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JD Transportphenomena 9

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< 9. Transport phenomena 2 Problems >
< Transport phenomena 2 by JD >
P.297 Problem 1) A 0.2m wide plate, 0.8m long, is placed on the bottom of a shallow tank. The plate is heated
and maintained at a constant surface temperature of 60โ„ƒ. Liquid water 12cm deep flows over the flat plate with
bulk volumetric flow rate of 2.4 × 10 ๐‘š /๐‘ ๐‘’๐‘. The bulk liquid water temperature may be assumed to remain
constant at 20โ„ƒ as it flows along the length of the plate.
0.2m
a) What are the values of ๐‘น๐’†๐‘ณ and ๐‘ท๐’“ for this convective heat transfer process?
Is the flow laminar or turbulent at the end of the plate?
๐‘…๐‘’ =
๐œŒ๐‘ˆ๐ฟ 992.2 × 0.1 × 0.8
=
= 120,632
๐œ‡
658 × 10
Laminar flow (๐‘…๐‘’ <200,000)
๐‘ƒ๐‘Ÿ =
๐œ‡๐ถ
658 × 10 × 4175
=
= 4.14
๐‘˜
0.663
< Transport phenomena 2 by JD >
P.297 Problem 1) A 0.2m wide plate, 0.8m long, is placed on the bottom of a shallow tank. The plate is heated
and maintained at a constant surface temperature of 60โ„ƒ. Liquid water 12cm deep flows over the flat plate with
bulk volumetric flow rate of 2.4 × 10 ๐‘š /๐‘ ๐‘’๐‘. The bulk liquid water temperature may be assumed to remain
constant at 20โ„ƒ as it flows along the length of the plate.
0.2m
b) What is the local heat flux at a distance 0.5 m from the leading edge of the plate?
Use the Blasius solution based Nusselt number equation.
๐‘˜
/
โ„Ž = 0.332 ๐‘…๐‘’ ๐‘ƒ๐‘Ÿ /
๐‘ฅ
0.663
992.2 × 0.1 × 0.5
= 0.332 ×
×
0.5
658 × 10
๐‘„
=โ„Ž ๐‘‡ −๐‘‡
๐ด
/
× 4.14
/
= 194.1
= 194.1 × 40 = 7764๐‘Š/๐‘š
< Transport phenomena 2 by JD >
P.297 Problem 1) A 0.2m wide plate, 0.8m long, is placed on the bottom of a shallow tank. The plate is heated
and maintained at a constant surface temperature of 60โ„ƒ. Liquid water 12cm deep flows over the flat plate with
bulk volumetric flow rate of 2.4 × 10 ๐‘š /๐‘ ๐‘’๐‘. The bulk liquid water temperature may be assumed to remain
constant at 20โ„ƒ as it flows along the length of the plate.
0.2m
c) What is the total heat flux rate from the surface?
โ„Ž๐ฟ
/
= 0.664๐‘…๐‘’ ๐‘ƒ๐‘Ÿ /
๐‘˜
992.2 × 0.1 × 0.8
= 0.664 ×
658 × 10
๐‘๐‘ข =
โ„Ž = 370.3 ×
.
.
๐‘„ = โ„Ž๐ด ๐‘‡ − ๐‘‡
/
× 4.14
/
= 370.3
= 307
= 307 × 0.16 × 40 = 1965๐‘Š/๐‘š
< Transport phenomena 2 by JD >
Problem 2) From an experiment of fluid flow over a heated flat plate, it has been shown that the temperature
profile within the TBL is given as below. (Here, flow velocity is assumed to depend only on y coordinate)
๐‘ฃ
๐‘‡ = 400โ„ƒ
๐‘ฆ
๐‘‡ = 300โ„ƒ
๐‘‡−๐‘‡
๐‘ฃ
= 1 − (0.7 +
๐‘ฆ)
๐‘‡ −๐‘‡
๐œˆ
/
๐‘ฅ
Q. If ๐‘‡ = 400โ„ƒ, ๐‘‡ = 300โ„ƒ, ๐‘˜ = 0.037 ๐‘Š/๐‘š๐พ, and
= 1000/๐‘š,
what is the heat flux and heat transfer coefficient on the surface?
