Спутник технологический. KJ 1B. [Редактировать]

KJ 1 (Kongjian 1) это малый космический аппарат который был создан для целей орбитальной отработки технологий . Более подробные детали об аппарате не раскрывались.

Дополнительная классификация

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

Найдено 1000 документов по запросу «KJ 1B». [Перейти к поиску]


Дата загрузки: 2016-12-24
Скачать документ
Скачать текст
0.05/5
...&-&, 8*$#, &O*6 $#0%(K, O$0T/0, '(0-&, 1#, 90%2$0, )&1(%* %.*$*$.JP*0,('&+(%'#, (&'$08 016 1&+, 7)&1&8 *)*n G?,' &@$#(%#1*0, $&4*, *1B&$8#3**:,*1B&$8#3*&112Q, *@/0$T0):, )&-1*6 %*'12Q, >O&@1#'#%04512Q?,&-$#1*901...(% '#A,g/1&'$08 011&, *1(%*% .3*&1#4 56 120, B&$82, (&(% #'4 KJ% ,*1B &$8 #3*&11.J, (% $.)% .$. &U0(O0901*K, 7BB0)%*'1&+ ,T* @10...'#0%,9$0@'29#+1&,O$*U4*6 @*%0451&+ A, D@80$01*0,k,7%&, )'#1%*B*)#3*K, *1B&$8#3**A, N#)%*90()*, 1* &/*1, %&'#$:,'21&(*82+,1#,O$&/#T.:, 10,('&U&/01...*K,80T/.,(%&$&6 1#8*,.T0,/ &(%*-1.%2:, O$0/' #$*%0451&,'(|,O$&(9*%#1&: *,(%&$&12,',304&8,/&'0$KJ%,/$.-,/$ .-.A, D@/0$T)*, #1#6 4*@#,7BB0)%*'1&(%*,%$#1(#)3**,O$0/.(8#%$*'#J%, (1*6 T01*0,7BB0)%* '1&(%*, )&1% $#)%#,10, %&4 5 )&, O$*, ' @#*6 8&/0+(%'** ,(, %&$-&'2 8, O#$% 10...,1#,U&400,306 401#O$#'4011&0, $#(O$&(%$#101*0, *1B&$8#3**A m1#4*@ , $0@ .45%#%&', B.1)3 *&1*$ &'#1* K,#-$#$1&-& (0)%&$#,M0(O.U4* )*, a#%#$(%#1, 4*i1...#(%*:, O$&Q&/*'i08 ,W, k F, (01%KU$K, Gb;Y,-A,',"#8#$0:,O$0/(%#'*% 04*, a#%#$()&+ m""M, "#U*$&',*,q&T)*1,O$0/&(%#'*4*,.(%1.J,*1B&$8#6 3*J,&,O4#1*$.JP08(K,$#+&10A,",&O&$&+,1#,8101*0,O$&6 B0..., >Mp"R:, O&4)&'1*),pARA,")#4&1?, *, /$A,p# R#45108, l&(% &)0:, ',S#$ U*10:, #)%*'1&, /0+(%'&'#4#, hM.(()#K, B#i*(% 6 ()#K,O#$%*Kj, VAlA, M&/@#0'()&-&A, l&, '%&$&+, O&4&'*10, GbH<6Q, --A, $.)&'&/*6 %04...&/, *@ %.O*)#jGYA M0#451# K, O$#)% *) #,B#i* (% ()&-&, (%$ &*% 045(%'#: h#8O.%#3*Kj,u0Q&(4&'#)**, *, /#4510+i*0,% 0$$ *%&$*6 #45120,@#Q'#% 2,d0$8 #1**, &)#@#4*,(.P0...()&+, *, O&4*%*90()&+, $0B&$8 #8 ', M&((**:, O$ 0/&O$0/04*4&,h$#@ $.i*% 045120, O&(40/6 (%'*Kj, $&((*+()*Q, O&4*%*90()* Q:, 7)&1&8*90()*Q,*, O$#6 '&'2Q, $0B&$8,',Gbb...*0,#-01%&', -0$8#1()*Q, (O03(4.TU, *@, 9*(4#, (&'0%6 ()*Q,-$#T/#1, *,O&4.901*0, $#@'0/2'#%0451&+,*1B&$8#3** ,O&,d0$8#1**A l,GbHH,k,GbHI, --A,$0@)&,*@801KJ% (K,O$*&$*%0%2,',&$-#1*@#3**,)&1%$$#@6 '0/2'#%0451...*K,iO*&1()&+, * /*'0$(*&11&+, /0K% 0451&(%*A, M.)&'&/(%'&,(%$#12,O$*6 /#'#4&,*1B&$8#3** ,&$-#1&', -&(U 0@&O#(1&(%* ,(0$50@6 1&0, ,@1#901*0A "&%$./1*)*,)&1% $$#@'0/2'#%04512Q,O&/$ #@/040...#(%*A, l,$0@.45%#%0, ,U24*, .(%#1&'4012,W,-0$6 8#1()*Q,O&//#112Q:,$#U&%#'i*Q,1#,@#'&/#Q:,&%)./#,O$&6 *(Q&/*4#,.%09)#,*1B&$8#3**A,V&1(.4,',q01*1-$#/0,c&86 80$,$0-.4K$1&, '(%$09#4(K, (, -0$8#1()*8*,(O03*#4*(%#6 8*, *,O&4.9#4,&%, 1*Q,(0)$0%1.J,*1B&$8#3*JA l,'2K'401**,$#@'0/2 '#%04512Q,.(%$08401...,('K@*, 80T/.,O$&*@6 '&/*%04K8*:,%&$-&'20,30O&9)*, *,(Q082,/&(%#')*,%&'#$&'A R&, )$*@*(#,, *1B&$8#3 *K, &, O$0/O$*K%*KQ,, $#@ 80P#4#(5,, ',"LD, U0(6 O4...)4#8#, (%#4#,*8011&, /'*6 -#%0408, %&$-&'4*:,(O&(&U&8, /&10(%*,/&, O&%$0U*%04K, *1B&$8#3*J, &, %&'#6 $0, *4*, .(4.-0:,1#+%*, /04&'&-&, O#$%10$#A, l, GbbG, -A, 8&T 1&, U24&, O&/#%5, ' -#@0%.,8#4015...)4#8#,1#, $#/*&,1#,O4#%1&+,&(1&6 '0,O&K'*4#(5,',GbbG, -A,','*/0,8#4015)*Q, O0$0/#90), *1B&$8#3*&11&-&, Q#6 $#)%0$#A, g(1&'128*, $0)4#8&/#% 04K8*, $#/*&, U24*, 9#(% 120...&-&, &Uf06 8#:, )#), -&'&$*%(K, h'O$&)jGYA m1#4&-*9128,&U$#@&8, (%$&*4#(5, *,$#@'*'#4#(5,(&6 3*#451#K, *1B$#(%$.)%.$#, ',(&'Q&@0,h"2@$#1()*+j,M#6 /*P0'()&-&,$#+&1#,`45K1&'()&+ ,&U4#(%*A,[4#-&/#$K,8#6 %0$*#45128,'4*'#1*K8...,$*()&'#11&-&,@08406 /04*K, (%#1&'* 4*(5,'2(&)&O$&/.)