. 11: Introduction to Compressible Flow When a fixed mass of air is heated from 20 o C to 100 o C, what is change in…. p 1 , h 1 , s 1 , 1 , u 1 , Vol 1 20 o C p 2 , h 2 , s 2 , 2 , u 2 , Vol 2 100 o C nstant s? constant p? constant volume?… STATE 1 STATE 2
Ch. 11: Introduction to Compressible Flow. When a fixed mass of air is heated from 20 o C to 100 o C, what is change in…. p 2 , h 2 , s 2 , 2 , u 2 , Vol 2 100 o C. STATE 2. p 1 , h 1 , s 1 , 1 , u 1 , Vol 1 20 o C. STATE 1. …. Constant s? constant p? constant volume?…. - PowerPoint PPT Presentation
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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
Ch. 11: Introduction to Compressible Flow
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?
When a fixed mass of air is heated from 20oC to 100oC –
What is the change in enthalpy?
h2 – h1 = Cp(T2- T1)
Ch. 11: Introduction to Compressible Flow
When a fixed mass of air is heated from 20oC to 100oC –
Change in entropy (constant volume)?
s2 – s1 = Cvln(T2/T1)
Ch. 11: Introduction to Compressible Flow
When a fixed mass of air is heated from 20oC to 100oC –
Change in entropy (constant pressure)?
s2 – s1 = Cpln(T2/T1)
Ch. 11: Introduction to Compressible Flow
When a fixed mass of air is heated from 20oC to 100oC –
If isentropic change in density?
2/1 = (T2/T1)1/(k-1)
Ch. 11: Introduction to Compressible Flow
When a fixed mass of air is heated from 20oC to 100oC –
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)
Steady, no body forces, one-dimensional, frictionless, ideal, calorically perfect,
adiabatic, isentropic
p/p0 = (T/To)k/(k-1)
To/T = 1 + {(k-1)/2} M2
p/p0 = (1 + {(k-1)/2} M2)k/(k-1)
Steady, no body forces, one-dimensional, frictionless, ideal, calorically perfect,