Control Volume and Reynolds Transport Theorem
10. 11. 2013
Hyunse Yoon, Ph.D. Assistant Research Scientist
IIHR-Hydroscience & Engineering University of Iowa
57:020 Fluids Mechanics Fall2013 1
Reynolds Transport Theorem (RTT)
• An analytical tool to shift from describing the laws governing fluid motion using the system concept to using the control volume concept
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System vs. Control Volume
• System: A collection of matter of fixed identity – Always the same atoms or fluid particles – A specific, identifiable quantity of matter
• Control Volume (CV): A volume in space
through which fluid may flow – A geometric entity – Independent of mass
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Examples of CV
Fixed CV Moving CV Deforming CV
CV fixed at a nozzle CV moving with ship CV deforming within cylinder
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Laws of Mechanics 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
= 0
𝐹𝐹 = 𝑑𝑑𝑎𝑎 = 𝑑𝑑𝑑𝑑𝑉𝑉𝑑𝑑𝑑𝑑 =
𝑑𝑑𝑑𝑑𝑑𝑑
𝑑𝑑𝑉𝑉
𝑀𝑀 =𝑑𝑑𝐻𝐻𝑑𝑑𝑑𝑑
1. Conservation of mass:
2. Conservation of linear momentum:
3. Conservation of angular momentum:
𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = �̇�𝑄 − �̇�𝑊 4. Conservation of Energy:
• The laws apply to either solid or fluid systems • Ideal for solid mechanics, where we follow the same system • For fluids, the laws need to be rewritten to apply to a specific region in
the neighborhood of our product (i.e., CV)
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Extensive vs. Intensive Property Governing Differential Equations (GDE’s):
𝑑𝑑𝑑𝑑𝑑𝑑
𝑑𝑑,𝑑𝑑𝑉𝑉,𝑑𝑑𝐵𝐵
= 0,𝐹𝐹, �̇�𝑄 − �̇�𝑊
𝑑𝑑 𝑒𝑒
• 𝐵𝐵 = The amount of 𝑑𝑑, 𝑑𝑑𝑉𝑉, or 𝑑𝑑 contained in the total mass of a system or a CV; Extensive property – Dependent on mass
• 𝛽𝛽 (or 𝑏𝑏) = The amount of 𝐵𝐵 per unit mass; Intensive
property – Independent on mass
𝛽𝛽 or 𝑏𝑏 = 𝐵𝐵/𝑑𝑑 (=𝑑𝑑𝐵𝐵𝑑𝑑𝑑𝑑 for nonuniform 𝐵𝐵)
𝐵𝐵 = 𝛽𝛽 ⋅ 𝑑𝑑 = � 𝛽𝛽 𝜌𝜌𝑑𝑑𝑉𝑉�=𝑑𝑑𝑑𝑑𝑉𝑉
for nonuniform 𝛽𝛽
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Fixed CV
At time 𝑑𝑑: SYS = CV
