Energy equation for a one-dimensional control volume

Figure 3.22 : Control Control Volume for an one-dimensional steady flow

 

Consider the one-dimensional control volume that we have analysed before and shown in Fig.3.22. If we interpret the velocity, density, pressure and other variables to be uniform across the ends or that they are the averaged values we have for a steady flow 

$$\dot{Q}~-~\dot{W_s} =~-(\rho~V~A)_1~(h~+gz~+~{1 \over 2} V^2)_1+(\rho~V~A)_2~(h~+gz~+~{1 \over 2} V^2)_2$$ (3.64)

$$\noalign{Noting that for continuity $(\rho~v~A)_1~=~(\rho~v~A)_2~=~\dot{m}$} \texttt{we have}$$

$$\dot{Q}~-~\dot{W_s} =~-\dot{m}~(h~+gz~+~{1 \over 2} V^2)_1+\dot{m}~(h~+gz~+~{1 \over 2} V^2)_2$$ (3.65)

On division by $\dot{m}$ and denoting $\dot{Q} \over \dot{m}$ by q and $\dot{W} \over \dot{m}$ by $w_s$ we have after rearrangement of terms,

$$(h~+gz~+~{1 \over 2} V^2)_1~=~(h~+gz~+~{1 \over 2} V^2)_2~-q~+w_s$$ (3.66)

Note that the term $(h~+gz~+~{1 \over 2} V^2)$ is equal to the Total enthalpy denoted by H₀. Accordingly the Eqn.3.66 becomes

$$H_{01}~=~H_{02}~-q~+w_s$$ (3.67)

That is to say that the total enthalpy of a control volume is conserved unless heat or work is added to / taken out of the control volume.

(c) Aerospace, Mechanical & Mechatronic Engg. 2005
University of Sydney