Question 5.3: Figure 5.13 shows a single-stage turbine and its map.The cru......

Part Ia: In order to calculate the cruise-operation power delivered, we proceed as follows:

w_{s}=U(V_{\theta\ 1}-V_{\theta\ 2})=U^{2}\left(\mathrm{because}\,V_{\theta\ 2}=0\right)

U=447.2\ {\mathrm{m/s}}

N={\frac{U}{r_{m}}}{\bigg(}{\frac{60}{2\pi}}{\bigg)}=33,892\ \mathrm{rpm}

{\frac{N}{\sqrt{\theta_{1}}}}\approx18,000\,\mathrm{rpm}

However,

w_{s}=\eta_{T} c_{p}T_{t\ 1}\biggl[1-\biggl({\frac{p_{t\ 2}}{p_{t\ 1}}}\biggr)^{\frac{\gamma-1}{\gamma}}\biggr]

which yields

{\frac{p_{t\ 1}}{p_{t\ 2}}}=2.6

We are now in a position to use the given turbine map to attain the following:

\frac{\dot{m}{\sqrt{\theta_{1}}}}{\delta_{1}}=2.55\ \mathrm{kg/s}

which, upon substitution, gives the physical mass-flow rate:

\dot{m}_{c r u i s e}=11.52\ \mathrm{kg/s}

Now, the cruise-operation power (P ) delivered by the turbine stage can be computed:

P=\dot m\eta_{T}c_{p}T_{t\ 1}\biggl[1-\left(\frac{p_{t\ 2}}{p_{t\ 1}}\right)^{\frac{\gamma-1}{\gamma}}\biggr]=2303\,\mathrm{kW}

Part Ib: With the magnitude of γ being 1.365, let us calculate the c_{p} magnitude:

c_{p}=\left({\frac{\gamma}{\gamma-1}}\right)R=1073.3\ \mathrm{J/}(\mathrm{kg}\ \mathrm{K})

Now, we calculate the stage-exit total temperature and pressure:

p_{t\ {2}}={\frac{8.5}{2.6}}=3.27\,\mathrm{bars}

T_{t\ 2}=T_{t\ 1}-{\frac{w_{s}}{c_{p}}}=833.7\,\mathrm{K}

Applying the continuity equation at the stage exit station, we have

\dot m=\rho_{2}V_{z}(2\pi r_{m}h_{2})

The simplification in the problem statement – that the stage-exit flow stream is incompressible – means that

\rho_{2}\approx\rho_{t\ 2}={\frac{p_{t\ 2}}{R T_{t\ 2}}}=1.37\ \mathrm{kg/m}^{3}

which yields

V_{z} = 177.4 m/s

In computing the inlet and exit total relative pressures, it is important to recall that in an axial-flow adiabatic rotor (such as the current rotor), the total relative temperature T_{t\ r} remains constant. Now, let us proceed with the aim being the inlet and exit total relative pressures:

V_{1}={\sqrt{V_{\theta\ 1}{}^{2}+V_{z}{}^{2}}}=481.1\ {\mathrm{m/s}}

T_{t,r,1}=T_{t\ 1}-\left(\frac{V_{1}{}^{2}-W_{1}{}^{2}}{2c_{p}}\right)=926.8\,\mathrm{K}=\,T_{t,r,2}

p_{t,r,1}=p_{t\,1}\bigg(\frac{T_{t,r,1}}{T_{t\,1}}\bigg)^{\frac{\gamma}{\gamma-1}}=5.94\,\mathrm{bars}

p_{t,r,2}=p_{t\,1}\left(\frac{T_{t,r,2}}{T_{t\ 2}}\right)^{\frac{\gamma}{\gamma-1}}=4.86\,\mathrm{bars}

\frac{\Delta p_{t_{r}}}{p_{t,r,1}}=18.2%

Part IIa: Equality of the corrected flow rate under the cruise and rig operating conditions will now enable us to compute the rig “physical” mass-flow rate. The rig speed, however, will be the result of equating the corrected speed, as follows:

{\dot{m}}_{{rig}}={\dot{m}}_{c r u i s e}\sqrt{\frac{T_{t\ 1{cr u i s e}}}{T_{t\ 1{rig}}}}\times\frac{p_{t\ 1ri g}}{p_{t\ 1cr u i s e}}=26.06\,\mathrm{Kg/s}

N_{r i g}=N_{c r u i s e}\sqrt{\frac{T_{t\ 1r i g}}{T_{t\ 1c r u i s e}}}=26,530\,\mathrm{rpm}

Thus

\tau_{r i g}=\left(\frac{\dot m w_{s}}{\omega}\right)_{r i g}=1148.9\,\mathrm{N}\cdot\mathrm{m}

Part IIb:

\left(\frac{V_{2}}{V_{c r_{2}}}\right)_{rig}=\left(\frac{V_{2}}{V_{c r_{2}}}\right)_{c r u i s e}=0.338

(\eta_{t-t})_{r i g}=(\eta_{t-t})_{c ru i s e}=0.81

With this total-to-total efficiency magnitude, and substituting in equation (3.59), we obtain the exit total temperature in the rig:

T_{t\ r}=T+{\frac{W^{2}}{2c_{p}}}   (3.59)

(\,T_{t\ 2})_{r i g}=510.9\,\mathrm{K}

The exit magnitude of critical velocity can now be calculated:

(V_{c r_{2}})_{ri g}={\sqrt{\left({\frac{2\gamma}{\gamma+1}}\right)R T_{t\ 2ri g}}}=411.4\ m/s

Finally, the exit velocity (V_{2}) in the test rig can now be calculated:

(V_{2})_{r i g}=\left(\frac{V_{2}}{V_{c r_{2}}}\right)_{r i g}\times V_{c r_{2}}=139.1\,\mathrm{m/s}