第三讲闭口、开口系统能量方程及例题
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1
2
U Q W t t t
Then, in the limit as Δt approaches zero dU W Q net net dt
例题2.1 • 气缸内储有定量的CO2气体，初态p1＝300kPa，T1＝ 200℃，V1＝0.2m3。经历一可逆过程后温度下降至T2＝ 100℃。如果过程中压力和比容间的关系满足pv1.2＝常 数，试确定该过程中CO2气体所作的功、比功、热力学 能变化量，气体与外界之间的热量交换。 • Solution： • Given: initial and end states, process, CO2 in a cylinder • Find: W, w, ΔU, Q • Model: closed system • Strategy: apply the basic closed system energy balance to solve State1 State 2 for the Q and ΔU, apply the definition of work to calculation W and w
Eout
c2 m( h gz) out 2
The control volume energy equation on one dimensional assumption:
For mass rate: m kg/s
2 2 dUCV c c W m in (h out (h Q gz) in m gz) out dt 2 2
For
a reversible process, the work can be calculated as:
w pdv
这里功的数量并不等于可以利用的功 For a closed system, heat transfer and work transfer are the only mechanisms by which energy can be transferred across the boundary. If we need to express the general energy balance on a rate basis by a finite time interval. This yields:
•W＝mw＝0.671×94500＝63425 J 因气体体积变化，故此功称为膨胀功。W>0，为气体对 外做功。在终态，气体体积为V2＝0.656m3 •2)系统内热力学能的变化：(ΔE)sys＝ΔU＝mcV(T2－T1) •查表取cv＝0.656kJ/kg· K
• 所以ΔU＝0.671×103×(100－200)＝－44018J （说明系统的热力学能是减少的） • 3)Q 的计算： Q ＝ ΔU ＋ W ＝ 63425 ＋（－ 44018 ）＝ 19407 J • Comment:该例题中气体对外作出的功量大于气体热力学 能的减少量，减少的部分由外界对气体的加热量所补充。 另外此处能量的单位用焦耳显得不方便，用 kJ 则好些。
in m out 0 m
in out
• Case 2 For one-dimensional flow:
dmCV in m out ( Ac) in ( Ac) out m dt in out in out
• Case 3 for both steady state and one-dimensional flow:
(Ac)
in
in
( Ac)out 0
out
•2nd Conservation of Energy Principle for A Open System or Control Volume •It’s clear when regardless the kinetic and potential energy of a system , we’ll have: dU CV W (m e) in ( m e) out Q dt •If we take the assumption- Steady state: W (m e)in (m e)out 0Q •If we take the assumption- One-dimensional flow: dU CV W m ( ein eout ) Q dt •If we take both the assumptions- Steady state and Onedimensional flow:
2.3 热力学第一定律的一般表达式
• 当系统由状态1经历一系列状态变化达到终状态2时，系 统总能量的变化为 • Δ Esys＝Ein－Eout＝（minein＋Q）－（mouteout＋W）
mout min ΔEsys Q W 能量守恒
• 当系统处于宏观上静止时 • Δ U＝（minein＋Q）－（mouteout＋W）
w 94.5 kJ/kg
U 44 .018 kJ
W 63 .425 kJ
Q 19.407 kJ
2.5开口系统的能量方程 Conservation of Energy Principle for A Open System or Control Volume
• Two steps of analysis – 1st Conservation of Mass Principle for A Open System or Control Volume – 2nd Conservation of Energy Principle for An Open System or Control Volume • 1st: Conservation of Mass Principle for A Open System or Control Volume
W m (ein eout ) 0Q
The energy in or out of the control volume ：
•When mass gets in the system, it will carry the several items of energy into the system
dmCV in m out m dt in out
m
Is defined as the mass flow rate
kg/s
•Steady state: any system in steady state if the system properties are constant with time at every position within and on the boundaries of the systems. •One-dimensional flow: if the properties at a flow boundary are uniform over the cross-sectional area. • Case 1 For steady state:
If we take another assumption of steady state:
2 2 c c W m in (h out (h 0Q gz) in m gz) out 2 2 There are numerous applications of the steady flow and steady state conservation of energy principle for which there is only one inlet (position 1) and one outlet (position 2) . In this circumstance above equation will be expressed as: 2 2 c c W m 1 (h 2 (h 0Q gz)1 m gz) 2 2 2 On a unit mass basis, it is convenient to express as: c2 c2 q w (h gz)1 ( h gz) 2 2 2 2 2 c2 c1 ( h2 h1 ) g ( z 2 z1 ) 2
W pAL pV mpv
We give this work the name of flow work or push work. When m kg of mass enters in a control volume :
W m pv
As 1) and 2) discussion, we have:
c2 Ek m 2
•4) and potential energy of the mass (When m kg of mass enters in a control volume)
EP mgz
From above discussing, for each m kg mass enters in the system, the total energy enters the control volume should be: c2 Ein m(h gz) in 2 the total energy leaves the control volume should be:
mu mpv m(u pv)
A definition:h u pv Nhomakorabeaor
H m h m(u pv)
We give h the name of enthalpy.
