ch. 11: introduction to compressible flow

Ch. 11: Introduction to Compressible Flow

When a fixed mass of air is heated from 20oC to 100oC, what

is change in….

p1, h1, s1, 1, u1, Vol1

20oC

p2, h2, s2, 2, u2, Vol2

100oC

…. Constant s? constant p? constant volume?…

STATE1

STATE2


When a fixed mass of air is heated from 20oC to 100oC –

What is the change in enthalpy? Change in entropy (constant volume)?Change in entropy (constant pressure)? If isentropic change in pressure? If isentropic change in density?

p = RT [R=Runiv/mmole] (11.1)

du = cvdT (11.2)

u2- u1 = cv(T2 – T1) (11.7a)

dh = cpdT (11.3)

h2- h1 = cp(T2 – T1) (11.7b)

IDEAL, CALORICALLY PERFECT GAS

h = u + pv IDEAL GAS

h = u + RT dh = du + RdT

IDEAL GAS

du = cvdT & dh = cpdT cpdT = cvdT + R dT

cp – cv = REq. (11.4)

cp - cv = R (11.4)

k cp/cv ([k=]) (11.5)

cp = kR/(k-1) (11.6a)

cv = R/(k-1) (11.6b)

IDEAL GAS

Ideal calorically perfect gas – constant cp, cv

p = RT; cp = dh/dT; cv = du/dT

s2 – s1 = cvln(T2/T1) - Rln(2/1)

s2 – s1 = cpln(T2/T1) - Rln(p2/p1)

always truedq + dw = du ds = q/T|rev

Tds = du - pdv = dh – vdp

s2 – s1 = cvln(T2/T1) - Rln(2/1)

s2 – s1 = cpln(T2/T1) - Rln(p2/p1)

Ideal / Calorically Perfect Gas

Handy if need to find change in entropy

Ideal / Calorically Perfect GasCv = du/dT; Cp = dh/dT; p = RT = (1/v)RT

Tds = du + pdv = dh –vdpds = du/T + RTdv/T

ds = cvdT/T + (R/v)dv

s2 – s1 = cvln(T2/T1) + Rln(v2/v1)

s2 – s1 = cvln(T2/T1) - Rln(2/1)


Tds = du + pdv = dh –vdpds = du/T + RTdv/T

ds = cvdT/T + (R/v)dv

Note: don’t be alarmed that cv and dv in same equation! cv = du/dT is ALWAYS TRUE for ideal gas

Tds = du + pdv = dh –vdpds = dh/T – vdp/T

ds = CpdT/T - (RT/[pT])dp

s2 – s1 = Cpln(T2/T1) - Rln(p2/p1)


Tds = du + pdv = dh –vdpds = dh/T – vdp/T

ds = CpdT/T - (RT/[pT])dp


Note: don’t be alarmed that cp and dp are in same equation! cp = dh/dT is ALWAYS TRUE for ideal gas

IsentropicIdeal / Calorically Perfect Gas

Handy if

isentropic

2/1 = (T2/T1)1/(k-1)

p2/p1 = (T2/T1)k/(k-1)

(2/1)k = p2/p1; p2/2k = const

c = kRT

s2 – s1 = Cvln(T2/T1) - Rln(2/1)

If isentropic s2 – s1 = 0 ln(T2/T1)Cv = ln(2/1)R

cp – cv = R; R/cv = k – 1

2/1 = (T2/T1)cv/R = (T2/T1)1/(k-1)

assumptions

ISENROPIC & IDEAL GAS& constant cp, cv

s2 – s1 = cpln(T2/T1) - Rln(p2/p1)If isentropic s2 – s1 = 0ln(T2/T1)cp = ln(p2/p1)R

cp – cv = R; R/cp = 1- 1/k

p2/p1 = (T2/T1)cp/R = (T2/T1)k/(k-1)

assumptionsISENROPIC & IDEAL GAS

& constant cp, cv

2/1 = (T2/T1)1/(k-1)

p2/p1 = (T2/T1)k/(k-1)

assumptionsISENROPIC & IDEAL GAS

& constant cp, cv

(2/1)k = p2/p1

p2/2k = p1/1

k = constant



What is the change in enthalpy?

h2 – h1 = Cp(T2- T1)



Change in entropy (constant volume)?

s2 – s1 = Cvln(T2/T1)



Change in entropy (constant pressure)?

s2 – s1 = Cpln(T2/T1)



If isentropic change in density?

