Спутник технологический. KS 1Q. [Редактировать]

KS 1Q это кубсат который был разработан коммерческой компанией из КНР с целью отработки технологий создания солнечных сенсоров, систем связи и т.п. Помимо технологической полезной нагрузки на нем установлена широкоугольная камера. Согласно заявлению компании производителя она планирует при помощи данного КА получить достаточно данных для создания более совершенных космических аппаратов.

Дополнительная классификация

#Наименования
1Страна оператор(владелец) - Китай
2Страна производитель - Китай
3Тип оператора(владельца) - государственный
4Тип орбиты - НОО

Информация об удачном запуске

#ХарактеристикаЗначение
1Космодром Цзюцюань
2Дата пуска2016-11-10
3Полезная нагрузка 1xLishui 1-01
4Полезная нагрузка 1xCAS-2T
5Полезная нагрузка 1xKS 1Q
6Полезная нагрузка 1xXiaoxiang 1
7Полезная нагрузка 1xXPNAV-1
8Ракета-носитель 1xВеликий поход 11

Найдено 198 документов по запросу «KS 1Q». [Перейти к поиску]


Дата загрузки: 2017-02-22
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0.07/5
...,4 94,7 491 511 Технологический КА [皮纳二号, KS-1Q]. 97,4 94,7 491 511 98...



Дата загрузки: 2017-04-23
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0.14/5
... innovation in firms en f ir (p (p l( ks ita ar ) b) )( P) P) DP GD GD... to the OECD median (Panel 1q). International co-patenting (Panel 1 r ) is... in firms en f ir ita ks (p P) P) b) (a ) )( DP GD GD rG er... innovation in firms en f ir (p (p l( ks ita ar ) b) )( P) P) DP GD GD... are well integrated internationally (Panel 1q, r ) and foreign affiliates account for... innovation in firms en f ir (p (p l( ks ita ar ) b) )( P) P) DP GD GD... innovation in firms en f ir (p (p l( ks ita ar ) b) )( P) P) DP GD GD... innovation in firms en f ir (p (p l( ks ita ar ) b) )( P) P) DP GD GD... innovation in firms en f ir (p (p l( ks ita ar ) b) )( P) P) DP GD GD... in firms en f ir ita ks (p P) P) DP GD GD (a ) )( P) DP rG...-authorship and co-invention (Panel 1q, r). The government seeks to improve... in firms en f ir ita ks (p ) b) )( P) P) DP GD GD rG er... in firms en f ir ita ks (p ) b) )( P) P) DP GD GD rG er... above the OECD median (Panel 1q, r). However, this also reflects the... below the OECD median (Panel 1q, r). The Interdepartmental Policy of International... innovation in firms en f ir (p (p l( ks ita ar ) b) )( P) P) DP GD GD... innovation in firms en f ir (p (p l( ks ita ar ) b) )( P) P) DP GD GD... knowledge and innovation networks (Panel 1q, r). However, industry-science linkages are... innovation in firms en f ir (p (p l( ks ita ar P) P) DP GD GD... and innovation is mixed (Panel 1q, r ). To exceed the EU average... innovation in firms en f ir (p (p l( ks ita ar ) b) )( P) P) DP GD GD... innovation in firms en f ir (p (p l( ks ita ar ) b) )( P) P) DP GD GD... innovation in firms en f ir (p (p l( ks ita ar ) b) )( P) P) DP GD GD... innovation in firms en f ir (p (p l( ks ita ar ) b) )( P) P) DP GD GD... in firms en f ir ita ks (p P) P) b) (a ) )( DP GD GD rG er... er pe (p l( m en f ir ita ks (p ) b) )( P) P) DP GD GD rG er... innovation in firms en f ir (p (p l( ks ita ar ) b) )( P) P) DP GD GD...’s indicator of co-authorship (Panel 1q). In 2013 leading Indonesian researchers... innovation in firms en f ir (p (p l( ks ita ar ) b) )( P) P) DP GD GD... applications involved international collaboration (Panel 1q, r), and funding from abroad accounted... innovation in firms en f ir (p (p l( ks ita ar ) b) )( P) P) DP GD GD... innovation in firms en f ir (p (p l( ks ita ar ) b) )( P) P) DP GD GD... innovation in firms en f ir (p (p l( ks ita ar ) b) )( P) P) DP GD GD... er pe s( eu m en f ir (p (p l( ks ita ar ) b) )( P) P) DP GD GD... international S&T co-operation networks (Panel 1q, r) and attracts few international R&D investments... innovation in firms en f ir (p (p l( ks ita ar ) b) )( P) P) DP GD GD... below the OECD median (Panel 1q, r). A traditionally strong focus on applied... innovation in firms en f ir (p (p l( ks ita ar ) b) )( P) P) DP GD GD...-authorship and co-invention (Panel 1q, r). Since 2007, it has addressed... innovation in firms en f ir (p (p l( ks ita ar ) b) )( P) P) DP GD GD... innovation in firms en f ir (p (p l( ks ita ar ) b) )( P) P) DP GD GD... of international co-authorship (Panel 1q) and international co-invention (Panel...



