ideal gas: p = rt (11.1) du = c v dt (11.2) dh= c p dt (11.3)
DESCRIPTION
1 st and 2 nd LAWS: Q-W = U Tds = du +pdv (11.10a) Tds = h –vdp (11.10b). IDEAL GAS: p = RT (11.1) du = c v dT (11.2) dh= c p dT (11.3). h = u + RT; dh = du RdT; c p dT = c v dT + RdT; c p = c v + R (11.4). Ideal Gas and s=0. IDEAL GAS + 1 st + 2 nd LAWS - PowerPoint PPT PresentationTRANSCRIPT
IDEAL GAS:p = RT (11.1)du = cvdT (11.2)dh= cpdT (11.3)
1st and 2nd LAWS:Q-W = UTds = du +pdv (11.10a) Tds = h –vdp (11.10b)
IDEAL GAS + 1st + 2nd LAWSds = du/T + pdv/T = cvdT/T + Rdv/v s2 – s1 = cv ln (T2/T1) + R ln (v2/v1) (11.11a)ds = dh/T - vdp/T = cpdT/T + Rdp/p s2 – s1 = cp ln (T2/T1) - R ln (p2/p1) (11.11b)
s2 – s1 = cv ln (T2/T1) + R ln (v2/v1) = cv ln(p2 1/p12) + (cp-cv) ln (v2/v1)s2 – s1 = cv ln(p2/p1) + cv ln(v2/v1) + cp ln (v2/v1) - cv ln (v2/v1)s2 – s1 = cv ln(p2/p1) + cp ln (v2/v1) (11.11c)
h = u + RT; dh = du RdT; cpdT = cvdT + RdT; cp = cv + R (11.4)
(11.20a)
(11.20b)
(11.20c)
Ideal Gas and s=0
5kg; s;p1=300kPaT1=60oC
5kg; s;p2=150kPa
T2= ?
Ideal Gas
Ideal Gas
IDEAL GAS + ADIABATIC + REVERSIBLE
Tvk-1 = T/(k-1) = c (11.12a)
Tp(1-k)/k = c (11.12b)
pvk = p/k = c (11.12c)
5kg; s;p1=300kPaT1=60oC
5kg; s;
p2=150kPa
T2= ?
isentropic
Tp(1-k)/k = c (11.12b)
T1(Ko)p1(1-k)/k = T2(Ko)p2
(1-k)/k
?
T1 = 333K; p1 = 300,000 Pa T2 = 273K; p2 = 150,000 Pa
Know: p1, T1, p2, T2
irreversibleWhat is s2-s1?
100,000 Pa273K
s1
200,000 Pa388K
s2
irreversible
Valid for any process between equilibrium states
dQ + dW = dE Tds = du + vdp
IDEAL GAS & cv and cp = const
s2-s1 = cvln(T2/T1) + Rln(v2/v1) (11.11a)
s2-s1 = cpln(T2/T1) - Rln(p2/p1) (11.11b)
s2-s1 = cvln(p2/p1) + cpln(v2/v1) (11.11c)
Know: p1, T1, p2, T2 & irreversibleWhat is s2-s1?
?
s2-s1 = cpln(T2/T1) - Rln(p2/p1) (11.11b)
= 103[J/kg-K] ln(388/273) – 287[J/kg-K] ln(200,000/100,000)= 134 J/(kg-K)
Know T1, p1= p2,T2
IDEAL GASs2-s1 = ?p1 = 4.5 MPa
p2 = p1
T1 = 858K
T2 = 15Cs1
s2
s2-s1 = cpln(T2/T1) - Rln(p2/p1) (11.11b)
s2-s1 = 1000 ln([273+15]/858)s2-s1 = -1.09 kJ/(kg-K)
0
T
s
1
2
IDEAL GAS & cv and cp = const
s2-s1 = cvln(T2/T1) + Rln(v2/v1) (11.11a)
s2-s1 = cpln(T2/T1) - Rln(p2/p1) (11.11b)
s2-s1 = cvln(p2/p1) + cpln(v2/v1) (11.11c)
s2 – s1 = cp ln (T2/T1) - R ln (p2/p1) (11.11b)
What equation has q in it?
q = dh = cpdT
q = cp(T2-T1) q = -572kJ/kg
Find po
Find po From Table A-3, pg 719z =12.5 km; p/pSL = 0.1776
/ SL = 0.2361SL = 1.225 kg/m3
pSL = 101.3 kPak = 1.4
po = 28.85 kPa
V from B.E. = ?V for compressible (=Mc) = ?
