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Week 4A Design Project 2: Ulysses - Starprobe (NASA-JPL) The primary objective of the Starprobe
mission was to investigate the properties of the solar wind during the fly-by mission within the 4 - 12 R.
distance from the Sun. NEAR-PERIHELION TRAJECTORY (VIEW PERPENDICULAR TO O... 

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The Intense Radiation from the Sun The protection of the Starprobe from the intense radiation from the
Sun was the key design challenge. FULL SCIENCE SPACECRAFT CONFIGURATION PERSPECTIVE
VIEW Sun OPTICAL BAFFLES & CONTAMINATION BARRIER RTG (3) RADIATOR Sol... 

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Calculation of Solar Flux on the Surface of the Sun Objective: to estimate the solar flux that Ulysses
spacecraft will experience during the close encounter. t = 12 hr The surface temperature of Sun is
approximately Starprobe 5,778 K (5,505 . C). The solar flux on the surface of the Sun: Enc... 

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Estimation of the Solar Flux around the Sun Energy balance equation: The total radiation energy
remains constant. qat sun * (4TR.) = qat4Ry * [4n(4R.)2] 4R 64,000,000 = 4,000,000 W Sun Starprobe
94RO 16 m2 Starprobe Ro 12R. 64,000,000 W Ro = radius of the Sun 912RO= = 444,000 ... 

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Calculation of Solar Flux on the Earth Orbit Estimation of the solar flux around the Earth orbit Earth
216R. 64,000,000 qEarth = 1370 - WV W (216)2 m2 - = 0.1370 cm 2 Ro R. = Radius of the Sun 10 Drexel
UNIVERSITY

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Equilibrium Temperature for a Primary Shield Objective: to estimate the equilibrium temperature Teq for
the primary shield with 30-deg cone (diameter = 2 m) at 4-solar-radius distance. (L,(m) For 2-m
diameter cone, we have r = 1m sin(15 deg) = 1 meter/L 15 deg - L = 1/sin 15 deg =3.8637m P... 

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Equilibrium Temperature for a Primary Shield at 4R. distance Cone surface area = n*L *r = 1*3.8637*1 =
12.1382 m2 Cone bottom surface = n*r2 L =3.8637 m r = 1 m D 15 deg - - Ja's 4) = 20 (12.1382 + - 4) TA
Surface area = Using qs = 4,000,000 W/m-, we get ## Teq Show more

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Energy Exchange between the Primary Shield and Cone Base Shape factors between the cone and
patch: To reach from patch to cone, the emission needs to go through Ab. Solar flux Fp1 = Fpb
Reciprocity relation 56.88 - A 1 , T, Primary shield cone A1 Fib = Ab Fb1 Enclosure relation ... 

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Energy Exchange between the Primary Shield and Payload Disk Radiation shields to further reduce
solar radiation for the protection of on-board instruments : Zero secondary shield case Energy balance
on the primary shield Solar flux Energy in = Energy out 56.88 A1 , T1 Energy in = Radiation ... 

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Calculation of Temperature of the Primary Shield T1 From the previous page, Surface area = Ab
(4,000,000) = [A1 + AbloTA Eq. * 1 GIVEN: b = 2 m (radius of cone base) 1 = 2/sin(28.44 deg) = 4.20 m
Ab = ? m2 A1 = ?m2 Thus, we get T1 = ? K Task #1: 0 = (5.67 * 10-8) W m2. K4 ## Stefan - ... 

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Energy Exchange between the Primary Shield and Payload Disk Calculation of B Solar flux Ap = B2 Ab
56.88 - A1 , TI Primary shield cone Ip = ? m Ap = ? m2 Ap = Base area of 3.013 cone end (d = 4 m) m Ap
= ? m2 Ap, To Payload B = ? 4.733 (perfect insulation) 28.950 # # 21 Drexel UNIVERSITY

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Energy Exchange between the Primary Shield and Payload Disk Now we have Solar flux B = ? Fop = ? =
F From the Energy balance at the payload, we had 56.88 -A1 , T1 Primary shield cone 4 Ap = Ap = (82) _
Ab Eq. * 1 AbFbp Fbp F Ab = Base area of 3.013 cone end (d = 4 m) Task #1: to find Tp. m A... 

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The Primary Shield and Payload with no Secondary Shield at 4 R. distance Task #1: Objective is to find
T, and Tp Solar flux Given: A1, Ab, R, Ro, and To (sun) = 5800 K 56.88 A1 , T1 Primary shield cone No
secondary shield Ab = Base area of T1 = temperature of the primary shield (cone) 3.013 co...