Дата загрузки: 2017-03-18
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0.17/5
... these features: – Jumbo frames – 802.1Q tunneling Release Notes for Cisco... per EtherChannel – Jumbo frames – 802.1Q tunneling – Traffic storm control WS... 15.1SY supports IEEE 802.1Q trunking. • NAC - L2 IP NAC...-mpls 6500: SXJ5 no 802.1Q VLAN TAG in vc type... session. CSCtu54300 ipsec-getvpn fn_VRFAwareGM: KS crashed while running getvpn unconfig... retransmitted CSCuc3176 1 ipsec-isakmp XE3.9 - KS crashes when removing GDOI groups...-mpls 6500: SXJ5 no 802.1Q VLAN TAG in vc type... session. CSCtu54300 ipsec-getvpn fn_VRFAwareGM: KS crashed while running getvpn unconfig... SA CSCuc31761 ipsec-isakmp XE3.9 - KS crashes when removing GDOI groups... failure CSCub4292 0 ipsec-getvpn GETVPN: KS fails to validate hash in... policy not downloaded to COOP-KS Sec CSCua5199 1 ipsec-ikev2 Inconsistency... 30 chars for variables CSCtb13421 — KS registration fails if one of... in HA pair CSCtc02012 — GETVPN: KS sends port 500 in his... instead of 848 CSCtc03011 — GETVPN KS Crash in unicast_rekey CSCtc04351 — RP... tunnel is configured CSCtc52655 — GetVPN KS/GM report sequence number failures... when doing "show crypto gdoi ks members" CSCtc78951 — C2W2C: port's not... traffic breaks crypto engine CSCte01303 — KS Policy Change not allowed on new Primary KS after a failover CSCte05199 — EEM syslog... (SPI leak) CSCtg08496 — After merge KS deletes all GMs, send rekey... GM after modifying ACL on KS CSCtg09619 — Web Auth host gets... multiple subnet support CSCtg55447 — Secondary KS TEK Seq number out of synch after primary KS failure CSCtg60424 — Fast-UDLD:Some... crypto gdoi" on KS does not clear the KS Policies CSCtg75452 — SDH... GM with psuedotime configured on KS CSCth61317 — Noc Payload Crc Error....1SY Identifier Technology Description CSCth85618 — KS Trace@%SYS-3-MGDTIMER@Process= "Crypto... — ASR1K: GM re-registers with KS when ACL is add/remove in KS. CSCti23872 — traceroute double hop with... with different type CSCtk84116 — GETVPN ks crash during split and merge...:having icmp/ip acl's in KS, ping is not working IN... manage the router CSCto89922 — GetVPN KS sends a Rekey ,even when the KS ACL is un-supported CSCto90252 — ... GM tries to register to KS and KS has issued rekey CSCtz41048...