2/1 = (T2/T1)1/(k-1)



If isentropic change in pressure?

p2/p1 = (T2/T1)k/(k-1)

Stagnation Reference (V=0)

(refers to “total” pressure (po), temperature (To) or density (o) if flow brought isentropically to rest)

11-3 REFERENCE STATE: LOCAL ISENTROPIC STAGNATION PROPERTIES

Since p, T, , u, h, s, V are all changing along the flow, the concept of stagnation conditions is extremely useful inthat it defines a convenient reference state for a flowing fluid. To obtain a useful final state, restrictions must be put on the deceleration process. For an isentropic (adiabatic and no friction) deceleration there are unique stagnation To, po, o, uo, so, ho (Vo=0) properties .

1-D, energy equation for adiabatic and no shaft or viscous work Eq. (8.28); hlT = [u2-u1] - Q/m

(p2/2) + u2 + ½ V22 + gz2 = (p1/1) + u1 + ½ V1

2 + gz1

Isentropic process

0

Definition: h = u + pv = u + p/;

assume z2 = z1

h2 + ½ V22 = h1 + ½ V1

2

= ho + 0

ho – h1 = ½ V12

1-D, energy equation for adiabatic and no shaft or viscous work (8.28, hlT = [u2-u1] - Q/m)

ho - h1 = ½ V12

ho – h1 = cp (To – T1)

½ V12 = cp (To – T1)

½ V12 + cpT1 = cp To

To = {½ V12 + cpT1}/cp

T0 = ½ V12/cp + T1 = ½ V2/cp + T

T0 = ½ V12/cp + T = T (1 + V2/[2cpT])

cp = kR/(k-1)

T0 = T (1 + V2/[2kRT/{(k-1)})

T0 = T (1 + (k-1)V2/[2kRT])

c2 = kRT

T0 = T (1 + (k-1)V2/[2c2])

M = V2/ c2

T0 = T (1 + [(k-1)/2] M2)

To/T = 1 + {(k-1)/2} M2

Steady, no body forces, one-dimensional, frictionless, ideal, calorically perfect,

adiabatic, isentropic

/o = (T/To)1/(k-1)

To/T = 1 + {(k-1)/2} M2

/o = (1 + {(k-1)/2} M2 )1/(k-1)



p/p0 = (T/To)k/(k-1)

To/T = 1 + {(k-1)/2} M2

p/p0 = (1 + {(k-1)/2} M2)k/(k-1)



p = RT; cp = dh/dT; cv = du/dT

s2 – s1 = cvln(T2/T1) - Rln(2/1)

s2 – s1 = cpln(T2/T1) - Rln(p2/p1)

2/1 = (T2/T1)1/(k-1); p2/p1 = (T2/T1)k/(k-1); p2/2

k = const; c = kRT

p0/p = (1 + {(k-1)/2} M2)k/(k-1); o/ = (1 + {(k-1)/2} M2 )1/(k-1)

To/T = 1 + {(k-1)/2} M2

Ideal & constant cp & cv

Ideal & constant cp & cv & isentropic

Ideal & constant cp & cv & isentropic + …

• Stagnation condition not useful for velocity• Use critical condition – when M = 1, V* = c*

(critical speed is the speed obtained when flow is isentropically accelerated or decelerated until M = 1)

• At critical conditions, the isentropic stagnation quantities become:

p0/p* = (1+{(k-1)/2} 12)k/(k-1) = {(k+1)/2}k/(k-1) o/ = (1+{(k-1)/2} 12 )1/(k-1) = {(k+1)/2}1/(k-1)

To/T = 1 + {(k-1)/2} 12 = (k+1)/2

p0/p = (1 + {(k-1)/2} M2)k/(k-1); o/ = (1 + {(k-1)/2} M2 )1/(k-1)

To/T = 1 + {(k-1)/2} M2

ch. 11: introduction to compressible flow

Documents

oc change

perfect gas constant

cp cv

dudt cp

dudt s2 s1

isentropic change

cvs2 s1

krts2 s1