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)

What equation has q and reversibility in it?

If reversibleQ/m = q = TdsTds = dh – vdpq = dh = cpdTq = cp(T2-T1) q = -572kJ/kg

0

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