At time 𝑑𝑑 + 𝛿𝛿𝑑𝑑: SYS = (CV – I) + II
𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 𝑑𝑑 = 𝐵𝐵𝐶𝐶𝑉𝑉(𝑑𝑑)
𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 𝑑𝑑 + 𝛿𝛿𝑑𝑑= 𝐵𝐵𝐶𝐶𝑉𝑉 𝑑𝑑 + 𝛿𝛿𝑑𝑑 − 𝐵𝐵𝐼𝐼 𝑑𝑑 + 𝛿𝛿𝑑𝑑+ 𝐵𝐵𝐼𝐼𝐼𝐼 𝑑𝑑 + 𝛿𝛿𝑑𝑑
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Time Rate of Change of 𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠
𝛿𝛿𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠𝛿𝛿𝑑𝑑 =
𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 𝑑𝑑 + 𝛿𝛿𝑑𝑑 − 𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠(𝑑𝑑)𝛿𝛿𝑑𝑑
=𝐵𝐵𝐶𝐶𝑉𝑉 𝑑𝑑 + 𝛿𝛿𝑑𝑑 − 𝐵𝐵𝐼𝐼 𝑑𝑑 + 𝛿𝛿𝑑𝑑 + 𝐵𝐵𝐼𝐼𝐼𝐼 𝑑𝑑 + 𝛿𝛿𝑑𝑑 − 𝐵𝐵𝐶𝐶𝑉𝑉 𝑑𝑑
𝛿𝛿𝑑𝑑
∴𝛿𝛿𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠𝛿𝛿𝑑𝑑 =
𝐵𝐵𝐶𝐶𝑉𝑉 𝑑𝑑 + 𝛿𝛿𝑑𝑑 − 𝐵𝐵𝐶𝐶𝑉𝑉 𝑑𝑑𝛿𝛿𝑑𝑑
1) Change of 𝐵𝐵 within CV over 𝛿𝛿𝛿𝛿
+𝐵𝐵𝐼𝐼𝐼𝐼(𝑑𝑑 + 𝛿𝛿𝑑𝑑)
𝛿𝛿𝑑𝑑2) Amount of 𝐵𝐵flowing outthrough CSover 𝛿𝛿𝛿𝛿
−𝐵𝐵𝐼𝐼 𝑑𝑑 + 𝛿𝛿𝑑𝑑
𝛿𝛿𝑑𝑑3) Amountt of 𝐵𝐵
flowing inthrough CSover 𝛿𝛿𝛿𝛿
Now, take limit of 𝛿𝛿𝑑𝑑 → 0 to Eq. (1) term by term
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Eq. (1)
LHS of Eq. (1)
lim𝛿𝛿𝛿𝛿→0
𝛿𝛿𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠𝛿𝛿𝑑𝑑 = lim
𝛿𝛿𝛿𝛿→0
𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠 𝑑𝑑 + 𝛿𝛿𝑑𝑑 − 𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠(𝑑𝑑)𝛿𝛿𝑑𝑑 =
𝑑𝑑𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠𝑑𝑑𝑑𝑑
Time rate ofchange of 𝐵𝐵within thesystem
or, =𝐷𝐷𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠𝐷𝐷𝑑𝑑 ; material derivative
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First term of RHS of Eq.(1)
lim𝛿𝛿𝛿𝛿→0
𝐵𝐵𝐶𝐶𝑉𝑉 𝑑𝑑 + 𝛿𝛿𝑑𝑑 − 𝐵𝐵𝐶𝐶𝑉𝑉(𝑑𝑑)𝛿𝛿𝑑𝑑
=𝑑𝑑𝐵𝐵𝐶𝐶𝑉𝑉𝑑𝑑𝑑𝑑 =
𝑑𝑑𝑑𝑑𝑑𝑑� 𝛽𝛽𝜌𝜌𝑑𝑑𝑉𝑉𝐶𝐶𝑉𝑉
Time rate of change of𝐵𝐵 withich CV
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2nd term of RHS of Eq.(1) 𝛿𝛿𝑑𝑑𝑜𝑜𝑜𝑜𝛿𝛿 = 𝜌𝜌𝛿𝛿𝑉𝑉
and 𝛿𝛿𝑉𝑉 = 𝛿𝛿𝛿𝛿 ⋅ 𝛿𝛿ℓ𝑛𝑛 = 𝛿𝛿𝛿𝛿 ⋅ 𝛿𝛿ℓ�
=𝑉𝑉𝛿𝛿𝛿𝛿cos𝜃𝜃 = 𝛿𝛿𝛿𝛿 ⋅ (𝑉𝑉𝛿𝛿𝑑𝑑 cos𝜃𝜃)
Thus, the amount of 𝐵𝐵 flowing out of CV through 𝛿𝛿𝛿𝛿 over a short time 𝛿𝛿𝑑𝑑:
∴ 𝛿𝛿𝐵𝐵𝑜𝑜𝑜𝑜𝛿𝛿 = 𝛽𝛽𝛿𝛿𝑑𝑑𝑜𝑜𝑜𝑜𝛿𝛿 = 𝛽𝛽𝜌𝜌𝑉𝑉 cos𝜃𝜃 𝛿𝛿𝑑𝑑𝛿𝛿𝛿𝛿
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𝑎𝑎 ⋅ 𝑏𝑏 = 𝑎𝑎 𝑏𝑏 cos𝜃𝜃
2nd term of RHS of Eq.(1) – Contd.
