MCP๋กœ ์—ฐ๊ฒฐ โ†’

๊ณ„์‚ฐ ์ž…๋ ฅ

๊ณต์‹

๊ด‘๊ณ 

๊ฒฐ๊ณผ

์ถ”์ • ๋ฌผ ์œ ๋Ÿ‰
1,666.08
๋ถ„๋‹น ๋ฆฌํ„ฐ (L/min)
์œ ๋Ÿ‰ (mยณ/s) 0.027768
์œ ๋Ÿ‰ (mยณ/h) 99.96
์œ ์† (m/s) 14.142
๋ฐฐ๊ด€ ๋‹จ๋ฉด์  (mยฒ) 0.001963

์ด ๊ณ„์‚ฐ๊ธฐ์˜ ๊ธฐ๋Šฅ

์ด ๋„๊ตฌ๋Š” ๋ฐฐ๊ด€์—์„œ ํ˜๋Ÿฌ๋‚˜์˜ค๋Š” ๋ฌผ(๋˜๋Š” ๋‹ค๋ฅธ ์œ ์ฒด)์˜ ์ฒด์  ์œ ๋Ÿ‰์„, ์œ ์ฒด๋ฅผ ๋ฐ€์–ด๋‚ด๋Š” ์••๋ ฅ๊ณผ ๋ฐฐ๊ด€ ์•ˆ์ง€๋ฆ„์„ ๋ฐ”ํƒ•์œผ๋กœ ์ถ”์ •ํ•ฉ๋‹ˆ๋‹ค. ์ด์ƒ์ ์ธ ๋ฒ ๋ฅด๋ˆ„์ด/ํ† ๋ฆฌ์ฒผ๋ฆฌ ๊ด€๊ณ„์‹์„ ์ ์šฉํ•ด ์••๋ ฅ ์—๋„ˆ์ง€๋ฅผ ์šด๋™ ์—๋„ˆ์ง€๋กœ ๋ณ€ํ™˜ํ•จ์œผ๋กœ์จ ์ถœ๊ตฌ ์œ ์†์„ ๊ตฌํ•œ ๋’ค, ์—ฌ๊ธฐ์— ๋ฐฐ๊ด€ ๋‹จ๋ฉด์ ์„ ๊ณฑํ•ด ์œ ๋Ÿ‰์„ ์‚ฐ์ถœํ•ฉ๋‹ˆ๋‹ค.

์‚ฌ์šฉ ๋ฐฉ๋ฒ•

๊ฒŒ์ด์ง€ ์••๋ ฅ์„ ํŒŒ์Šค์นผ(Pa) ๋‹จ์œ„๋กœ, ๋ฐฐ๊ด€ ์•ˆ์ง€๋ฆ„์„ ๋ฐ€๋ฆฌ๋ฏธํ„ฐ(mm) ๋‹จ์œ„๋กœ, ์œ ์ฒด ๋ฐ€๋„๋ฅผ ์„ธ์ œ๊ณฑ๋ฏธํ„ฐ๋‹น ํ‚ฌ๋กœ๊ทธ๋žจ(kg/mยณ) ๋‹จ์œ„๋กœ ์ž…๋ ฅํ•˜์„ธ์š”. ์ƒ์˜จ์˜ ๋ฌผ์€ ์•ฝ 1000 kg/mยณ์ž…๋‹ˆ๋‹ค. ๊ณ„์‚ฐ๊ธฐ๋Š” ์œ ๋Ÿ‰์„ ๋ถ„๋‹น ๋ฆฌํ„ฐ(L/min), ์ดˆ๋‹น ์„ธ์ œ๊ณฑ๋ฏธํ„ฐ(mยณ/s), ์‹œ๊ฐ„๋‹น ์„ธ์ œ๊ณฑ๋ฏธํ„ฐ(mยณ/h)๋กœ ๋ณด์—ฌ ์ฃผ๋ฉฐ, ํ•จ๊ป˜ ์œ ์†๊ณผ ๋ฐฐ๊ด€ ๋‹จ๋ฉด์ ๋„ ์•Œ๋ ค ์ค๋‹ˆ๋‹ค.

