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ProblemSet5_2011

# ProblemSet5_2011 - ES 300 Principles of Naval Weapon...

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Unformatted text preview: ES 300 Principles of Naval Weapon Systems Problem Set 5 — AY2011 NAME: Section: 1) Fill in the following tables. EXAMPLE Mum) WEAPON EO SPECTRUM TYPE OF E0 SENSOR BAND SYSTEM USING DIVISION Mum) BAND SENSOR UV 4 _ / V 34-0,),ﬂ v 515% I" * V1) ‘ 5 ‘ ‘ ' B’"°°“/a’ VISIBLE LIGHT - . A .3 I k. 5 —L. t ,1 ;, NEARIR U ’3 55 a A I. ’r I‘ MID IR 5/(v i I, FAR IR . c» . L» —I > ,k . Image lntensifier [I I L/ . -,,, EXTREMEIR I; “(A ‘ 2) Using the Universal Blackbody Curve, calculate the percentage of total energy radiated by a blackbody at 750°C... a) In the transmission window between 7pm and mum: »- .. 1‘ :I‘eg’i" 7:3” 0 Pitfall Prevention: 7” I1“ "*3 5 ‘ {L A' l Calculate fupper and ﬂower \ (f: y,“ x 131;; : 7Ié I x I7r< IL : :J. 3 1 separately as functions ‘ _ H3 g .2 a; INK a H oftemperature and A. )MT ""- ‘ lea > IL = 5' H 'L- .‘1'3 1 C 42 b) What is another name for this value (i.e., what term in the max IR range)? 10> kl I’M; t4) 4‘0 H L; 4.? {n c) What detector and target parameters affect this value? DC ’19.; A i " VIVUI I; _,I I I- d) What about the detector physically determines the detector parameter for part (c)? II 0 Hint: This isthesamething KI. I’. ~: I") I»! w" 4”: that E5 on the final exam is asking. 1 ES 300 Principles of Naval Weapon Systems 3) Given the following: Tar et Data IJE_ Environmental Data S urface Area = 25 m Mower = 6 microns Air Density = 1.25 x10-3 Reflectivit = 0.20 £4043 Effective Detector Aerture = 0.15 m2 Air Tem = 23° Exhaust Tern = 37°C—‘3/c/ MDS = -63 dBm Emissivity = 0.9 a) Calculate the Bandwidth Factor (F) required to achieve a max IR range of 55 km: ,. #40311»; , i ,g, 7(13/OK >I’Vli1: A17: v f ﬁ—a/rx!‘5«0.§<w;{w’ -0.X ‘ ’tc [CY .— lid / v1 / I /v T— to C/_ I ' ‘I /I" , ‘f I ‘1 \ V, q we"?! ;Zr'l‘ a]: Semc'i‘l 3'0 05 ’276 ‘Orﬂ-‘ZZ’?’ bi" MngF/‘b I/Ll 4? wk I’” '1 IL ’ a ﬁxation" »r ,h 5 :(5w ; OJX¥ V} ‘5". mic/(VAC 27.17sz '(Q If) b) What Aupper is necessary in order to achieve the above Bandwidth Factor? Fgﬁ :OJX7 I; “-1[(/\Ll/I) #76614”. r'S’Ok) rays/0'0) '7 0.06 51f”; :O‘IXLaoS -’ 0-2‘37 #Hsom) Aﬂékn) M:32P;[t74ﬂ 2/0 l 4) Calculate the Minimum Discernable Signal (MDS) required to detect a destroyer (DD)-sized target at a range of 30km given the following parameters: . / .5. '/6‘ Environmental Temperature: 12 °C 5 1’5 7 K : T DD Skin Temperature: 33 °C 7 '30 6 K ’ DD Skin Emissivity: 0.92 i 2:7 Environmental Emissivity 0.98 r be DD Surface Area: ' 600 m2 , Ar Detector Opening: .11 m wide, .09 m tall (rectangular) Detector Efficiency: 0.90 1 Detector Bandwidth: 8.5— 11 microns ‘ Hl M , ,. Joe“ 40‘] AO’OJI « I \\ l ‘1 \ “mm Mosatolﬁ(3*13 tub ES 300 Principles of Naval Weapon Systems / .5) 5) You are once again using the Uncooled Manportable Forward Looking Infra-Red (UMFLIR) to search for Fedayeen Saddam. Relevant data are as follows: Estimated surface area of Fedayeen Ar = 0.5 m2 0 Pitfall Prevention: Reflectivity of Fedayeen £1 = l - 0.20 =08 Don’t forget to convert Emissivity of environment 5 5,: 0.95 reflectivity to emissivity Personnel temperature rr : 38 °C 2 Wars for use in the Rmale Night air temperature Tc. = 12 °C was l‘ equation. UMFLIR Bandwidth 8-12 pm UMFLIR effective detector area A: r 0.005 m2 UMFLIR detector minimum signal for detection 50 pW a) Are the Fedayeen in the UMFLIR bandwidth? [(#75821: 07/) K N” I ' [5-10 5: 625’s: '2 yes, sully +9: moonlit. b) With these Fedayeen targets, what is the bandwidth factor for the UMFLIR? _ ﬂ _ _ g A ’CJ’I'A ’; O! .-, L. FrfCN-l/r)~i(>t,Trirférzaui-Rx4/0”!“1313"“1"“)’0'”; 5 c) What is the maximum IR range at which your UMFLIR can image these targets? , , V n l I 2 ,. A’LATACF Vb [éW43/L: ‘5] 660-“ 5 km MA" I v 7 ( .1 “’th “511’ H W SMM VI m‘, w,‘1 , \ bl ’L‘ :ég-Ci 'nL Ii . it I( lrplbi‘lo‘s low/ta] ’O'L/S 3‘ J m” d) Explain why you can or cannot see near-IR moonlight energy reflecting off the Fedayeen through your UMFLIR: l/amL HZ =03 km 2 410T Lani/W )55“’D‘»’107H um”; raw * 84va e) On the graph below, draw two approximate, scaled exitance curves: one for Fedayeen exitance and one for environment exitance. Indicate the peak wavelength of each exitance on the x-axis. (Spectral radiant values are not important except to indicate relative values based upon temperature.) indicate by two labeled vertical lines the bandwidth of the UMFLIR (on the x-axis). Shade in the area difference between the two curves corresponding to the contrast in exitance seen by the UMFLIR and residing within its bandwidth. [\WW ._ ZL‘M __ 43/», ,0 I , lc L55 Spectral Radiant Emittance (w/mzk) ...
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