Design and Stress Analysis of Extraterrestrial ... - The Black Vault

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In the cross section of pipes and radiator elements a certain temperature gradient is established which leads to the appearance of thermal stresses. The following is an approach to solving this problem. We shall fix the element rigidly in the axial direction. Then elastic and thermal strains will equal each other, i.e., AL =A. Let us find the stresses in the element in this position. The application of bonds leads to a disruption of the boundary conditions of a free system. We restore the boundary conditions by applying to the system a force equal to the reaction arising in the attachment, but opposite in direction. Let us find the stresses from this force. They will only be elastic. Y 3' Fig. 4.12. Symmetrical temperature gradient. It is completely obviously that the sum of the stresses of the first and second calcolations gives us the real thermal stresses in the element. Examples of gradients can be very different. Let us examine the first case when the temperature gradient is symmetric (Fig. 4.12) relative to the axis. In the presence of such a temperature gradient throughout the cross section in a free, unattached elemen't, thermal stresses arise. We shall evaluate these stresses. 1450
+ LI£L 6 Fig. 4.13. Determining the stresses with a symmetric temperature gradient'. Let us assume that this cross-sectional temperature gradient will be preserved for the entire length of the element (Fig. 4.13). The following are known: E, a, At; find stress a. If we represent such an element in the free state, its elongation as a result of healing will be proportional to the temperature gradient; c -- aAt, where At = t - t 0 is the temperature variation with heating. (1 = -EaAt arise. If attachment is effected, compressive stresses The diagram of compressive stresses will accurately trace the temperature gradient and will be identical in any section of the fixed element. Let us find the stresses arising in the element if we free it from attachment. For this we apply tensile stresses to the ends of the rod, which ensure zero stresses on the ends. 451
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FTD:-MT-24-1462-71 FOREIGN TECHNOLO
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tUNCLASSIFIED - Security Ct-9-1uifl
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t IABLE OF CONTENTS U. S. Board on
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5.2. Ion motors ...................
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II jah _X JOLIGAETECbtSODN USA N NL
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,.is the calculation Of thermal str
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S1 In stress analysis the calculati
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"I 3 f !7 2 II 3 I 44; a ' 3 I 4 2
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)I requires special protection of t
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electrical power of the power insta
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Extraterrestrial motors in power fr
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S1' , . that a part operatesbefore!
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Fig. 1.3. Diagram of an ERE with me
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Table 1.3 shows the working media,
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I I The schematic of a nuclear ther
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Fig. 1.5. converter. Diagram of an
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The advantage of this arrangement i
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A complex system or any part of it,
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I I Stages and content of operation
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'the use of earlier systems and uni
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Structural diagram of operations Th
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. . . . . __.. . . . . .. .. . . .
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-- indication of state, transmissio
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Stages ERE, I. Analysis of the tech
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Space vehicle (SV) jParts jettisone
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Spc veil (SV)!- 0 4. ho 0. 4)Z - H4
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~tto 4- jo 0o (utd~nb 'Sa vtho- 4)i
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'Li However, 'this list can be chan
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- S developed. Or course, there are
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Sthat From the design and operation
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As the number or tests builds up-an
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II P 'Po. = -ý SFig. 1.20. Curves
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Stress is the intensity of the inte
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strength criterion only when even s
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JV Strength- reserve is the ratio o
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'2 1 I III I, . - I - S We know tha
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i J However, when the operating tim
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Plastic deformations in the latter
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Let us examine the basic design rel
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I I The constants B and n are deter
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Along with equation (1.18) we use e
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Iftesting continues, as shown in Fi
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I I Based on the known deformation
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6%Fx (a) Fig. 1,31., Uniaxial (a) a
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The element is in plastic state (1
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Total operating time of the sample
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I:! Substituting expression (1.35)
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Function B(T) is found after determ
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first We shall express stresses in
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etween it and the stressed state of
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a) b) c) Fig. 1.36. Determining the
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other conditions being equal, the c
