Question 5.1: The design-point operation of a single-stage axial-flow turb......
The design-point operation of a single-stage axial-flow turbine (Fig. 5.11) is simulated in a cold rig. The two sets of operating conditions are as follows:
Design-point operation mode
• Stage-inlet total pressure p_{t\ 0} = 12 bars
• Stage-inlet total temperature T_{t\ 0} = 1420 K
• Stage-exit total pressure p_{t\ 2} = 5.4 bars
• Rotational speed (N) = 58, 000 rpm
• Mass-flow rate {\dot{m}} = 3.8 kg/s
• Rotor-inlet relative flow angle (β_{1}) = 0.0°
• Stage efficiency (η_{t−t} ) = 86%
• Stator-exit critical Mach number (V/V_{cr} )_{1} = 0.96
Cold-rig operation mode
• Stage-inlet total pressure = 2.6 bars
• Stage-inlet total temperature = 388 K
The stage has a mean radius r_{m} of 0.1 m, and the axial velocity component across the stage is assumed constant, which applies to both sets of operating conditions.
With an isentropic stator assumption and an average specific-heat ratio γ of 1.365 (to apply to both sets of operating conditions), calculate:
a) The stage-exit total temperature T_{t\ 2} in the cold rig;
b) The power produced by the turbine stage in the cold rig.
Part a: According to the rules of dynamic similarity, we have
\left({\frac{p_{t\ 0}}{p_{t\ 2}}}\right)_{rig}= \left({\frac{p_{t\ 0}}{p_{t\ 2}}}\right)_{d e s.~p t.}=2.22
Also
\eta_{rig}=\eta_{d e s.\;p t.}=0.86=\frac{1-(T_{t\ 2}/T_{t\ 0})}{1-(p_{t\ 2}/p_{t\ 0})^{\frac{\gamma-1}{\gamma}}}
which yields
(T_{t\ 2})_{rig}=323.9\,\mathrm{K}
Part b: The dynamic similarity rules also require equality of the corrected flow rate, meaning that
\left({\frac{\dot m {\sqrt{\theta_{0}}}}{\delta_{0}}}\right)_{r i g} =\left({\frac{\dot m\sqrt{\theta_{0}}}{\delta_{0}}}\right)_{d e s.~p t.}
which, upon substitution, gives the “physical” mass-flow rate in the rig as follows:
{\dot{m}}_{r i g}=1.58,\ \mathrm{kg/s}
Finally, we can calculate the stage-supplied power (P ) in the test rig as follows:
P_{rig}=\dot m_{r i g}c_{p}(T_{t\ 0}-T_{t\ 2})=108.4\ \mathrm{kW}