The system on a modern engine can be modeled as a turbine which extracts energy
ID: 1841724 • Letter: T
Question
The system on a modern engine can be modeled as a turbine which extracts energy from the hot exhaust connected compressor, which compresses the ambient air to higher fixed shaft which powers pressure (shown in figure). The air volume the compressor niet is 2.14m3/min. mass of exhaust gas through the turbine is the sum of the air mass flow and fuel mass flow into the engine, but, mar P, 160 kPa Exh Gas and therefore mcomp as mturb). The air can be modeled as an ideal gas with c 1.006 k kg-k, and the gas be modeled as an ideal gas with c 1.070 kJ/kg-K Assume R 0.287 both gases. The turbine is well insulated, but the compressor s not. Determine a) (5 pts The mass of air through the system in [kg/sg 120kPa P 100kPa (4) 350°C b (5 pts) The shaft power (hy,) delivered by the turbine in [kWn.(Don't forget to draw your system!)Explanation / Answer
a)
Air density at compressor inlet, rho1 = P1 / (R*T1)
= 100 / (0.287*(20+273))
= 1.189 kg/m3
Mass flow rate = density*volume flow rate
= 1.189*2.14
= 2.545 kg/min
= 2.545/60 kg/s
= 0.0424 kg/s
b)
Applying first law to turbine, Q - W = dH
Q = 0 since turbine is insulated.
W = -dH
W = m*Cp*(T3 - T4)
Turbine power = m*Cp*(T3 - T4)
= 0.0424*1.070*(400 - 350)
= 2.269 kW
c)
For turbine entropy generation,
ds = m*Cp ln (T4 / T3) - m*R ln (P4 / P3)
= 0.0424*1.070 ln ((350+273) / (400+273)) - 0.0424*0.287 ln (120 / 220)
= 0.00387 kW/K
d)
For compressor, Q - W = dH
dH = m*Cp*(T2 - T1)
= 0.0424*1.006*(70 - 20)
= 2.133 kW
Q - (-2.269) = 2.133......W will be negative for compressor as work is done on the air.
Q = -0.136 kW........negative sign indicates heat lost to surroundings.
e)
Sgen = Q / T
= -0.136 / 300
= -0.00045 kW/K
f)
T2 / T1 = (P2 / P1)(k-1)/k
T2is / (20+273) = (160/100)(1.4-1)/1.4
T2is = 335 K = 62 deg C
g)
Eff = (T2is - T1) / (T2 - T1)
= (62 - 20) / (70 - 20)
= 0.842 or 84.2%
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