CUVX Design Report - the AOE home page - Virginia Tech

CUVX Design – VT Team 2 Page 112Module 4 - Space Available4a. Underwater available hull volume: V HUW V FL4b. Above waterline available hull volume:For sheer line, choose from 3 criteria: - keep deck edge above water at 25 heel (DDS-079-1).- ensure longitudinal strength.- contain machinery box (in height).0.21 B T13.185MLWL15M = 14.22 m D 10MIN max( M) D 10MIN = 26.59 m D 10H MBreq N . HANGDK H HANGDK 1m .26.59D 0MIN2.011827.T6.36215.10 6. LWL 2 2.780649. 10 2 . LWL Dft0MIN = 19.235 m D 0 D 10 D 0 = 29.624 mD 20MIN0.014. 1.25.10 3LWL. 2.125.LWL T D 20MIN = 16.043 m D 20 D 10 D 20 = 29.624 mftF 0 D 0 T F 10 D 10 T F 20 D 20 T (Assuming zero trim)A PROF AVLWL F 0 4F . 10 F. 20(Trapezoidal rule)0.98 6A PROLWLflare10. deg if HullForm 60. deg otherwiseF AV = 22.999 m D AV F AV T D AV = 30.084 m F AV = 75.457 ftC W = 0.8632Above water hull volume: V HAW F . AV tan( flare). LWL LWL. B. C . W F AV V HAW = 1.031. 10 5 m 3 V HAW = 3.641. 10 6 ft 3Total hull volume: V HT V HUW V HAW V HT = 1.323. 10 5 m 3V HUW = 1.031. 10 6 ft 3Cubic number:CNV HT10 5 . ft 3CN = 46.723Total ship volume: V T V HT V D V T = 135369.256 m 3Machinery boxH MBH MB = 10.624 m V . MB VH MBreqMBreqV MB = 1.558. 10 4 m 3Module 5 - Electrical LoadV TBased on DDS 310-1. Estimate maximum functional load for winter cruise condition:(SWBS 200, propulsion). KW P 0.00323. kW . Php BPENGTOT KW P 168.kW(SWBS 561, steering). KW S 0.00826. kW . LWL TKW S 29.9 kWft 2= 4.781. 10 6 ft 3(SWBS 300, electric plant, lighting). KW E 0.000213. kW . V TKW E 500.8.kWft 3(SWBS 430+475, miscelaneous). KW M 101.4. kWKW M 111.5.kW(Collective Protection System) KW CPS 0.000135. kW . V TKW CPS = 645.37 kWft 3(SWBS 517, aux boiler). KW B 0.235. N T kWKW B 3913.5 kW(SWBS 521, firemain). KW F 0.000097. kW . V TKW F 257.6.kWft 3(SWBS 540, fuel handling). KW HN 0.000177. kW . V HTKW HN 113.4 kWft 3(SWBS 530+550, misc aux). KW A 0.65. N . T kW KW fins KW A 626.kW(SWBS 600, services). KW SERV 0.395. N . T kWKW SERV 342 . kWThe calculations are iterative, because KWH, KWV and KWAC depend on VAUX, which depends on the maximalfunctional load. Non payload functional load (without the above mentioned loads):KW NP KW P KW S KW E KW M KW B KW F KW HN KW A KW SERV(non-Payload)Maximum Functional Load: (Summer AC assumed worse case) KW NP = 6.063. 10 6 WKW MFL KW MFL 1000.kWKW X0kW .KW MFL KW Xwhile> 0.01KW MFLKW XKW MFLV AUX 40000 ft3 KW. . XkW 3411KW AC 0.67.0.1. kW. N T 0.00067. kW . V T V MB V AUX 0.1.BATKW PAYft 3KW V 0.103. KW AC BATKW PAY KW CPSKW MFL KW NP KW V KW AC BATKW PAYKW MFL EDMF = 1.1 EFMF = 1.1KW MFL = 9.633. 10 3 kWKW MFLM EDMF. EFMF. KW MFLKW MFLM = 1.166. 10 4 kW (MFL w/margins)The iterative process yields: TRANKW PAY = 636.27 kWV AUX 40000 ft3 KW. . MFLkW 3411KW AC0.67. 0.1. kW. N T 0.00067. kW . V T V MB V AUX 0.1.BATKW PAY KW AC = 1.955. 10 3 kWft 3KW V 0.103. KW AC BATKW PAY KW CPS(Ventilation) KW V = 918.498 kWPower required per generator for mechanical drive system, with one in stand-by:KW MFLMKW GREQmechif PSYSN SSG 1 .TYP 1 KW GREQmech = 0kW0.90.0. kW otherwise