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