Control Volume and Reynolds Transport Theoremfluids/Posting/Lecture_Notes... · 57:020 Fluids...

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


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


Examples of CV

Fixed CV Moving CV Deforming CV

CV fixed at a nozzle CV moving with ship CV deforming within cylinder


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)


Extensive vs. Intensive Property Governing Differential Equations (GDE’s):


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


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


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


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


First term of RHS of Eq.(1)


𝐵𝐵𝐶𝐶𝑉𝑉 𝑑𝑑 + 𝛿𝛿𝑑𝑑 − 𝐵𝐵𝐶𝐶𝑉𝑉(𝑑𝑑)𝛿𝛿𝑑𝑑

=𝑑𝑑𝐵𝐵𝐶𝐶𝑉𝑉𝑑𝑑𝑑𝑑 =

𝑑𝑑𝑑𝑑𝑑𝑑� 𝛽𝛽𝜌𝜌𝑑𝑑𝑉𝑉𝐶𝐶𝑉𝑉

Time rate of change of𝐵𝐵 withich CV


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


𝑎𝑎 ⋅ 𝑏𝑏 = 𝑎𝑎 𝑏𝑏 cos𝜃𝜃

2nd term of RHS of Eq.(1) – Contd.


𝐵𝐵𝐼𝐼𝐼𝐼(𝑑𝑑 + 𝛿𝛿𝑑𝑑)𝛿𝛿𝑑𝑑 = lim


1𝛿𝛿𝑑𝑑 𝛿𝛿𝑑𝑑� 𝛽𝛽𝜌𝜌𝑉𝑉 cos𝜃𝜃 𝑑𝑑𝛿𝛿

𝐶𝐶𝑆𝑆𝑜𝑜𝑢𝑢𝑢𝑢

∴ �̇�𝐵out = � 𝛽𝛽𝜌𝜌𝑉𝑉 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆𝑜𝑜𝑢𝑢𝑢𝑢


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


3rd term of RHS of Eq.(1) – Contd.


𝐵𝐵𝐼𝐼(𝑑𝑑 + 𝛿𝛿𝑑𝑑)𝛿𝛿𝑑𝑑 = lim


1𝛿𝛿𝑑𝑑 𝛿𝛿𝑑𝑑� −𝛽𝛽𝜌𝜌𝑉𝑉 cos𝜃𝜃 𝑑𝑑𝛿𝛿

𝐶𝐶𝑆𝑆𝑖𝑖𝑛𝑛

∴ �̇�𝐵in = −� 𝛽𝛽𝜌𝜌𝑉𝑉 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆𝑖𝑖𝑛𝑛


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 =


+ � 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,



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


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


Special Case: 𝑉𝑉= constant over discrete CS’s

∴𝐷𝐷𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠𝐷𝐷𝑑𝑑 =


𝐶𝐶𝑉𝑉+ � 𝛽𝛽𝑗𝑗 𝜌𝜌𝑗𝑗𝑉𝑉𝑗𝑗𝛿𝛿𝑗𝑗

�̇�𝑑𝑗𝑗 𝑜𝑜𝑜𝑜𝛿𝛿𝑗𝑗

−� 𝛽𝛽𝑖𝑖 𝜌𝜌𝑖𝑖𝑉𝑉𝑖𝑖𝛿𝛿𝑖𝑖�̇�𝑑𝑖𝑖 𝑖𝑖𝑛𝑛𝑖𝑖


�̇�𝐵𝑖𝑖𝑛𝑛 = � 𝛽𝛽𝜌𝜌 𝑉𝑉 ⋅ 𝒏𝒏��constant

𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆𝑖𝑖𝑛𝑛

= � 𝛽𝛽𝑖𝑖𝜌𝜌𝑖𝑖𝑉𝑉𝑖𝑖𝛿𝛿𝑖𝑖 𝑖𝑖𝑛𝑛𝑖𝑖

�̇�𝐵𝑜𝑜𝑜𝑜𝛿𝛿 = � 𝛽𝛽𝜌𝜌 𝑉𝑉 ⋅ 𝒏𝒏��constant

𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆𝑜𝑜𝑢𝑢𝑢𝑢

= � 𝛽𝛽𝑗𝑗𝜌𝜌𝑗𝑗𝑉𝑉𝑗𝑗𝛿𝛿𝑗𝑗 𝑜𝑜𝑜𝑜𝛿𝛿𝑗𝑗


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


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 = ?


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



𝐶𝐶𝑉𝑉+ � 𝛽𝛽𝜌𝜌𝑉𝑉𝑟𝑟 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿

𝐶𝐶𝑆𝑆

(i.e., relative velocity 𝑉𝑉𝑟𝑟)


RTT for Moving and Deforming CV

*Ref) Fluid Mechanics by Frank M. White, McGraw Hill

𝑑𝑑𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠𝑑𝑑𝑑𝑑 =


+ � 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


RTT Summary (1) 𝑑𝑑𝐵𝐵𝑠𝑠𝑠𝑠𝑠𝑠𝑑𝑑𝑑𝑑 =


� 𝛽𝛽𝜌𝜌𝑑𝑑𝑉𝑉𝐶𝐶𝑉𝑉

+ � 𝛽𝛽𝜌𝜌𝑉𝑉𝑟𝑟 ⋅ 𝒏𝒏�𝑑𝑑𝛿𝛿𝐶𝐶𝑆𝑆


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:


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


= 0

2. If 𝑉𝑉 = constant over discrete CS’s (i.e., one-dimensional flow)


= �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)


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.


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

Control Volume and Reynolds Transport Theoremfluids/Posting/Lecture_Notes... · 57:020 Fluids...

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