๊ณต์‹ ์ดํ•ดํ•˜๊ธฐ

์ถœ๊ตฌ ์œ ์†์€ ์••๋ ฅ ์—๋„ˆ์ง€์™€ ์šด๋™ ์—๋„ˆ์ง€๋ฅผ ๊ฐ™๋‹ค๊ณ  ๋†“์•„ ๊ตฌํ•ฉ๋‹ˆ๋‹ค:

$$v = \sqrt{\frac{2P}{\rho}}$$

์›ํ˜• ๋ฐฐ๊ด€์˜ ๋‹จ๋ฉด์ ์€

$$A = \frac{\pi D^{2}}{4}$$

์ž…๋‹ˆ๋‹ค. ์œ ์†์— ๋‹จ๋ฉด์ ์„ ๊ณฑํ•˜๋ฉด ์ฒด์  ์œ ๋Ÿ‰

$$Q = A \cdot \sqrt{\frac{2P}{\rho}}$$

๊ฐ€ ๋ฉ๋‹ˆ๋‹ค. ์ด๋Š” ์–ด๋””๊นŒ์ง€๋‚˜ ์ด์ƒ์ ์ธ ๊ฐ’์œผ๋กœ, ๋งˆ์ฐฐ ์†์‹คยท์ด์Œ๊ด€ยท์ ์„ฑยท์œ ์ถœ ๊ณ„์ˆ˜๋ฅผ ๋ฌด์‹œํ•œ ๊ฒฐ๊ณผ์ž…๋‹ˆ๋‹ค. ๋”ฐ๋ผ์„œ ์‹ค์ œ ์œ ๋Ÿ‰์€ ์ด๋ณด๋‹ค ์ž‘๊ฒŒ ๋‚˜ํƒ€๋‚ฉ๋‹ˆ๋‹ค.

Circle representing pipe cross-sectional area with diameter D and area formula relationship
The cross-sectional area A depends on the inner diameter D.
Cross-section of a pipe showing pressure pushing water through a circular area of diameter D, with flow exiting as a velocity arrow
Pressure P drives water through the pipe's circular cross-section (area A = ฯ€Dยฒ/4), producing flow velocity and rate Q.

๊ณ„์‚ฐ ์˜ˆ์‹œ

\(P = 100{,}000\ \text{Pa}\), \(D = 50\ \text{mm}\), \(\rho = 1000\ \text{kg/m}^3\) ์ธ ๊ฒฝ์šฐ:

$$A = \frac{\pi \cdot (0.05)^{2}}{4} = 0.0019635\ \text{m}^2$$

์œ ์†

$$v = \sqrt{\frac{2 \cdot 100000}{1000}} = \sqrt{200} \approx 14.142\ \text{m/s}$$

๋”ฐ๋ผ์„œ

$$Q = 0.0019635 \times 14.142 \approx 0.02777\ \text{m}^3\text{/s}$$

์ด๋ฉฐ, ์ด๋Š” ์•ฝ 1,666 L/min ์— ํ•ด๋‹นํ•ฉ๋‹ˆ๋‹ค.

์ž์ฃผ ๋ฌป๋Š” ์งˆ๋ฌธ

์ด ๊ฐ’์ด ์ •ํ™•ํ•œ๊ฐ€์š”? ์•„๋‹™๋‹ˆ๋‹ค. ๋งˆ์ฐฐ์ด ์—†๊ณ  ์ ์„ฑ์ด ์—†๋Š” ํ๋ฆ„์„ ๊ฐ€์ •ํ•œ ์ด๋ก ์  ์ตœ๋Œ“๊ฐ’์ž…๋‹ˆ๋‹ค. ์‹ค์ œ ์ถ”์ •์—๋Š” ์œ ์ถœ ๊ณ„์ˆ˜(๋ณดํ†ต 0.6~0.9)๋ฅผ ๊ณฑํ•ด ๋ณด์ •ํ•˜์„ธ์š”.

์–ด๋–ค ์••๋ ฅ์„ ์ž…๋ ฅํ•ด์•ผ ํ•˜๋‚˜์š”? ํ๋ฆ„์„ ๋ฐ€์–ด๋‚ด๋Š” ์••๋ ฅ(๊ฒŒ์ด์ง€ ์••๋ ฅ)์„ ์ž…๋ ฅํ•˜์„ธ์š”. 1 bar = 100,000 Pa, 1 psi โ‰ˆ 6,895 Pa ์ž…๋‹ˆ๋‹ค.

๋‹ค๋ฅธ ์œ ์ฒด์—๋„ ์“ธ ์ˆ˜ ์žˆ๋‚˜์š”? ๋„ค โ€” ํ•ด๋‹น ์œ ์ฒด์˜ ๋ฐ€๋„๋งŒ ์ž…๋ ฅํ•˜๋ฉด ๋ฉ๋‹ˆ๋‹ค. ๋ฌผ๋ฆฌ ์›๋ฆฌ๋Š” ๋™์ผํ•ฉ๋‹ˆ๋‹ค.

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