%*'128*, $#+ &1#8* A l80(%0, (,1*8*,i*$*4#(5,*,h-0&-$#B*Kj,(%$&*%045(%'# (0401*+A,p#, ($0/(%'#,804*&$#%*'12Q, &$-#1*@#3*+, O$&6 '&/*4#(5, O0...)%2, (&' $08011&-&,-.8#1*%#$1&-&, @1#1*KA, "#$#%&' :, ;< laC?,%#)T0,10,/#4*,1* )#)&+ *1B&$8#3** ;A,d4#'12+,#$Q*':,-/0,Q$#1K%(K,8#%0$* #42,%&$-&'&6 O$ &8 2i6..., &UP06 (%'#, *, O$0/&(% #'4K%5(K, ', /&(%.O1&8,/4K, O&4.901*K,* #1#4*@ #, *1B&$8#3**, '*/0A GGA, ƒ5*:E*B, ,A53;:/;K*! -&( ./#$(%'0112Q, $#(Q&/&' O$0/O&4#-#0%:,9%&, UJ...)0,7)&1&8*90()&-&,O&'0/01*K O0$ (&1#4#,&$-#1*@ #3**, U#@*$.J%(K,1#, O$0/O&4&T01** &,%&8:,9%&,1#,0-&, O&'0/01*0,'4*KJ%,*,10)&%&$20,)&1%#6 8*1#1%2A,p#O$*80$:, (#8,Q#$#)%0$, %$./#,8&T0...,R4K 7%&-&,O$0/(%&K4&, $0i*%5,@#/#9*,O&,&$-#1*@#3**,(&'$06 8011&+,*1B$#(%$.)%.$2:, B&$8*$&'#1*J, @/&$&'&-&,&U6 $#@#,T*@1*:,O&,.4.9i01*J,(&3*#451&+,*,O$*$&/1&+,7)&6 4&-**A,l#T10+ i*8*,@#/#9#8*, %#)T0..., O$&30((2,',$0-*&1#Q, Q#$#)%0$*6 @&'#4*(5,O&%0$0+, .O$ #'4K08&(% *:,9%&, '2@2'#4&,U&45i*0,(U&*,', (&3*&6 ).45%.$1&8, $#@'*%**, *,(&3*#451&+,*1B$#(%$.)%.$0, (04#A,g(&U011&, U&456 i&0, )&4*90(%'&, 10...* -&%&'2,'@K%5,1#,(0UK,&%'0%(%'011&(%5,@#,$0i01*0,(&3*6 #4512Q,'&O$&( &'A,l, *%&-0,(&3*#451#K,* 1B$#(%$.)% .$# (04#:,',%&8,9*(40,(B0$#,&U$#@&'#1*K:,O$*i4*,),)$*@*(6 1&8. ,(&(%&K1...$0/0401&, '4*K1*0, )&8O5J%0$ 12Q, % 0Q1&4&-*+ ,1#, O&%&)*, *1B&$ 8#3** , 1#, O$&82 i4011&8 O$0/O$ *K%** A kTQTORb-N+l+,, \Ÿ´•,Ÿ¦,š–¦Ÿ\••”šŸ–,š–,••‘š–¯,Ÿ¦,•©•š–š’”\•”š¢•,©•“š’šŸ–’, –©•\,”™•,š–¦´ •–“•,Ÿ¦ ”™•, •v”•\–•´,–Ÿ–•› š´š•\š •,••©š • ”‡]^,x€z]‚„w,]^, †w†]‚xzw†,z_..., 1#O$ #'401* K, )&$O&$#%* '1&-&, .O$#'401*K, ',(B0$0, B.1)3* &1* $&'#1* K, U#1)&'()*Q, (%$.)6 %.$ A, M#()$2%#, O$&U408#,% $#1(O&$01%1&(%*, *1B&$ 8#3** ,> O$ &@$#91&(%*? :, #,% #)T0, &O$ 0/0401#, 10&UQ...#K,(%$.)%.$#, U*@10(60/*6 1*3:, )&%&$#K, '4* K0%, 1#,B&$8*$&'#1*0, .90%1&6#1#4*%*90()&+,*1B&$8#3**, &,@#%$#%#Q, )&8O#1*+, (&%&'&+, ('K@*A D((40/&'#12, &(&U011&(%*, *Q, /0K..., g[6 MmcglmpD• l,(%#%50,$ #((8&%$012, 7%#O2, *, 1#O$ #'401*K, *1B&$ 8#%*@#3** , (&3*#451&67)&1&8* 90( )&-&,*, *(%&$*90()&-& &U$#@&'#1*K:,O$#4*@*$&'#1, &O2...



Дата загрузки: 2017-03-18
Скачать документ
Скачать текст
0.07/5
... ΔE ~ ΔH, effect of PΔV usually small ΔE = -2045 kJ ΔH = - 2043 kJ PΔV = +2 kJ C3H8(g) + 5 O2(g) –> 3 CO2(g) + 4 H2O... C3H8(g) + 5 O2(g) –> 3 CO2(g) + 4 H2O(l) ΔH = - 2219 kJ (End Test 2) Effect of quantity (coefficients) is significant ΔH = - 4438 kJ 2 C3H8(g) + 10 O2(g) –> 6 CO2(g) + 8 H2O...(g) –> 2.25 CO2(g) + 3 H2O(l ) ΔH = - 2219 kJ] ΔH = - 1664 kJ Ch 8-CH1100-Fall 2005 Start... Eqns 2 NH3(g) ΔH˚ = + 92.38 kJ ΔH˚ = – 92.38 kJ N2(g) + 3 H2(g) Enthalpy ΔH magnitude... through). NH3(g) ==> 1/2 N2(g) + 3/2 H2(g) ΔH˚ = + 46.2 kJ If you want the reverse.... 4 x [ 1/2N2(g) + 3/2 H2(g) ==> 1 NH3(g) – 46.2 kJ ] ΔH˚ = -184.8 kJ Ch 8-CH1100-Fall 2005 12... ΔHsub Ex: If ΔHfus = 10 kJ/mol. How much heat to... 0.7 mol? Sol’n: (0.7 mol) x (10 kJ/mol) = 7 kJ ΔHfusion solid ΔHsub = ΔHfusion + ΔHvap Ex: ΔH sub = 83 kJ/mol ΔHvap = 53 kJ/mol. What is ΔH to... of 100 g/mol? Ans:15 kJ Ch 8-CH1100-Fall 2005 13... ˚C) q = (350 g) (4.184 j/g˚C)(–23 ˚C) q = – 33681.2 J = – 33.7 kJ Ch 8-CH1100-Fall 2005 16... q p = ΔE + PΔV = ΔH (for liqs ΔV ~ 0) q rxn = ΔH = – 2790.2 J ~ – 2.79 kJ Ch 8-CH1100-Fall 2005 18... shown: 1 N2(g)+ 3 H2(g) => 2 NH3(g) ΔH˚ = - 92 kJ Use above rxn to find... NH3(g) : N2(g)+ 3H2(g) => 2 NH3(g) ΔH˚= - 92 kJ) ΔH˚ = - 92 kJ 2 NH3(g) => 1 N2(g)+ 3 H2(g) ΔH˚= +92...