ChE 304 Final Exam. ü Mark your answers Put your name on the back

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1 ChE 304 Final Exam ü Mark your anwer Put your name on the back (5). A piton having a cro-ectional area of 0.07 m i located in a cylinder containing water. An open U-tube manometer i connected to the cylinder a hown below. For h l = 60 mm and h = 00 mm, what i the value of the applied force, P A, acting on the piton? The weight of the piton i negligible. ü Write the manometer equation and olve for p eq = p + h gw - h ghg ã 0; eq = p ê. Flatten@Solve@eq, pd, D; Print@"p = ", eqd p = h ghg - h gw ü Subtitute an = eq êê. 9h Ø 0. m, h Ø 0.06 m, A Ø 0.07 m, gw Ø 980 Nt ë m 3, ghg Ø gw 3.56=; PrintA"Preure = ", an, "\nforce = ", an 0.07 m êê Framed, "\ng Hg Preure = Force = g Hg = = ", gw 3.56 êê. 9gw Ø 980 Nt ë m 3, ghg Ø gw 3.56=E 73.8 Nt m Nt Nt m 3

2 FinalExam.nb (5). Water flow teadily with negligible vicou effect through the pipe hown below. It i known that the 4-indiameter ection of thin walled tubing will collape if the preure within it become le than 0 pi below atmopheric preure. Determine the maximum value that h can have without cauing collape of the tubing. From the top of the water in the tank to the centerline i 4 ft. ü Write the BE and et 4 + h = z = z and zero out p, p, z, and v g + z + v g + z + v eq = p g == p g ; eq = eq êê. 8p Ø 0, p Ø 0, z Ø 0, z Ø z, v Ø 0< 0 ã z + v g vel = Flatten@Solve@eq, v D, D@@, DD -Â g z ü Rewrite the BE from the center of the contraction to the end. eq3 = p 3 g + z 3 + v 3 g == p g + z + v g ; eq4 = eq3 êê. 8p Ø 0, z 3 Ø 0, z Ø -h, v Ø vel< p 3 g + v 3 g ã -h - z Set p 3 = -0 lbf ë in 44 in ë ft 0 gage = 4.7 ab ü Expre v 3 in term of v an = eq4 êê. 9p 3 Ø -0 lbf ë in 44 in ë ft, g Ø 6.4 lbf ë ft 3, v 3 Ø vel a@0.5 ftd ê a@ ê 3 ftd, g Ø 3. ft ë = ft z ã -h - z h = ft eqa = eq ê. :p Ø 0, p -> -0 lb f in 44, z in ft Ø 0, z Ø -4 ft, v Ø 0, g Ø 6.4 lb f ë ft 3, g Ø 3. ft ë > 0 ã ft v ft

3 FinalExam.nb 3 vela = Flatten@Solve@eqa, v D, D@@DD êê Lat ft eq g + v g + z ã p g + v g + z p eqa = eq ê. :p Ø 0, p -> -0 lb f in 44, z in ft Ø 0, z Ø z, v Ø vela a@ ê 3D ê a@0.5d, v Ø vela, g Ø 6.4 lb f ë ft 3, g Ø 3. ft ë > ft ã 4. ft + z Solve@eqa, zd êê Flatten 8z Ø ft< vela a@ ê 3D ê a@0.5d ft