lim𝛿𝛿𝛿𝛿→0
𝐵𝐵𝐼𝐼𝐼𝐼(𝑑𝑑 + 𝛿𝛿𝑑𝑑)𝛿𝛿𝑑𝑑 = lim
𝛿𝛿𝛿𝛿→0
1𝛿𝛿𝑑𝑑 𝛿𝛿𝑑𝑑� 𝛽𝛽𝜌𝜌𝑉𝑉 cos𝜃𝜃 𝑑𝑑𝛿𝛿
𝐶𝐶𝑆𝑆𝑜𝑜𝑢𝑢𝑢𝑢
∴ �̇�𝐵out = � 𝛽𝛽𝜌𝜌𝑉𝑉 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆𝑜𝑜𝑢𝑢𝑢𝑢
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𝐵𝐵𝐼𝐼𝐼𝐼 𝑑𝑑 + 𝛿𝛿𝑑𝑑 = � 𝑑𝑑𝐵𝐵𝑜𝑜𝑜𝑜𝛿𝛿𝐶𝐶𝑆𝑆𝑜𝑜𝑢𝑢𝑢𝑢
= � 𝛽𝛽𝜌𝜌𝑉𝑉 cos𝜃𝜃 𝛿𝛿𝑑𝑑𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆𝑜𝑜𝑢𝑢𝑢𝑢
Thus,
By integrating 𝛿𝛿𝐵𝐵𝑜𝑜𝑜𝑜𝛿𝛿 over the entire outflow portion of CS,
= � 𝛽𝛽𝜌𝜌𝑉𝑉 cos𝜃𝜃=𝑉𝑉𝑛𝑛
𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆𝑜𝑜𝑢𝑢𝑢𝑢
≡ �̇�𝐵𝑜𝑜𝑜𝑜𝛿𝛿
Note that 𝑉𝑉 cos𝜃𝜃 = 𝑉𝑉 ⋅ 𝒏𝒏�,
i.e., Out flux of 𝐵𝐵 through CS
3rd term of RHS of Eq.(1) 𝛿𝛿𝑑𝑑𝑖𝑖𝑛𝑛 = 𝜌𝜌𝛿𝛿𝑉𝑉
and
𝛿𝛿𝑉𝑉 = 𝛿𝛿𝛿𝛿 ⋅ 𝛿𝛿ℓ𝑛𝑛 = 𝛿𝛿𝛿𝛿 ⋅ 𝛿𝛿ℓ�=𝑉𝑉𝛿𝛿𝛿𝛿
− cos𝜃𝜃< 0
= 𝛿𝛿𝛿𝛿 ⋅ (−𝑉𝑉𝛿𝛿𝑑𝑑 cos𝜃𝜃)
Thus, the amount of 𝐵𝐵 flowing out of CV through 𝛿𝛿𝛿𝛿 over a short time 𝛿𝛿𝑑𝑑:
∴ 𝛿𝛿𝐵𝐵𝑖𝑖𝑛𝑛 = 𝛽𝛽𝛿𝛿𝑑𝑑𝑖𝑖𝑛𝑛 = −𝛽𝛽𝜌𝜌𝑉𝑉 cos𝜃𝜃 𝛿𝛿𝑑𝑑𝛿𝛿𝛿𝛿
57:020 Fluids Mechanics Fall2013 13
3rd term of RHS of Eq.(1) – Contd.