Дата загрузки: 2017-01-30
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0.2/5
... круге U функций f , для которых ( ∫1 ∥f ∥Ap,q = Ωn ( ∫π Ωn (1 − r) 0 ) 1q ) pq |f (rw)|p dm(w) dr < +∞, 0 < p, q < +∞. −π В первом... βn ( ( − ) 2 ( = ) βn ( − )(1 (+ 1) ) = βn− p1 + p1′ + 1q + q1′ = βn− p1 + 1q − p1′ + q1′ + p1 + 1q n− p1 + 1q n ) ( ) ( ) ( 1 1 1 1 1 1 = βn+ p + q (n − 1)− p′ + q′ − 1 − p′ + 1 − q′ n ) ( ) ( ) ( 1 1 1 1 1 1 = βn − 2n+ p + q (n − 1)− p′ + q′ + p′ + q′ n = = = = 29...θ )||D(F )(ρeiθ )|dθdρ ≤ 1 1 p+q × ) 1 + q1′ p′ (n−1)+ (n−1) × 0 ( ∫π 1 q × ωΠ (1 − ρ) |g(ρe )| dθ 1 q ωΠ (1 − ρ) −π ( ( ∫1 × iθ ( (β−2)nq ′ + (1 − ρ) ( ∫π ωΠ (1 − ρ) ) q′ +1 p′ × ) q′′ ) 1′ p ′ |D(F )(ρe )| dθ p q dρ = −π (n−1) ( ∫π ωΠ (1 − ρ) ) pq |g(ρeiθ )|p dθ ) 1q dρ × −π 0 ( ∫1 ) 1q dρ |g(ρeiθ )|p dθ iθ q p +1 (1 − ρ) = ) pq (n−1) ( ∫π ωΠ (1 − ρ) ( ) ( ∫1 dρ ≤ −π q′ q 0 p p′ −π (n−1) 0 ) 1′ |D(F )(ρe )| dθ p ) q p +1 (1 − ρ) ≤ ) p1 ( ∫π iθ ( ∫1 × Π (1 1 q 0 × 1 q ( ′ (β−2)nq + (1 − ρ) ) q′ +1 p′ (n−1) 1−q ′ ωΠ (1 ( ∫π − ρ) ) q′′ −π 0 = ∥g∥Ap... ωΠ (1 − |z|) = ω(1 − |z|)(1 − |z|) нормой 1 (∫ ∥f ∥A1,q∗ = ω ωΠ (1 − r)(1 − r)q( p −1) 1 ( ωΠ (1 − r) (1 − r) = (∫ ωΠ (1 − r) Qn |f (rζ)|dmn (ζ) ) 1q rdr 1 p −1 )q ∫ |f (rζ)|dmn (ζ) ) 1q rdr Tn Qn (∫ )q Tn... (1.1) получим |D α+1 ωΠ∗ (1 − |z|) ∥ψ∥L∞ (U n ) . (1 − |z|)α g(z)| Тогда (1 − |z|)α− p − q +2 1 = sup ∥g∥λp,q ω 1 q 1 ωΠ (1 − |z|) z∈U n |Dα+1 g(z)| (1 − |z|)α− p − q +2 ωΠ∗ (1 − |z|) sup ∥ψ∥L∞ (U n ) = 1 α (1 − |z|) q z∈U n ω (1 − |z|) 1 1 Π α− p1 − 1q +2 = sup z∈U n (1 − |z|) 1 ωΠq (1 − |z|) 1 q ωΠ (1 − |z|)(1 − |z|) p + q −2 ∥ψ∥L∞ (U n ) = (1 − |z|)α 1 1 = ∥ψ∥L∞ (U n ) = ∥Φ∥. Итак, доказано, что ∥g∥λp,q ∥Φ∥. Это... ) p1 p ) 1′ 1 (1 − r)α− q +1 1 q ωΠ (1 − r) ) 1′ } g(rτ )| dmn (τ ) p × × × 72 (∫ × ) pq (∫ |f (rτ )|p dmn (τ ) ωΠ (1 − r) ) 1q rdr = ∥g∥λ˜˜p,q ∥f ∥Ap,q (ω) . ω Tn Qn Если... (τ ) τ ∈T n Qn (∫ q′ (1 − r) sup |Dα+1 g(rτ )|q ) 1′ ( ∫ ′ q (∫ |f (rτ )|p dmn (τ ) ωΠ (1 − r)× rdr q′ q ωΠ (1 − r) ) pq ) 1q Qn Qn = ∥g∥λ˜p,q ∥f ∥Ap,q (ω) . ω rdr Tn... z∈U n (∫ ωΠ (1 − r) ) pq |f (rτ )|p dmn (τ ) Tn Qn α+2− p1 − 1q (1 − |z|) 1 q ωΠ (1 − |z|) ) 1q rdr = ] |Dα+1 g(z)| ∥f ∥Ap,q (ω) = ∥g∥λp,q ∥f ∥Ap,q (ω) . ω Итак, |Φ(f )(z)| ограничен... ea,p (z) = p ∏ ( s=1 m zks ks ∂ m ∂zks ks (1 − aks zks )2 mks ) n ∏ ( s=p+1 ) mks z ∂ ks , mk s ∂zks (1 − aks... mj = αj + 1, j = 1, ..., n мы имеем ∥ea,p ∥Aω (α,m) ∫1 p n ∏ ωks (1 − |aks |) ∏ ωks (u) du. 2 2 (1 − |a |) u k s s=1 s=p+1 1−|aks | (2.32) 100 Найдем... ∫ 1 Th (ea,p )(z) = (2πi)n h(ζ) p ∏ Tn h(ζ) n ∏ (ζks − aks )2 Tn n ∏ ζks |dζ| s=p+1 s=1 p ∏ n ∏ (ζks − aks ) s=p+1 ∫ p ∏ ζk2s s=1 n ∏ (ζks − aks )2 s=1 1 = (2πi)n p ∏ = (1 − ζs z s ) s=1 ζks s=1 (ζks − aks ) s=p+1 n ∏ dζ, (1 − ζs z s ) s=1 ζ = (ζ1 , ..., ζn ), z = (z1... zkjss (1 − aks zks )js +1 s=1 ] 1 n ∏ = (1 − ζks aks ) s=p+1 1 ∂ p hp (a) ∏ = + ∂ζ1 ...