< Transport phenomena 2 by JD >
Problem 2) From an experiment of fluid flow over a heated flat plate, it has been shown that the temperature
profile within the TBL is given as below. (Here, flow velocity is assumed to depend only on y coordinate)
๐‘ฃ
๐‘‡ = 400โ„ƒ
๐‘ฆ
๐‘˜ = 0.037๐‘Š/๐‘š๐พ
๐‘‡−๐‘‡
๐‘ฃ
= 1 − (0.7 +
๐‘ฆ)
๐‘‡ −๐‘‡
๐œˆ
๐‘‡ = 300โ„ƒ
/
๐‘ฅ
๐œ•๐‘‡
๐œ•๐‘ฆ
=−
1
๐‘‡ −๐‘‡
7
๐‘ž = −๐‘˜
๐œ•๐‘‡
๐œ•๐‘ฆ
๐‘ฃ
๐œˆ
0.7 +
=โ„Ž ๐‘‡ −๐‘‡
๐‘ฃ
๐‘ฆ
๐œˆ
=−
= 3947๐‘Š/๐‘š
1
๐‘‡ −๐‘‡
7
๐‘ฃ
๐œˆ
0.7 +
๐‘ฃ
๐‘ฆ
๐œˆ
= −106,669๐พ/๐‘š
โ„Ž = 39.5๐‘Š/๐‘š ๐พ
< Transport phenomena 2 by JD >
Problem 2) From an experiment of fluid flow over a heated flat plate, it has been shown that the temperature
profile within the TBL is given as below. (Here, flow velocity is assumed to depend only on y coordinate)
๐‘ฃ
๐‘‡ = 400โ„ƒ
๐‘ฆ
๐‘‡ = 300โ„ƒ
๐‘˜ = 0.037๐‘Š/๐‘š๐พ
๐‘‡−๐‘‡
๐‘ฃ
= 1 − (0.7 +
๐‘ฆ)
๐‘‡ −๐‘‡
๐œˆ
/
๐‘ฅ
Further discussion) What if, velocity is dependent both on ๐’™
and ๐’š ??
Hint. Energy balance analysis with control volume integral (P.286~287)
< Transport phenomena 2 by JD >
Problem 3) Consider a human body as a cylinder with 40cm diameter and 1.8m height in still air. If the body
temperature is 2 โ„ƒ lower than that of the air, how much of energy will be required to maintain body
temperature(37 โ„ƒ)? Generally, a human intakes about 8400KJ (2000kcal)/day. Compare it with calculated value.
1) Vertical cylinder problem ๏ƒ 
(๐บ๐‘Ÿ =
2)
=
.
.
= 0.222 ≥
= 1.16 × 10 × 2 × 1.8
=
.
= 0.182
= 191.8)
< Churchill and Chu’s Experiments >
0.387 × 1.16 × 10 × 2 × 1.8 × 0.705
= 0.825 +
/
1 + 0.492/0.705 /
= 0.825 + 10.17
= 121 ๏ƒ  = โ„Ž = 121 × = 121 ×
.
.
= 1.82
/
= 121
< Transport phenomena 2 by JD >
Problem 3) Consider a human body as a cylinder with 40cm diameter and 1.8m height in still air. If the body
temperature is 2 โ„ƒ lower than that of the air, how much of energy will be required to maintain body
temperature(37 โ„ƒ)? Generally, a human intakes about 8400KJ (2000kcal)/day. Compare it with calculated value.
3) ๐‘„ = โ„Ž๐ดโˆ†๐‘‡ = 1.82 × 0.4 × ๐œ‹ × 1.8 × 2 = 8.23๐‘Š
Total heat loss = 8.23๐‘Š × 3600 × 24 = 711072
711072
= 0.084
8400 × 1000
About 8% of energy is used to maintain body temperature, even in the very hot weather
< Transport phenomena 2 by JD >
Problem 4)
Problem 4)
Problem 4)
๐‘จ
ODE to be solved
Boundary conditions
Problem 5) JD left a cup of coffee for two weeks in his office. Initially, height of coffee was 7cm.
