Question 5.2: Figure 5.12 shows the map of a compressor in a turboprop eng......
Figure 5.12 shows the map of a compressor in a turboprop engine. The cruise operation of the compressor is defined as follows:
• The power supplied P_{alt} = 836.0 kW
• The mass-flow rate \dot m_{alt} = 3.8 kg/s
• The inlet total pressure (p_{t\ in})_{alt} = 0.235 bars
• The inlet total temperature (T_{t\ in})_{alt} = 216.0 K
• Total-to-total efficiency η_{C} = 80%
These operating conditions are simulated in a test rig utilizing air at inlet total pressure and temperature of 1.0 bar and 288 K, respectively. Assuming an average specific-heat ratio of 1.4 for both sets of operating conditions, calculate the following variables in the test rig:
a) The “physical” shaft speed (N) in rpm.
b) The torque ( \tau ) transmitted to the compressor.
Part a: Let us first calculate the cruise-operation total-to-total pressure ratio (Pr), starting with the supplied-power expression, namely
P={\frac{\dot m c_{p}T_{t\ i n}}{\eta_{C}}}\biggl[\biggl({\frac{p_{t\ e x}}{p_{t\ i n}}}\biggr)^{\frac{\gamma-1}{\gamma}}-1\biggr]
which, upon substitution, yields
(P r)_{c r ui s e}={\frac{P_{t\ e x}}{P_{t\ i n}}}\approx8.0
Also,
\frac{\dot m\sqrt{T_{t\ i n}}}{\delta_{ i n}}\approx14.0\,\mathrm{kg/s}
These newly computed variables allow us to place the point signifying the cruise operation mode on the compressor map, as shown in Figure 5.12. Now we can read off the cruise-operation corrected-speed magnitude:
\left(\frac{N}{\sqrt{\theta_{i n}}}\right)_{c r u i s e}=36,000\ \mathrm{rpm}
The foregoing three variables correspond not only to the compressor cruise operation but also its rig operation because the two operating modes are dynamically similar. Let us now use the equality of corrected speeds:
\left(\frac{N}{\sqrt{\theta_{i n}}}\right)_{r i g}=\left(\frac{N}{\sqrt{\theta_{i n}}}\right)_{c r u i s e}=36,000\ \mathrm{rpm}
where
(\theta_{i n})_{r i g}=\frac{T_{t\ i n}}{T_{S T P}}=\frac{288.0}{288.0}=1.0
which means that the “physical” speed in the test rig is
N_{r i g}=36,000\ \mathrm{rpm}
Part b: Next, let us implement the equality of the corrected flow rate under both sets of operating conditions:
\left(\frac{\dot m{\sqrt{\theta_{i n}}}}{\delta_{i n}}\right)_{r ig}=14.0\ \mathrm{kg/s}
where
(\delta_{i n})_{r i g}={\frac{p_{t\ i n}}{p_{ S T P}}}={\frac{1.0}{1.0}}=1.0
The rig-operation “physical” mass-flow rate can now be determined:
(\dot m)_{r i g}=14.0\ \mathrm{kg/s}
Finally, we can calculate the shaft-transmitted torque (\tau ) in the rig as follows:
\tau_{r i g}={\frac{(P o w e r)_{r i g}}{\omega_{r i g}}}=1089.6 N · m