ÀÎÅͳÝÈÇаøÇÐ+±¸±Û+³×À̹ö
ÀÚµ¿·Î±×ÀÎ
ÄÁÅÙÃ÷ | contents
ÀϹݰԽÃÆÇ
Q&A°Ô½ÃÆÇ
°øÇÐ S/W
½ºÅ͵ð | study
´ÜÀ§È¯»ê
³óµµ°è»ê
ÀϹÝÈÇÐ
È°øÀϹÝ
È°ø½Ç¹«
¾Ð·Âȯ»ê
±Û¾´ÀÌ
:
¾ç¹Îµ¿
°íÀ¯ID
: ¾ç¹Îµ¿
³¯Â¥
: 00-00-00 00:00
Á¶È¸
: 7135
Ç¥ÁØ´ë±â¾Ð 1atmÀ϶§ ¿¡º£·¹½ºÆ®»ê 8800m´Â ¸î atm Àΰ¡¿ä?
Çؼ³ ºÎŹµå¸³´Ï´Ù.
Stefano
Stefano
39-08-02 00:00
¾Õ¼ ¼Ò°³ÇÑ Âü°í ½ÎÀÌÆ®¿¡¼´Â ´ÙÀ½°ú °°´Ù°í ÇÕ´Ï´Ù. ´Ü ¿©±â¼ÀÇ h´Â ft ´ÜÀ§ (ÀÌ°ªµµ °è»êÇؼ ºñ±³ÇØ ¸Â´ÂÁö È®Àιٶø´Ï´Ù.)
Àý´ë¿Âµµ = (To)*(1-h/145,442)
¾Ð·Â = (Po)*(1-h/145442)^5.255876
¹Ðµµ = (Do)*(1-h/145442)^4.255876
¾Õ¼ ¼Ò°³ÇÑ Âü°í ½ÎÀÌÆ®¿¡¼´Â ´ÙÀ½°ú °°´Ù°í ÇÕ´Ï´Ù. ´Ü ¿©±â¼ÀÇ h´Â ft ´ÜÀ§ (ÀÌ°ªµµ °è»êÇؼ ºñ±³ÇØ ¸Â´ÂÁö È®Àιٶø´Ï´Ù.) Àý´ë¿Âµµ = (To)*(1-h/145,442) ¾Ð·Â = (Po)*(1-h/145442)^5.255876 ¹Ðµµ = (Do)*(1-h/145442)^4.255876
Stefano
Stefano
39-08-02 00:00
(0) ¹Ù·Î ¸çÄ¥Àü °Ô½Ã¹° #2377¿¡¼ Áú¹®Çß´ø ³»¿ë°ú À¯»çÇÑ Áú¹®ÀÔ´Ï´Ù. ¸ÕÀú ã¾Æº¸µµ·Ï Çϼ¼¿ä.
(1) °íµµ¿¡ µû¶ó ¿Âµµ°¡ Å©°Ô º¯È°¡ ¾ø´Ù°í °¡Á¤Çϸé
´ë±â¿Âµµ(T)·Î À¯ÁöµÇ´Â ³ôÀÌ(h)¿¡¼´Â
-dP = ¥ñ g dh = (PM/RT) g dh R=±âü»ó¼ö,M=ºÐÀÚ·®.....(1)
¿Âµµ°¡ ÀÏÁ¤ÇÏ´Ù°í ÇÏ°í À̸¦ ÀûºÐÇÏ¸é ³ôÀÌ¿¡ µû¸¥ ±â¾ÐÀÌ ¾ò¾îÁý´Ï´Ù.
