P09-163

Equations

Thermodynamics - An Engineering Approach (5th Ed) - Cengel, Boles - Mcgraw-Hill (2006) - pg. 549

Ciclul Brayton pentru turbina cu gaze, cp=ct: influente pratio, T[3], ηC , ηT...

Sa se determine influenta raportului de comprimare, a temperaturii maxime a ciclului si a randamentelor turbineisi compresorului asupra lucrului mecanic specific net si a randamentului unui ciclu Brayton simplu cu aer. Laintrarea in compresor aerul are 100 kPa si 300 K. Se considera ptr aer calduri specifice constante egale cu celede la temperatura ambianta.

$UnitSystem kPa K

procedure ConstPropResult(T1, P1, rcomp, T3 : ηth,ConstProp, ηth,easy) (1)

For Air:

cv = 0.718 [kJ/kg ·K] (2)

k = 1.4 (3)

T2 = T1 · rk−1comp (4)

P2 = P1 · rkcomp (5)

qin,23 = cv · (T3 − T2) (6)

T4 = T3 · (1/rcomp)k−1 (7)

qout,41 = cv · (T4− T1) (8)

ηth,ConstProp = (1− qout,41/qin,23) · |100 %| [%] (9)

The Easy Way to calculate the constant property Otto cycle efficiency is:

ηth,easy =

(1− 1

rk−1comp

)· |100 %| [%] (10)

end (11)

Marimi de intrare - Input Data:

T1 = 300 [K] (12)

1

P1 = 100 [kPa] (13)

rcomp = 12 (14)

T3 = 1000 [K] (15)

Rezolvare:

R = 0.287 [kJ/kg ·K] (16)

Process 1-2 is isentropic compression

s1 = s (air, T = T1, P = P1) (17)

P1 · v1 = R · T1 (18)

s2 = s1 (19)

V2 =V1rcomp

(20)

T2 = T (air, s = s2, P = P2) (21)

P2 ·v2T2

= P1 ·v1T1

(22)

Conservation of energy for process 1 to 2

q12 − w12 = ∆u12 (23)

q12 = 0 isentropic process (24)

∆u12 = u (air, T = T2)− u (air, T = T1) (25)

Process 2-3 is constant volume heat addition

v3 = v2 (26)

s3 = s (air, T = T3, P = P3) (27)

P3 · v3 = R · T3 (28)

Conservation of energy for process 2 to 3

q23 − w23 = ∆u23 (29)

w23 = 0 constant volume process (30)

∆u23 = u (air, T = T3)− u (air, T = T2) (31)

2

Process 3-4 is isentropic expansion

s4 = s3 (32)

s4 = s (air, T = T4, P = P4) (33)

P4 · v4 = R · T4 (34)

Conservation of energy for process 3 to 4

q34 − w34 = ∆u34 (35)

q34 = 0 isentropic process (36)

∆u34 = u (air, T = T4)− u (air, T = T3) (37)

Process 4-1 is constant volume heat rejection

V4 = V1 (38)

Conservation of energy for process 4 to 1

q41 − w41 = ∆u41 (39)

w41 = 0 constant volume process (40)

∆u41 = u (air, T = T1)− u (air, T = T4) (41)

qin,total = q23 (42)

qout,total = −q41 (43)

wnet = w12 + w23 + w34 + w41 (44)

ηth = wnet/qin,total · |100 %| Thermal efficiency, in percent (45)

call ConstPropResult(T1, P1, rcomp, T3 : ηth,ConstProp, ηth,easy) (46)

PerCentError =ABS (ηth − ηth,ConstProp)

ηth· |100 %| [%] (47)

Data$ = Date$ (48)

SolutionVariables in Main program

Data$ = ‘2014-02-06’ ∆u12 = 361.8 [kJ/kg] ∆u23 = 183.1 [kJ/kg] ∆u34 = −473.2 [kJ/kg]∆u41 = −71.79 [kJ/kg] ηth = 60.8 [%] ηth,ConstProp = 62.99 [%] ηth,easy = 62.99 [%]PerCentError = 3.605 [%] q12 = 0 [kJ/kg] q23 = 183.1 [kJ/kg] q34 = 0 [kJ/kg]q41 = −71.79 [kJ/kg] qin,total = 183.1 [kJ/kg] qout,total = 71.79 [kJ/kg] R = 0.287 [kJ/kg-K]rcomp = 12 w12 = −361.8 [kJ/kg] w23 = 0 [kJ/kg] w34 = 473.2 [kJ/kg]w41 = 0 [kJ/kg] wnet = 111.3 [kJ/kg]

3

Variables in Procedure ConstPropResultT [1] = 300 [K] P [1] = 100 [kPa] rcomp = 12 T [3] = 1000 [K]ηth,ConstProp = 62.99 [%] ηth,easy = 62.99 [%] cv = 0.718 [kJ/kg-K] k = 1.4T2 = 810.6 [K] P2 = 3242 [kPa] qin,23 = 136 [kJ/kg] T4 = 370.1 [K]qout,41 = 50.34 [kJ/kg]

Arrays

Row Pi Ti si vi[kPa] [K] [kJ/kg-K] [m3/kg]

1 100 300 5.705 0.8612 3119 779.7 5.705 0.071753 4000 1000 5.912 0.071754 133.2 399.5 5.912 0.861

Percenterror

4

T[3]

5