ChE 304 Exam 3 Put your name on the back of the exam. Do your own work. 1. Water i pumped through a 4-inch in diameter pipe (ee the figure (a) below). The pump characteritic (pump head veru flow rate) are given in part (b) of the figure. Determine the flow rate if the head lo in the pipe i given by h L = 8 v g. Note: Q = V. In addition, you have to determine the unit on the 16 and the 5 in the equation h p = 16-5 V.
Exam 3v8_1.nb Set up the BE eq = p 1 g + z 1 + v 1 g + h p ã p g + z + v g + h L; Subtitute known value eq1 = eq êê. :z 1 Ø 0, z Ø 1 ft, p 1 Ø 0, p Ø 0, v 1 Ø 0, v Ø v, h p Ø 16 ft - 5 ft v, v Ø v A, h L Ø 8 v 16 ft - 0.43633 v ã 1 ft + 0.13975 v Solve for velocity v = 4.01198 ft ft g, A Ø p ê 4. H1 ê 3 ftl, g Ø 3. ft ë > Convert to volumetric flow rate v p ê 4 H1 ê 3 ftl 0.35011 ft 3 v = 0.00991089 m 3
Exam 3v8_1.nb 3. Gaoline (g = 0.68) flow through a pump at 0.1 m 3 ë a indicated in the figure. The lo between ection (1) and () i given by h L = 0.3 v. What will the difference in preure between ection (1) and g () be if 0 kw i delivered by the pump to the fluid? Ueful function a@d_d := p ê 4. d ; BE h p + p 1 g + v 1 g + z 1 ã h L + p g + v g + z Solve for p - p 1 g v 1 g - v g - h L + h p + z 1 - z Put the power into a form that can be ubtituted into the BE P = v h p g h p =.99814 m4 vdot
4 Exam 3v8_1.nb Fill in the value to be ubtituted into eq v = v 1 ub = : vdot Ø 0.1 m3 v Ø 0.1 m3, v 1 Ø 0.1 m3 ì a@d 1D, ì a@d D, d 1 Ø 0.1 m, d Ø 0. m, h p Ø hp, v Ø v 1, A Ø a@d 1 D, g Ø 9.81 m ë, z Ø 3 m, z 1 Ø 0, h L Ø 0.3 v g >; Subtitute 9.5697 m g Subtitute for g 197 53. Nt m Fill in the value to be ubtituted into eq v = v eq1 = eqa êê. :vdot Ø 0.1 m3, v 1 Ø 0.1 m3 ì a@d 1D, v Ø 0.1 m3 ì a@d D, d 1 Ø 0.1 m, d Ø 0. m, h p Ø hp, v Ø v, A Ø a@d 1 D, g Ø 9.81 m ë, h L Ø 0.3 v g, z 1 Ø 0, z Ø 3 m> 3.916 m g 19 576. Nt m Fill in the value to be ubtituted into eq v = (v 1 + v L ê eq1 = eqa êê. :vdot Ø 0.1 m3, v 1 Ø 0.1 m3 ì a@d 1D, v Ø 0.1 m3 ì a@d D, d 1 Ø 0.1 m, d Ø 0. m, h p Ø hp, v Ø Hv 1 + v L ê, A Ø a@d 1 D, g Ø 9.81 m ë, h L Ø 0.3 v g, z 1 Ø 0, z Ø 3 m> 31.7448 m g
Exam 3v8_1.nb 5 11 763. Nt m Fill in the value to be ubtituted into eq h L = 0.3 v 1 + 0.3 v g g for the h p term eq1 = eqa êê. :vdot Ø 0.1 m3 v Ø 0.1 m3, v 1 Ø 0.1 m3 ì a@d 1D, ì a@d D, d 1 Ø 0.1 m, d Ø 0. m, h p Ø hp, A Ø a@d 1 D, g Ø 9.81 m ë, h L Ø 0.3 v 1 g + 0.3 v 9.3466 m g g, z 1 Ø 0, z Ø 3 m> 195 765. Nt m eq1 = eqa êê. :vdot Ø 0.1 m3 v Ø 0.1 m3, v 1 Ø 0.1 m3 ì a@d 1D, ì a@d D, d 1 Ø 0.1 m, d Ø 0. m, h p Ø hp, A Ø a@d 1 D, g Ø 9.81 m ë, h L Ø 0.5 µ 0.3 v 1 g + 0.5 µ 0.3 v 31.48 m g g, z 1 Ø 0, z Ø 3 m> 08 415. Nt m
6 Exam 3v8_1.nb h L = 0.3 v 1 g h L = 0.3 v g 197 53. Nt m 19 576. Nt m h L = 0.3 I v 1+v g L 11 763. Nt m h L = 0.3 v 1 g +0.3 v g 195 765. Nt m h L = 0.5 0.3 v 1 g +0.5 0.3 v g 08 415. Nt m v 1 = 15.789 m v = 3.8197 m h p = 4.9845 m
Exam 3v8_1.nb 7 3. Air dicharge from a - inch in diameter nozzle and trike a curved vane, which i in a vertical plane (hown below). A tagnation tube connected to a water U-tube manometer i located in the free air jet. Determine the horizontal component of the force that the air jet exert on the vane. Neglect the weight of the air and all friction. Ue the BE to find the air velocity eq = p 1 g + z 1 + v 1 g ã p g + z + v g ; Ueful function a@d_d := p 4. d ; Eliminate variable v 1 g ã p g Solve for v 1 the air velocity v 1 = g p g Subtitute for p = hg w g h gw g