(g)+ 3 H2(g) ΔH˚= +92 kJ] 5 NH3(g) => 2.5 N2(g)+ 7.5 H2(g) ΔH˚ = - 92 kJ) ΔH˚ = -184 kJ (reverse) (multiply to get correct coef.) (5/2 x 92 kJ) = ΔH˚= +230 kJ Ch 8-CH1100-Fall 2005 21... C3H8(g) + 5 O2(g) => 3 CO2(g) + 4 H2O(g) ΔH˚ = -2219 kJ b) Draw PE diagram of that... tell you conversion factor of kJ ΔH rxn for the # mol in... you know: - 2219 kJ/ 1 mol C3H8 . Go to kJ for mols you... mol C3H8 x (- 2219 kJ/ 1 mol C3H8) = 777 kJ Ch 8-CH1100-Fall 2005...)+ 2 NH3(aq) + 10 H2O(l) ΔH˚ = +80.3 kJ Ch 8-CH1100-Fall 2005 8.7 23...: 1 C3H8(g) + 5 O2(g) => 3 CO2(g) + 4 H2O(g) ΔH˚ = - 2219 kJ Solution: Convert g to moles: (15... x (– 2219 kJ/1mol C3H8) => ΔH˚= - 776.7 kJ 1b) 0.35 C3H8(g)+ 1.75 O2(g) ΔH˚= - 776.7 kJ 1.05... rxn. 3 H2(g) + N2(g) ==> 2 NH3(g) ΔH˚ = -92.2 kJ Intermediates H2 + N2H4 Reactants 3H2 + N2 Step 1 ΔH˚ = +95.4 kJ Step 2 ΔH˚ = -187.6 kJ Final ΔH ˚ = -92.2 kJ Product 2 NH3 Hess...(g) --> CO2(g) ΔH = - 393.5 kJ • CO2(g) --> C(s) + O2(g) ΔH = + 393.5 kJ • 2 CO2(g) --> 2 C(s) + 2 O2(g) ΔH = +787 kJ • This MUST...(l) ΔH˚ = ? CH4(g) + O2(g) ==> CH2O(g) + H2O(g) ΔH˚ = - 284 kJ CH2O(g) + O2 (g) ==> CO2(g) + H2O(g) ΔH˚ = - 518 kJ H2O(l) ==> H2O(g) ΔH˚ = + 44 kJ Ch 8-CH1100-Fall 2005... (g) ==> CO2(g) + H2O(g) ΔH˚ = - 518 kJ 2 [ H2O(g) ==> H2O(l) ΔH˚ = + 44 kJ ] ΔH˚ = - 88 kJ CH4(g) + 2 O2(g) ==> CO2(g) + 2 H2O(l) ΔH˚ = - 890 kJ Ch 8-CH1100... data to find ΔH˚rxn (in kJ) for: C2H4(g) + 3 O2(g) ==> 2 CO2(g) + 2 H2O(g) Substance ΔHf˚ (kJ/mol) H2O(l) H2O(g) CO2(g) - 285... ΔG = ΔH - TΔS For a given reaction ΔH is +45 kJ and ΔS is +199 J/K. At what...? (226.1 K) A reaction has ΔG of - 55 kJ and ΔS of +145J/K at 25... is non-spontaneous? Answers: a) - 11.8 kJ b) Y c) Y d) Y Ch 8-CH1100-Fall 2005 e) None...



Дата загрузки: 2017-12-26
Скачать документ
Скачать текст
0.03/5
... Гессе H обеспечи2 вает отыскание минимума x ∗ = H −1b не более чем за n шагов... Лагранжа: ( k k L x, λ , μ , r + ( где λk = λk1 , … , λkm ) T 1 2r k k m ) = f ( x ) + ∑j =1 p λ kj rk g j ( x) + 2 { } m ∑ [ g j ( x) ] 2 + j =1 ( ) 2 ⎧⎡ k k k ⎤ max 0, ( ) r g x μ + ∑ ⎨ ⎣⎢ j j ⎥ − μj ⎦ j = m +1 ⎩ ( , μ k = μ km +1 , … , μ kp ) T 2 ⎫ ⎬, ⎭ – векторы... Шаг 2. Составить модифицированную функцию Лагранжа: ( k k L x, λ , μ , r + k m ) = f ( x) + ∑j =1 p λ kj rk g j ( x) + 2 { m ∑ j =1 } [ g j ( x) ] 2 + ( ) 2 ⎧⎡ k k k ⎨ ⎢⎣ max 0, μ j + r g j ( x) ⎤⎥⎦ − μ j j = m +1 ⎩ 1 ∑ 2r k ( 2⎫ ⎬. ⎭ ) Шаг 3. Найти...-равенств); (x (λ , μ , r )) } (пересчет множителей для ограничений- λk +1 = λk + r k g x ∗ λk , μ k , r k { μ kj +1 = max 0, μ kj + r k g j неравенств); ( ∗ k k k ) x k +1 = x ∗ λk , μ k , r k , k = k + 1 , и перейти к шагу 2. Сходимость... достаточные условия. Если среди множителей λkj есть отрицательные, то это означает... по абсолютному значению отрицательный множитель λkj . Такая процедура поиска позволяет отыскать... условия f x k + t k∗ d k = min f x k + t k d k , а величина { } ( tk ≥ 0 ) t k ∗∗ = min t kj , t kj удовлетворяет условиям g j x k + t k d k = 0 , t k ≥ 0 . Направление спуска d k удовлетворяет... вектор базисных переменных: xBk = Bk−1b − Bk−1N k xN k . (11.28... вектора небазисных переменных: xBk = Bk−1b ≥ 0, xN k = 0 . (11.29) Допустимое базисное... x , Bk B k Nk Nk ⎟ ⎠ −1 −1 T T f ( x ) = cTBk Bk−1b − cTBk Bk−1N k xN k + cTN... b + (cN k − cBk Bk N k ) x N k . Обозначая β k = Bk−1b – матрицу-столбец значений базисных переменных...) где α k = B k− 1 N k – матрица размеров (m × (n − m)) со столбцами α kj = B k− 1 ( N k ) j , ( j = 1, ... , n − m ; Δk = (Δk1 ,..., Δkn − m ) ; x N k = ( x