∂ζp s=1 (1 − ζks aks ) n [ 1 ∑ + j1 , ..., jp = 0, |j| ̸= 0 ∂ p−|j| hp... 1 ∂ζk1−j ...∂ζ k 1 p p ∏ zkjss (1 − aks zks )js +1 s=1 ] 1 n ∏ (1 − ζks aks ) s=p+1 Следовательно, учитывая лемму 2.8, получаем... aks (1 − aks zks )mks +1 ( −1 + a z + 1 + a z ) 2zks ks ks ks ks × = . 1 − aks zks (1 − aks zks )mks... +1+js Следовательно, [ 1 ∂ p hp (a) ∏ 1 + D (Th (ea,p )(z)) = m! ∂ζ1 ...∂ζp s=1 (1 − ζks aks ) m n . 102 1 ∑ + p ( ∏ ∂ p−|j| hp (a) 1−jp j1...+1 (a) ∏ ≤ |Dm Th (ea,p+1 )(z)|+ m +1 k ∂ζk1 ...∂ζkp+1 s=1 |1 − ζks aks | s + 1 ∑ p+1 ( ∏ ∂ p+1−|j| hp+1 (a) 1−j j1 , ..., jp+1 = 0, |j| = j1... ≤ p + 1. Учитывая индукционное предположение получим ( p+1 ∏ ∂ hp+1 (a) ωks (1 − |aks |) ≤ 1−j 1 (1 − |aks |)2 ∫1 ∂ζk1−j ...∂ζkp p+1 s=1 1 p+1−|j| 1−|aks... ∂ζk1 ...∂ζkp+1 s=1 u2 p+1 1−|as | ∫1 p+1 n ∏ ωks (1 − |aks |) ∏ ωks (u) du+ 2 2 (1 − |a |) u k s s=1 s=p+2 1−|aks | ( 1 ∑ + ) p+1 ∫1 αks −1 ∏ ωks (1 − ρ)(1 − ρ) dρ ∂ hp+1 (a) × 1−j |1 − aks |js... Но, учитывая, что mks = αks + 1, s = 1, ..., n имеем ∫1 0 αks −1 ωks (1 − ρ)(1 − ρ) dρ ≈ j +1+m ks |1 − aks | s { ωks (1−|aks |) (1−|aks |)2 , если js = 1, ∫1 ωks (u) u2 du, если js = 1−|aks... ...∂ζkp+1 s=1 u2 1−|as | + ( p+1 ∏ s=1 ∫1 p+1 n ∏ ωks (1 − |aks |) ∏ ωks (u) du+ 2 2 (1 − |a |) u k s s=1 s=p+2 ωks (1 − |aks |) ∫1 ωks (u) (1 − |aks |)2 u2 du 1−|aks... p+1( )j ∏ ωk (1 − |ak |) s s s × 2 (1 − |a |) k s s=1 (2.39) 1−|aks | ( ∫1 × 1−|aks | )1−js ωks (u) du + u2 1 ∑ ∂ p+1−|j| hp+1 (a) 1−j j1 , ..., jp...−j ...∂ζkp+1p+1 1 × 105 ( )js ( ∫1 )1−js p+1 ∏ ωks (1 − |aks |) ωks (u) . × du 2 2 (1 − |a |) u k s s=1 1−|aks | Теперь, еще... 1−jp+1 , |j| = 1−j1 ∂ζk 1 ...∂ζkp [ p+1 ∏ ∂ p+1−|j| hp+1 (a) ωks (1 − |aks |) ∫1 ωks (u) 2 (1 − |aks |) u2 du 1−jp+1 1 ∂ζk1...+1 s=1 u2 1−|as | + [ p+1 ∏ s=1 p+1 ∏ ωks (1 − |aks |) n ∫1 ∏ s=1 (1 − |a |)2 ks s=p+2 1−|aks | ωks (1 − |aks |) ∫1 ωks (u) 2 (1 − |aks |) u2 du + ωks (u) u2 du ]1−js p+1( )1−js )j ( ∫1 ∏ ωk (1 − |ak |) s ω (u) ks... ∂ζk1 ...∂ζkp+1 s=1 u2 1−|as | ∫1 p+1 n ∏ ωks (1 − |aks |) ∏ ωks (u) du+ 2 2 (1 − |a u |) k s s=1 s=p+2 1−|aks | p+1 ∏ ωks (1 − |aks |) . + 2 (1 − |a |) k s s=1 В итоге, из..., что ∂ p+1 hp+1 (a) ∂ζk1 ...∂ζkp+1 p+1 ∏ ωks (1 − |aks |) . ∫1 ωks (u) s=1 (1 − |a |)2 ks u2 du 1−|aks | Требуемая оценка...



