Thermofluid Processes
Reversible Non-Flow Process
Constant volume process => no work (as rigid vessel) Q1-2 = (u2 –u1) + W1-2 = u2 –u1 => (U2 –U1)
= m cv (T2 – T1)
Constant pressure process: boundary moves (isobaric)
1 1
2 2
v v
1-2 2 1
v v
1-2 2 1 1-2 2 1 2 1
2 2 1 1 2 1 2 1
p 2 1
W = pdv= p dv= p (v - v )
Q = (u -u ) + W = (u -u ) + p (v - v ) = (u - p v ) - (u + p v ) = h -h = H -H = m C (T -T )
v p
v H2O
PG
1
v2 – v1
2
p 0
v2 – v1 v2 = v1
2
1
W1-2 = 0
p
1 2
p2 – p1
v1 v2 v
1 2
W1-2 = p (v2 – v1)
v1 v2
v p
.05 kg heated at 2 bar until V2 = .0658 m3 (a) Steam initially dry saturated
(b) Air initially at 130◦C
Const temperature (isothermal) process p
1 2
heat supplied
0.8856 v 1.316
[m3/ kg]
h1 = hg = 2707 kJ/kg at 2 bar
v2 = V2/ m = .0658/ .05 = 1.316 m3/ kg p2 = 2 * 105 N/m2
T2 = 300◦C, h2 = 3072 kJ/kg Q1-2 = H2 – H1 = m(h2 - h1)
= .05 (3072 - 2707) = 18.25 kJ
W1-2 = p(v2 – v1) =2 * 105(1.316 - .8856) Nm/kg = .05*2 * 105(1.316 - .8856)* 10-3
= 4.304 kJ [N/ m2]
2 * 105
p
1 2
[m3/ kg]
T2 = (p2 V2)/ (mR)
= (2 * 105* .0658) / (.05* .287 *103) = 917K Q1-2 = mCp(T2 - T1)
= .05* 1.005(917 - 403) = 25.83 kJ
W1-2 = p (v2 – v1) R(T2 - T1) = 0.287 *(917 - 403) kJ/kg = .05 * .287 * 514 = 7.38 kJ
[N/ m2]
2 * 105
v
1
2
v
Q1-2 = (u2 – u1) + W1-2
Tset @ 7 bar T2 = T1 = 165◦C
Steam @ 7 bar, x = .9, (T), rev 1.5 bar
∆u, ∆h, W1-2 ?
7 * 105
1.5 * 105
p [N/ m2]
p1
p2
Q1-2
W1-2
Isothermal process for a perfect gas: pv = RT
Joule’s Law: U2 –U1 = m cv (T2 – T1) = 0 for (T) isothermal process 1st Law: q = (u2 – u1) + w = W
Reversible adiabatic non-flow process q = (u2 – u1) + w W = u1 – u2
Perfect gas: dq = du + dw, dw = pdv dq = du + pdv = 0 adiabatic h = pv + u, dh = du + pdv + vdp
dq = dh – vdp = 0 p = RT/v, du + RTdv/v = 0 u = Cv T du = Cv dT
v
C dT RTdv 0
v
Cv ln T + R ln v = const p2
p1
pv = const 1
2
p1v1 = p1v1 = C
2
2 1 1
2 1 1
1 2 2 2
1 2 1 1
1
2 1 1
1 1
1 2 2
1 1 1
2 1 2
v W = C dv = Cln
v v
=p v lnv v
=p v ln v v W = p V ln v
v
v p p
= , W = p V ln
v p p
p v = RT, w = RT ln p p = W = mRT ln p
p
p
0 as adiabatic
γ-1
γ-1
γ
γ
const
γ
γ-1 γ-1
γ γ
γ
γ-1 1-1
γ
1 2
2 1
ln pv + lnv = const R
ln pv v = const R
ln p v = const R
p v = e =const R
p v =const p = RT
v
RTv = const Tv = const
RT RT
v = => pv = p = const
p p
T T
= const or = const
p p
p v
p = v
γ 1 2 γ-1 1 1 1-1γ2 1 2 2
T v T p
, = , =
T v T p
v
v
v
pv R
ln + lnv = const
R C
R R
C = or = γ-1
γ-1 C
ln pv+ (γ-1) lnv = const R
v1 v2
1 2 v 1 2 v
1 2
1 1 2 2
W = u - u = C (T -T ) C = R γ-1 R(T -T )
= pv = RT γ-1
p v -p v W =
γ-1
Polytropic Process:
2
1
2 n
1
2 1-n v 1-n 1-n
2 1
n
1 v
n 1-n n 1-n
1 1 1 2 2 2
1 1 2 2
n
1 2
2 1
pv = const w = pdv v -v
dv v
= C = C = C
v 1-n 1-n
p v v - p v v
= n-1
p v - p v
= n-1
p v
p = v
In a polytropic process the index n depends only on the heat and work quantities during the process.