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The fuel elements also rest on a li
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As the reflector and moderator in t
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Let us examine the examples of weld
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Between these shells passes a fluid
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The procedure for welding a two-lay
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A fuel element is a closed pressuri
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1 2A S(a) II • • /Fig. 2.8. Con
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The reflector material 5 (beryllium
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Figure 2.12 is a sketch of a fast n
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I. ! ~ ~~. . ........... li .... I
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I jZ ICIS qo 'Im klol
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THERMAL STRESSES IN FUEL ELEMENTS T
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Ora-O"when r=a; when r=b. (2.8) Sub
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I The exponential logarithmic depen
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I I of the cylinder between two adj
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EuI~, 21,2 Eat, • bb 2-(1 ---xInb
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-a - Fig. 2.18. Thermal stresses in
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I a It is necessary that n > 1l. If
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The locus of points equidistant fro
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Equations of equilibrium for an axi
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Fig. 2.23. Equilibrium of a shell e
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:I I I I The first equilibrium equa
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The wall stress of this sphere is d
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I These formulas enable us to plot
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. . . S_.. I i. stresses. Maximum i
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Shells in the reactors which we sha
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axial stresses. Simplicity fully Ju
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abl-ab (R+AR)d-•Rdf AR ab Rd? R (
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V 1! II Point 3 corresponds to the
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SI Now we seek the pressure of the
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!I As is apparent from Fig. 2.39, t
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I I I I '' 9 capacity of a shell in
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SI a i f fit~ Z.rl -= •.r nl--~ ~
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We plot the unknown curve p = f(AR)
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-.. -, 7 ''°-^ ...... I}" L7,50s.
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I" Table 2.2. (Cont d) (1) V(. )3 (
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W/ dr (2.48) "The sign is negative
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Ve designate th6 relaitie -momenits
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a a I a A4l "• dr; P= pr dr idy;
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'(2.61) or in shortened form, V 2 V
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4i In this case, the constant C2 in
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will be If the external force is co
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I W -P [(a2 -I a +(•1 -r2)In a ("
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+) If we find that Wmax • h, the
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I I W c I , I- a , I. ' We cut the
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- (0,,i66.0,239. 10-6-0,484.0,33. 1
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-n I I I I Pig, 2.56. Determining t
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In some cases, it is assumed that t
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6e, I Smat a a SI r.+0P6h2a .. 'fro
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In designing a source it is necessa
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removal, for example, in a containe
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- 2d I I I ______ -. - I 2o ( 'I Z
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Ir dp;- N,= N, .(-N,,lr d$ + d (3,I
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It is perfectly obvious that the el
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dr dr, dMr, dr2 d, d2or 1 3 dr dr "
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I ! i I I The quality of the device
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The mirror surface, in this case, i
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Fig. 2.69. Thermal trap with contro
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i ffased on this factor, fuel eleme
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In extremely small pores (pore c) t
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why fuel elements with a solid elec
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Peazeemnl (l) Fig. 2.73. A regenera
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I. II As a result' of the- process,
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'4I '21
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U k IIl + (n2lpiim)() Fig. 2.78. Co
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I I I I Figure 1.3 shows one of the
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• C a is the axial velocity of th
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Figure 3.1 shows working blades of
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Fig. 3.11. Union of disk with shaft
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I *1 warping, the material of the h
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(a) (b) Fig. 3.6. Dry-fri'ction bea
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ing groove, and through the incline
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I '' S Albng with the variation in
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Friction and wear in such a bearing
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The transition from the oversaturat
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O a) CO- 4) U'% 0 4Z Oo .~ 0 t'24a)
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I _____ ', "I I The Na-K pump consi
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E- W2l.i rx.