This is a property of matter. (Combined property) •3) kinetic energy of the mass (When m kg of mass enters in a control volume )
•1) internal energy of the mass; •2) ? •3) kinetic energy of the mass; •4) and potential energy of the mass. •1) internal energy of the mass (if m kg mass enters)
•Analysis: •1)work：可逆δw＝pdv 2 2 1 1.2 dv 1.2 dv w pv 1.2 p1v1 1.2 ( p1v1 p2 v2 ) 0.2 v v 1 1
Rg 189 (T1 T2 ) (200 100 ) 94500 J/kg 0.2 0.2 p1V1 300 103 0.2 m 0.671kg RT1 189 (200 372)
U in (mu)in
•2) how to let the mass across the boundary into a system?
According to the fig. shown below:
For one-dimensional assumption, Work should be done as:
• 看问题的角度:进入系统的能量为正,离开为负
2.4闭口系统能量方程 A Conservation of Energy Principle for Closed Systems
A closed system undergoing a process between state 1 and state 2, the energy equation will be: Δ U＝ Q － W or Q＝Δ U＋W or for 1kg of mass q ＝ Δ u＋ w • for a differential basis • δ Q＝dU＋δ W or δ q＝du＋δ w
2
U Q W t t t
Then, in the limit as Δt approaches zero dU W Q net net dt
例题2.1 • 气缸内储有定量的CO2气体，初态p1＝300kPa，T1＝ 200℃，V1＝0.2m3。经历一可逆过程后温度下降至T2＝ 100℃。如果过程中压力和比容间的关系满足pv1.2＝常 数，试确定该过程中CO2气体所作的功、比功、热力学 能变化量，气体与外界之间的热量交换。 • Solution： • Given: initial and end states, process, CO2 in a cylinder • Find: W, w, ΔU, Q • Model: closed system • Strategy: apply the basic closed system energy balance to solve State1 State 2 for the Q and ΔU, apply the definition of work to calculation W and w
Eout
c2 m( h gz) out 2
The control volume energy equation on one dimensional assumption:
For mass rate: m kg/s
2 2 dUCV c c W m in (h out (h Q gz) in m gz) out dt 2 2
For
a reversible process, the work can be calculated as:
w pdv
这里功的数量并不等于可以利用的功 For a closed system, heat transfer and work transfer are the only mechanisms by which energy can be transferred across the boundary. If we need to express the general energy balance on a rate basis by a finite time interval. This yields:
•W＝mw＝0.671×94500＝63425 J 因气体体积变化，故此功称为膨胀功。W>0，为气体对 外做功。在终态，气体体积为V2＝0.656m3 •2)系统内热力学能的变化：(ΔE)sys＝ΔU＝mcV(T2－T1) •查表取cv＝0.656kJ/kg· K
• 所以ΔU＝0.671×103×(100－200)＝－44018J （说明系统的热力学能是减少的） • 3)Q 的计算： Q ＝ ΔU ＋ W ＝ 63425 ＋（－ 44018 ）＝ 19407 J • Comment:该例题中气体对外作出的功量大于气体热力学 能的减少量，减少的部分由外界对气体的加热量所补充。 另外此处能量的单位用焦耳显得不方便，用 kJ 则好些。
in m out 0 m
in out
• Case 2 For one-dimensional flow:
dmCV in m out ( Ac) in ( Ac) out m dt in out in out
• Case 3 for both steady state and one-dimensional flow:
(Ac)
in
in
( Ac)out 0
out
•2nd Conservation of Energy Principle for A Open System or Control Volume •It’s clear when regardless the kinetic and potential energy of a system , we’ll have: dU CV W (m e) in ( m e) out Q dt •If we take the assumption- Steady state: W (m e)in (m e)out 0Q •If we take the assumption- One-dimensional flow: dU CV W m ( ein eout ) Q dt •If we take both the assumptions- Steady state and Onedimensional flow:
2.3 热力学第一定律的一般表达式
• 当系统由状态1经历一系列状态变化达到终状态2时，系 统总能量的变化为 • Δ Esys＝Ein－Eout＝（minein＋Q）－（mouteout＋W）