0
-
γ
n = 0 pv = p = const : isobaric
n = pv = const or p v = const, v = const
n = 1 pv = const or T = const : isothermal ( pv =const) T
n = γ pv = const : reversible, adiatic
p2
p1
pvᵧ
= const 1
2
2
1
2 2 1-γ v
γ
1 1 v
1-γ 1-γ 1-γ 1-γ
2 1 1 2
2 1-γ 2 1-γ
γ γ
1 1 1 2 2 2
1 1 2 2
1 1 2 2
Cdv v
W = pdv = = C
v 1-γ
v -v v -v
= C = C
1-γ γ-1
p v v - p v v
= C = p v = p v
γ-1 p v - p v W = γ-1
p
v2
v1 v
Irreversible processes
lagged (perfectly thermally insulated)
(1) Unrestricted, or free, expansion
Irreversible: extenal work must be done to restore the fluid to its initial condition q = (u2 – u1) + w q = 0 adiabatic
w = 0 the system boundary does not move The process is adiabatic, but irreversible. u2 = u1
Perfect gas: u= Cv T CvT1 = CvT2 or T1 = T2 For a perfect gas undergoing a free expansion (2) Throttling
1 - A': (P) heating 1 - B': (T) expansion 1 - C': rev. adiabatic expansion
1 - D': (V) cooling
1 - A: (P) cooling 1 – B: (T) compression 1 - C: rev. adiabatic compression 1 - D': (V) heating
Vessel A
Vessel B
t = 0: A filled with a fluid at a certain pressrue B is evacuated
t > 0: A & B filled at a lower pressure
lagging 2 2
1 2
1 2
V V
h + + q = h + + w
2 2
When V1 & V2 are small or close to each other, then h1 = h2
Perfect gas: h = CpT orifice
0 0
CpT1 = CpT2 or T1 = T2
(3) Adiabatic mixing
-highly irreversible due to the large amount of eddying and churning of the fluid.
Reversibele flow processess
2 2 2 2
1 2 1 2
1 2 1 2
V V V -V
h + + q = h + + w => w = h -h +
2 2 2
Non-steady flow processes
E: total energy of the system within the boundary
δm1: mass enetering the system during a small time interval δm2: mass leaving the system during a small time interval δq: heat transferred during the same time
δw: work done during the same time δm1 p1v1: work done at inlet
δm2 p2v2: work done at outlet u1 + V12/2 + gz1: energy at inlet u2 + V22/2 + gz2: energy at outlet
q = 0, w = 0 H1 + H2 = H3
neglecting KE
. . . .
1 1 2 2 1 2 3
. . . .
1 2 1 2
p p 1 p 2 p 3
. . . .
1 1 2 2 1 2 3
m h + m h = m + m h
pG: h = C T, m C T + m C T = m + m C T m T + m T = m + m T
. 1
1
m T
. 3
3
m T
. 2
2
m T
. .
1 2
mm
0 0
1st law:
(Energy entering the system) – (Energy leaving the system) = (Change in energy of the system) δm1 p1v1 + δm1(u1 + V12/2 + gz1) + δQ - δw - δm2 p2v2 - δm2 (u2 + V22/2 + gz2) = δE
:
Q
Q total heat transferred during a finite time :W
w total work done during a finite time m': initial mass within the system boundariesu': initial internal energy within the system boundaries m'': final mass within the system boundaries
u'': final internal energy within the system boundaries
2 2
1 2
1 1 1 1 1 2 2 2 2 2
δE=m u -m u +(m u -m u )
V V
δm u +p v + +gz +Q= δm u +p v + +gz +W
2 2
Continuity of mass: (Mass entering) – (Mass leaving) = (change in mass of the system)
1 2
δm - δm =m -m
Vessel fluid
escaping valve
system boundary
W = 0,
δm
1= 0 (no mass enter) gz2 = 0 (negligiable)2 2
2 2
Q=δm h + V +(m u -m u ) 2
where h2 = u2 + p2v2
