Model

The mass (continuity equation) balance can be written as

$$\dot{m} = \rho A U = constant$$ (9.1)

$$\hookrightarrow \rho_1 U_1 = \rho_2 U_2$$

The energy conservation (under the assumption that this model is adiabatic flow and the friction is not transformed into thermal energy) reads

$${T_{0}}_1 = {T_{0}}_2$$

$$\hookrightarrow T_1 + { {U_1}^2 \over 2c_p} = T_2 + { {U_2}^2 \over 2c_p}$$ (9.2)

Or in a derivative form

$$C_p dT +d \left( U^2 \over 2 \right) = 0$$ (9.3)

Again for simplicity, the perfect gas model is assumed^9.3.

$$P = \rho R T$$ (9.4)

$$\hookrightarrow {P_1 \over \rho_{1} T_{1}} = {P_2 \over \rho_{2} T_{2}}$$

It is assumed that the flow can be approximated as one-dimensional. The force acting on the gas is the friction at the wall and the momentum conservation reads

$$-AdP -\tau_{w} dA_{w} = \dot{m} dU$$ (9.5)

It is convenient to define a hydraulic diameter as

$$D_{H} = {4 \times \hbox{Cross Section Area} \over \hbox{ wetted perimeter }}$$ (9.6)

Or in other words

$$A = {\pi {D_H}^2 \over 4}$$ (9.7)

It is convenient to substitute $D$ for $D_H$ and yet it still will be referred to the same name as the hydraulic diameter. The infinitesimal area that shear stress is acting on is

$$dA_w = \pi D dx$$ (9.8)

Introducing the Fanning friction factor as a dimensionless friction factor which is some times referred to as the friction coefficient and reads as the following:

$$f = { \tau_w \over \half \rho U^{2} }$$ (9.9)

By utilizing equation (9.2) and substituting equation (9.10) into momentum equation (9.6) yields

$$- \overbrace{\pi D^2 \over 4}^{A} dP - { \pi D dx } \overbrace{ f... ...\rho U^{2} \right) }^{\tau_w} = A { \overbrace{\rho U}^{\dot {m} \over A } dU }$$ (9.10)

Dividing equation (9.11) by the cross section area, $\mathbf{A}$ and rearranging yields

$$- dP + { 4 f dx \over D} \left( \half \rho U^{2} \right) = {\rho U dU }$$ (9.11)

The second law is the last equation to be utilized to determine the flow direction.

$$s_2 \geq s_1$$ (9.12)