Question 13.4: If the whirl velocities in an axial stage are expressed as C......
If the whirl velocities in an axial stage are expressed as
C_{\text{u1}}=aR^{n}-\frac{b}{R} , \quad C_{\text{u2}}=aR^n+\frac{b}{R}
Prove that if a constant specific work is assumed and the exponent, n = 2, then
C^2_{\text{a1}}-C_{\text{a2}}^2=12ab(R-1)
| TABLE 13.3 Governing Equations for Different Design Methods | |||
| C_{\text{u1}}=aR^n-\frac{b}{R} \\ a=U_{\text{m}}(1-\Lambda_{\text{m}}) | a=U_{\text{m}}(1-\Lambda_{\text{m}}) \\ b=Cp\Delta T_{0s}/2U_{\text{m}} | ||
| Blading | C_{\text{a}} | Λ | C_{\text{u}} |
| Free vortex
n = -1 |
C_{\text{a1}}=C_{\text{a1m}} \\ C_{\text{a2}}=C_{\text{a2m}}\\C_{\text{a1m}}=C_{\text{a2m}} | \Lambda=1-\frac{a}{U_{\text{m}}R^2} \\ \Lambda=1-\frac{(1-\Lambda_{\text{m}})}{R^2} | C_{\text{u1}}=\frac{a-b}{R} \\ C_{\text{u2}}=\frac{a+b}{R} |
| Exponential
n = 0 |
C_{\text{a1}}^2-C_{\text{a1m}}^2=-2\left[a^2 \ell nR+\frac{ab}{R}-ab \right] \\ C_{\text{a2}}^2-C_{\text{a2m}}^2=-2\left[a^2 \ell nR-\frac{ab}{R}+ab \right] \\ C_{\text{a1m}}=C_{\text{a2m}} \\ C_{\text{a2}}^2-C_{\text{a1}}^2=4ab\left(\frac{1}{R}-1 \right) | \Lambda=1+\frac{a}{U_{\text{m}}} -\frac{2a}{U_{\text{m}}R} \\ \Lambda =1+\left(1-\frac{2}{R} \right) (1-\Lambda_{\text{m}}) \\ \quad \\ \quad | C_{\text{u1}}=a-\frac{b}{R} \\ C_{\text{u2}}=a+\frac{b}{R} \\ \quad \\ \quad |
| First power
n=1 |
C_{\text{a1}}^2=C_{\text{a1m}}^2=-2[a^2(R^2-1)-2ab \ell nR] \\ C_{\text{a2}}^2-C^2_{\text{a2m}}=-2[a^2(R^2-1)+2ab \ell n R]\\ C^2_{\text{a2}}-C_{\text{a1}}^2=-8ab \ell n R \\ C_{\text{a1m}}=C_{\text{a2m}} | \Lambda=1+\frac{2a \ell nR}{U_{\text{m}}} -\frac{a}{U_{\text{m}}} \\ \Lambda=1+(2\ell nR-1)(1-\Lambda_{\text{m}}) \\ \quad \\ \quad | C_{\text{u1}}=aR-\frac{b}{R} \\ C_{\text{u2}}=aR+\frac{b}{R} |
The vortex energy equation is expressed by
\frac{\text{d}h_o}{\text{d}r} =C_{\text{a}}\frac{\text{d}C_{\text{a}}}{\text{d}r} +C_{\text{u}}\frac{\text{d}C_{\text{u}}}{\text{d}r} +\frac{C_{\text{u}^2}}{r}
Since the specific work is constant in the radial direction (or \text{d}h_0/\text{d}r=0), then the vortex energy equation is reduced to
-C_{\text{a}}\frac{\text{d}C_{\text{a}}}{\text{d}r} =C_{\text{u}}\frac{\text{d}C_{\text{u}}}{\text{d}r} +\frac{C_{\text{u}^2}}{r} \\ -C_{\text{a}}\text{d}C_{\text{a}}=C_{\text{u}}\text{d}C_{\text{u}}+\frac{C^2_{\text{u}}}{r} \text{d}r \\ R=\frac{r}{r_{\text{m}}} \\ -\int \limits^{C_{\text{a}}}_{C_{\text{am}}}C_{\text{a}}\text{d}C_{\text{a}}=\int \limits^{C_{\text{u}}}_{C_{\text{um}}}C_{\text{u}}\text{d}C_{\text{u}}+\int \limits^{R}_{1}C^2_{\text{u}}\frac{\text{d}R}{R} \\ -\frac{1}{2}[C_{\text{a}}^2-C_{\text{am}}^2] =\frac{1}{2}[C_{\text{u}}^2-C^2_{\text{um}}] +I \quad \quad \quad (1)
where I=\int \limits ^{R}_{I}C_{\text{u}}^2\frac{\text{d}R}{R}
n=2 \\ C_{\text{u1}}=aR^2-\frac{b}{R} \\ C_{\text{u2}}=aR^2+\frac{b}{R}
At station (1), then equation (1) has the form
-\frac{1}{2}[C_{\text{a1}}^2-C_{\text{a1m}}^2] =\frac{1}{2} [C_{\text{u1}}^2-C_{\text{u1m}}^2]+\int \limits ^R_1\left(aR^2-\frac{b}{R} \right) ^2\frac{\text{d}R}{R} \\ -\frac{1}{2}[C_{\text{a1}}^2-C_{\text{a1m}}^2] =\frac{1}{2} [(aR^2-\frac{b}{R} )^2-(a-b)^2]+\int \limits ^R_1 (a^2R^3-2ab+\frac{b^2}{R^3} )\text{d}R \\ -\frac{1}{2}[C_{\text{a1}}^2-C_{\text{a1m}}^2] =\frac{1}{2} [(aR^2-\frac{b}{R} )^2-(a-b)^2]+[\frac{a^2R^4}{4}-2abR-\frac{b^2}{2R^2} ]^R_1 \\ -\frac{1}{2}[C_{\text{a1}}^2-C_{\text{a1m}}^2] =\frac{1}{2} [(a^2R^4-2abR+\frac{b^2}{R^2} )-(a^2-2ab+b^2)] +[(\frac{a^2R^4}{4}-2abR-\frac{b^2}{2R^2} )-(\frac{a^2}{4}-2ab-\frac{b^2}{2} )] \\ C_{\text{a1}}^2-C_{\text{a1m}}^2 = (\frac{3}{2}a^2-6ab )-(\frac{3}{2}a^2R^4-6abR )
Similarly
C^2_{\text{a2}}-C^2_{\text{a2m}}=(\frac{3}{2}a^2+6ab )=(\frac{3}{2}a^2R^4+6abR )
Since C_{\text{a1m}}=C_{\text{a2m}}
C_{\text{a1}}^2-C_{\text{a2}}^2=12ab(R-1)
From the above equation, the following conclusions are deduced:
| R < 1 | R = 1 | R > 1 |
| C_{\text{a1}} \lt C_{\text{a2}} | C_{\text{a1m}} = C_{\text{a2m}} | C_{\text{a1}} \gt C_{\text{a2}} |