N k )1 ,..., ( x N k ) n − m )T ; (N k ) j – j -й стол- бец матрицы... q находится из условия Δkq = max Δkj ; 1≤ j ≤ n − m • число s находится из условия β ks... начальное базисное решение: x B0 = B0−1b = b = β 0 , xN 0 = 0 . Шаг 3. Вычислить матрицу относительных...: Δk = cTN − cTB B k−1N k . k k Если все Δ kj ≤ 0 , j = 1, ... , n − m , то условие окончания выполнено, а решение..., если относительные оценки строго отрицательные: Δ kj < 0 . Если имеется оценка, равная нулю... столбец, которому соответствует нулевая оценка Δ kj , и перейти к шагу 4. Если решается M -задача..., ... , xn . Если среди относительных оценок Δ kj есть положительные, то необходимо найти... максимальная положительная оценка: Δkq = max Δkj . 1≤ j ≤ n − m Если имеется неоднозначность выбора, то... N k . Поделить элементы матрицы-столбца βk = Bk−1b = xBk на соответствующие элементы матрицы... элементы матрицы β0 на соответβ 0 = B0−1b = ⎜⎜ ⎝0 1⎠⎝4⎠ ⎝4⎠ ствующие элементы матрицы α02 и определим... базисное решение: ⎛ x ⎞ ⎛0⎞ ⎛x ⎞ ⎛ 1 0⎞⎛ 2⎞ ⎛2⎞ ⎟⎟ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ = β1 ; xN 1 = ⎜⎜ 1 ⎟⎟ = ⎜⎜ ⎟⎟ . x B1 = ⎜⎜ 2 ⎟⎟ = B1−1b = ⎜⎜ ⎝ x3 ⎠ ⎝ 0 ⎠ ⎝ − 1 1⎠⎝4⎠ ⎝2⎠ ⎝ x4 ⎠ Положим k = 1 и перейдем к шагу... имеет вид: ⎛ 1 1⎞ ⎟ ⎛ 2 ⎞ ⎛3⎞ ⎜ ⎛ x4 ⎞ ⎛ 0 ⎞ ⎛ x2 ⎞ x B2 = ⎜⎜ ⎟⎟ = B 2−1b = ⎜ 2 2 ⎟ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ = β 2 ; xN 2 = ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ . ⎜ − 1 1 ⎟ ⎝ 4 ⎠ ⎝1⎠ ⎝ x3 ⎠ ⎝ 0 ⎠ ⎝ x1 ⎠ ⎟ ⎜ ⎝ 2 2⎠ Положим k = 2 и перейдем... матрицы β0 на соответствующие что β 0 = B0−1b = ⎜⎜ 0 1 ⎝ ⎠⎝ 5 ⎠ ⎝ 5 ⎠ элементы матрицы α02 и определим номер... элементы матрицы β0 на соответстчто β 0 = B0−1b = ⎜⎜ ⎝0 1⎠⎝4⎠ ⎝4⎠ вующие элементы матрицы α10 и определим... решение имеет вид: ⎛x ⎞ ⎛1 0⎞ ⎛ 2⎞ ⎛2⎞ ⎟⎟ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ = β1 ; x B1 = ⎜⎜ 2 ⎟⎟ = B1−1b = ⎜⎜ ⎝2 1⎠ ⎝4⎠ ⎝8⎠ ⎝ x1 ⎠ ⎛ x ⎞ ⎛0⎞ xN 1 = ⎜⎜ 3 ⎟⎟ = ⎜⎜ ⎟⎟ . ⎝ x4 ⎠ ⎝ 0 ⎠ Положим k = 1 и перейдем... матрицы β0 на соответствующие элеβ 0 = B0−1b = ⎜⎜ ⎝0 1⎠⎝ 2⎠ ⎝ 2⎠ менты матрицы α02 и определим номер... α10 = B0−1 ( N 0 )1 = ⎜ ⎟⎜ ⎟ = ⎜ ⎟ . Учитывая, что ⎝ 0 1 ⎠⎝ 1 ⎠ ⎝ 1 ⎠ ⎛1 0⎞ ⎛ 4⎞ ⎛ 4⎞ 0 β 0 = B0−1b = ⎜ ⎟ ⎜ ⎟ = ⎜ ⎟ = x B0 , поделим элементы матрицы β на...: Вычислим B1−1 : ⎛x ⎞ ⎛ 1 4 ⎞ ⎛ 4 ⎞ ⎛ 20 ⎞ ⎟⎟ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ = β1 ; x B1 = ⎜⎜ 3 ⎟⎟ = B1−1b = ⎜⎜ x 0 1 ⎠⎝4⎠ ⎝ 4 ⎠ ⎝ ⎝ 1⎠ ⎛ x ⎞ ⎛0⎞ xN 1 = ⎜⎜ 4 ⎟⎟ = ⎜⎜ ⎟⎟ . ⎝ x2 ⎠ ⎝ 0 ⎠ Положим k = 1 и перейдем к шагу... правилу и сформировать: [ ] [ ] а) два дополнительных ограничения: x j ≤ x kj ∗ , x j ≥ x kj ∗ + 1 ; б) две задачи ЗЛП- 2k + i , i = 1,2 : ЗЛП... к задаче ЗЛП-k дополни- [ ] тельного ограничения x j ≤ x kj ∗ ; ЗЛП- 2k + 2 , получаемую в результате добавления к задаче ЗЛП-k дополни- [ ] тельного ограничения x j ≥ x kj ∗ + 1 . Положить i = 1 и перейти к шагу 4. Шаг 4. Решить...



Дата загрузки: 2017-12-26
Скачать документ
Скачать текст
0.11/5
..., CRP2 EL2CR5, 2CR5 CR-V3/1B 6/24 6/24 6/24 6/24 6/24... EPX625, H-D, V625PX EPX675, KX675 539, KJ, 4018, 867 CART/CASE 6/24... P-P201 PA/1B BA P-P202 PA/1B BA P-P301 PA/1B BA P-P302 PA/1B BA P-P303 PA/1B BA P-P304 PA/1B BA P-P305 PA/1B BA P-P501 PA/1B BA P-P543 PA/1B BA P-P592 PA/1B BA P-P507 A/BA1 BA P-P508 A/1B BA PA HHR-P 505 A/1B BA PA HHR-P 401 A/1B BA PA HHR-P506 A/1B BA PA HHR-P509 A/1B TYPE VOLTS CAPACITY... 1604A — — — KC KD — K9V K9V — — KJ BA3042/U BA3030/U — BA3090/U BA3090/U — — — AM2... — — — PS-6360 26-04 — — — — — — 26-1B — — — 26-132 — — — — — 12M9 6V5K — LCR12V4BP...