Дата загрузки: 2017-03-18
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0.19/5
... ycTpaHeHyr nenocmrxoB; _ - hopc€ryy r6v€H^1q€fi tuapnapra y)qeer Be!rri...}lle oryy ycnyft c 6oryruy; 6aArSlaie{1q 'd rir'l paqnoHanrHoro Has'{ex!rcq...) rap!.scqqhi\ o6pa3oBaHuq rpar,!a!. (w[v.med.ks) x B o6rracrr rpaHcnopra (w\inv.mrc.sov..."y xapaErapa!H I B o6xacrr r?aHcflopra (q"w.edu.go!.ks). B o6,racrg oxpaHH 3aopoBb, 6pounopanapLrBaH...



Дата загрузки: 2017-03-09
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0.39/5
... OHNE ZUBEHOR ZUGLANGE TORNILLO FB KS 3.9X25 GRIS PARA SOLERA PICAPORTE....CINCADO IZQ. CERR.3625-153-1Q CINCADO DCH. SKG2 CERR.3625-153-1Q CINCADO IZQ. SKG2 CERR.3625....KS 25.ABE.KS 25.ADC.KS 25.AEV.KS 25.AFJ.KS 25.CAB.KS 25.CAE.KS 25.CAF.KS 25.CAJ.KS....010.00 26.AAV.KS 26.ADC.KS 27.007.01 30.... F/16 CINCADO PROLONGADOR KFV V2500-1Q F/16 CINCADO PROLONGADOR KFV V2500Q... JGO.2 UNID.) PROLONGADOR KFV V2600-1Q F/16 CINCADO PROLONGADOR KFV V2600Q...



Дата загрузки: 2017-02-06
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0.13/5
.../guide/nm_script_tcl.html Transport 802.1q and 802.1p Tags over... 3745 • The primary Key Server (KS) now displays a registered Group Encrypted... server key distribution occurs, one KS declares itself as the primary KS, creates a policy, and sends out ... to the other secondary KS. The secondary KS continues to wait before declaring the primary KS as the primary KS and continues to... both the primary and secondary KS have a policy, the GM registration... server key distribution occurs, one KS declares itself as primary, creates... other secondary KS. The secondary KS declares the primary KS as primary KS when... the election mode. The secondary KS now also blocks GM registration...:00:16: %GDOI-5-COOP_KS_BLOCK_NEW_GM_REGISTER_ELECTION: This KS temporarily blocks GM with ip... in group diffint as the KS election is underway Features and.../ip6-bfd.html Preserve 802.1q with 802.1p Marking over... no workaround. • CSCtk84116 Symptoms: A GETVPN ks crash may occur when split... Co-operative key server (COOP KS) split/merge corner cases. 2. After... in corner cases of COOP KS split/merge scenarios. Workaround: There... triggered to register with the KS at the same time. 3. While... observed on the key server (KS) when the show crypto gdoi ks members command is executed. As a ... session goes down between the KS and the iBGP neighbor. Conditions... Bugs • CSCte01303 Symptoms: New Primary KS after failover does not allow KS policy changes. Conditions: This symptom is observed when a KS failover occurs first, then the... applied on the new primary KS. Workaround: Apply the policy change in the primary KS once it comes up, then force a KS role re-election by entering the clear crypto gdoi ks role in the new primary KS. Once the previously primary KS is restored as the primary KS, apply the... configured on a subinterface and 802.1q trunking is used to connect... 349 Bugs • CSCtk84116 Symptoms: A GETVPN ks crash may occur when split... in crypto ACL downloaded from KS. Its pattern of deny alternating... a crash. • CSCth16962 Symptoms: The primary KS KEK timer will get stuck... configured on a subinterface and 802.1q trunking is used to connect... purge interface. • CSCtk84116 Symptoms: A GETVPN KS crash may occur when split... configured on a subinterface and 802.1q trunking is used to connect... registration: %GDOI-5-GM_REGS_COMPL: Registration to KS complete for group using address... in crypto ACL downloaded from KS. Workaround: Simplify and consolidate the... connected to the Key Server (KS) 4. clear crypto sessions 5. attach the... default manner and allow the KS to send the SAs. • CSCsx65975... workaround. • CSCth16962 Symptoms: The primary KS KEK timer gets stuck or... no workaround. • CSCtk84116 Symptoms: A GETVPN ks crash may occur. Conditions: This...