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w is the angular velocity of disk r
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; I ' | * * I I j I l7 I II I' *
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and, finally, dPa 2.ir( 2r a dr z z
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Flexural stresses at any point' of
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Another means of unloading consists
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0 I I I In cae wetakii CO/7eHU~e *1
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Determining the frequencies of inhe
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Differentiating this equation, we f
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Equation '(3.?5) is a differentiale
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Functions S, T, U and V during diff
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1 JVQP (3.30') Fig. 3.24. Forms of
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Then the bendinv moment is t ! ! EI
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'!..... " - ;-~ .- .-.. ~ - - J I T
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Figure 3.25 shows the operation of
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2 I ' whe, For plottfng the donvert
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satisfies the bounaary conditions o
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- o0 CC 0 CL P% c - t- -. C4 t C Ci
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The angular velocity w Aof natural
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I. Each of the harmonics of the ang
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At each of the points of interse-ti
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The actual frequency of natural ben
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1 I i , II ' t S S "•< i, I S, 3.
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(In view of the smallness of angle
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dr dr r d• _._._.• .d~t 1 •t-
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7WI - For conical disks equation (3
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Hyperbolic disks. If the, thickness
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o•/. or Oer 17,0 5,0 16,0-2 4,5 5
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- 0CX=, 9 - 280- 270- 110 0,15 __ -
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11 0,7 2,- 7,5 - '2,4- a'l 0,5 02 0
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55- . \, 105 50-100 45 ' 95 S85 "J
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Let us examine two cases: disk) and
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4. 100 0,6- WoO, ~: 1 0,94 493,8 3,
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0,7 3,3 -3,3 0,6 -3,2 0,15 3,0 2,9
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f~rc 200 -PC.; 190 40 170 30 I/t 0,
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a) arB 0 if the disk does not bear
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I I I I II I , Substituting this so
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Z= lz 1 x _-, 9 S1- 0,TF,1 o~ 5o o
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ii 0,1 42 0,3 ,4 0,50q6 0,7 08 0,9
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The expression for a* is obtained f
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, The'actual stresses on the bounda
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ý4.1 j L. - - c* m oc ~ .7 lira___
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C,4 C4C4 I IC4) CI tz +3Il + + CC.)
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C.. C4 0 C4 C I -T I -J . C :.d C?3
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Widely used methods are those based
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Solving the last equation relative
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Substituting the difference (awl -
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0 g h•con- - o7 g , 0,7 .Op • o
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iJ relaxation) occurs and with stea
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In the examired theory during plast
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Let, us p w I I ! • Let, us proce
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Rupture sets in only when plastic f
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£@ 1 hdr .1 (o) hr r(3.102) Usuall
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I Since the shaft is elastic, then,
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I I SI hence r M! rad/s ,(3.104) de
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y ii y0 \ 1 0-1 H t ieemNw idn 5ai7
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Critical angular velocities- of a s
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We assume that the disk and shaft a
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Expanding the determinant of (3.113
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y=aP-P+a, 2 M; a 21aiP+a22M# (3.119
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(1) ((2) O~7~C.• o~P'7t7- A O6/UC
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Wihen X = w, forward synchronous pr
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we find For forward synchronous pre
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•+ , , ; Im• S, -I i I I- I' su
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When analyzing the natural frequenc
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Vz 7M 7 ' z (KR--K-) z z (R (3.137)
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The concept of forced vibrations. t
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Resonance modes can be shown on the
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' C Yon no'. ern12 •. -p. 4 a..(a
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I ~ I 'Existing methods of damping
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Vibrations in rotors on hydrostatic
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SQTi •-fT 1 (O,0625VI)Y+1-0, 1-(p
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During the forward speed of the piv
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~t The coefficients of hydraulic fr
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of forces in the absence of pivot v
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To solve the problem of the vibrati
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I I k i = Ib the projections of dis
Page 414 and 415: 3 The general solution to (3.160) c
Page 416 and 417: (3.166) Thus, near the boundary of
Page 418 and 419: 2 1z or 2.. .• .. :. • C .2,e.