Дата загрузки: 2017-06-15
Скачать документ
Скачать текст
0.08/5
....29 ϩ ͸ N ⌬hƒk ⌬G ƒЊ(298.15 K) ϭ 53.88 ϩ k (kJ molϪ1) (3-3.1) k (kJ molϪ1) (3-3.2) k ƒ k C pЊ(T ) ϭ ͭ͸ ͭ͸ ͭ͸ k k k ͮ ͭ͸ ͮ ͮ NkCpAk Ϫ 37.93....15 K) ϭ 53.88 ϩ k ƒk ϭ Ϫ25.73 kJ molϪ1 kϭ1 C Њp(700) ϭ ͭ͸ ͭ͸ ͭ͸ ͮ ͭ͸ NkCpAk Ϫ 37.93... Base Property ⌬G Њƒ(298.15 K) kJ molϪ1 ⌬H Њƒ(298.15 K) kJ molϪ1 C Њp(100 K) J mol....5981ͬ ␪ ) ϩ W ͸ M (C ) Ϫ 10.7983ͬ ␪ (3-4.1) (kJ molϪ1) (3-4.2) j pA2j j Nk(CpB1k k ϩ (kJ molϪ) j Nk(CpA1k....15 K) ϭ 10.835 ϩ Ϫ1 k kϭ1 ͸ N (gƒk) ϭ Ϫ23.597 kJ mol 5 ⌬G ƒЊ(298.15 K ) ϭ Ϫ14.83.... % Err. (2nd-Order) ⌬G ƒЊ(298.15 K), kJ molϪ1 Experimental Calculated (First Order... tested with the method Property ⌬G Њƒ kJ molϪ1 ⌬H Њƒ kJ molϪ1 C Њƒ(100 K) J molϪ1 KϪ1 C Њp(298... tested with the method Property ⌬G Њƒ kJ molϪ1 ⌬H Њƒ kJ molϪ1 C Њƒ(100 K) J molϪ1 KϪ1 C Њp(298... C Њp 500K C Њp 600K C Њp 800K C Њp 1000K C Њp 1500K kJ molϪ1 J molϪ1KϪ1 J molϪ1K... C Њp 500K C Њp 600K C Њp 800K C Њp 1000K C Њp 1500K kJ molϪ1 J molϪ1KϪ1 J molϪ1K... C Њp 500K C Њp 600K C Њp 800K C Њp 1000K C Њp 1500K kJ molϪ1 J molϪ1KϪ1 J molϪ1K... C Њp 500K C Њp 600K C Њp 800K C Њp 1000K C Њp 1500K kJ molϪ1 J molϪ1KϪ1 J molϪ1K... C Њp 500K C Њp 600K C Њp 800K C Њp 1000K C Њp 1500K kJ molϪ1 J molϪ1KϪ1 J molϪ1K... C Њp 500K C Њp 600K C Њp 800K C Њp 1000K C Њp 1500K kJ molϪ1 J molϪ1KϪ1 J molϪ1K... C Њp 500K C Њp 600K C Њp 800K C Њp 1000K C Њp 1500K kJ molϪ1 J molϪ1KϪ1 J molϪ1K... Joback % Error C/G % Error Benson C Њp(700K) kJ molϪ1 % Error Joback % Error C/G % Error....22 kJ mol 7 ⌬H Њƒ(298.15 K) ϭ k Ϫ1 ƒk kϭ1 ͸ N (S Њ) ϭ 0.397 kJ mol 7 S Њ(298.15 K) ϭ k Ϫ1 k KϪ1 kϭ1 ͸ N (S Њ) ϭ 0.801 kJ mol...) Ϫ S el Њ (298.15K)] ϭ Ϫ24.92 kJ molϪ1 ͸ N (C Њ ) ϭ 303.6 J mol 7 C Њp(800 K) ϭ k pk...) standard enthalpy of reaction at T, kJ molϪ1; Eqs. (3-1.2) and (3-1.4) standard (lower... at T and 1.01326 bar (1 atm.), kJ molϪ1 KϪ1; Eq. (3-1.8), Sec. 3-5 entropy contribution... of T and V Zϵ PV ϭ ƒP (T, P ) RT (4-2.1a) ϭ ƒV (T, V ) (4-2.1b) where V is the molar volume... Zϭ1ϩB ϭ1ϩ ͩ ͪ ͩ ͪ P P ϩ (C Ϫ B2) RT RT B C ϩ ϩ⅐⅐⅐ V V2 2 ϩ⅐⅐⅐ (4-5.1a) (4-5.1b) where the coefficients B, C, . . . are.... Equation (4-6.1) in the form Eq. (4-2.1b) is Zϭ V (⌰ / RT )V(V Ϫ ␩) Ϫ V Ϫ b (V Ϫ b)(V 2 ϩ ␦V ϩ ␧) (4-6.2) When it is... P Ϫ P0 ϭ ƒ (T, ␳, ␳0) (4-12.1a) ␳ Ϫ ␳0 ϭ ƒ (T, P, P0) (4-12.1b) where the reference density, ␳0, is..., J molϪ1 KϪ1 enthalpy change of boiling, kJ molϪ1 reaction equilibrium constant molecular... Zϭ1ϩB ϭ1ϩ ͩ ͪ ͩ ͪ P P ϩ (C Ϫ B2) RT RT 2 ϩ ... B C ϩ ϩ ... V V2 (5-4.1a) (5-4.1b) where the coefficients B, C, . . . are... ϭ 0.282 and ZRA2 ϭ 0.247. Equation (5-8.1b) then gives ZRAm ϭ 0.7 ϫ 0.282 ϩ 0.3 ϫ 0.247..., Eq. (5-8.3d ) weight fraction, Eq. (5-2.1b) mole fraction, Eq. (5.2-1a) compressibility.... (4-5.1). Zϭ1ϩB ϭ1ϩ ͩ ͪ ͩ ͪ P P ϩ (C Ϫ B2) RT RT 2 ϩ... B C ϩ ϩ... V V2 (4-5.1a) (4-5.1b) This is the only case... 2 RT B 2 ϩ... ϩ... P (C Ϫ B ) P Ϫ RT 2 RT ϪB P RT 2 2 2 Ϫ... 2 ϩ... Z (T, V ), Eq. (4-5.1b) ͩͪ 1ϪZ U ig Ϫ U RT 2 P P Ϫ (C Ϫ B2) RT RT... (5-4.2) Virial Equation ͫ͸ ln ␾i n (5-4.1a) & (5-4.2a) 2 jϭ1 (5-4.1b) & (5-4.2a,b) 2 V ͸yB n j jϭ1 ͬ yj Bij(T ) Ϫ B(T, {y}) ij... ij j j ͸ xk⌳ki N xj⌳ij ϩ 1 Ϫ k ͸ x⌳ ͸x␶ G ␶ Ϫ ͸G x N j kj j NRTL ͸␶G x g ϭ͸x RT ͸G x ͸␶G x xG ln ␥ ϭ ϩ͸ ͸G x ͸G x N N E ji... ͩ͸ ␪ ␶ ͪ N ln ␥i ϭ ln i ij kj k k ⌽i xi ϩ ΂ Ϫ j ji ⌽i k kj N kj k k i j ji j i j N j j j N j ij j N k kj k where ⌽i ϭ ri xi ͸rx N k k k and ␪i ϭ qi xi ͸qx N k k k i ΃ kj... group in molecule j ͸x␯ X ϭ ͸x ͸␯ j kj j (8-10.56) k j j kj k where ␯kj is the number of... 1 Ϫ ln (8-10.59) ͸xl ͩ͸ ͪ ͸ ͸ ͬ ␪j ␶ij ␪j ␶j i Ϫ j ␪k␶kj j k z li ϭ (ri Ϫ qi) Ϫ (ri Ϫ 1) z ϭ 10... 1 Ϫ ln ␪j␶ji Ϫ jϭ1 ͸ 3 jϭ1 ΂͸ ΃΅ ␪j␶ij (8-16.16) 3 ␪k␶kj kϭ1 where qi is the UNIQUAC... factor Greek ␣ ␣ij ␤ ␥i ⌫k ␦ ␦ ␧ ␰ ␩ ␪ ⍜i ␬ ␭ ␭ij ⌳ij ␯ (i) k ␯kj ␲* ␳ ␴2 ␶ ␶ij ␾i ⌽i ␹ ⌿mn ␺12 ␻ ⍀ parameter in...-Nies: Dechema Chemistry Series, 1: Part 1b, 108 (1974). Gmehling, J., and U. Onken... viscosity equation is: ͸ K ͩ1 ϩ 2 ͸ H n ␩m ϭ iϪ1 i iϭ1 ij ͸ ͸ n Kj ϩ jϭ1 n ͪ Hij Hik Kj Kk jϭ1 i kϭ1 i (9-5.1) where ␩m is the.... Data Ser., Vol. I, Parts 1 and 1b, DECHEMA, Frankfurt, 1977. Gordon, A. R.: J. Chem... (methyl amine) CAS # DelHf0, kJ / mol DelGf0, kJ / mol 7440-37-1 7726...-19-7 431-89-0 DelHf0, kJ / mol DelGf0, kJ / mol Ϫ264.40 Ϫ83....31 13838-16-9 DelHb, kJ / mol DelHm, kJ / mol 14.70 38...-2-pentene 2-methyl-2-butene DelHf0, kJ / mol DelGf0, kJ / mol 554-12-1 141....08 106.90 DelHb, kJ / mol DelHm, kJ / mol 32.15 35...) CAS # DelHf0, kJ / mol DelGf0, kJ / mol DelHb, kJ / mol DelHm, kJ / mol 1.88... 2,2,3,4-tetramethylpentane 2,2,4,4-tetramethylpentane 2,3,3,4-tetramethylpentane CAS # DelHf0, kJ / mol 540-84-1 Ϫ224.01....31 Ϫ116.59 DelHb, kJ / mol DelHm, kJ / mol 30.79 32... CAS # DelHf0, kJ / mol DelGf0, kJ / mol DelHb, kJ / mol DelHm, kJ / mol 138...-17-8 7440-63-3 DelHf0, kJ / mol DelGf0, kJ / mol 0.00 0.00 Ϫ241... 0.00 0.00 0.00 DelHb, kJ / mol DelHm, kJ / mol 0.90 40.66... 0.0729 hv1k hƒ1k gƒ1k w1k kJ molϪ1 kJ molϪ1 0.296 0.147 Ϫ0.071... in App. A. hƒ1k gƒ1k w1k kJ molϪ1 kJ molϪ1 0.263 0.500 X X X X 0.503...Ϫ1 0.297 0.292 Ϫ0.399 Ϫ0.720 kJ molϪ1 kJ molϪ1 1.252 1.041 Ϫ2.792 Ϫ2.092... w2j X 0.03654 0.21106 gƒ2j kJ molϪ1 kJ molϪ1 Ϫ9.874 Ϫ3.887 Ϫ24.125...