8 Exam 3v8_1.nb Subtitute known value and evaluate to yield the air velocity v 1 = 176.8 ft Write the equation for the force in the x - direction F x ã - v1 v1x r A 1 g c + v vx r A g c Subtitute F x ã -.96371 lb f The force of the fluid on the vane i the (-) of the reaction force Force of the air on the vane =.96371 lb f Evaluate the piece - v1 v1x r A 1 g c êê. :v1 Ø eq4, v1x Ø eq4, g c Ø 3. lb m ft lb f, A 1 Ø a@ ê 1 ftd, v Ø eq4, vx Ø v Co@10 DegreeD, A Ø a@ ê 1 ftd, r Ø 0.075 lb m ë ft 3 > -1.5885 lb f v vx r A g c êê. :v1 Ø eq4, v1x Ø eq4, g c Ø 3. lb m ft lb f, A 1 Ø a@ ê 1 ftd, v Ø eq4, vx Ø v Co@10 DegreeD, A Ø a@ ê 1 ftd, r Ø 0.075 lb m ë ft 3 > -1.37546 lb f
Exam 3v8_1.nb 9 4. For the piping ytem hown, all pipe are concrete with a roughne of 0.04 inche. Neglecting minor loe, compute the overall preure drop Hp 1 - p L in lb f ë in if V = 0 ft 3 ë. The fluid i water at 0 C. A Label the firt node a A and the path a u upper, m middle, l lower eq = p 1 g + z 1 + v 1 g ã p A g + z A + v A g + h L; Ueful function a@d_d := p ê 4 d ; vel@fr_, d_d := fr ê a@dd; re@v_, d_, nu_d := v d ê nu; ff@re_, eod_d := 1.6363883057031 LogA0.34043317053636 eod 1.11 + 6.9 re E Set up the contant for ubtitution ub1 = 8z 1 Ø 0, z A Ø 0<; ub = 8v 1 Ø 0, v A Ø 0<; ub3 = 9g Ø 3. ft ë, g Ø 6.4 lb f ë ft 3, n Ø 1.05 µ 10-5 ft ë =; ub4 = :h L Ø fr L d v g >; ub5 = 8d u Ø 8 ê 1 ft, d m Ø 1 ê 1 ft, d l Ø 15 ê 1 ft, L u Ø 1500 ft, L m Ø 800 ft, L l Ø 100 ft<; Find the Dp between 1 and A p 1 g ã h L + p A g Rearrange equation and define p 1 - p A = Dp Dp ã g h L
10 Exam 3v8_1.nb Subtitute for h L = Jf L d Dp ã fr L v g d g Evaluate Dp ã 17 001.3 lb f ft Convert to pi v g N p 1 -p A = 118.1 lb f in Now we have the preure drop from point 1 to point A. Next we find the preure drop from point A to point. Write the equation for each branch hlu = fr u L u v u g d u hlm = fr m L m v m g d m hll = fr l L l v l g d l Expre the fourth equation in term of v A 1 v ã 4 p d l v l + 1 4 p d m v m + 1 4 p d u v u Set up the contant for ubtitution. Strip the unit and ue FindRoot ub1 = 8z 1 Ø 0, z A Ø 0<; ub = 8v 1 Ø 0, v A Ø 0<; ub3 = 9g Ø 3., g Ø 6.4, n Ø 1.05 µ 10-5 =; ub4 = :h L Ø fr L d v g >; ub5 = 8d u Ø 8 ê 1, d m Ø 1 ê 1, d l Ø 15 ê 1, L u Ø 1500, L m Ø 800, L l Ø 100<;
Exam 3v8_1.nb 11 Now we have 4 equation in three unknown, v u, v m, and v l. Subtituting value and olving any three of them imultaneouly hould produce value for the unknown. eq6 = hlu - hlm ã 0 êê. Join@ub3, ub5, 8fr u Ø ff@re@v u, d u, nd, H0.04 ê 1L ê d u D, fr m Ø ff@re@v m, d m, nd, H0.04 ê 1L ê d m D<D eq7 = hlm - hll ã 0 êê. Join@ub3, ub5, 8fr l Ø ff@re@v l, d l, nd, H0.04 ê 1L ê d l D, fr m Ø ff@re@v m, d m, nd, H0.04 ê 1L ê d m D<D eq8 = eq5 êê. ub5 - - 0.378 v m LogA0.000416573+ 0.00007588 v m E + 4.3934 v l + LogB0.00035178+ 0.0000580704 F v l 5 p v l v ã + p v m 64 4 + p v u 9 57.17 v u LogA0.00065336+ 0.00010888 v u E ã 0 0.378 v m LogA0.000416573+ 0.00007588 v m E ã 0 FindRoot find a numeric olution to the three equation in three unknown 85.4817, 9.35937, 8.81466< The equation for p A - p i Dp ã fr L v g d g Retore unit before ubtituting ub1 = 8z 1 Ø 0, z A Ø 0<; ub = 8v 1 Ø 0, v A Ø 0<; ub3 = 9g Ø 3. ft ë, g Ø 6.4 lb f ë ft 3, n Ø 1.05 µ 10-5 ft ë =; ub4 = :h L Ø fr L d v g >; ub5 = 8d u Ø 8 ê 1 ft, d m Ø 1 ê 1 ft, d l Ø 15 ê 1 ft, L u Ø 1500 ft, L m Ø 800 ft, L l Ø 100 ft<; Evaluate Dp ã 184.83 lb f ft
1 Exam 3v8_1.nb Convert to pi Dp = 1.7974 lb f in Find the total preure drop Total Dp = Hp 1 -p A L+Hp A -p L= p 1 -p = 131. lb f in v t = 5.4817 v m = 9.35937 v b = 8.81466 Dp 1 = 118.065 lb f Dp in = 1.7974 lb f in