Дата загрузки: 2016-12-04
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0.78/5
... inside the cylinder? Solution: Fs = ks∆x = ks ∆V/Ap ; π 0.12 = 0.007854 m2 4 T3... spring 1.5 Ap = (π/4) × 0.152 = 0.01767 m2 ; ks = 100 kN/m (matches P in kPa... = 163 W 4.3 A linear spring, F = ks(x − xo), with spring constant ks = 500 N/m, is stretched... input. Solution: F = ks(x - x0) = 500 × 0.1 = 50 N 2 W = ∫ F dx = ⌠ ⌡ ks(x - x0)d(x - x0) = ks(x - x0) /2 = 500.../s 4.64E A linear spring, F = ks(x − xo), with spring constant ks = 35 lbf/ft...) = 35 × 2.5/12 = 7.292 lbf W=⌠ ⌡Fdx = ⌠ ⌡ks(x - x0)d(x - x0) = = 1 k (x - x0)2 2 s 1 × 35 × (2.5/12... 1W2ÊÊelec = -power × ∆t = -Amp × volts × ∆t = 1Q 2 10 x 6 x 10 x 60 = -36 kJ... = AREA = 2 (300 + 400)(1.0 - 0.5) = 175 kJ 1Q 2 5.17 = m(u2-u1) + 1W2 = 1(3292... State 2: P2 , v2 = v1 T ≅ 110°C, 1Q 2 1 => superheated vapor u2 = h2 - P2v2... x 5000 x(0.0009995 - 0.0633) = -3115 kJ 1Q 2 = 10 x(88.65 - 3316.2) = -32276... – V1) = 400 (0.4625 – 0.1) = 145 kJ 1Q 2 = m(u2 - u1) + 1W2 = 1 (2553.6 – 87... x 0.02937 = = 12.1 kJ 1W2 = ∫ P dV = 1 - (-1/3) 1-n 1Q 2 = m(u2- u1) + 1W2 = 2(165.8 - 98... (0.4088 - 0.27) 2 1W2 = 36.9kJ 1Q 2 = m(u2 - u1) + 1W2 = 0.5(2749.7 - 1931...) = 400 x 5(0.46246 - 0.3354) = 254.1 kJ 1Q 3 = 5 (2553.6 - 836.4) + 254.1 = 8840 kJ... =  x 0.5 x (0.9924 - 0.89186) = 17.56 kJ 2   1Q 2 = m(u2 - u1) + 1W2 = 0.5 x (3668 - 2529... + 1600 =  (0.0102 - 0.008) = 2.76 kJ 2   1Q 2 = 0.502 x (285.5 - 166.87) + 2.76... x 0.00315 - 2000 x 0.005) = -2.106 kJ 4 1Q 2 = m(u2 - u1) + 1W2 = 0.005 (1203...×0.05416 - 201.7×0.09921) = -7.07 kJ 1Q 2 = m(u2 -u1) + 1W2 = 0.5(440.9 - 372...) = 500 × 0.1 (0.5226 - 0.4243) = 4.91 kJ 1Q 2 = m(u2 - u1) + 1W2 = 0.1(2802.9 - 939... 323.88 0.5 θ + θ ] = -1186.3 1.75 0.5 7.732 1Q 2 = m(u2 - u1) = 0.374(-1186.3) = -444...) = (200 - 100) 1-n 1-n 1 - 1.2 = -40.196 kJ 1Q 2 = m(u2 - u1) + 1W2 ≅ mCv(T2... – 150 x 0.001) = - 0.192 kJ n-1 n-1 1W2 = 1Q 2 = m(u2 – u1) + 1W2 = P1V1 RT1... and P2 << PC => State 2: 2 1W2 = 1Q 2 ∫ 1 P dV = P2 V 2 - P 1 V 1 1- n = Also Ideal Gas... - P1V1 mR(T2 - T1) ⌠ PdV = W = ⌡ = 1 2 1-n 1-n 1 = 1Q 2 1× 0.18855 × (700 − 40) = 41.48....8 = 694.5, uB1 = 2416.6 kJ/kg 1Q 2 5.86 = 9.121x10-3(694.5 - 2416.6) = -15... ⌡ PdV = m Process: PVn = const → 1W2 = ⌠ 1-n 1 1Q 2 = 52.5 = m(u2-u1) + m P2v2-P1v1... 78-100x0.001767)/(4/3) = 6.11 kJ 1 2 1-(-1/3) 1 1Q 2 = m2u2 - mA1uA1 - mB1uB1 + 1W2 = 0.851x182... - 0.1233) 2 2 = 12.6 + 17.6 = 30.2 kJ 1Q 3 = m(u3 - u1) + 1W3 = 3.981 (406... gives 1 1 ⌠ 2 1W2 = ⌡PdV = 2(P 1 + P2)(V2 - V 1) v 1Q 2 = m(u2 - u1) + 1W2 Equation of... - T1) = × 1-n 1-n 778 1 − 1.2 = - 56.12 Btu 1Q 2 = m(u2 - u1) + 1W2 ≅ mCv(T2....2 = 449.9 Btu/lbm; uB1 = 1040.2 1Q 2 = 0.02138 (449.9 - 1040.2) = -12.6 Btu... x 0.0654)/(0.287 x 300) = 0.152 kg 1Q 2 = 0.66x250.32 - 0.152x214.364 - 0.508x401... = mihi + 1Q2 - 1W2 P Process: P ~ D ~ V1/3 2 1Q 2 1 so P = P1(V/V1)1/3 v V1 = mRT1... the work becomes 1W3 = ∫ P dV = 1Q 3 P 3V 3 − P 1V 1 1−n = 703x1.4368 − 1200x4... USUF. Process (linear spring): P2 - P 1 = kS kS A A2 2 (V 2 - V 1) = (m2v2 - m1v1) 1   v2 - 1 x 0.029588... = 6.494 ; v2 = 1.5658 ; h2 = 1581.2 1Q 2 = 1(273 + 50)(6.494 - 5.265) = 396... = 1 × 1000(0.1955 - 0.145) = 50.5 kJ 1Q 2 = m(u2 - u1) + 1W2 = 1 × (1566.7 - 1391....3 - 1000×0.145 = 1391.3, s2 = 5.2654 P T 1 1 2 2 T v s 1Q 2 = mT(s2 − s1) = 2 × 323.15... W = 0, thus provides heat transfer as 1Q 2 2 1 s = m(u2 - u1) = 3213.7 kJ Entropy... Process: T = constant, reversible 1S 2 gen = 0 => P T 2 2 1 1 T v 1Q 2 ⌠Tds = mT(s2 - s1) = 0.1349.../v1 = 0.001/0.07029 = 0.0142 kg P 1 2 a 1Q 2 WHP H.P. V T v=C QL Tamb 1 2 a s State a: v = 11...) = 400 (9.316 - 1) × 0.001 = 3.33 kJ 1Q 2 = m(u2 − u1) + 1W2 = 0.0142 (2804... = 2.1607 m3, P u2 = 2654.4 T 1 2 1 2 T v 8.25 s 1Q 2 = mT(s2 − s1) = 2 × 473.15...? C.V. Water. T 1 o 130 C Energy Eq.: P1 1Q 2 u = 2540 7.0259 s = 0 = m(u2 - u1) + 1W2... 1Q2 = H2 - H1 = m(h2 - h1) P 1 T 2 v 2 1 v 1Q 2 = 2(3900.1 - 422.72) = 6954.76... + 150) 1 (0.35411 - 0.001002) = 203 kJ 1Q 2 = 1(3124.3 - 83.94) +203 = 3243... m(s2 - s1) = 1(7.7621 - 0.2968) = 7.4655 1Q 2 / Tsource = 3.7146 kJ/K (for source... + 2.339) × 3 × (0.6382 - 0.8041) = -125 kJ 2 1Q 2 = m(u2 - u1) + 1W2 = 3(109.46... kJ/kg-K P T 1 3 1 3 2 2 v 1st Law: 1‡2, 1W2 1Q 2 = m(u2 − u1) + 1W2 ; s 1Q2 =0 = m(u1... m3 m = ρV = 2300 × 12 = 27600 kg 1Q 2 = mC∆T = 27600 × 0.65(-5) = -89700 kJ... so ideal gas => u 2 = u1 ) T P 1 1 2 1 P2 2 v 1Q 2 s = 1W2 = ⌠ ⌡PdV = P1V1 ln (V2....3, P2 = 2.7 × P1 = 113.57 P T 2 v=C 2 1 P1 1 v 1Q 2 s = 1 × 0.3122 (818.3 - 303.15) = 160...) = 750 kPa State: 1 = initial, WHE A 1Q 3 1 -> 3 1Q 2 H.P. B 2 = final hot 1 -> 2 3 = final cold To... and mass State: 1 = initial, WHE 1Q 3 A 1 -> 3 1Q 2 H.P. B 2 = final hot 1 -> 2 3 = final cold P2...)/(1-n) =⌡ = [604.8 × 0.16914 - 1000 × 0.2]/[1-(-3)] = -24.4 kJ 1Q 2 = 1.847 × 0.6529(20 - 300) - 24...(2088.3 - 2410.2) = -1.90 kJ (2W3 = 0) 1Q 3 = 1Q2 + 2Q3 = m(u3 - u2) = 0.0059... R Process: T = constant, reversible 1S 2 gen = 0 => P T 2 2 1 1 T v 1Q 2 =⌠ ⌡Tds = mT(s2 - s1) = 0.3859... × 7) 144 ⌠ W = ⌡ PdV = = × = -24.3 Btu 1 2 778 1-n 1+3 1Q 2 = 4.064 × 0.158 × (530 - 1060) - 24... + minsin m2s2 - m1s1 = ⌡ Process: Adiabatic 1Q 2 = 0 , Process ideal 1S 2 gen = 0 , s1...