lim𝛿𝛿𝛿𝛿→0
𝐵𝐵𝐼𝐼(𝑑𝑑 + 𝛿𝛿𝑑𝑑)𝛿𝛿𝑑𝑑 = lim
𝛿𝛿𝛿𝛿→0
1𝛿𝛿𝑑𝑑 𝛿𝛿𝑑𝑑� −𝛽𝛽𝜌𝜌𝑉𝑉 cos𝜃𝜃 𝑑𝑑𝛿𝛿
𝐶𝐶𝑆𝑆𝑖𝑖𝑛𝑛
∴ �̇�𝐵in = −� 𝛽𝛽𝜌𝜌𝑉𝑉 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆𝑖𝑖𝑛𝑛
57:020 Fluids Mechanics Fall2013 14
𝐵𝐵𝐼𝐼 𝑑𝑑 + 𝛿𝛿𝑑𝑑 = � 𝑑𝑑𝐵𝐵𝑖𝑖𝑛𝑛𝐶𝐶𝑆𝑆𝑖𝑖𝑛𝑛
= � −𝛽𝛽𝜌𝜌𝑉𝑉 cos𝜃𝜃 𝛿𝛿𝑑𝑑𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆𝑖𝑖𝑛𝑛
Thus,
By integrating 𝛿𝛿𝐵𝐵𝑜𝑜𝑜𝑜𝛿𝛿 over the entire outflow portion of CS,
= −� 𝛽𝛽𝜌𝜌𝑉𝑉 cos𝜃𝜃=−𝑉𝑉𝑛𝑛
𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆𝑖𝑖𝑛𝑛
≡ �̇�𝐵𝑖𝑖𝑛𝑛
Note that 𝑉𝑉 cos𝜃𝜃 = 𝑉𝑉 ⋅ 𝒏𝒏�,
i.e., influx of 𝐵𝐵 through CS
RTT for Fixed CV
𝐷𝐷𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠𝐷𝐷𝑑𝑑 =
𝑑𝑑𝑑𝑑𝑑𝑑� 𝛽𝛽𝜌𝜌𝑑𝑑𝑉𝑉𝐶𝐶𝑉𝑉
+ � 𝛽𝛽𝜌𝜌𝑉𝑉 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆𝑜𝑜𝑢𝑢𝑢𝑢
�̇�𝐵𝑜𝑜𝑢𝑢𝑢𝑢
− −� 𝛽𝛽𝜌𝜌𝑉𝑉 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆𝑖𝑖𝑛𝑛
�̇�𝐵𝑖𝑖𝑛𝑛
Now the relationship between the time rate of change of 𝐵𝐵 for the system and that for the CV is given by,
57:020 Fluids Mechanics Fall2013 15
𝐷𝐷𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠𝐷𝐷𝑑𝑑 =
𝑑𝑑𝑑𝑑𝑑𝑑� 𝛽𝛽𝜌𝜌𝑑𝑑𝑉𝑉
𝐶𝐶𝑉𝑉+ � 𝛽𝛽𝜌𝜌𝑉𝑉 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿
𝐶𝐶𝑆𝑆
Time rate of change of 𝐵𝐵 within a system
Time rate of change of 𝐵𝐵 within CV
Net flux of 𝐵𝐵 through CS = �̇�𝐵𝑜𝑜𝑜𝑜𝛿𝛿 − �̇�𝐵𝑖𝑖𝑛𝑛
= +
With the fact that 𝐶𝐶𝐶𝐶 = 𝐶𝐶𝐶𝐶𝑜𝑜𝑜𝑜𝛿𝛿 + 𝐶𝐶𝐶𝐶𝑖𝑖𝑛𝑛,
Example 1 57:020 Fluids Mechanics Fall2013 16
Example 1 - Contd.