Find plane speed assuming incompressible* B.E.po = p + ½ V (p, T and are for z=12.5 km)po = 28.85kPa; p = 17.99kPa; = 0.2892kg/m3
V = 274.1 m/sFind plane speed assuming compressible flow.V =Mc = M(kRT)1/2 V = 250.8 m/s ~ 9% error
Find To and po
Find To and po
To = 996oF
p0 = 184 psia
dm/dt = VA = ?
Find mass flow rate
dm/dt = VA = 174 lbm/secA = 2 ft2
V = Mc = M(kRT)1/2 R = Ru/Mm for air = 1717 ft2(s2-R) = 8314 m2(s2-K) For R = 1717 ft2/(s2-R) , T must be in Rankine (460 +60 = 520R)
V = 3.0(1.4*1717*520*)1/2 = 3354 ft/sec
= p/(RT) = 5[lbf/in2][32.2lbm/lbf][144in2/ft2]/(1717[ft2/(s2-R)]520R) = 0.0260 lbm/ft3
Know p, T, M
M1
T1
p1
M2>M1
T2>T1
p1>p2
Q added flow
s2 – s1 = cv ln (T2/T1) + R ln (v2/v1) (11.11a)
s2 – s1 = cp ln (T2/T1) - R ln (p2/p1) (11.11b)
s2 – s1 = cv ln(p2/p1) + cp ln (v2/v1) (11.11c)
s2 – s1 = cp ln (T2/T1) - R ln (p2/p1) (11.11b)
KNOW
KNOW
= p/(RT)
po1 = 1.0MPa[1 + 0.2*(0.2)2]3.5 = 1.028 MPa
At location 1
po2 = 862.7kPa[1 + 0.2*(0.4)2]3.5 = 0.9632kPa
At location 2
At location 1
At location 2
To1 = 580K[1 + 0.2*(0.2)2] = 584.6K
To2 = 1727K[1 + 0.2*(0.4)2] = 1782K
s2 – s1 = cp ln (T2/T1) - R ln (p2/p1) (11.11b)
1004 J/kg-K 1727/580
0.8627/1.0
287 J/kg-k
s2 – s1 = 1138 J/kg-K
?
s2 – s1 = 1138 J/kg-K
po1 = 1.028 MPa
po2 = 0.9632kPa
To1 = 584.6K
To2 = 1782K
T1 = 1573oK; p1 = 2.0 MPaT2 = 773oK; p2 = 101 kPa
u = ?; h = ?; s = ?
Can consider ideal gas
Valid for any process between equilibrium states
dQ + dW = dE Tds = du + vdp
IDEAL GAS & cv and cp = const
s2-s1 = cvln(T2/T1) + Rln(v2/v1) (11.11a)
s2-s1 = cpln(T2/T1) - Rln(p2/p1) (11.11b)
=143 J/(kg-K)
s2-s1 = cvln(p2/p1) + cpln(v2/v1) (11.11c)
u = cVT (11.2)
h = cpT (11.3)
increasing pressure
Steady, adiabatic flow of air, dm/dt = 0.5 kg/sec, through a turbine.At inlet, V1 = 0, T1 = 1300C, p1 = 2.0 mPa (abs)At outlet, V2 = 200 m/s, T2 = 500C, p2 = 101 kPa
Label state points on a Ts diagram:
If isentropic: T2 = T1 (p2/p1)(k-1)/k = 670K (397C) 500CSo not isentropic!