4 4 FinalExam.nb 3 (5). Water flow through a -in-diameter pipe with a velocity of 5 ft/. The relative roughne of the pipe i and the lo coefficient (K) for the exit i.0. Determine the height h to which the water rie in the piezometer tube. eq = p g + z + v g == p g + z + v g + h L; eqt = eq êê. JoinB:p Ø h g, p Ø 0, z Ø 0, z Ø 8 ft, v Ø 0, v Ø 0, v Ø 5 ft ê, h L Ø f l d 9ke Ø, g Ø 3. ft ë, d Ø ê ft, l Ø 8 ft=, 9f Ø ffarea5 ft ê, ê 6 ft,.05 µ 0-5 ft ë E, 0.004E=F v g + v g ke>, 0. + h ã ft ü Write the BE from the top of the peizometer to the top of the liquid in the tank and ubtitute eq = p g + z + v g == p g + z + v g + h L; eq = eq êê. :p Ø 0, p Ø 0, z Ø h, z Ø 8 ft, v Ø 0, v Ø 0, v Ø 5 ft ê, h L Ø f l d v g + v g ke> h ã 8 ft + 5 ft ke + 5 f ft l g d g ü Subtitute the head lo value an = eq êê. 9ke Ø, g Ø 3. ft ë, d Ø ê ft, l Ø 8 ft= h ã.4938 ft f ft

5 FinalExam.nb 5 height = an ê. 9f Ø ffarea5 ft ê, ê 6 ft,.05 µ 0-5 ft ë E, 0.004E= êê Framed; PrintBheight, "\nfriction Factor = ", ffarea5 ft ê, ê 6 ft,.05 µ 0-5 ft ë E, 0.004E, "\nre = ", rea5 ft ê, ê 6 ft,.05 µ 0-5 ft ë E, "\nh L = ", f l d v g + v g ke êê. JoinA9ke Ø, g Ø 3. ft ë, d Ø ê ft, l Ø 8 ft, v Ø 5 ft ê =, 9f Ø ffarea5 ft ê, ê 6 ft,.05 µ 0-5 ft ë E, 0.004E=EF h ã ft Friction Factor = Re = h L = ft

6 6 FinalExam.nb 4 (5). The pump hown below deliver a head of 50 ft to the water. Determine the power that the pump add to the water. The difference in elevation of the two pond i 00 ft. ü Write the BE from the top of the lower reervoir to the top of the upper reervoir and ubtitute eq = p g + z + v g + h p == p g + z + v g + h L; eq = eq ê. 8p Ø 0, p Ø 0, z Ø 0, v Ø 0, v Ø 0< h p ã h L + z ü Enter the head lo equation eq = eq ê. h L Ø v h p ã g v f l ü Subtitute and check unit g f l d + K ent + 4 K elb + K ext + K v d + 4 K elb + K ent + K ext + K v + z eq3 = eq ê. 9h p Ø 50 ft, K elb Ø.5, K ent Ø 0.8, K ext Ø, K v Ø 5.0, g Ø 3.3 ft ë, l Ø 500 ft, d Ø 0.75 ft, z Ø 00 ft= êê Simplify ft I. ft + H fl v M ã 0 eq4 = Solve@eq3, vd êê Lat 9v Ø H. ftl ë H, H fl L= ü Strip unit and ubtitute eq3 = FlattenASolveAeq ê. 9h p Ø 50, K elb Ø.5, K ent Ø 0.8, K ext Ø, K v Ø 5.0, g Ø 3.3, l Ø 500, d Ø 0.75, z Ø 00=, ve, E êê Lat v Ø ë H, H fll ü Find velocity and print the pump power vel = Iv ê. FindRootAv == eq3@@dd ê. 9f Ø ffareav, 0.75,.05 µ 0-5 E, 0E=, 8v, 0<EM ft ê ; flowrate = vel a@0.75 ftd; power = vel a@0.75 ftd 6.4 lb f ë ft 3 50 ft; Power = ft lb f = Watt = 56.6 HP