�̇�𝐵𝑜𝑜𝑜𝑜𝛿𝛿 = � 𝜌𝜌𝛽𝛽𝑽𝑽 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆𝑜𝑜𝑢𝑢𝑢𝑢
(4.16)
With 𝛽𝛽= 1 and 𝑽𝑽 ⋅ 𝒏𝒏� = 𝑉𝑉 cos𝜃𝜃 ,
�̇�𝐵𝑜𝑜𝑜𝑜𝛿𝛿 = � 𝜌𝜌𝑉𝑉 cos𝜃𝜃 𝑑𝑑𝛿𝛿𝐶𝐶𝐷𝐷
= 𝜌𝜌𝑉𝑉 cos𝜃𝜃� 𝑑𝑑𝛿𝛿𝐶𝐶𝐷𝐷=𝐴𝐴𝐶𝐶𝐶𝐶
= 𝜌𝜌𝑉𝑉 cos𝜃𝜃 𝛿𝛿𝐶𝐶𝐷𝐷
where
𝛿𝛿𝐶𝐶𝐷𝐷 = ℓ × (2 𝑑𝑑)
=0.5 𝑑𝑑cos𝜃𝜃 2 𝑑𝑑 =
1cos𝜃𝜃 𝑑𝑑2
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Example 1 - Contd. Thus, with 𝜌𝜌 = 1,000 kg/m3 for water and 𝑉𝑉 = 3 m/s,
�̇�𝐵𝑜𝑜𝑜𝑜𝛿𝛿 = 1,000 kgm3 3
ms
cos𝜃𝜃1
cos𝜃𝜃 m2 = 3,000 kg/s
With 𝛽𝛽 = 1/𝜌𝜌,
�̇�𝐵𝑜𝑜𝑜𝑜𝛿𝛿 = � 𝑉𝑉 cos𝜃𝜃 𝑑𝑑𝛿𝛿𝐶𝐶𝐷𝐷
= 𝑉𝑉 cos𝜃𝜃 � 𝑑𝑑𝛿𝛿𝐶𝐶𝐷𝐷=𝐴𝐴𝐶𝐶𝐶𝐶
=1/ cos 𝜃𝜃
= 𝑉𝑉 cos𝜃𝜃 𝛿𝛿𝐶𝐶𝐷𝐷
= 3ms cos𝜃𝜃
1cos𝜃𝜃 m2 = 3 m3 s⁄ (𝑖𝑖. 𝑒𝑒. , volume flow rate)
Note: These results are the same for all 𝜃𝜃 values
57:020 Fluids Mechanics Fall2013 18
Special Case: 𝑉𝑉= constant over discrete CS’s
∴𝐷𝐷𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠𝐷𝐷𝑑𝑑 =
𝑑𝑑𝑑𝑑𝑑𝑑� 𝛽𝛽𝜌𝜌𝑑𝑑𝑉𝑉
𝐶𝐶𝑉𝑉+ � 𝛽𝛽𝑗𝑗 𝜌𝜌𝑗𝑗𝑉𝑉𝑗𝑗𝛿𝛿𝑗𝑗
�̇�𝑑𝑗𝑗 𝑜𝑜𝑜𝑜𝛿𝛿𝑗𝑗
−� 𝛽𝛽𝑖𝑖 𝜌𝜌𝑖𝑖𝑉𝑉𝑖𝑖𝛿𝛿𝑖𝑖�̇�𝑑𝑖𝑖 𝑖𝑖𝑛𝑛𝑖𝑖
57:020 Fluids Mechanics Fall2013 19
�̇�𝐵𝑖𝑖𝑛𝑛 = � 𝛽𝛽𝜌𝜌 𝑉𝑉 ⋅ 𝒏𝒏��constant
𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆𝑖𝑖𝑛𝑛
= � 𝛽𝛽𝑖𝑖𝜌𝜌𝑖𝑖𝑉𝑉𝑖𝑖𝛿𝛿𝑖𝑖 𝑖𝑖𝑛𝑛𝑖𝑖
�̇�𝐵𝑜𝑜𝑜𝑜𝛿𝛿 = � 𝛽𝛽𝜌𝜌 𝑉𝑉 ⋅ 𝒏𝒏��constant
𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆𝑜𝑜𝑢𝑢𝑢𝑢
= � 𝛽𝛽𝑗𝑗𝜌𝜌𝑗𝑗𝑉𝑉𝑗𝑗𝛿𝛿𝑗𝑗 𝑜𝑜𝑜𝑜𝛿𝛿𝑗𝑗
Example 2 57:020 Fluids Mechanics Fall2013 20
Given: • Water flow (𝜌𝜌 = constant) • 𝐷𝐷1 = 10 cm; 𝐷𝐷2 = 15 cm • 𝑉𝑉1 = 10 cm/s • Steady flow
Find: 𝑉𝑉2 = ?