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... Spectre fan ) Bitfenix BFA-RCN-KS-RP Recon Black system status... , support WoL , 802.3x , 802.1Q, 802.1P Intel PWLA8391GT , Retail... , support WoL , 802.3x , 802.1Q, 802.1P E OE , price VAT... trunking + mirroring , port-based 802.1Q VLAN , IGMP snooping , SNMP management... trunking + mirroring , port-based 802.1Q VLAN , IGMP snooping , SNMP management... MAC address , port trunking , 802.1Q VLAN , 802.1p priority tagging... address , port trunking + mirroring , 802.1Q VLAN , 802.1p priority tagging... address , port trunking + mirroring , 802.1Q VLAN 802.1p priorityWeb tagging... address , port trunking + mirroring , 802.1Q VLAN , 802.1p priority tagging... trunking + mirroring , port-based 802.1Q VLAN , IGMP snooping , SNMPor management... trunking + mirroring , port-based 802.1Q VLAN , IGMPL2 snooping D-Link DGS... trunking + mirroring , port-based 802.1Q VLAN , IGMPL2 snooping , rackmount ready... trunking + mirroring , port-based 802.1Q , IGMP snooping , SNMP , rackmount D-Link... trunking + mirroring , port-based 802.1Q , IGMP snooping , SNMP , rackmount D-Link... SDRAM , 128k packet buffer , 802.1Q VLAN , 802.1p priority tagging... SDRAM , 768k packet buffer , 802.1Q VLAN , 802.1p priority tagging... flash , 2mb packet buffer , 802.1Q DGS-3120-24Pc VLAN , 802... flash , 2mb packet buffer , 802.1Q VLAN , 802.1p priority tagging... flash , 2mb packet buffer , 802.1Q VLAN , 802.1p priority tagging... flash , 2mb packet buffer , 802.1Q VLAN , 802.1p priority tagging... flash , 750k packet buffer , 802.1Q VLAN , 802.1p priority tagging... flash , 750k packet buffer , 802.1Q VLAN , 802.1p priority tagging... kvm switch cables KS-CBP5 KS-CBU2 KS-CBU5 KS-CBUP2 KS-CBUP3 KS-CBUP5 R 212... optional Sunix KS-SKMC102PV KS-SKMS124U D-Link KS-DKVM121 KS-DKVM222 KS-DKVM4 KS-DKVM4U KS-DKVM440 R 410 R 589 R 604 R 852 R 3,517 KS...