Page 420 and 421: I 3.2. THERMOEMISSION ENERGY CONVER
Page 422 and 423: STRUCTURAL DIAGRAMS AND DESIGN OF C
Page 424 and 425: In the creation of the design serio
Page 426 and 427: I £ The nortcoming in this design
Page 428 and 429: 'I, Fig. 3.92. Overall view of reac
Page 430 and 431: Figure 3.93 is a diagram of an elec
Page 432 and 433: We shall set up one of 'he shell se
Page 434 and 435: E3 e 2 , --e 1 , o A= B 3 , 0 --E 1
Page 436 and 437: worst efficiency. In the table the
Page 438 and 439: 7,7 417 1 2 Fig. 3.96. Diagram of a
Page 440 and 441: A thermoelectric converter consists
Page 442 and 443: I 1A 7ý,, -- Fig. 3.101. Controlla
Page 444 and 445: Photoelectric Converters In a photo
Page 446 and 447: of 140 W with allowance for all los
Page 448 and 449: Radiator coolers are divided into s
Page 450 and 451: 2 A Fig. 4.2. A conical flexible un
Page 452 and 453: Folding conical, radiators. As seen
Page 454 and 455: * ,. I- ] 2 : I KEY: ()aA (V)w °+
Page 456 and 457: point does not exceed the temperatu
Page 458 and 459: The subscript "cep" e ~'~;+ f~dx. 0
Page 460 and 461: -- - ' P l -- u -A - a .' • U1 1.
Page 462 and 463: AL Fig. 11.11. Determining thermal_
Page 466 and 467: o> . .. Fig. 4.14. Stress diagrams
Page 468 and 469: 1 I I I Thermal stresses in a a, fl
Page 470 and 471: We shall use Hooke's law for a two-
Page 472 and 473: i.e., the plate bends along a spher
Page 474 and 475: then "- D(0 + ) B& 3 ( +,u) aAt EAh
Page 476 and 477: Then instead of At(z) we shall writ
Page 478 and 479: Heat exchangers in which the heat t
Page 480 and 481: which is done by introducing into t
Page 482 and 483: -.............. . a U-shaped jacket
Page 484 and 485: The mechanical strength of heat exc
Page 486 and 487: All the welding seams on the pipes
Page 488 and 489: I I! .jn the design of a heat excha
Page 490 and 491: .The following types of corrosion a
Page 492 and 493: To •clean lithium q the oxygen an
Page 494 and 495: t I * I where p is the permissible
Page 496 and 497: Additional tensile (compressive) st
Page 498 and 499: Calculation formulas are valid unde
Page 500 and 501: The compliance parameter is determi
Page 502 and 503: -KI 1- SI ' 2I -log VJ L0 Fig. 4.38
Page 504 and 505: Case 3. The shell is loaded simulta
Page 506 and 507: The value of n for thin cylindrical
Page 508 and 509: In the general case, critical time
Page 510 and 511: when ).
Page 512 and 513: Axial displacement of compensator:
Page 514 and 515:
CHAPTER V MOTORS 5.1. PLASMA MOTORS
Page 516 and 517:
Pulsed pinch plasma motor This inot
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In the plasma generator unit the ev
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If an external magnetic field is cr
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forces Fig. 5.5. acting Simplified
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Z I . . .. . . Pulsed plasma face-t
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The grain of the working medium is
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J 4J 0 0 514 ~
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I I Forces T e, while acting normal
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where D - 12 "). Let us return to e
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It is significant also that when n
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where B is a coefficient; n is the
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Bend w is an elastic bend of the sh
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Thus, a linear law of temperature g
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;' -- e--- 1= - (20MO cosVxJr- Q 0
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Table 5. 1. 3x __ _ _ V I '0 I ! 0
