历届亚运会MECH 225 Engineering Science 2 3 Energy
3.5A. Steady Flow Energy Equation (SFEE)
(a) Recap:
Clod system:no mass flow across the boundary
(control mass considered)
NFEE Q + W = ∆U
Open system:mass flow across the boundary
(control volume considered)
Will the kinetic energy of the fluid affect the energy balance? What other energy changes will there be?
Q
1’站立的近义词
(b) The First Law of Thermodynamics: Q + W = ∆E
Q, W are energy transfers by heating and working, respectively
Q = heating
W consists of two parts:
(i) shaft work, work which is transferred to the surroundings对中国免签
(ii) flow work, the work required to push the fluid through the control volume, where the control volume is shown above between 1 and 2.
果的偏旁
∆E is the change in the total energy, which will be made up of changes in: (i) internal energy (due to changes in temperature)
(ii) kinetic energy (due to changes in flow velocity)
(iii) gravitational potential energy (due to changes in height of the fluid)
(c) Energy transfers – the flow work term
Work done in getting the mass ∆m of fluid into the system, from 1 to 1’:
= force * distance有声胎教故事大全
齐庄公
Flow work = (pA)*L Flow work per unit mass = AL
pAL )V (pAL m pAL ρ=∆ρ=∆∴ flow work per unit mass =
pv p =ρ
where ρ is the density of the fluid ,and v is its specific volume (volume of 1kg).At the inlet, flow work = p 1v 1 (work done on the system to get fluid in).At outlet, flow work = -p 2V 2 (work done by the s
ystem, as fluid exits).The shaft work is denoted by W s .
(d) Changes in total energy of the control mass, ∆m
As the mass ∆m moves into the control volume, at flow station 1, for steady flow, an equal mass ∆m will move out of the control volume, at flow station 2.There will in general be changes in its:
(i) kinetic energy:∆E k = )c c )(m (2
12122−∆where c is the flow velocity (a change in symbol, so as not to
confu it with “v” for specific volume or “u” for specific internal
energy)
(ii) gravitational potential energy:∆E p = (∆m)g(z 2 – z 1)∆m
(iii) Internal energy: ∆U = (∆m)(u 2 – u 1)
Therefore, when we apply the First Law of Thermodynamics to the fluid in the control volume, we get:
Q + W s + p 1V 1 – p 2V 2 = )c c )(m (2
12122−∆ + (∆m)g(z 2 – z 1) + (∆m)(u 2 – u 1)Note that becau the flow is steady, the properties of the fluid within the volume between 1’ and 2 do not change. All that has changed are the properties of the small mass ∆m entering and leaving the control volume.For unit mass of fluid entering the control volume, divide by ∆m:任何人英语
q + w s + p 1v 1 – p 2v 2 = )c c (212122
− + g(z 2 – z 1) + (u 2 – u 1)which can be written as:
q + w s = )c c (212122
− + g(z 2 – z 1) + (u 2 – u 1) + p 2v 2 - p 1v 1We introduce a new term (e also 1.6), enthalpy
Becau the combination of energy terms (u + pv) occurs so frequently in thermodynamic analys, it has been given a special name and symbol,enthalpy, h.
Thus h = u + pv is the specific enthalpy
While H = U + pV is the enthalpy of a mass, m, say.
So we can rewrite the energy balance above as:
This is the steady flow energy equation (SFEE)
What are the units of each term?
祖国在我心中手抄报内容For a mass flow rate, m
, the SFEE becomes:úû
ùêëé−+−+−=+)z z (g )c c (21h h m W Q 12212212s Where Q
is the rate of heating applied to the control volume s
W is the rate of shaft work input Since the are rates of energy transfer, the units are watts (Js -1)
NOTE: Key assumptions are:steady flow
properties constant with time.
Further reading:
Bacon and Stephens, Mechanical Technology 23.9
Rogers, G and Mayhew, Y,
Engineering Thermodynamics Work and Heat Transfer Ch 4 The Open University, T236 Introduction to thermofluid mechanics Block 6
