(^_^)
06-11-2010, 04:02 AM
السلام عليكم يا باشمهندسين :]
الTopic ده علشان اللى عنده نصيحة او حاجة ناخد بالنا منها علشان منقعش فيها فى الامتحان او فكرة مهمة نركز عليها و يا ريت تكون المشاركات لاضافة معلومة و الردود مش هنا و لا الاسئلة بعد اذنكوا :
1st Heating part:
the sheet idea and tricks:
Pb1: we 've to know that Power(parallel)=n^2 P(series) where n :no of heating resistances
Pb2:as it is 3 pahse à P(elect.)=3*(v^2)/R & P(heat)=3*H*A
V must be Phase while the given in problem is V(L-L) but this is because the material is connected as Y, if connected as delta then V(phase) shall be the same as V(L_L)
Width , length and thickness are substituted all in( meter)
Units : P(watt )=H(watt/m^2)* A(m^2) ,P is given in Kw but substit. With watt
Pb:3 & Pb:4 as Pb:2(Hint Pb3 geh zayha f 2007)
Pb:5 is as Pb:2 except that it is single phase so (mnadrbsh f 3) à P(elect.)=(v^2)/R &
P(heat)=H*A
Pb:6 its idea is that P=Q(heat energy)/T(time)
For the units watt=joule/sec
Kilowatt= (Mjoule/3.6) / hour à (Mjoule/3.6)=KWhr
Ana sha5sayn 7a7fz en (watt=joule/sec) w lamma ygely “Mjoule” a3wad b (10^6 joule)
W lw galy hours a7ott-ha (3600 sec) w aray7 nfsy
Pb:7 in problem (hearth is full) and later it became (half full)!!Dh yfr2 fel current y3ny f I1 w I2
Remember: R=Z cos (phay)
X=Z sin (phay)
Pb8: differentiate between (furnace efficiency)=P(useful)/P(arc)
(electric efficiency )=P(arc)/P(i/p)
From phasor and the circuit : V=Varc +I(Req+Xeq)
lVl=square root of( (Varc+I Req)^2 + (I Xeq)^2 )
cos (phay)=Xeq/Req
Pb:6,7,8homma el mas2al elly fel sheet 3la el goz2 elly el nazary bta3o mal3’ay bas homma 3laina
Pb:9 (Dielectric)
Mo3zam afkar el Dielectric enny a7seb el C ,w a3wad f mo3dlet el P to get el V
Then I=P/V cos (phay)
Eo=8.85*10^-12 (mesh given fel emt7an w lazm nef7zha aw elly m3ah calculator fx-570 7yl2y el constants 3la el cover mn gowa (shift 7 à 32 )
Pb:10 na2sa w el sa7 bt3aha in Pb5 in evaluation P.165
Evaluation of Heating part:
Pb1: (geh zayaha 2006 )
2complete changes of air per hour àtime=0.5 hr
(y3ny el cycle bt7sal marratin fel sa3a,yb2a wa2t el cycle aw el operation ad eh!!nos sa3a)
(in 2006 was :assume the air is filtered and changed every 20 min)àtime=20 min
P(rated)=P(air )+P(losses)
P(air ) n7sebha mn el: heat energy(Q)/time(t)
P(losses) mn kanon el conduction w 7sab el Area na5od balna en el walls netr7 menha el window el doors, w el floor w el ceiling lehom ta2ser
Pb2:t3wed mobashr
take care T in "Kelvin" and w ndrb 10^4 bas ne2sm el T/1000 or mndrbsh feha bas ne2sm el T/100
Pb3: just know the law of “FORCED CONVECTION”
Take care the rate of flow is given (m^3)/min n3wad beh (m^3)/sec
Pb4: take care e7na 7asbna el P(o/p)=Q/t & from efficiency we got P(i/p)
To get V, use P(i/p)=(v^2) c ..…..