mout min ΔEsys Q W 能量守恒
• 当系统处于宏观上静止时 • Δ U＝（minein＋Q）－（mouteout＋W）
w 94.5 kJ/kg
U 44 .018 kJ
W 63 .425 kJ
Q 19.407 kJ
2.5开口系统的能量方程 Conservation of Energy Principle for A Open System or Control Volume
• Two steps of analysis – 1st Conservation of Mass Principle for A Open System or Control Volume – 2nd Conservation of Energy Principle for An Open System or Control Volume • 1st: Conservation of Mass Principle for A Open System or Control Volume
W m (ein eout ) 0Q
The energy in or out of the control volume ：
•When mass gets in the system, it will carry the several items of energy into the system
dmCV in m out m dt in out
m
Is defined as the mass flow rate
kg/s
•Steady state: any system in steady state if the system properties are constant with time at every position within and on the boundaries of the systems. •One-dimensional flow: if the properties at a flow boundary are uniform over the cross-sectional area. • Case 1 For steady state:
If we take another assumption of steady state:
2 2 c c W m in (h out (h 0Q gz) in m gz) out 2 2 There are numerous applications of the steady flow and steady state conservation of energy principle for which there is only one inlet (position 1) and one outlet (position 2) . In this circumstance above equation will be expressed as: 2 2 c c W m 1 (h 2 (h 0Q gz)1 m gz) 2 2 2 On a unit mass basis, it is convenient to express as: c2 c2 q w (h gz)1 ( h gz) 2 2 2 2 2 c2 c1 ( h2 h1 ) g ( z 2 z1 ) 2
W pAL pV mpv
We give this work the name of flow work or push work. When m kg of mass enters in a control volume :
W m pv
As 1) and 2) discussion, we have:
c2 Ek m 2
•4) and potential energy of the mass (When m kg of mass enters in a control volume)
EP mgz
From above discussing, for each m kg mass enters in the system, the total energy enters the control volume should be: c2 Ein m(h gz) in 2 the total energy leaves the control volume should be:
mu mpv m(u pv)
A definition:h u pv Nhomakorabeaor
H m h m(u pv)
We give h the name of enthalpy.
This is a property of matter. (Combined property) •3) kinetic energy of the mass (When m kg of mass enters in a control volume )
•1) internal energy of the mass; •2) ? •3) kinetic energy of the mass; •4) and potential energy of the mass. •1) internal energy of the mass (if m kg mass enters)
•Analysis: •1)work：可逆δw＝pdv 2 2 1 1.2 dv 1.2 dv w pv 1.2 p1v1 1.2 ( p1v1 p2 v2 ) 0.2 v v 1 1
Rg 189 (T1 T2 ) (200 100 ) 94500 J/kg 0.2 0.2 p1V1 300 103 0.2 m 0.671kg RT1 189 (200 372)
U in (mu)in
•2) how to let the mass across the boundary into a system?
According to the fig. shown below:
For one-dimensional assumption, Work should be done as:
• 看问题的角度:进入系统的能量为正,离开为负
2.4闭口系统能量方程 A Conservation of Energy Principle for Closed Systems
A closed system undergoing a process between state 1 and state 2, the energy equation will be: Δ U＝ Q － W or Q＝Δ U＋W or for 1kg of mass q ＝ Δ u＋ w • for a differential basis • δ Q＝dU＋δ W or δ q＝du＋δ w