Comparison with Incompressible Flow

The Mach number of the flow in some instances is relatively small. In these cases, one should expect that the isothermal flow should have similar characteristics as incompressible flow. For incompressible flow, the pressure loss is expressed as follows

$$P_{1} - P_{2} = \frac{4fL}{D} { U ^{2} \over 2}$$ (8.39)

Now note that for incompressible flow $U_{1} = U_{2}= U$ and $\frac{4fL}{D}$ represent the ratio of the traditional $h_{12}$ . To obtain a similar expression for isothermal flow, a relationship between $M_{2}$ and $M_{1}$ and pressures has to be derived. From equation (8.39) one can obtained that

$$M_{2} = M_{1} {P_{1} \over P_{2}}$$ (8.40)

Substituting this expression into (8.40) yields

$$\frac{4fL}{D} = {1 \over k{M_{1}}^{2}} \left( 1 - \left( {P_{2} \over P_{1}} \right)^{2} \right) - \ln \left( {P_{2} \over P_{1}} \right)^{2}$$ (8.41)

Because $f$ is always positive there is only one solution to the above equation even though M2.

Expanding the solution for small pressure ratio drop, $P_{1} - P_{2} /P_{1}$ , by some mathematics.

denote

$$\chi = {P_{1} -P_{2} \over P _{1} }$$ (8.42)

Now equation (8.41) can be transformed into

$$\frac{4fL}{D} = { 1 \over k {M_{1}}^{2}} \left( 1 - \left( { {P_{... ...}} \right)^{2} \right) -\ln \left( { 1 \over { P_{2} \over P_{1} }} \right)^{2}$$ (8.43)

$$\frac{4fL}{D} = { 1 \over k {M_{1}}^{2}} \left( 1 - \left( { 1 -\chi } \right)^{2} \right) -\ln \left( { 1 \over 1 -\chi } \right)^{2}$$ (8.44)

$$\frac{4fL}{D} = { 1 \over k {M_{1}}^{2}} \left( 2\chi - \chi^{2} \right) -\ln \left( { 1 \over 1 -\chi } \right)^{2}$$ (8.45)

now we have to expand into a series around $\chi=0$ and remember that

$$f(x) = f(0) + f'(0) x + f''(0) {x^2 \over 2} + 0\left(x^{3}\right)$$ (8.46)

and for example the first derivative of

$$\left. {d \over d\chi } \ln \left( { 1 \over 1 -\chi } \right)^{2} \right\vert _{\chi= 0} =$$

$$\left. \left( 1 - \chi \right)^{2} \times {\left[ (-2) (1-\chi)^{-3} \right] } {(-1)} \right\vert _{\chi=0} = 2$$ (8.47)

similarly it can be shown that $f''(\chi=0)=1$ equation (8.45) now can be approximated as

$$\frac{4fL}{D} = {1 \over k{M_{1}}^{2}} (2\chi - \chi^{2} ) - \left( 2 \chi - \chi^{2}\right) + f \left( \chi^{3} \right)$$ (8.48)

rearranging equation (8.48) yields

$$\frac{4fL}{D} = {\chi \over k{M_{1}}^{2}} \left[ (2 - \chi ) - k{M_{1}}^{2} \left( 2 - \chi \right) \right] + f \left( \chi^{3} \right)$$ (8.49)

and further rearrangement yields

$$\frac{4fL}{D} = {\chi \over k{M_{1}}^{2}} \left[ 2(1 - k{M_{1}}^{2} ) - \left( 1 + k{M_{1}}^{2} \right) \chi \right] + f \left( \chi^{3} \right)$$ (8.50)

in cases that $\chi$ is small

$$\frac{4fL}{D} \approx {\chi \over k{M_{1}}^{2}} \left[ 2(1 - k{M_{1}}^{2} ) - \left( 1 + k{M_{1}}^{2} \right) \chi \right]$$ (8.51)

The pressure difference can be plotted as a function of the $M_{1}$ for given value of $\frac{4fL}{D}$ . Equation (8.51) can be solved explicitly to produce a solution for

$$\chi = { 1 -k{M_{1}}^{2} \over 1 +k{M_{1}}^{2}} - \sqrt{{ 1 -k{M_... ...} \over 1 +k{M_{1}}^{2}} - {k{M_{1}}^{2} \over 1 +k{M_{1}}^{2} } \frac{4fL}{D}}$$ (8.52)

A few observations can be made about equation (8.52).