Question 3.12: Compressibility Factor Compare incompressible vs. compressib......

(A) Incompressible Flow: Applying Bernoulli between pointⓢ 0 anⓓ s with Δz = 0 and $u _ s =0$ at the stagnation point,

$\mathrm{p}_{s}=\mathrm{p}_{0}+\frac{\mathsf{\rho}}{2}\mathrm{u}_{0}^{2}$                               (E.3.12.1)

Thus, with $ρ_0 = ρ_s = ρ$ for incompressible flow (see given β – definition):

To express the stagnation point temperature, we use (see ideal gas):

$\rm {\frac{T_{s}}{T_{0}}}={\frac{\bf p_{s}}{\bf p_{0}}}$                       (E.3.12.2a)

so that with Eq. (E.3.12.1):

$\rm {T}_{{s}}={ T}_{0}\bigl({ p}_{{{s}}}/{ p}_{0})={T}_{0}\Biggl[{ l}+\frac{{u}_{0}^{2}}{2}\left\lgroup\frac{ \rho}{ p}_{0}\right\rgroup \Biggr]$                         ( E.3.12.2b)

Employing the ideal gas law in the form $\rm \rho/p_0=(RT_0)^{-1}$, we rewrite
Eq. (E.3.12.2b) as:

$\rm T_s=t_0+\frac{u_0^2}{2R}$                         (E.3.12.2c)

(B) Compressible Flow: For this situation, Bernoulli’s equation reads:

$\rm \int\frac{dp}{\rho}+\frac{u^2}{2}=¢$       (E.3.12.3a)

Expressing ρ as ρ(p) = (p / C)$^{1/k}$, C ≡ p$^{1/k}$ / ρ , we can integrate the first term, i.e.,

$\rm \int \frac{dp}{\rho} =\frac{k}{k-1} \left\lgroup\frac{p}{\rho} \right\rgroup$ + constant

Hence, Eq. (E.3.12.3a) now reads:

$\rm \frac{k}{k-1} \frac{p}{\rho}+\frac{u^2}{2} =¢$                       (E.3.12.3b)

Applying Eq. (E.3.12.3b) to points ⓪ and ⓢ yields:

$\rm \frac{k}{k-1} \left\lgroup\frac{ps}{\rho_s} \right\rgroup =\frac{k}{k-1} \left\lgroup\frac{p_0}{\rho_0} \right\rgroup +\frac{u_0^2}{2}$                 (E.3.12.3c)

Expressing Eq. (3.12.3c) again in terms of $\rm T_s$ with the equation of state p / ρ = RT , we have

$\rm \mathrm{T}_{s}=\mathrm{T}_{0}+\frac{\mathrm{k}-1}{\mathrm{k}}\frac{\mathrm{u}_{0}^{2}}{2\mathrm{R}}$                     (E.3.12.4a)

Comparing (E.3.12.4a) with (E.3.12.2c) it is already evident that the second-term coefficient (k − 1)/k (≈0.3 for air) captures the difference between incompressible and compressible flows. In order to express β in terms of the Mach number, we use Eq. (3.42b) which provides $\rm kR=c^2/T_0~\text{and}~u_0^2/c^2\equiv M_0^2$, i.e.,

$\rm c=\sqrt{kRT}$                              (3.42b)

$\rm T_{s}={\mathrm T_{0}}\!\left[1+\frac{1}{2}({ k}-1){ M}_{0}^{2}\right]$                            (E.3.12.4b)

Now, with Eq. (3.39) we obtain

$\rm\frac{T_2}{T_1}=\left\lgroup\frac{p_2}{p_1} \right\rgroup^{(k-1)/k} \qquad\text{and}\qquad\frac{p_2}{p_1}=\left\lgroup\frac{\rho_2}{\rho_1} \right\rgroup ^k$                                          (3.39/40)

$\rm { p}_{s}={ p}_{0}\left\lgroup\frac{{T}_{z}}{{ T}_{0}}\right\rgroup ^{\frac{k}{{ k}-1}}={ p}_{0}\left\lgroup1+\frac{{ k}-1}{2}{ M}_{0}^{2}\right\rgroup ^{\frac{k}{{ k}-1}}$                            (E.3.12.5a)

For subsonic flow $\rm \frac{k-1}{2}M_0^2\lt 1$ and Eq. (E.3.12.5a) can be expanded with the binominal theorem to (see App. A):

In order to form the compressibility factor, we rearrange (E.3.12.5b) as

$\rm \mathrm{p}_{s}-{ p}_{0}=\frac{ k}2{p}_{0}{ M}_{0}^{2}\biggl[1+\frac{1}{4}{ M}_{0}^{2}+\frac{2-{\bf k}}{24}{ M}_{0}^{4}+…\biggr]$                        (E.3.12.5c)

However, by definition the factor

so that

$\rm \mathrm{p}_{s}-\mathrm{p}_{o}=\frac{ \rho_{o}}2u_{0}^{2}\bigg[1+\frac14{ M}_{0}^{2}+\frac{2- k}{24}{ M}_{0}^{4}+…\bigg]$                         (E.3.12.5d)

Finally,

Graph:

Comment: The graph β(M) indicates that for M < 0.3 in air (being an ideal gas), the error in assuming incompressible flow is less than 2%; however, at M = 1 the error is 28%.