dot Q =\dot W = 0 can be written for two points on a horizontal streamline as:
(Energy Conservation) {\rm\dot Q-\dot W=\frac{\dot m}{2} (v_2^2-v_1^2)+\dot H_1-\dot H_2}\\{\rm +\dot mg\Delta z} (3.27)
\mathrm{h}_{1}+{\frac{\mathrm{u}_{1}^{2}}{2}}=\mathrm{h}_{2}+{\frac{\mathrm{u}_{2}^{2}}{2}}=¢ (E.3.11.1)
Recall from Sect. 3.2 that for steady inviscid flow, the Euler equation can be written as
\rm udu+\frac{dp}{\rho} =0 (E.3.11.2a)
or in integrated form as the Bernoulli equation, i.e.,
\rm \frac{u^2}{2}+\int\frac{dp}{\rho} =¢ (E.3.11.2b)
Now, for isentropic flow, i.e., zero entropy changes, Eq. (3.35a) reduces to ( v = 1/ρ ):
\rm Tds = dh – \nu dp (3.35a)
\rm 0=dh-vdp\,~~or~~\,dh=\frac{dp}{\rho} (E.3.11.3a,b)
Using (E.3.11.3b) in (E.3.11.2a) we obtain
\rm udu+dh=0 (E.3.11.4)
which is exactly the differential form of the energy equation (E.3.11.1).