Mass conservation: • 𝐷𝐷𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠/𝐷𝐷𝑑𝑑 = 0 • 𝛽𝛽 = 1 • 𝜌𝜌1 = 𝜌𝜌2 = 𝜌𝜌
0 =𝑑𝑑𝑑𝑑𝑑𝑑� 𝜌𝜌𝑑𝑑𝑉𝑉
𝐶𝐶𝑉𝑉+ 𝜌𝜌2𝑉𝑉2𝛿𝛿2 − 𝜌𝜌1𝑉𝑉1𝛿𝛿1
Steady flow
or, 𝜌𝜌1𝑉𝑉1𝛿𝛿1 = 𝜌𝜌2𝑉𝑉2𝛿𝛿2
∴ 𝑉𝑉2 =𝜌𝜌1𝛿𝛿1𝜌𝜌2𝛿𝛿2
𝑉𝑉1 =𝜌𝜌𝜌𝜌
𝐷𝐷1𝐷𝐷2
2
𝑉𝑉1 = 110 cm15 cm
2
10 cms = 4.4 cm/s
Example 3 57:020 Fluids Mechanics Fall2013 21
Given: • 𝐷𝐷1 = 5 cm; 𝐷𝐷2 = 7 cm • 𝑉𝑉1 = 3 m/s • 𝑄𝑄3 = 𝑉𝑉3𝛿𝛿3 = 0.01 m3/s • ℎ = constant (i.e., steady flow) • 𝜌𝜌1 = 𝜌𝜌2 = 𝜌𝜌3 = 𝜌𝜌water
Find: 𝑉𝑉2 = ?
0 =𝑑𝑑𝑑𝑑𝑑𝑑� 𝜌𝜌𝑑𝑑𝑉𝑉
𝐶𝐶𝑉𝑉+ 𝜌𝜌2𝑉𝑉2𝛿𝛿2 − 𝜌𝜌1𝑉𝑉1𝛿𝛿1 − (𝜌𝜌3𝑉𝑉3𝛿𝛿3)
= 0; steady flow
or, 𝑉𝑉2𝛿𝛿2 = 𝑉𝑉1𝛿𝛿1 + 𝑉𝑉3𝛿𝛿3�= 𝑄𝑄3
∴ 𝑉𝑉2 =𝑉𝑉1𝛿𝛿1 + 𝑄𝑄3
𝛿𝛿2=
3 𝜋𝜋 0.05 2/4 + (0.01)𝜋𝜋 0.07 2/4 = 4.13 m/s
Moving CV 57:020 Fluids Mechanics Fall2013 22
RTT for Moving CV
𝐷𝐷𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠𝐷𝐷𝑑𝑑 =
𝑑𝑑𝑑𝑑𝑑𝑑� 𝛽𝛽𝜌𝜌𝑑𝑑𝑉𝑉
𝐶𝐶𝑉𝑉+ � 𝛽𝛽𝜌𝜌𝑉𝑉𝑟𝑟 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿
𝐶𝐶𝑆𝑆
(i.e., relative velocity 𝑉𝑉𝑟𝑟)
57:020 Fluids Mechanics Fall2013 23
RTT for Moving and Deforming CV
*Ref) Fluid Mechanics by Frank M. White, McGraw Hill
𝑑𝑑𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠𝑑𝑑𝑑𝑑 =
𝑑𝑑𝑑𝑑𝑑𝑑� 𝛽𝛽𝜌𝜌𝑑𝑑𝑉𝑉𝐶𝐶𝑉𝑉
+ � 𝛽𝛽𝜌𝜌 𝑽𝑽𝒓𝒓 ⋅ 𝒏𝒏� 𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆
∗
𝑽𝑽𝑟𝑟 = 𝑽𝑽(𝒙𝒙, 𝑑𝑑) − 𝑽𝑽𝑆𝑆(𝒙𝒙, 𝑑𝑑) • 𝑉𝑉𝑆𝑆(𝑥𝑥, 𝑑𝑑): Velocity of CS • 𝑉𝑉(𝑥𝑥, 𝑑𝑑): Fluid velocity in the coordinate
system in which the 𝑉𝑉𝑠𝑠 is observed • 𝑉𝑉𝑟𝑟: Relative velocity of fluid seen by an
observer riding on the CV
Both CV and CS change their shape and location with time
57:020 Fluids Mechanics Fall2013 24
RTT Summary (1) 𝑑𝑑𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠𝑑𝑑𝑑𝑑 =
𝑑𝑑𝑑𝑑𝑑𝑑
� 𝛽𝛽𝜌𝜌𝑑𝑑𝑉𝑉𝐶𝐶𝑉𝑉
+ � 𝛽𝛽𝜌𝜌𝑉𝑉𝑟𝑟 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆
57:020 Fluids Mechanics Fall2013 25
General RTT (for moving and deforming CV):
𝑑𝑑𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠𝑑𝑑𝑑𝑑 = �
𝜕𝜕𝜕𝜕𝑑𝑑 𝛽𝛽𝜌𝜌 𝑑𝑑𝑉𝑉
𝐶𝐶𝑉𝑉+ � 𝛽𝛽𝜌𝜌𝑉𝑉𝑟𝑟 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿
𝐶𝐶𝑆𝑆
1) Non-deforming (but moving) CV
𝑑𝑑𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠𝑑𝑑𝑑𝑑 = �
𝜕𝜕𝜕𝜕𝑑𝑑 𝛽𝛽𝜌𝜌 𝑑𝑑𝑉𝑉
𝐶𝐶𝑉𝑉+ � 𝛽𝛽𝜌𝜌𝑉𝑉 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿
𝐶𝐶𝑆𝑆
2) Fixed CV
𝜕𝜕𝜕𝜕𝑑𝑑 = 0
3) Steady flow:
Special Cases:
� 𝛽𝛽𝜌𝜌𝑉𝑉 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆
= � 𝛽𝛽�̇�𝑑 𝑜𝑜𝑜𝑜𝛿𝛿 −� 𝛽𝛽�̇�𝑑 𝑖𝑖𝑛𝑛
4) Flux terms for uniform flow across discrete CS’s (steady or unsteady)
RTT Summary (2)
Parameter (𝐵𝐵) 𝛽𝛽 = 𝐵𝐵/𝑑𝑑 RTT Remark
Mass (𝑑𝑑) 1 0 =𝑑𝑑𝑑𝑑𝑑𝑑� 𝜌𝜌𝑑𝑑𝑉𝑉𝐶𝐶𝑉𝑉
+ � 𝜌𝜌𝑉𝑉 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆
Continuity eq.
(Ch. 5.1)
Momentum (𝑑𝑑𝑉𝑉)
𝑉𝑉 �𝐹𝐹 =𝑑𝑑𝑑𝑑𝑑𝑑� 𝑉𝑉𝜌𝜌𝑑𝑑𝑉𝑉𝐶𝐶𝑉𝑉
+ � 𝑉𝑉𝜌𝜌𝑉𝑉 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆
Linear momentum eq.
(Ch. 5.2)
Energy (𝑑𝑑) 𝑒𝑒 �̇�𝑄 − �̇�𝑊 =𝑑𝑑𝑑𝑑𝑑𝑑� 𝑒𝑒𝜌𝜌𝑑𝑑𝑉𝑉𝐶𝐶𝑉𝑉
+ � 𝑒𝑒𝜌𝜌𝑉𝑉 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆
Energy eq. (Ch. 5.3)
For fixed CV’s:
57:020 Fluids Mechanics Fall2013 26
Continuity Equation (Ch. 5.1) 57:020 Fluids Mechanics Fall2013 27
RTT with 𝐵𝐵 = mass and 𝛽𝛽 = 1,
0 =𝐷𝐷𝑀𝑀𝑠𝑠𝑠𝑠𝑠𝑠
𝐷𝐷𝑑𝑑mass conservatoin
=𝑑𝑑𝑑𝑑𝑑𝑑� 𝜌𝜌𝑑𝑑𝑉𝑉𝐶𝐶𝑉𝑉
+ � 𝜌𝜌𝑉𝑉 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆
or
� 𝜌𝜌𝑉𝑉 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆
Net rate of outflowof mass across CS
= −𝑑𝑑𝑑𝑑𝑑𝑑� 𝜌𝜌𝑑𝑑𝑉𝑉
𝐶𝐶𝑉𝑉Rate of decrease of mass within CV
Note: Incompressible fluid (𝜌𝜌 = constant)
� 𝑉𝑉 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆
= −𝑑𝑑𝑑𝑑𝑑𝑑� 𝑑𝑑𝑉𝑉
𝐶𝐶𝑉𝑉 (Conservation of volume)
