mech3020-s11-hw02-soln5

# mech3020-s11-hw02-soln5 - = P 3 r c r k = 378 kPa The net...

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MECH 3020 Thermodynamics II Spring 2011: Homework set #2 5. Say a 4–stroke Diesel engine, with r = 18, q H = 1500 kJ/kg, and V D = 12 L, is operating at 1200 RPM. (a) Calculate the power output from the engine. Use the usual assumptions for the initial state. (b) Calculate the mass ﬂow rate through the engine. This will be ˙ m = m RPM 60 · 2 , with m = P 1 V 1 RT 1 (c) Calculate the temperature and pressure at the end of the power stroke, state 4. (d) Say now the exhaust stream leaves the engine at temperature T 4 and pressure P 4 with mass ﬂow ˙ m . How much power could be derived from this ﬂow via a reversible expansion across an adiabatic turbine to 1 atm? What would the net thermal ef- ﬁciency of the engine now be (including the engine and the turbine)? The mass of air in the engine is m = P 1 V 1 RT 1 = P 1 V D RT 1 (1 - 1 /r ) = 0 . 0149 kg The cycle is now analyzed step–by–step, using a con- stant C P assumption: T 2 = T 1 r k - 1 = 953 K , P 2 = P 1 r k = 5794 kPa T 3 = T 2 + q H C P = 2441 K , P 3 = P 2 r c = T 3 T 2 = 2 . 56 T 4 = T 3 ± V 4 V 3 ² k - 1 = ³ r c r ´ k - 1 = 1119 K P 4
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Unformatted text preview: = P 3 r c r k = 378 kPa The net work per unit mass for the Diesel cycle alone is w d = q H-q L = q H-C V ( T 4-T 1 ) = 910 . 2 kJ / kg Alternatively, we could calculate the thermal eciency using d = 1- r 1-k r k c-1 k ( r c-1) = 0 . 607 and w d = d q H = 910 . 2 kJ / kg The mass ow rate is the mass per cycle times cycles per second: m = m RPM 60 2 = 0 . 149 kg / s so the power is W D = mw d = 135 . 6 kW An isentropic turbine expanding to P 5 = 101 . 3 kPa would exit at T 5 = T 4 P 5 P 4 ( k-1) /k = 768 K and w t = C P ( T 4-T 5 ) = 353 . 7 kJ / kg So the turbine adds over 1/3 extra power to the cycle, yet the heat input is the same. The overall thermal eciency is = w d + w t q H = 0 . 84 The total power produced is W = m ( w d + w t ) = 188 . 9 kW 1...
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