7 FinalExam.nb 7 velocity = flow rate = f = Re = ft 5.57 ft3

8 8 FinalExam.nb 5 (5). Etimate the dicharge J ft3 N of oil through the pipe hown below. ü Write the BE and ubtitute eq = p g + z + v g + h p == p g + z + v g + h L; eq = eq ê. 9p Ø 0, p Ø 0, z Ø 00 ft, z Ø 64 ft, v Ø 0, v Ø v, h p Ø 0= êê Simplify 7 ft ã v g + h L ü Solve with exit lo eq = eq êê. :h L Ø v 7 ft ã g f L d + Ke + Kv, L Ø 300 ft, d Ø ft, g Ø 3. ft ë, Ke Ø 0.5, Kv Ø 5.0> êê Simplify ft H fl v ü Strip unit o FindRoot can be ued eq = eq êê. :p Ø 0, p Ø 0, z Ø 00, z Ø 64, v Ø 0, v Ø v, h p Ø 0, h L Ø v g f L + Ke + Kv, L Ø 300, d Ø, g Ø 3., Ke Ø 0.5, Kv Ø 5.0> êê Simplify d H fl v ã 7.78 ü Solve an = Iv ê. FindRootAeq ê. 9f Ø ffareav,, 0-4 E, E=, 8v, 0<EM ft ê ; Print@"velocity = ", and velocity = ft dicharge = an a@ ftd; PrintA"dicharge = ", dicharge êê Framed, "\nvelocity = ", an, "\nf = ", ffareaan@@dd,, 0-4 E, E, "\nre = ", reaan@@dd,, 0-4 EE dicharge =.038 ft 3 velocity = f = Re = ft

9 FinalExam.nb 9 ü No minor loe ü Subtitute the head lo equation v eq = eq êê. :h L Ø f L d g, L Ø 300 ft, d Ø ft, g Ø 3. ft ë > êê Simplify 7 ft ã ft H fl v eq = eq êê. :p Ø 0, p Ø 0, z Ø 00, z Ø 64, v Ø 0, v Ø 0, h p Ø 0, h L Ø f L d v, L Ø 300, d Ø, g Ø 3.> êê Simplify g. f v ã 7.78 ü Solve an = Iv ê. FindRootAeq ê. 9f Ø ffareav,, 0-4 E, E=, 8v, 0<EM ft ê ; Print@"velocity = ", and velocity =.748 ft dicharge = an a@ ftd; PrintA"dicharge = ", dicharge êê Framed, "\nveloity = ", an, "\nf = ", ffareaan@@dd,, 0-4 E, E, "\nre = ", reaan@@dd,, 0-4 EE dicharge = ft 3 veloity = f = Re = ft

10 0 FinalExam.nb 6 (5). A gate i m wide and. m tall and hinged at the bottom. On one ide the gate hold back a -m-deep body of water. On the other ide, a 5-cm diameter water jet hit the gate at a height of m. What jet peed V i required to hold the gate vertical? ü Firt find the center of preure eq = ycp ã y + ibar y a ;

11 FinalExam.nb eq = eq êê. 9ibar Ø b h 3 ë, y Ø 0.5 m, a Ø b h, b Ø m, h Ø m= ycp ã m ü Evaluate the force at the midpoint eqf = h g a êê. 9h Ø 0.5 m, g Ø 980 Nt ë m 3, a Ø m = Nt ü The force due to the water jet i given by F x = r ê g c v x v A Co@pD eqfx = F x ã r ê g c v x v A Co@pD; eqfx = IeqFx êê. 9r Ø 000 kg ë m 3, g c Ø kg m ë INt M, v x Ø v, A Ø a@0.05 md=m@@dd Nt v m ü Set up the moment equation eqm = F L ã F L; ü Subtitute taking care to get the ign correct. anvel = v ê. Flatten@Solve@eqM êê. 8F Ø -eqf, L Ø eq@@dd, F Ø eqfx, L Ø m<, vd, D@@DD; PrintA"velocity = ", anvel êê Framed, "\n\ncheck the moment\nf L = ", -HeqFx ê. v Ø anvell m, "\nf L = ", -eqf ê 3 m, "\ny cp = ", eq@@dd, "\nforce at midpoint = ", eqfe velocity = m Check the moment F L = 370. m Nt F L = m Nt y cp = m Force at midpoint = Nt