Simplifications 57:020 Fluids Mechanics Fall2013 28
1. Steady flow
� 𝜌𝜌𝑉𝑉 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆
= 0
2. If 𝑉𝑉 = constant over discrete CS’s (i.e., one-dimensional flow)
� 𝜌𝜌𝑉𝑉 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆
= �𝜌𝜌𝑉𝑉𝛿𝛿𝑜𝑜𝑜𝑜𝛿𝛿
−�𝜌𝜌𝑉𝑉𝛿𝛿𝑖𝑖𝑛𝑛
3. Steady one-dimensional flow in a conduit
𝜌𝜌𝑉𝑉𝛿𝛿 𝑜𝑜𝑜𝑜𝛿𝛿 − 𝜌𝜌𝑉𝑉𝛿𝛿 𝑖𝑖𝑛𝑛 = 0
or
𝜌𝜌2𝑉𝑉2𝛿𝛿2 − 𝜌𝜌1𝑉𝑉1𝛿𝛿1 = 0
For 𝜌𝜌 = constant
𝑉𝑉1𝛿𝛿1 = 𝑉𝑉2𝛿𝛿2 (or 𝑄𝑄1 = 𝑄𝑄2)
Some useful definitions 57:020 Fluids Mechanics Fall2013 29
Mass flux �̇�𝑑 = � 𝜌𝜌𝑉𝑉 ⋅ 𝑑𝑑𝛿𝛿𝐴𝐴
Volume flux 𝑄𝑄 = � 𝑉𝑉 ⋅ 𝑑𝑑𝛿𝛿𝐴𝐴
Average velocity �̅�𝛿 =𝑄𝑄𝛿𝛿 =
1𝛿𝛿� 𝑉𝑉 ⋅ 𝑑𝑑𝛿𝛿
𝐴𝐴
Average density �̅�𝜌 =1𝛿𝛿� 𝜌𝜌𝑑𝑑𝛿𝛿
𝐴𝐴
Note: �̇�𝑑 ≠ �̅�𝜌𝑄𝑄 unless 𝜌𝜌 = constant
(Note: 𝑉𝑉 ⋅ 𝑑𝑑𝛿𝛿 = 𝑉𝑉 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿)
Example 4 57:020 Fluids Mechanics Fall2013 30
Estimate the time required to fill with water a cone-shaped container 5 ft hight and 5 ft across at the top if the filling rate is 20 gal/min.
0 =𝑑𝑑𝑑𝑑𝑑𝑑� 𝜌𝜌𝑑𝑑𝑉𝑉
𝐶𝐶𝑉𝑉+ � 𝜌𝜌𝑉𝑉 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿
𝐶𝐶𝑆𝑆
Apply the conservation of mass (𝛽𝛽 = 1)
For incompressible fluid (i.e., 𝜌𝜌 = constant) and one inlet,
0 =𝑑𝑑𝑑𝑑𝑑𝑑� 𝑑𝑑𝑉𝑉
𝐶𝐶𝑉𝑉=𝑉𝑉
− 𝑉𝑉𝛿𝛿 𝑖𝑖𝑛𝑛=𝑄𝑄𝑖𝑖𝑛𝑛
Example 4 – Contd. 57:020 Fluids Mechanics Fall2013 31
Volume of the cone at time t,
𝑉𝑉 𝑑𝑑 =𝜋𝜋𝐷𝐷2
12 ℎ 𝑑𝑑
Flow rate at the inlet,
𝑄𝑄 = 20galmin 231
in3
gal 1,728in3
ft3� = 2.674 ft3/min
The continuity eq. becomes
𝜋𝜋𝐷𝐷2
12 ⋅𝑑𝑑ℎ𝑑𝑑𝑑𝑑 = 𝑄𝑄 or
𝑑𝑑ℎ𝑑𝑑𝑑𝑑 =
12𝑄𝑄𝜋𝜋𝐷𝐷2
Example 4 – Contd. 57:020 Fluids Mechanics Fall2013 32
ℎ 𝑑𝑑 = �12𝑄𝑄𝜋𝜋𝐷𝐷2
𝛿𝛿
0𝑑𝑑𝑑𝑑 =
12𝑄𝑄 ⋅ 𝑑𝑑𝜋𝜋𝐷𝐷2
Solve for ℎ(𝑑𝑑),
Thus, the time for ℎ = 5 ft is
𝑑𝑑 =𝜋𝜋𝐷𝐷2ℎ12𝑄𝑄 =
𝜋𝜋 5 ft 2(5 ft)(12)(2.674 ft3/min) = 12.2 min