Upload
others
View
56
Download
0
Embed Size (px)
Citation preview
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