Pb5: (geh zayaha 2008)
Rem: P directly proportional to (V^2)*F
Pb6:the answer is not completed in papers, but I guess the answer is
b) AT=N*I
as N not given take =1
to get I à P=(I^2)*R & R=(resistivity*pi*d)/(delta*L) see papers P.26 & 27
c) velocity=(length/time)
Take care:
on lecture of date (15/5/2010) Dr. Mahmoud mentioned
1-causes of failure
2-modes of temp, control
3- control adjustment to vary the temp (makmlsh el point deh
Bas el klam dah mesh fl wr2 so plz return to the lecture or the book for this points (homma bel tarteb kano Question (2) 2008 )
arkam el ketab feha 7agat kteer 3'alt,f 3ady en arkmna tetl3 mo5talefa ,so follow eng.Mostafa's results l2n 7alloh sa7 3la el arakm elly howa by7l beha w en kan sa3at el arkam el given btkon 3ando 3'air el mas2al (zay mas2la 8,)
Lw feh ay 7aga 3’alt plz forgive me,and say the correction
الTopic ده علشان اللى عنده نصيحة او حاجة ناخد بالنا منها علشان منقعش فيها فى الامتحان او فكرة مهمة نركز عليها و يا ريت تكون المشاركات لاضافة معلومة و الردود مش هنا و لا الاسئلة بعد اذنكوا :
1st Heating part:
the sheet idea and tricks:
Pb1: we 've to know that Power(parallel)=n^2 P(series) where n :no of heating resistances
Pb2:as it is 3 pahse à P(elect.)=3*(v^2)/R & P(heat)=3*H*A
V must be Phase while the given in problem is V(L-L) but this is because the material is connected as Y, if connected as delta then V(phase) shall be the same as V(L_L)
Width , length and thickness are substituted all in( meter)
Units : P(watt )=H(watt/m^2)* A(m^2) ,P is given in Kw but substit. With watt
Pb:3 & Pb:4 as Pb:2(Hint Pb3 geh zayha f 2007)
Pb:5 is as Pb:2 except that it is single phase so (mnadrbsh f 3) à P(elect.)=(v^2)/R &
P(heat)=H*A
Pb:6 its idea is that P=Q(heat energy)/T(time)
For the units watt=joule/sec
Kilowatt= (Mjoule/3.6) / hour à (Mjoule/3.6)=KWhr
Ana sha5sayn 7a7fz en (watt=joule/sec) w lamma ygely “Mjoule” a3wad b (10^6 joule)
W lw galy hours a7ott-ha (3600 sec) w aray7 nfsy
Pb:7 in problem (hearth is full) and later it became (half full)!!Dh yfr2 fel current y3ny f I1 w I2
Remember: R=Z cos (phay)
X=Z sin (phay)
Pb8: differentiate between (furnace efficiency)=P(useful)/P(arc)
(electric efficiency )=P(arc)/P(i/p)
From phasor and the circuit : V=Varc +I(Req+Xeq)
lVl=square root of( (Varc+I Req)^2 + (I Xeq)^2 )
cos (phay)=Xeq/Req
Pb:6,7,8homma el mas2al elly fel sheet 3la el goz2 elly el nazary bta3o mal3’ay bas homma 3laina
Pb:9 (Dielectric)
Mo3zam afkar el Dielectric enny a7seb el C ,w a3wad f mo3dlet el P to get el V
Then I=P/V cos (phay)
Eo=8.85*10^-12 (mesh given fel emt7an w lazm nef7zha aw elly m3ah calculator fx-570 7yl2y el constants 3la el cover mn gowa (shift 7 à 32 )
Pb:10 na2sa w el sa7 bt3aha in Pb5 in evaluation P.165
Evaluation of Heating part:
Pb1: (geh zayaha 2006 )
2complete changes of air per hour àtime=0.5 hr
(y3ny el cycle bt7sal marratin fel sa3a,yb2a wa2t el cycle aw el operation ad eh!!nos sa3a)
(in 2006 was :assume the air is filtered and changed every 20 min)àtime=20 min
P(rated)=P(air )+P(losses)
P(air ) n7sebha mn el: heat energy(Q)/time(t)
P(losses) mn kanon el conduction w 7sab el Area na5od balna en el walls netr7 menha el window el doors, w el floor w el ceiling lehom ta2ser
Pb2:t3wed mobashr
take care T in "Kelvin" and w ndrb 10^4 bas ne2sm el T/1000 or mndrbsh feha bas ne2sm el T/100
Pb3: just know the law of “FORCED CONVECTION”
Take care the rate of flow is given (m^3)/min n3wad beh (m^3)/sec
Pb4: take care e7na 7asbna el P(o/p)=Q/t & from efficiency we got P(i/p)
To get V, use P(i/p)=(v^2) c ..…..
Pb5: (geh zayaha 2008)
Rem: P directly proportional to (V^2)*F
Pb6:the answer is not completed in papers, but I guess the answer is
b) AT=N*I
as N not given take =1
to get I à P=(I^2)*R & R=(resistivity*pi*d)/(delta*L) see papers P.26 & 27
c) velocity=(length/time)
Take care:
on lecture of date (15/5/2010) Dr. Mahmoud mentioned
1-causes of failure
2-modes of temp, control
3- control adjustment to vary the temp (makmlsh el point deh
Bas el klam dah mesh fl wr2 so plz return to the lecture or the book for this points (homma bel tarteb kano Question (2) 2008 )
arkam el ketab feha 7agat kteer 3'alt,f 3ady en arkmna tetl3 mo5talefa ,so follow eng.Mostafa's results l2n 7alloh sa7 3la el arakm elly howa by7l beha w en kan sa3at el arkam el given btkon 3ando 3'air el mas2al (zay mas2la 8,)
Lw feh ay 7aga 3’alt plz forgive me,and say the correction