10.07.2015 Views

Thermodynamics

Thermodynamics

Thermodynamics

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

502 | <strong>Thermodynamics</strong>P, psia14.7q in2 3IsentropicIsentropic41q outV 2 = V 1 /18 V 3 = 2V 2V 1 = V 4VFIGURE 9–24P-V diagram for the ideal Diesel cyclediscussed in Example 9–3.the end of each process, (b) the net work output and the thermal efficiency,and (c) the mean effective pressure.Solution An ideal Diesel cycle is considered. The temperature and pressureat the end of each process, the net work output, the thermal efficiency, andthe mean effective pressure are to be determined.Assumptions 1 The cold-air-standard assumptions are applicable and thusair can be assumed to have constant specific heats at room temperature.2 Kinetic and potential energy changes are negligible.Properties The gas constant of air is R 0.3704 psia · ft 3 /lbm · R and itsother properties at room temperature are c p 0.240 Btu/lbm · R, c v 0.171 Btu/lbm · R, and k 1.4 (Table A–2Ea).Analysis The P-V diagram of the ideal Diesel cycle described is shown inFig. 9–24. We note that the air contained in the cylinder forms a closedsystem.(a) The temperature and pressure values at the end of each process can bedetermined by utilizing the ideal-gas isentropic relations for processes 1-2and 3-4. But first we determine the volumes at the end of each process fromthe definitions of the compression ratio and the cutoff ratio:Process 1-2 (isentropic compression of an ideal gas, constant specific heats):Process 2-3 (constant-pressure heat addition to an ideal gas):Process 3-4 (isentropic expansion of an ideal gas, constant specific heats):(b) The net work for a cycle is equivalent to the net heat transfer. But firstwe find the mass of air:m P 1V 1RT 1P 3 P 2 841 psiaP 2 V 2T 2 P 3V 3T 3S T 3 T 2 a V 3V 2b 11716 R2 122 3432 RT 4 T 3 a V k1 1.41313 in3b 13432 R2aV 4 117 in b 1425 R3P 4 P 3 a V k 1.4313 in3b 1841 psia2aV 4 117 in b 38.8 psia3V 2 V 1r117 in318V 3 r c V 2 12216.5 in 3 2 13 in 3V 4 V 1 117 in 3T 2 T 1 a V k11b 1540 R2 1182 1.41 1716 RV 2P 2 P 1 a V k1b 114.7 psia2 1182 1.4 841 psiaV 2114.7 psia2 1117 in 3 2 6.5 in 310.3704 psia # ft 3 >lbm # R21540 R2a1 ft 3b 0.00498 lbm31728 in

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!