# 11.7.1: The Control Volume Analysis/Governing equations

Figure 11.17 describes the flow of gas from the left to the right. The heat transfer up stream (or down stream) is assumed to be negligible. Hence, the energy equation can be written as the following:

$\dfrac{d\, Q }{ \dot{m} } = c_p dT + d \dfrac{U^2 }{ 2} = c_p dT_{0} \label{isothermal:eq:CV} \tag{1}$
The momentum equation is written as the following
$-A\, dP - \tau_{w}\, dA_{\text{wetted area}} = \dot{m}\, dU \label{isothermal:eq:momentum} \tag{2}$

where $$A$$ is the cross section area (it doesn't have to be a perfect circle; a close enough shape is sufficient.). The shear stress is the force per area that acts on the fluid by the tube wall. The $$A_{wetted\;\;area}$$ is the area that shear stress acts on. The second law of thermodynamics reads

${s_2 - s_1 \over C_p} = \ln {T_2 \over T_1 } - {k -1 \over k} \ln {P_2 \over P_1} \label{isothermal:eq:2law} \tag{3}$
The mass conservation is reduced to
$\dot {m} = \text{constant} = \rho\, U\, A \label{isothermal:eq:mass} \tag{4}$
Again it is assumed that the gas is a perfect gas and therefore, equation of state is expressed as the following:
$P = \rho\, R\, T \label{isothermal:eq:state} \tag{5}$

### Contributors

• Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.