Question

Problem 3. (20) An air Otto cycle has compression ratio of 10. At the beginning of the compression process, pressure is 1 bar and temperature is 300 K. The maximum temperature in the cycle is 2500 K. Cp=718 J/kg/K. k=1.4. Air is ideal gas. Find: (1) Sketch the process on a T-s diagram and a P-v diagram. (2+2) (2) Temperature at cach point from Ti to T4 in K. (12) (3) The net work per unit of air in kJ/kg. (4')

Answer 1

(1) The pV diagram and the TS diagram can be drawn respectively as

.

ס 3 2 4 Ibar 1 V/10 T 2500k 3 AV = 0 4 2 A 3ook AV = 0 +S

Here
42
and $3\rightarrow 4$ are adiabatic processes while $2\rightarrow 3$ and $4\rightarrow 1$ are isochoric processes. Since the compression ratio is 10,
10/i= ܕ
.

(2) Based on the given information and the diagram shown in (a), we have already seen that
T1 300K
and
T3 - 2500K
. Now, we have to find
T
and
TA
.

For an adiabatic process,

$TV^{\gamma -1}=constant$,

where $\gamma = \frac{C_p}{C_p-k}$ .

Consider the adiabatic process for which

42

.

- ) 8 - ( Ti VI — Ta va - | VI Та = т (v V2 VI - 1 0 От now, V2=villo V2 Ср and 8-1= с -) Cp-k ep-k — 0.002 — 718 - 1. 4 О. OOR , TI Та – lo – 10 О. Ооа Х ЗООК = 301 . 4 К

Similarly, for the adiabatic process $3\rightarrow 4$ ,

.

8-1 Tz vz = TAVA 81=TAVI cr, T3 va g-| 8-1 = , TA T3 ( 22 ) (á) 0:00 on, TA = T3 2500 : TA= K 100.002 =2488.5K

(3) The work done in the entire cycle can be obtained as

4 2 2 3 W = Spar t Sb dr + Sb dut sb dv+Sb dv 3 4 2 2 2 now, spdv S p dv=-ev бат since pdvtedt Tds =0 =eu (Ti-T2) for adiabete process since ev=ep-k = G-&)(T-T2) = (718-1.4) (30 300 - 301-4) 3/za 1003.24 J/kg - Speves Spav=o since dr=0 for 233 Гь b dv = (ep-k) (T3-Tq) just like 17일 SI (718-1.4) (2500-2488.5) J 28240.9 J 31 kg since dr=0 for 471. p dv =0 4 W = ( 8240.9 -1003.2915 =7237-66 e]