Estimate the height of coffee after two weeks.
Temperature is 293K and diffusion coefficient (D) of coffee (water) is ๐ท = 2.5 × 10 ๐‘š /๐‘ ๐‘’๐‘
and pressure is 2.34 × 10 ๐‘ƒ๐‘Ž.
z
1) Pseudo steady state assumption & Mass balance
๐‘†๐‘
8cm
− ๐‘†๐‘
๐‘
= ๐‘ฆ (๐‘
=0
๐‘
+ ๐‘ ) − cD
๐‘‘๐‘ฆ
= constant
๐‘‘z
7cm
๐‘
=
= ๐ถ๐‘œ๐‘›๐‘ ๐‘ก๐‘Ž๐‘›๐‘ก, ๐‘
cD
1−y
ln(
)
z -z
1−y
=0
Problem 5) JD left a cup of coffee for two weeks in his office. Initially, height of coffee was 7cm.
Estimate the height of coffee after two weeks.
Temperature is 293K and diffusion coefficient (D) of coffee (water) is ๐ท = 2.5 × 10 ๐‘š /๐‘ ๐‘’๐‘
and pressure is 2.34 × 10 ๐‘ƒ๐‘Ž.
z
๐‘ฆ =
๐‘ A=
๐œŒd(z − z )
A
Mdt
๐‘
8cm
=
๐‘ฆ =0
cD
1−y
๐œŒd(z − z )
ln(
)=
z -z
1−y
Mdt
7cm
C=
*
z − z d(z − z ) =
cMD 1 − y
ln(
)dt
๐œŒ
1−y
< Open position >
Complex fluid and Soft matter physics lab
์—ฐ๊ตฌ ๋ถ„์•ผ: 1) Rheology (๋ฌผ์งˆ์˜ ๋ณ€ํ˜•๊ณผ ํ๋ฆ„ ์—ฐ๊ตฌ)
2) Soft matter (Colloid & Polymer) 3) Process modeling & analysis
์ง€์› ์ž๊ฒฉ: ๋Œ€ํ•™์› ์ง„ํ•™์„ ๊ณ ๋ คํ•˜๋Š” ์ฐฝ์˜์ ์ด๊ณ  ์ ๊ทน์ ์ธ 3,4ํ•™๋…„ (Attitude is all!!)
๋ชจ์ง‘ ์ธ์›: 2๋ช…
Colloidal gel
Colloidal glass
์ง„ํ–‰ ์˜ˆ์ • ์—ฐ๊ตฌ: 1) Temperature-Responsive polymer์˜ ์œ ๋ณ€๋ฌผ์„ฑ ์ œ์–ด๋ฅผ ํ†ตํ•œ ๋ฌผ์งˆ ์ „๋‹ฌ ์กฐ์ ˆ
(Sun-screen ๋ฐ Drug & Cosmetics์—์˜ ์‘์šฉ)
2) Li-ion battery paste์˜ rheological property์— ๋”ฐ๋ฅธ ๊ณต์ • ๋ฐ ์„ฑ๋Šฅ ๋ถ„์„
3) Liquid metal์˜ ์œ ๋ณ€ํ•™์  ํŠน์„ฑ ๋ถ„์„ ๋ฐ ์‘์šฉ
4) ์‹ํ’ˆ์— ์‚ฌ์šฉ๋˜๋Š” Hydrocolloids์˜ ์œ ๋ณ€ํ•™ ํŠน์„ฑ ๋ถ„์„
5) ์ด๋ก ์  ์ ‘๊ทผ (๋ชจ๋ธ๋ง ๋ฐ ์ž…์ž ์‹œ๋ฎฌ๋ ˆ์ด์…˜)์„ ์ด์šฉํ•œ Colloid suspension์˜ ์œ ๋ณ€ํ•™ ๊ฑฐ๋™ ๋ถ„์„
6) ์ˆ˜์น˜ํ•ด์„์„ ์ด์šฉํ•œ ์œ ๋™ ๋ถ„์„
7) ๊ธฐํƒ€ ๋ณธ์ธ์ด ํ•˜๊ณ  ์‹ถ์€ ์—ฐ๊ตฌ ์ œ์•ˆ ์–ผ๋งˆ๋“ ์ง€ ํ™˜์˜!!