Дата загрузки: 2017-12-10
Скачать документ
Скачать текст
0.27/5
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g("'(*<#$.$(0.*a>>*8$-#.*'?*d$-%,"($. * ?#'7*37$#"5-x.*M"($.0*Z$J.*+'9#5$I$V81$,/@P`$)@/13$a@;:89AD8I [/;9/@/1B$)/@AI$KLTTI$Q+@88$.28<8@`$)56D5243...$a1B$53=$ #7/07234$V/92/3C$/E$H/?@3502C6IR...$V891/@P8=$&3$53=$&!0238IR$ @'(0$74'#-#H*+'5"',')H$Kj$GUN`$UYjOUdYI F8348@B$#928338I$KLL...



Дата загрузки: 2017-06-15
Скачать документ
Скачать текст
0.51/5
... from a transient formaldehyde complex. Ph3P 1b H H H O + H2 PPh3 H Ru Ph3P CO... Scheme 1: Alcohol dehydrogenation catalysed by 1b. The decarbonylation side-reaction affords... mechanism for dehydrogenation catalysed by 1b, comparing methanol, ethanol and 2-propanol... H Dehydrogenation Pathway B H Ru Ph3P Ph3P 1b Initiation H PPh3 H H H O Ph3P - H2 - HCHO... transition states; (c) free energy profile (kJ/mol), with calculations carried at... driving force for 1b + MeOH  1b + H2CO + H2, G = + 50.5 kJ/mol),[6] formation of..., cf. the reaction: 1b + MeOH  22 + 3 H2, G = -84.5 kJ/mol (4) Table 1: Reaction... (G) for pathways E-G (all data in kJ/mol).a E EBSSE Esolvb EG G Overall... + H2 -14.4 0.0 11.9 -81.9 -84.5 1b + MeOH  22 + 3 H2 Pathway E (methanol) 69.9 -4.2 -0.2 -48.5 20.2 1b  2 + H2 -17.6 14.2 16.4 16... Pathway F (methanol) -28.0 -5.9 -87.7 94.6 1b + MeOH  13H + H2 + PPh3 216... small activation barriers (below 22.4 kJ/mol), a feature which is consistent... transition states; (c) free energy profile (kJ/mol), with calculations carried at... the overall barrier of G = 51.7 kJ/mol relative to 31 (Table 3, 18.8 kJ/mol + 32.9 kJ/mol) is easily overcome... transition states; (c) free energy profile (kJ/mol), with calculations carried at... H Ru H Ph3P 13H' H Me H O H - MeCHO 1b + PPh3 14H' Decarbonylation Pathway F - H2... transition states; (c) free energy profile (kJ/mol), with calculations carried at... F with ethanol (all data in kJ/mol).a E EBSSE Esolvb EG G Overall... + H2 -71.8 0.0 10.2 -76.8 -138.4 1b + EtOH  22 + 2 H2 + CH4 Pathway F (ethanol) 1b + EtOH  13H’ + H2 + PPh3 203... Bb Cb Cneut. Db F G Methanol 1b + MeO-  TS3-4 + H2 9 + MeOH  TS3Hb-7 + HCHO 1b + MeO-  TS13-14 + H2 + PPh3... + MeO1b + MeOH  TS9perp-19 + 2 H2 1b + MeOH  TS13H-23perp + H2 + PPh3...-  TS13-28 + H2 + PPh3 Ethanol f 1b + EtO-  TS’3-4 + H2 2 + EtOH  TS...’13-28 + H2 + PPh3 2-propanol h 1b + iPrO-  TS”3-4 + H2 2 + iPrOH  TS”3Hb-7 1b + iPrO-  TS”13-14 + H2... this case, as G = 116.8 kJ/mol for: 1b + EtOH  TS’23-24... barriers are lowered by 6-28 kJ/mol for dehydrogenation and by up to 9 kJ/mol for decarbonylation. This result... decarbonylation path (G), by ca. 30 kJ/mol). Given typical uncertainties of... XTOF (TDTS) Methanol 95 100 A 1b TS3-4 5 4 2 100 B 7 TS3Hb-7 98 9 100 92 C 1b TS13-14 TS14-15 a 8 100... overall free energy barrier (148.9 kJ/mol) than previously reported (134... (0.999) -H transfer 1b + MeO-  TS3-4 + H2 1b + EtO-  TS’3-4 + H2 1b + iPrO-  TS...  TS3Hb-7 + HCHO    flipping of HCHO 1b + MeO-  TS13-14 + H2 + PPh3 2  1 HCHO and rotation of H2 1b + MeOH  TS13H-14H + H2 + PPh3...-17 +HCHO+ MeO2 1 H-agostic interaction 1b + MeOH  TS9perp-19 + 2 H2 7  TS9perp-19 H2 dissociation 1b + MeOH  TS13H-23perp + H2 + PPh3... 1.88 (4.21) 1.012 (1.008) C-H +0.03 1b + MeO  TS13-28 + H2 + PPh3... intermediates, bottom: free energy profile (kJ/mol, B97-D2/ECP2 level... 20-perp, but only by +2.1 kJ/mol. [14] e.g.: a) B. B. Wayland, B. A. Woods and... to afford 1b. This process is very demanding thermodynamically (ΔG = +135 kJ/mol... 1b, EG = -95.8 kJ/mol at P = 1354 atm, whereas it is -120.9 kJ... the ECP3 level, in kJ/mol.(a) Global 1b (b) Methanol 2 (b) TS9perp-19 TS13H... B, D and E. Methanol E E’BSSEa Esolv EG G b B 1b + MeOH  TS3Hb-7 + H2 91.3 26...-7 21.8 28.9 21.3 62.3 134.3 D 1b + MeOH  TS16Hb-17 + PPh3 185.4 -18.5 -2.0 -63.4 101.5 E 1b + MeOH  TS9perp-19 + 2 H2 152....4 10.8 13.3 15.7 123.1 Ethanol c B 1b + EtOH  TS’3Hb-7 + H2 84...-7 + MeCHO 105.9 10.5 1.4 -22.5 95.3 D 1b + EtOH  TS’16Hb-17 + PPh3 178.6 -18.6 -2.7 -56.2 101.1 2-propanol d B 1b + iPrOH  TS”3Hb-7 + H2 74...-7 + Me2CO 51.3 13.8 -0.1 -38.6 26.4 D 1b + iPrOH  TS”16Hb-17 + PPh3... difference of intramolecular BSSE in 1b and the transition state (see... of Pathway F. E E’BSSEa Esolv EG G 1b + MeOH  TS13H-23perp + H2 + PPh3 251.2 -18.7 -7.8 -97.6 127.1 1b + MeOH  TS23-24 + 2 H2 + PPh3 270.2 -21.5 -7.9 -120.7 120.1 1b + MeOH  TS24-25 + 2 H2 + PPh3 276.8 -21.1 -7.7 -130.0 118.0 1b + EtOH  TS’13H-23 + H2 + PPh3 236.8 -18.0 -5.3 -95.6 117.8 1b + EtOH  TS’23-24 + 2 H2... 