-dP/P= (gM/RT) dh --> ln(P)= ln(Po) - (gM/RT)h ......(2)
(2) ´ë·ù±Ç Áß(Çؼö¸é~ÇØ¹ß 11km Á¤µµ)¹üÀ§¿¡¼´Â °íµµ 1km ¿Ã¶ó°¥ ¶§¸¶´Ù -6.5 K¾¿ ¶³¾îÁø´Ù°í ÇϹǷÎ
T= To - (6.5/1000)h..................................(3)
ÀÌ ½ÄÀ» (1)ÀÇ T¿¡ ´ëÀÔÇؼ ÀûºÐÇϸé
-dP = ¥ñ g dh = (P*M/R/(To-6.5/1000 h)*g*dh
-dP/P = (Mg/R)*dh/(To-0.0065 h)
ln (Po/P) = (Mg/R) (1/0.0065) ln(To/(To-0.0065 h)
P= Po * {To/(To-0.0065 h)}^(-Mg/0.0065R)..............(4)
À§ÀÇ (4)½Ä¿¡´Ù Po= 1atm, h=8800 m, To = 288.15 K, M =29 kg/kg-mole, g = 9.8 m/s^2, R = 8314.47 Pa-m3/K/kg (Pa = kg-m/s2/m2). ±×¸®°í »ó¼ö 0.0065ÀÇ ´ÜÀ§´Â K/mÀÓÀ» °í·ÁÇÏ¿© °è»êÇؼ ´ä±Û·Î ¿Ã·ÁÁÖ¼¼¿ä. (¼ö½ÄÀÌ ¸Â³ª È®Àεµ ÇØÁÖ¼¼¿ä)
(3) ¸¶Ä§ ÀÌ°÷¿¡¼ À¯»çÇÑ ³»¿ëÀ» ã¾Æ º¼ ¼ö ÀÖ½À´Ï´Ù.
http://www.atmosculator.com/The%20Standard%20Atmosphere.html?
(0) ¹Ù·Î ¸çÄ¥Àü °Ô½Ã¹° #2377¿¡¼ Áú¹®Çß´ø ³»¿ë°ú À¯»çÇÑ Áú¹®ÀÔ´Ï´Ù. ¸ÕÀú ã¾Æº¸µµ·Ï Çϼ¼¿ä. (1) °íµµ¿¡ µû¶ó ¿Âµµ°¡ Å©°Ô º¯È°¡ ¾ø´Ù°í °¡Á¤ÇÏ¸é ´ë±â¿Âµµ(T)·Î À¯ÁöµÇ´Â ³ôÀÌ(h)¿¡¼´Â -dP = ¥ñ g dh = (PM/RT) g dh R=±âü»ó¼ö,M=ºÐÀÚ·®.....(1) ¿Âµµ°¡ ÀÏÁ¤ÇÏ´Ù°í ÇÏ°í À̸¦ ÀûºÐÇÏ¸é ³ôÀÌ¿¡ µû¸¥ ±â¾ÐÀÌ ¾ò¾îÁý´Ï´Ù. -dP/P= (gM/RT) dh --> ln(P)= ln(Po) - (gM/RT)h ......(2) (2) ´ë·ù±Ç Áß(Çؼö¸é~ÇØ¹ß 11km Á¤µµ)¹üÀ§¿¡¼´Â °íµµ 1km ¿Ã¶ó°¥ ¶§¸¶´Ù -6.5 K¾¿ ¶³¾îÁø´Ù°í ÇϹǷΠT= To - (6.5/1000)h..................................(3) ÀÌ ½ÄÀ» (1)ÀÇ T¿¡ ´ëÀÔÇؼ ÀûºÐÇϸé -dP = ¥ñ g dh = (P*M/R/(To-6.5/1000 h)*g*dh -dP/P = (Mg/R)*dh/(To-0.0065 h) ln (Po/P) = (Mg/R) (1/0.0065) ln(To/(To-0.0065 h) P= Po * {To/(To-0.0065 h)}^(-Mg/0.0065R)..............(4) À§ÀÇ (4)½Ä¿¡´Ù Po= 1atm, h=8800 m, To = 288.15 K, M =29 kg/kg-mole, g = 9.8 m/s^2, R = 8314.47 Pa-m3/K/kg (Pa = kg-m/s2/m2). ±×¸®°í »ó¼ö 0.0065ÀÇ ´ÜÀ§´Â K/mÀÓÀ» °í·ÁÇÏ¿© °è»êÇؼ ´ä±Û·Î ¿Ã·ÁÁÖ¼¼¿ä. (¼ö½ÄÀÌ ¸Â³ª È®Àεµ ÇØÁÖ¼¼¿ä) (3) ¸¶Ä§ ÀÌ°÷¿¡¼ À¯»çÇÑ ³»¿ëÀ» ã¾Æ º¼ ¼ö ÀÖ½À´Ï´Ù. http://www.atmosculator.com/The%20Standard%20Atmosphere.html?
Copyright 1999.07.10-Now ChemEng.co.kr & 3D System Engineering. (mail : ykjang@naver.com, call 010-4456-8090)