< Open position >
Complex fluid and Soft matter physics lab
๊ทผ๋ฌด ์กฐ๊ฑด: ์ถœํ‡ด๊ทผ ์ž์œ , ์ฃผ1ํšŒ ๊ฐœ์ธ ๋ฏธํŒ…, ์ฃผ1ํšŒ ์—ฐ๊ตฌ์‹ค ๋ฏธํŒ… (Paper & Research presentation)
์—ฐ๊ตฌ ๋ฐ ํ•™์—…์—๋งŒ ์ง‘์ค‘ํ•  ์ˆ˜ ์žˆ๋Š” ์ธ๊ฑด๋น„, ์ž์œ ๋กœ์šด & ํ•™๋ฌธ์ ์ธ ๋ถ„์œ„๊ธฐ (๊ณผ์ œ ๋ถ€๋‹ด X)
์‹œ์ž‘ ์‹œ๊ธฐ: 2021.1์›” ํ˜น์€ 3์›” ~
์ด๋Ÿฐ ๋ถ„์—๊ฒŒ ์ถ”์ฒœ
- ๋ฌผ๋ฆฌํ™”ํ•™, ๊ณตํ•™์ˆ˜ํ•™, ์ „๋‹ฌํ˜„์ƒ 1&2 ์— ํฅ๋ฏธ๋ฅผ ๋Š๋‚€๋‹ค (ํ™”ํ•™ ๋ณด๋‹ค๋Š” ๋ฌผ๋ฆฌ & ์ˆ˜ํ•™์ด์ง€!)
Colloidal gel
Colloidal glass
- ์‹คํ—˜์„ ๋งŽ์ด ํ•˜๋Š”๊ฒŒ ์‹ซ๋‹ค & ๋ชธ์— ์•ˆ ์ข‹์€ ํ™”ํ•™์•ฝํ’ˆ์„ ์“ฐ๊ณ  ์‹ถ์ง€ ์•Š๋‹ค
- ์ปดํ“จํ„ฐ ์‹œ๋ฎฌ๋ ˆ์ด์…˜์ด๋‚˜ ์ด๋ก ์ด ์ข‹๋‹ค (๋‚˜๋Š” ์ „์ƒ์— ์„ ๋น„์˜€๋‹ค)
- ๊ถ๊ธˆํ•œ ๊ฑด ๋ชป ์ฐธ์•„! ์•Œ ๋•Œ๊นŒ์ง€ ๊ณ ๋ฏผํ•œ๋‹ค!
- ์–ด๋ ค์šด ๋ฌธ์ œ๋ฅผ ํ•ด๊ฒฐํ•˜๋Š” ๊ฒƒ์—์„œ ์žฌ๋ฏธ๋ฅผ ๋Š๋‚€๋‹ค
- ๋‚จ๋“ค ๋‹คํ•˜๋Š” ์—๋„ˆ์ง€, ๋ฐ”์ด์˜ค๋Š” ์‹์ƒํ•ด!!
- ENTJ ๋ž‘ ์ž˜ ๋งž๊ฒ ๋Š”๋ฐ?
์ง€์› ๋ฐฉ๋ฒ•: jdpark@sookmyung.ac.kr ๋กœ ๊ฐ„๋‹จํ•œ ์ž๊ธฐ์†Œ๊ฐœ & ์ง€์›๋™๊ธฐ ์ ์–ด์„œ ๋ณด๋‚ด์ฃผ์„ธ์š”~
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