268.0 -17.5 -13.5 -120.2 116.8 1b + iPrOH  TS”13H-23perp + H2 + PPh3 245.9 -16.5 -3.8 93.3 132.3 1b + iPrOH  TS”23-24 + 2 H2... difference of intramolecular BSSE in 1b and the transition state (see... EH EG 13.5 -5.4 12.7 -6.2 13.5 A 1b + MeO-  TS3-4 + H2 -6.2 B 9 + MeOH  TS3Hb... -14.4 -11.0 -10.1 -7.1 -14.4 -11.0 C 1b + MeO-  TS13-14 + H2 + PPh3....5 -24.2 -84.4 -34.2 -91.5 Cneut 1b + MeOH  TS13H-14H + H2 + PPh3....8 15.7 -13.4 22.4 -22.8 15.7 F 1b + MeOH  TS13H-23perp + H2 + PPh3....5 -97.6 -22.7 -90.7 -33.5 -97.5 G 1b + MeO-  TS13-28 + H2 + PPh3....90 0.00 0.00 12.10 (1b) (2) (3) (4) (5) (1b) (TS1b-2) (no TS) (TS3-4) (no... 0.00 5.10 5.10 0.00 0.00 (1b) (2) (TS1b-2) (no TS) - 22 - Bühl... 2013 PPh3 H Ru Ph3P Ph3P 1b Dehydrogenation Pathway B H H Initiation PPh3 H H H Ru...



Дата загрузки: 2017-12-26
Скачать документ
Скачать текст
0.17/5
... K x (303-298) K = 604.531,5844 kJ Ll.Hmasuk = H selulosa + H hemiselulosa + H Lignin... K x (363-298) K = 479.736,0003 kJ LiHke1uar = Q selulosa + Q hemiselulosa + Q Lignin + Q air...,80 kg = 2.l91.694,963 kJ x 1,250 kJ/kgOC x (90-25tC =mxcpx... - (5% x Qin) = Qout 5.611.824,3383 kJ = H Selulosa + H Hemiselulosa + H Lignin + H air T2... - (5% x Qin) = Quut 5.397.129,2980 kJ = H Selulosa + H Hemiselulosa + H Lignin + Hair + H H2S04....831.442,7 J= 1.692.831,4427 kJ 2) H herniselulosa = m x cp x D.T 3) H Lignin = 26.952...): Hidrolisa (R-210): Remahan tongkol , Jumlah(kJ) Buburlslurry: i Jagung: Selulosa 1.604.054... kJ/mo! = , = -285 840 kJ/mol = , g/ -212,02 kJ/ / mol x 1000g - -1413 47 kJ...,19 kJ + 35122463,21 kJ + 58365063,27 kJ = = 52.793.031,29 kJ AHa... = 34.545.053,04 kJ = 2.120,448 kJ/kg Pada suhu 148... tangki impregnator (R-210): Keluar Jumlah (kJ) Komponen I filter (H-220): Bubur/sluny... x L'l.T = 24.368,75 = 2319905 kJ x 1,12 kJ/kg °C x (110-25) °C =mxcpxL'l.T = 9.264...% x 33.334.214,5649 kJ = 3.333421,4565 kJ Entalpi keluar Qin - (5% x Qin... °C x 001,6-25) °C = 2.090.676,4418 kJ =m x cp x ~T Prarencana Pabrik Bioetanol dari....783,9543 kJ = 18.423.474,2197 kJ x 6,039 kJ/kg Panas Reaksi...--= g/ kg 74,I/rnol -13313,9 kJ/kg = -13313,9 kJ/kg x 6.995,94 kg = -93143245,57 kJ = -811 32 kJ/mo] = , -811,32 kJ/ I lOOOg / rno...,02 kJ/kg = -76639438,1 kJ = -1432,7 kJ/g mol = x 9.264,9 kg -1432,7 lnmoI kJ 13615 g/ , lmol -10522 kJ/kg... kg/kgmol = -135252036 kJ = -285 840 kJ/mol = , = -15880 kJ/kg x -285.840... kJ MIrx = (-135252036 kJ)+(2x -26986767,74 kJ)+(-93143245,57 kJ)+(76639438,lkJ) = -359008255,2 kJ (eksoterrnis) Panas Total = 14842222,61 kJ +359008255,2 kJ = 373850477,8 kJ... = 4.128,4906 kg x 141,3198 kJ/kg = 3 1 4 4 -T., )+-.c.(To -T., )+-d.(T, -T] ) 583.437,2911...,2197 kJ + 359008255,2 kJ = 153.036.937,0162 kJ +92l.l73,7110 kJ + Q yang... (F- (F-221): 231): Filtrat: Filtra1:: Jumlah (kJ) Glukosa 1.986.949,8787 Glukosa...(SO)4 = 33,1 kkaVkgoC x 4,184 kJ = 138,49 kJ/kgOC keal = m x cp x L'l.T = 12857... = 7.311.808,5134 kJ 2) H Glukosa 3) H Xylosa x 129,681 kJ/kg = m x cp x t.T = 24... kkal x 4,184 kJlkkal = 564,6341 kJ 2. H Magnesium sulfat = m x cp x d T = 0,8701 kg...) °c = 0,9623 kkal x 4,184 kJlkkal =4,0264 kJ 3. H Kalium klorida =m x cp x dT = 0,5220...) °c = 0,4280 kkal x 4,184 kJlkkal = 1,7909 kJ 4. H antifoam = m x cp x dT = 0,2871.10...) °c = 8,4407.10- 4 kkal x 4,184 kJ/kkal = 3,5316.10-3 kJ 5. H Amonium Hidroksida =m x cp....766,088 kJ + 52.521,6602 kJ = -309.244,4278 kJ (eksotermis) Prarencana...,049 kg = 20.873,105 kJ x 1,45 kJ/kg °C x (30-25) °C 30... UK Komponen --- KELUAR Jumlah (kJ) Komponen Jumlah(kJ) Dari tangki Starter Ke... °C x (30-25) °C = 187.859,2762 kJ 4) Hyeast extract = m x cp x L'l.T = 6,6347 x 8,6131... Jagung 3 - APPENDIXB. NERACAPANAS =-25621 kJ/g mol = , B-57 - 25621 kJ / ' /molxl000g I 180'~ kg... kg =-31217611,56 kJ = -277,63 kJ/g mol = -277 63 kj" ' mol x 1000g... 180kg I kgmol = -33827464,96 kJ = -393,51 kJ/g mol = - 393,51 ~/ I 1000... I = -31217611,56 kJ + -33827464,96 kJ + -47946738,94 kJ = -112991815,5 kJ Reaksi kedua CSHIOOS... kJ/kkal = -887,092 kJ/gmol - 887 092 kJ/ , / mol x 1000 g = -5913,95 kJ... kJ/kg = x 25911,44 kg -153238960,6 kJ Panas reaksi II = -182737078,8 kJ + -20729,4 kJ + -153238960,6 kJ = -335996768,8 kJ Panas Reaksi Total = -112991815,5 kJ + -335996768,8 kJ = -448.988.584,3 kJ... kg = 474.363,1887 kJ = m x cp x x 9,348 kJ/kg ~T = 2591,1750 kg... APPENDIX