Equations - University of GalațiCiclul Otto ideal Un ciclu Otto ideal are un raport de comprimare...
Transcript of Equations - University of GalațiCiclul Otto ideal Un ciclu Otto ideal are un raport de comprimare...
P09-035
Equations
Thermodynamics - An Engineering Approach (5th Ed) - Cengel, Boles - Mcgraw-Hill (2006) - pg. 540
Ciclul Otto ideal
Un ciclu Otto ideal are un raport de comprimare de 8. La inceputul procesului de comprimare aerul are 95 kPasi 27◦C, iar in timpul procesului de primire de caldura la v=ct. primeste o cantiate de caldura de 750 kJ/kg. Dacase considera variatia caldurii specifice cu temperatura, sa se calculeze:(a) presiunea si temperatura la sfarsitul procesului la v=ct.(b) lucrul mecanic net(c) randamentul termic(d) presiunea medie efectiva in ciclu
Sa se traseze ciclul in diagramele T-s si p-V, si sa se studieze influenta raportului de comprimare asupra lu-crului mecanic net si a randamentului termic
$UnitSystem K kPa
Marimi de intrare:
T1 = 300 [K] ; (1)
P1 = 95 [kPa] ; (2)
q23 = 750 [kJ/kg] ; (3)
rcomp = 8; (4)
Rezolvare:
Process 1-2 is isentropic compression
s1 = s (air, T = T1, P = P1) ; (5)
s2 = s1; (6)
T2 = T (air, s = s2, P = P2) ; (7)
P2 ·v2T2
= P1 ·v1T1
; (8)
P1 · v1 = 0.287 [kJ/kg ·K] · T1; (9)
1
V2 =V1rcomp
; (10)
Conservation of energy for process 1 to 2
q12 − w12 = ∆u12; (11)
q12 = 0 isentropic process (12)
∆u12 = u (air, T = T2) − u (air, T = T1) ; (13)
Process 2-3 is constant volume heat addition
v3 = v2; (14)
s3 = s (air, T = T3, P = P3) ; (15)
P3 · v3 = 0.287 [kJ/kg ·K] · T3; (16)
Conservation of energy for process 2 to 3
q23 − w23 = ∆u23; (17)
w23 = 0 constant volume process (18)
∆u23 = u (air, T = T3) − u (air, T = T2) ; (19)
Process 3-4 is isentropic expansion
s4 = s3; (20)
s4 = s (air, T = T4, P = P4) ; (21)
P4 · v4 = 0.287 [kJ/kg ·K] · T4; (22)
Conservation of energy for process 3 to 4
q34 − w34 = ∆u34; (23)
q34 = 0 isentropic process (24)
∆u34 = u (air, T = T4) − u (air, T = T3) ; (25)
Process 4-1 is constant volume heat rejection
V4 = V1; (26)
Conservation of energy for process 4 to 1
q41 − w41 = ∆u41; (27)
2
w41 = 0; constant volume process (28)
∆u41 = u (air, T = T1) − u (air, T = T4) ; (29)
qin,total = q23; qout,total = −q41; (30)
wnet = w12 + w23 + w34 + w41; (31)
ηth = wnet/qin,total · 100; Thermal efficiency, in percent (32)
The mean effective pressure is:
MEP =wnet
V1 − V2; [kPa] (33)
Data$ = Date$; (34)
Solution
Data$ = ‘2014-02-04’ ∆u12 = 277.4 [kJ/kg] ∆u23 = 750 [kJ/kg] ∆u34 = −670.2 [kJ/kg]
∆u41 = −357.3 [kJ/kg] ηth = 52.36 MEP = 495.2 [kPa] q12 = 0 [kJ/kg]
q23 = 750 [kJ/kg] q34 = 0 [kJ/kg] q41 = −357.3 [kJ/kg] qin,total = 750 [kJ/kg]qout,total = 357.3 [kJ/kg] rcomp = 8 w12 = −277.4 [kJ/kg] w23 = 0 [kJ/kg]
w34 = 670.2 [kJ/kg] w41 = 0 [kJ/kg] wnet = 392.7 [kJ/kg]
Arrays
Row Pi Ti si vi[kPa] [K] [kJ/kg-K] [m3/kg]
1 95 300 5.72 0.90632 1706 673.4 5.72 0.11333 3899 1539 6.427 0.11334 245.1 774 6.427 0.9063
3
P-v: Air
T-s: Air
4
Net Cycle Work vs Compression Ratio
MEP vs Compression Ratio
5
Cycle Thermal Efficiency vs Compression Ratio
6