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... езы 8 Сп е ц и ф и к а ц и я 9 О б о р уд о в а н и е А ккум уляторного 1Q Обо руд о в ан и е бака 41... Сп ециф икация 9 Об о р у д о в а н и е Аккумуляторного 1Q Об о р уд о в а н и е 14 Ос т а н о в к а 12... КРАСКОЙ ПОВЕРХНОСТЬ грунтом БАКОВ Xtc - 0 1Q в ТОЛЩИНОЙ ДЛЯ 130 микрон ТО... 19903- 74 ДнИШЕ £?1В 25G-1Q ф25 7 2SG-1,0 3 10,65 5 ГОСТ... 26 -О—1—0 33 i J1 ~ - 2г о в КвМАНДА V ■+kS° 1ШТП65 63 53. 002. Т у с т = 5А... ЕТ EAkCkUU 1ЯЗСП11Ш91$П$Я Л И Г гЗ 3KV KS-3 d 33 S T з4 зкк ! k l 63 Ццу... Р 2 А Е v^npРчБДЕНИЯ •О— 24 Ве н т и л е м на п о д п итке Ь KS-11 С Х Е М уЭ М А И С тЗ 64 Ь СХЕМ... 1КЗЗ- 2-1-3 1 3300 кч 1 К 33-2-1-4 1 3300 KS ТК 3 9- 2- 2.1 1 4000 Кб П 3 3 .2 - 2 .2 1БСТ6-2А... 39 3. 2ЧО -3 ПС20 3.000 3 ^ ПС 1Q ПИ ПС 9 ПСЙ ПС 1В... ЭПЗ-400ИЗ Стадия Лист Аистов 1Q р Са р м и н а С ХЕМ А РАСПОЛОЖЕНИЯ ОПОр ПОД...