B. NERACA PANAS = (0;) ~ bottom = Ki Kj K etanol = Kair B-72 1,1508 0.507...,73 kJ + m x 589,13 754.801,8755 kJ + m x 2.144,77 kJ/kg kJ/kg 1.170.457,855 kJ m= 1.170.457,855 kJ 1.555,64 kJ/kg = 1.555,64 kJ/kg...,73 kJ + m125, 79 kJ/kg = 1.583.858,6190 kJ + m205,l54 kJ/kg....259,73 kJ - 1.583.858,6190 kJ = 341.401,1113 kJ ep air... kJ + 12.188.159,8586 kJ + m x632,20 kJ/kg + 0,05 x m x 2.120,448 kJ/kg mx2.120,448 kJ... = 12.293.749,3500 kJ 1.382,2256 kJ/kgxm = 12.293.749....771,86 kJ - 5622.892,314 kJ = 18782.879,55 kJ Prarencana Pabrik...,8586 kJ + m x 125,79 kJ/kg = 8.437.909,9419 kJ + m x 209,33 kJ/kg 3.750249,9090 kJ = 83,54 kJ/kg....ft 7073,8263 = 104 8444 1b material kering = ~x(9.268S)2' 4 . . f 0 Time 0 passage... = 4 . H P operas! = p 144 P design = \ = '!'X(3+ ~kr) 4 633786 1b/ ' / jt3 x 14J673jt . . 144 90in 1= 1,76...- - ' - - - - ' - - - - - - - - - - - - - 69.266,7866 6,1658.10- 3 1b /ft .s (turbulen) Prarencana Pabrik Bioetanol...= 1 7246 1 ' 7 ~-in 2 . H P operas! = p - = 144 758859 1b/ ' / jt3 x 19,6406jt 144 10...



Дата загрузки: 2017-03-26
Скачать документ
Скачать текст
0.23/5
...) voor een paling zijn 0.68 kJ.kg-1.km-1 terwijl de COT voor forel 2.73 kJ.kg-1.km-1 is. Het geschatte... (COT) for eel was 0.68 kJ.kg-1.km-1 while the COT for trout was 2.73 kJ.kg-1.km-1. The estimated fat... larvae (Knights 2003). From figure 1B we can see that recruitment... much fluctuations and were < 0° C (Figure 1B). Recruitment then declined markedly during..., University of Westminster, UK. Figure 1B (Bottom): The ‘Sargasso Sea Surface... and oxygen signal, see Figure 1B) they had a significantly higher SMR... in the range of 433 kJ.mol-1 which corresponds to a mixed... the range of 426-433 kJ.mol-1 for all four groups... total energy consumption was 2367 kJ kg-1 for a 3 months period, and... fish eggs of 23.48 kJ per g dry weight11, a 2-kg female... for respirometry of 18.89 kJ ml-1 O2 (Elliot & Davison 1975... energy consumption of 2316.58 kJ kg-1 fish over 5533.2 km or a COT of 0.42 kJ.kg-1.km-1. Alternatively the energy... mass fish-1) 3. Initial energy content (kJ g-1 dry mass) * initial dry matter... energy content begin (kJ fish –1). 4. End energy content (kJ g-1 dry mass) * end... fish-1) = total energy content end (kJ fish –1). 5. total energy content begin (kJ fish –1 ) - total energy content end (kJ fish –1 ) = total energy difference (kJ fish –1). 6. Total energy difference (kJ fish –1 ) / geometric bodyweight (g) * 1000 = total energy usage (kJ kg-1). With: geometric bodyweight = EXP... usage (= total energy consumption in kJ kg-1 fish, Table 1) multiplied by... bomb-calorimetry of 33.73 kJ/g dry mass (Brafield and Llewellyn... h-1) Total energy consumption (kJ kg-1 fish) COT (kJ /kg /km) Oxygen consumption...), we obtained a value of 2317 kJ kg-1 fish for the energy... to a COT value of 0.42 kJ kg-1 km-1. The swimming group... used for swimming was 3450 kJ kg-1 fish, corresponding to a COT value of 0.62 kJ kg-1 km-1. The two COT... trout of 0.68 and 2.73 kJ kg-1 km-1 respectively. Our video... analyses are 0.42 and 0.62 kJ kg-1 km-1, respectively, whereas trout... values (respirometry only) of 2.73 kJ kg-1 km-1. The COT in... them to stop swimming (Fig. 1B). The hemorrhage took the form... after 5500km to 42.3% (±2.1) (Fig. 1B). Normal hematocrit values for healthy... their unexposed controls (Table 1, figure 1b). This difference was significantly different... the PCB rest group (Figure 1B). So an important conclusion from... (COT) for eel was 0.68 kJ.kg-1.km-1 while the COT for trout was 2.73 kJ.kg1 .km-1. The estimated fat... the total amount of energy (kJ) it takes to transport one... the range of 2.52-2.58 kJ/kg/km (Brett 1973). The... 0.68 (eel) and 2.73 (trout) kJ/kg/km. Video films of...



Дата загрузки: 2017-12-26
Скачать документ
Скачать текст
0.06/5
.... Наиболее эффектной она выглядела в λ=Hα +0.75A 1b). Позднее ее интенсивность уменьшилась и в 0530... видимая в Hα − 0.75˚ A , хорошо видна в Hα + 1˚ A (рис. 1b). Возможно различие между картиной скоростей... − рентгеновские петли. На рис.1a и 1b приводятся зарисовки рентгеновских петель и арок...′ 10 R3 = η2 2304µ6 Здесь (4′ ) 5e′2 ; 0}. 3 (3) (3) Kj Dj ; Φ3 = 0. j=1 ∂A r2 + 3Ae2 sin v2... K1(4) = A2, K4(4) = 9B 2, 18 (4) (4) Kj Dj ; j=1 K2 (4) = 3AB, K5(4) = 9BC... : γ 2u′8 L′ 16η2 Φ5 = 552960µ10 48 (5) Kj Dj (5) ∗ j=1 γ 2u′8 L′ 16η2 и R5 = 552960µ10 48 Kj (5) Dj (5) , j=1 (12) где (опять отбрасывая...