What is power produced by turbine?
dW/dt + dQ/dt = (dm/dt) [(h2 + (V2)2
/2 + gz2) - (h1 + (V1)2 /2 + gz1)]
0
z2 = z1
h2 – h1 = cp (T2 – T1)
Steady, adiabatic flow of air, dm/dt = 0.5 kg/sec, through a turbine. At inlet, V1 = 0, T1 = 1300C, p1 = 2.0 mPa (abs); At outlet, V2 = 200 m/s, T2 = 500C, p2 = 101 kPa
Can speed of car at 60 mph and 120 mph be considered incompressible?
M = ? < 0.3 the answer is yes!
[0 - 1]/0 = ?< 5% then we consider incompressible
0 = 1{ 1 + [(k-1)/2]M12}1/(k-1)
M1 = V1/c1
c1 = (kRT1)1/2
[0 - 1]/0 = 0.3%M = 0.0782
V1 = 60 mph = 26.8 m/s; R = 287 J/(kg-K); 1 = p1/(RT1) = 1.201 kg.m3;
0 = 1{ 1 + [(k-1)/2]M12}1/(k-1)
M1 = V2/c1
c1 = (kRT1)1/2
[0 - 1]/0 = 1.21%M = 0.156
V1 = 120 mph = 53.6 m/s; R = 287 J/(kg-K); 1 = p1/(RT1) = 1.201 kg.m3;
Know p0, p and T and are asked to find
V of aircraft.
Know p0, p and T and are asked to find V of aircraft.
M = V/c so V = Mc
c = (kRT)1/2
po/p = (1 + [(k-1)/2] M2)k/(k-1)
Know p0, p and M and are asked to find
maximum temperature and pressure on aircraft.
po/p = {1 + [(k-1)/2]M2}k/(k-1)
To/T = 1 + [(k-1)/2]M2
Maximum pressure and temperature will be local isentropic stagnation conditions.
Know p0, p and T and are asked to find M and V of aircraft.
Otto Cycle Diesel cycle
pv diagrams for the internal combustion engine
0 0 0
1-2 doing work on system + work2-3 & 3-4 system doing work - work
0
Net W, , is negative so net Q, , is positive
pv = c
(TdS = Q)rev.
s2-s1 = cvln(T2/T1) + Rln(v2/v1)s2-s1 = cpln(T2/T1) - Rln(p2/p1)
s2-s1 = cvln(T2/T1) + Rln(v2/v1) (11.11a) ; Constant v, T = Toexp(s-so)/cv
s2-s1 = cpln(T2/T1) - Rln(p2/p1) (11.11b); Constant p, T = Toexp(s-so)/cp
Because cp>cv for all gasses, slope of const p <constant v
s2-s1 = cvln(T2/T1) + Rln(v2/v1); s2-s1 = cpln(T2/T1) - Rln(p2/p1)
s2-s1=0
s2-s1=0
V from B.E. = ?V for compressible (=Mc) = ?
Find plane speed assuming incompressible* B.E.po = p + ½ V (p, T and are for z=12.5 km)po = 28.85kPa; p = 17.99kPa; = 0.2892kg/m3
V = 274.1 m/sFind plane speed assuming compressible flow.V =Mc = M(kRT)1/2 V = 250.8 m/s ~ 9% error
Find To and po
Find To and po
To = 996oF
p0 = 184 psia
dm/dt = VA = ?
