Question

Problem 9.003 SI At the beginning of the compression process of an air standard Otto cycle, p1 = 1 bar, T1 = 300 K. The maximum temperature in the cycle is 2250 K and the compression ratio is 9.8. The engine has 4 cylinders and an engine displacement of Vo = 2.4 L. Determine per cylinder: a) e) f) the volume at state 1. the air mass per cycle. the heat addition per cycle, in kJ. the heat rejection per cycle, in kJ. the net work per cycle, in kJ. the thermal efficiency. the mean effective pressure, in bar. Develop a full exergy accounting per cycle, in kJ. Let To = 300 K, PO = 1 bar. Devise and evaluate the exergetic efficiency for the cycle. h) i) Step 1 Your answer is correct. Determine per cylinder the volume at state 1, in liters. v-7.660 .

Answer 1

Otto cycle is an ideal cycle which is used to describe the processes taking place in a spark ignition internal combustion engine. PV diagram for the cycle is: It consists of two isentropic processes and two constant volume processes. Step 1: Isentropic compression process Step 2: Constant volume process Step 3: Isentropic expansion process Step 4: Constant volume process Given that: P1 =1 bar = 100 kPa T1 = 300 K Compression ratio, y = 9.8 number of cylinders = 4 Engine displacement. Va = 2.4 L

The displacement for 4 cylinders is 2.4 L. Then, for 1 cylinder: _ 2.4L =0.6 L And the volume displacement is the difference between initial and final volumes. V =V-V Calculate the initial volume using the relation of compression ratio: The expression for compression ratio is: 9.8= 1,-0.6 9.8(V; -0.6) =V 9.8V, -5.88=V V(9.8-1)=5.88 Vi = 0.668 L Now calculate the initial mass of air (mass of air at initial conditions) assuming the gas as ideal gas: PV = mRT kJ Use the value of Ras specific gas constant R=0.287 kg-K Then,

PV = MRT ma PV RT 100 kPax0.668 Lx 1m 0.287.2K X300 K 100 kPa 0.668 L *1000 x 1x 0 kPa 0.668 23 W 0.287 ke K* 1 Pax 300 K 1 bed m=7.76x10* kg Heat addition per cycle is calculated by the formula: Q =m(U-U) Here, U is the internal energy and Q is the amount of heat added. Read the values internal energy at point 3 and 2. At point 3, turbine inlet temperature is given as, T, = 2250 K. At, Ty = 2250 K, specific internal energy of air is, Ug = 1921.3 To get the internal energy at point 2, apply isentropic relation at process 1-2: Here, V, and V, are the relative volumes at step 1 and 2 respectively. At initial temperature of. T1 = 300 K relative volume is. V,1 = 621.2. We know that y= -9.8 1712 then,

Then, relative volume at step 2 will be, 12.1 621.2 9.8 V,2 = 621.2*9.8 1,2 = 63.4 Now read the value of internal energy corresponding to this relative volume value. And the value will be In property table of air, no direct value is there for v,2 = 63.4. Interpolate the given values and get the internal energy value at v,2 = 63.4. at, v, = 62.13+U=536.07 kJ at, V, = 64.53+U=528.14 Then, V,2 = 63.4: U=(a = 63.4)=536.07 +(63.4–62. 13) 523.59 = 53687) U (1,2 = 63.4) Now substitute the known values and calculate the heat absorbed as shown below: Q=m(U, -U) = 7.76x104 kg (1921.3-531.87) Q=1.08 kJ