第三讲闭口开口系统能量方程及例题 PPT
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d d C m t Vin m in om u o t u tin (A )in c o( uA t )oc ut
• Case 3 for both steady state and one-dimensional flow:
(A)ic n (A)o cu t0
•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。
• 看问题的角度:进入系统的能量为正,
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:
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
• ΔEsys＝Ein－Eout＝（minein＋Q）－（mouteout＋W）
mout min
ΔEsys
W Q
• 当系统处于宏观上静止时
• ΔU＝（minein＋Q）－（mouteout＋W）
•One-dimensional flow: if the properties at a flow boundary are uniform over the cross-sectional area.
• Case 1 For steady state:
0 m in m out
in
out
• Case 2 For one-dimensional flow:
• 1st: Conservation of Mass Principle for A Open System or Control Volume
dm CV
dt
in
m in m out
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.
ΔU＝Q －W
or
Q＝ΔU＋W
or for 1kg of mass
q＝Δu＋w
Baidu Nhomakorabea• for a differential basis
•
δQ＝dU＋δW
or
δq＝du＋δw
For a reversible process, the work can be calculated as:
2
w pdv
这里功的数量并不等于可1 以利用的功
U Q W t t t
Then, in the limit as Δt approaches zero
ddUt QnetWnet
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 for the Q and ΔU, apply the
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:
State1
State 2
definition of work to calculation W and w
•Analysis: •1)work：可逆δw＝pdv
w 1 2p1 .2 v v d 1 .2 v1 2p 1 v 1 1 .2v d 1 .2 v0 1 .2(p 1 v 1p 2 v 2)
w94.5kJ/kg W6.342k5J
U4.0 41k8J Q19.40k7J
• 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
0 R .2 g(T 1 T 2)1 0 .28 (2 90 10 )0 9 04J5/0 k0 g mp1V13010300.2 0.6k7g1
R1T18(9 203 07) 2
•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)
• Case 3 for both steady state and one-dimensional flow:
(A)ic n (A)o cu t0
•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。
• 看问题的角度:进入系统的能量为正,
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:
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
• ΔEsys＝Ein－Eout＝（minein＋Q）－（mouteout＋W）
mout min
ΔEsys
W Q
• 当系统处于宏观上静止时
• ΔU＝（minein＋Q）－（mouteout＋W）
•One-dimensional flow: if the properties at a flow boundary are uniform over the cross-sectional area.
• Case 1 For steady state:
0 m in m out
in
out
• Case 2 For one-dimensional flow:
• 1st: Conservation of Mass Principle for A Open System or Control Volume
dm CV
dt
in
m in m out
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.
ΔU＝Q －W
or
Q＝ΔU＋W
or for 1kg of mass
q＝Δu＋w
Baidu Nhomakorabea• for a differential basis
•
δQ＝dU＋δW
or
δq＝du＋δw
For a reversible process, the work can be calculated as:
2
w pdv
这里功的数量并不等于可1 以利用的功
U Q W t t t
Then, in the limit as Δt approaches zero
ddUt QnetWnet
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 for the Q and ΔU, apply the
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:
State1
State 2
definition of work to calculation W and w
•Analysis: •1)work：可逆δw＝pdv
w 1 2p1 .2 v v d 1 .2 v1 2p 1 v 1 1 .2v d 1 .2 v0 1 .2(p 1 v 1p 2 v 2)
w94.5kJ/kg W6.342k5J
U4.0 41k8J Q19.40k7J
• 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
0 R .2 g(T 1 T 2)1 0 .28 (2 90 10 )0 9 04J5/0 k0 g mp1V13010300.2 0.6k7g1
R1T18(9 203 07) 2
•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)