Find mass flow rate
dm/dt = VA = 174 lbm/secA = 2 ft2
V = Mc = M(kRT)1/2 R = Ru/Mm for air = 1717 ft2(s2-R) = 8314 m2(s2-K) For R = 1717 ft2/(s2-R) , T must be in Rankine (460 +60 = 520R)
V = 3.0(1.4*1717*520*)1/2 = 3354 ft/sec
= p/(RT) = 5[lbf/in2][32.2lbm/lbf][144in2/ft2]/(1717[ft2/(s2-R)]520R) = 0.0260 lbm/ft3
Know p, T, M
M1
T1
p1
M2>M1
T2>T1
p1>p2
Q added flow
s2 – s1 = cv ln (T2/T1) + R ln (v2/v1) (11.11a)
s2 – s1 = cp ln (T2/T1) - R ln (p2/p1) (11.11b)
s2 – s1 = cv ln(p2/p1) + cp ln (v2/v1) (11.11c)
s2 – s1 = cp ln (T2/T1) - R ln (p2/p1) (11.11b)
KNOW
KNOW
= p/(RT)
po1 = 1.0MPa[1 + 0.2*(0.2)2]3.5 = 1.028 MPa
At location 1
po2 = 862.7kPa[1 + 0.2*(0.4)2]3.5 = 0.9632kPa
At location 2
At location 1
At location 2
To1 = 580K[1 + 0.2*(0.2)2] = 584.6K
To2 = 1727K[1 + 0.2*(0.4)2] = 1782K
s2 – s1 = cp ln (T2/T1) - R ln (p2/p1) (11.11b)
1004 J/kg-K 1727/580
0.8627/1.0
287 J/kg-k
s2 – s1 = 1138 J/kg-K
?
s2 – s1 = 1138 J/kg-K
po1 = 1.028 MPa
po2 = 0.9632kPa
To1 = 584.6K
To2 = 1782K
V = c*M; c=(kRT)1/2;
before shock
(pobeforeshock – poaftershock)/ pobeforeshock = 0.326 changes across shock
same across shock
V = c*M; c=(kRT)1/2
k=1.4; R=286.9 N-m/kg-KT = ?
z = 3600ft = 36000*0.3048 ~ 11kmFrom Table A-3, pg 719; T = 216.8K
V = 620 m/s
before shock
z ~ 11km; from Table A-3, pg 719; p/pSL=0.224; p=0.224*101.3kPa
k =1.4; M=2.1po=208 kPa
(pobeforeshock – poaftershock)/ pobeforeshock = 0.326
p0 changes across shock
208 kPa = 140 kPa
T0 does not change across shock.
z ~ 11km; from Table A-3, pg 719; T = 216.8oK
To =408oK
p2 < p1; M2<M1; T2<T1
s2 – s1 = cp ln (T2/T1) - R ln (p2/p1) (11.11b)
Know p1, M1, T1
Calculate po1= 101.3kPaTo1 = 288.3K
Know p2, M2, T2
Calculate po2= 39.92kPaTo2 = 288.3K
s2 – s1 = 267 J/kg-K
p2 < p1; M2<M1; T2<T1
po1 > p1 > po2 > p2
To1 = To2 s is positive
Suppose I asked what isso2 – so1 ?
po1= 101.3kPa; To1 = 288.3Kpo2= 39.92kPa; To2 = 288.3K
s2 – s1 = 267 J/kg-K
s2 – s1 = so2 – so1 = 267 J/kg-K !!
s2 – s1 = cp ln (T2/T1) - R ln (p2/p1)
so2 – so1 = cp ln (To2/T01) - R ln (p2/p1)
cp ln (T2/T1) - R ln (p2/p1)?=?cpln(To2/T01) - R ln(p2/p1)
1004ln(1727/580) - 286ln(0.8627/1.0) = 1138 J/kg-K1004ln(1782/584.6) -287ln(0.9632/1.028) = 1136J/kg-K
s2 – s1 = s02 – s01
s2 – s1 = cp ln (T2/T1) + R ln (p2/p1) (11.11b)
DRAW s-T DIAGRAM
s2 – s1 = cp ln (T2/T1) + R ln (p2/p1)
always >1
always >1
s2 – s1 = 59.6 J/kg-K
If p2 > p1 then:
(a) po2 > po1 (b) po2 = po1 (c) po2 < po1 (d) po2 ? po1
po2 = p2[f(M2)]po1 = p1[f(M1)]
po2/p01 = p2/p1 only if M1=M=2
Even through a shockT01= T02
p1<p2
T1<T2
po1<po2