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Hence generalized stress is SV2= f
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Let us 3train a free shell loaded w
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tt Solution is sought from formula
Page 552 and 553:
The value of wp is found from formu
Page 554 and 555:
We determine 1.29 1.29 D' Eh3 1,45
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Table 5.2. CT- p ,INI'•I.R 1. po-
Page 558 and 559:
. sin ir P ( 2P2C 4 ) + eP cos ,x (
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,_ . l90 _. h-ID NI -rcz Lit 'w. a)
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Computation of w'" is made in ,line
Page 564 and 565:
Table I 5.3. (2(2) (1 l10 _ _ _ _ _
Page 566 and 567:
We integr'ate this equation once mo
Page 568 and 569:
Table 5.4. _ _ _ () Kit)I ) (3) ( U
Page 570 and 571:
These formulas show that on the fre
Page 572 and 573:
pR _prtgii h (5.39) where R is the
Page 574 and 575:
K i... Linear shearing forces, obta
Page 576 and 577:
1.1: deformation correspondz to-the
Page 578 and 579:
The assumed law of At enables us to
Page 580 and 581:
'I Thus, the solution to the nonhom
Page 582 and 583:
The diagram of the shell is shown i
Page 584 and 585:
Tne z?.ell dimension~, in our probl
Page 586 and 587:
Stresses in a circular diret.ion or
Page 588 and 589:
w lW 2- W- =--, (); r- '=r g "I ; r
Page 590 and 591:
I, * I I Suzt-ttuting conditions (5
Page 592 and 593:
-7 The obtained equation is a homog
Page 594 and 595:
PRINCIPAL AND OF MOTORS STRUCTURAL
Page 596 and 597:
The ionizer unit consists of the io
Page 598 and 599:
-ii 4 I medium 6 and heater 7. The
Page 600 and 601:
!8 41 C Fig. 5. 37. The positive io
Page 602 and 603:
Soldering the porous plate with the
Page 604 and 605:
I I I . 1 2 Fig. 5.39. Design of io
Page 606 and 607:
i,, 1B) 6".R1 Fig. 5.41. Plate elec
Page 608 and 609:
electrons from the cathode and then
Page 610 and 611:
Stress analysis of motor elements A
Page 612 and 613:
A- Fig. 5.47. Motor unit design. 4z
Page 614 and 615:
Let us project all forces acting on
Page 616 and 617:
I- (' l1dx dy '-7dx =0; Ox=O Rý tg
Page 618 and 619:
I The quaotity Scx consists of thre
Page 620 and 621:
The angle of shift is the variation
Page 622 and 623:
V Iv Yvox a, R&tgo Ox Oy z 2Ow+ 2 0
Page 624 and 625:
7 3- h 2 h 2 iY dz; 1Y=- -h'2 -h 2
Page 626 and 627:
F 2Ea / 2r 3 + a3 , A G _ L(aa3) 0
Page 628 and 629:
I dimension r, dr. We apply to the
Page 630 and 631:
We drop • from equations (5.86) -
Page 632 and 633:
1-IL'L + z A. 2 o,-1_€ ('~-1,2-lm
Page 634 and 635:
Function F(r) can be found from the
Page 636 and 637:
We integrate this equation twice. T
Page 638 and 639:
Here A 1 = (2)/(3,)(z + 1)(z + 3);
Page 640 and 641:
- Absolutely flexible membrane If a
Page 642 and 643:
2500- 2000 1500 -/ 1000 Y - 500 4j-
Page 644 and 645:
Movement of the bellows also depend
Page 646 and 647:
Is The coefficient kla! calculated
Page 648 and 649:
Fig. 5.65. Stress analysis of a bel
Page 650 and 651:
Table A.l. Mechanical properties of
Page 652 and 653:
.(ii (2) A, 6r,/M,.ajpa £'40- 6 dH
Page 654 and 655:
A shortcoming of graphite is the no
Page 656 and 657:
0 (1) (2) 40 6-16- i!• 2 [K1/MM 2
Page 658 and 659:
E ( dafi/MM 1 2 U) 2_ou . Iv 5000 1
Page 660 and 661:
I I I3 Table A.3. (Continued)' (1a)
Page 662 and 663:
Table A.11. Characteristics of niob
Page 664 and 665:
Table A.5. Mechanical properties of
Page 666 and 667:
Tungsten and its alloys Figure A.16
Page 668 and 669:
, - ,, I Table A.8. Modulus of elas
Page 670 and 671:
Bibliography 1. A it;ý' Ip e e Bn.
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Design and Stress Analysis of Extraterrestrial ... - The Black Vault

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