Pipe flow with friction losses solutions using HP and TI calculators By Gilberto E. Urroz, October 2005

Transcription

1 Pipe low with riction losses solutions using HP and TI calculators By Gilberto E. Urroz, October arcy-weisbach Equation and riction actor The basic equation governing riction losses in a pipeline is the arcy-weisbach equation: L V 8 L Q h =, or h = (1) 5 g π g where is the riction actor, e is the absolute roughness (or equivalent sand roughness) o the pipe, is the pipe diameter, R is the Reynolds number, L is the pipe length, V is the low velocity, Q is the discharge, and g is the acceleration o gravity. The discharge Q and the low velocity V are related by the continuity equation, namely, V 4 Q =, or π π Q = V 4. (). The Reynolds number is deined as ρv V 4 ρ Q 4 Q R = =, or R = = μ ν π μ π ν. (3) where ρ is the luid density, μ is its dynamic (or absolute) viscosity, and ν = μ/ρ is its kinematic viscosity [units = m /s or t /s]. The riction actor is a unction o the relative roughness e/ and o the Reynolds number R. Values o can be obtained rom the Moody diagram that shows curves o constant relative roughness or a range o values o the Reynolds number and the corresponding riction actors. The diagram includes also the laminar riction actor given by Stokes equation: 64 =. (4) R Function ARCY in HP calculators The HP 48 G and HP 49 G series calculators provide unction ARCY(ee/,R) 1 to calculate the riction actor or turbulent lows. In the HP 48 G, unction ARCY is available by using the keystrokes: The resulting menu will show the unctions: [ ][ EQ LIB ][UTILS] 1 Here we use ee instead o e because the HP calculators would interpret e as exp(1). 1

2 MINE: ZFACT: FANNI: ARCY: F0λ: SIEN: Minesweeper game Gas compressibility Z actor unction Fanning s riction actor (4 times -W s riction actor) The unction o interest (arcy-weisbach s riction actor) Black-body emissive power unction Silicon intrinsic density In the HP 49 G, HP 49 G+, and HP 48GII, unction ARCY is available through the unction catalog, [CAT], or you could simply type the name o the unction. To calculate the riction actor with unction ARCY using the HP 48 G series calculators, or the HP 49 G, HP 49 G+, and HP 48GII calculators in RPN (Reverse Polish Notation), enter the values o e/ and R in the stack, and then invoke unction ARCY. I using the HP 49 G, HP 49 G+, and HP 48GII calculators in ALG (algebraic) mode, enter the expression ARCY(value o e/, value o R). As an example, calculate the riction actors or the ollowing combinations o riction actors and Reynolds numbers. Veriy the values listed in Table 1. Table 1. Friction actors calculated with HP s unction ARCY e/ R x x x x x Function ARCY implements the Coolebrook-White equation, as well as the laminar equation, to calculate the riction actor. Using unction ARCY in the HP calculators is equivalent to reading out o the Moody diagram knowing the values o e/ and R. Equations or the riction actor The Coolebrook-White equation, shown below, is an implicit equation in, and, thereore, not amenable to direct solution. 1 e.51 = log R (5) Modern alternatives or the Coolebrook-White equation that are explicit in include Haaland s equation and Swamee-Jain s equation. Haaland s equation is: 1 e = 1.8 log , (6) R

3 or, equivalently, =. (7) 1.11 e 6.9 log R Swamee-Jain equation is given by: 0.5 = (8) e 5. log R In equations (5) through (8), log stands or the logarithm o base 10, and log () [log()]. Implementing Haaland s and Swamee-Jain equations in HP calculators As alternatives or the unction ARCY in the HP calculators, one can deine unctions HA and SJ to implement the explicit orm o the Haaland s and Swamee-Jain s equations, respectively. In order to keep all the pipe-related unctions and equations together, I suggest creating a sub-directory, call it PIPES, within the HOME directory o your calculator. In order to deine the unctions you need to use the key EF with the ollowing arguments: HA(e,R)=0.3086/(LOG((e/3.7)^ /R))^ SJ(e,R)=0.5/(LOG(e/ /R^0.9))^ In these deinitions e stands or the relative roughness (e/) and R stands or the Reynolds number R. Ater deining these unctions there will be sot-menu keys labeled [ HA ] and [ SJ ] in your calculator. To see the variables available in your PIPES subdirectory you may have to press the [VAR] key. The operation o these two user-deined unctions, namely, HA and SJ, is very similar to the operation o unction ARCY. To veriy the implementation o these unctions in HP calculators check the ollowing values returned by the unctions or the parameters e/ and R as given in Table. Table. Values o calculated with user-deined unctions HA and SJ e/ R HA SJ x x x x x Implementing Haaland s and Swamee-Jain equations in TI calculators In the TI-89 or TI-9 calculators, you can program unctions ha(k,r) and sj(k,r) to calculate the riction actor using Haaland s and Swamee-Jain s equations, respectively. 3

4 In these deinitions, k represents the relative roughness (e/) and r represents the Reynolds number. To program unction ha, select the Program Editor in your calculator, and then select the option 3:New to enter a new unction. Select: Type: Folder: Variable: :Function (your avorite older, e.g., one called luids ) ha and press [Enter]. Edit the unction such that the editor s window looks as ollows: :ha(k,r) :Func :.3086/(log((k/3.7)^ /r)^ :EndFunc Similarly, to implement unction sj select the Program Editor in your calculator, and then select the option 3:New to enter a new unction. Select: Type: Folder: Variable: :Function (your avorite older, e.g., one called luids ) sj and press [Enter]. Edit the unction such that the editor s window looks as ollows: :sj(k,r) :Func :.5/(log(k/(3.75)+5.74/r^(.9))^ :EndFunc Press [HOME] to return to the HOME screen. At this point you are ready to use unctions ha and sj to calculate riction actors. To load the unction name in the HOME screen entry line, you can either type the unction name ha or sj, or use [N][VAR-LINK] and select the unction rom the list thus produced. The unction name must be ollowed by a set o parentheses including the values o k = e/ and Re separated by commas. Press [ENTER] to evaluate the unction. Ater implementing unctions ha and sj in your TI calculator, veriy the results shown in Table.. Types o problems involving the arcy-weisbach equation or riction losses The textbook by Finnemore and Franzini identiies three types o problems using the arcy-weisbach equation, namely: 1. Head loss problem: calculate h given, Q or V, and g, L, e, ν.. ischarge problem: calculate Q or V, given, h and g, L, e, ν. 3. Sizing problem: calculate, given Q, h and g, L, e, ν. Examples o the three types o problems are shown next: Problem [1]. Given = 0.3 t, Q = 0.0 cs, g = 3. t/s, L = 1000 t, e = 0.00 in = t, and ν = 1.13x10-5 t /s, ind h. 4

5 Problem []. Given = 0.7 t, h = 15 t, g = 3. t/s, L = 750 t, e = in = t, and ν = 1.x10-5 t /s, ind Q. Problem [3]. Given Q = 3 cs, h = 10 t, g = 3. t/s, L = 1500 t, e = 0.01 in = t, and ν = 1.5x10-5 t /s, ind. These problems are solved next using a variety o approaches. Solution o sample problems using the Moody diagram Solution to Problem [1]: For e/ , R = 4Q/(πν) = 7.5x10 4, the Moody diagram shows 0.0, thereore, h = 8LQ /(π g 5 ) = 9.1 t. Solution to Problem []: e/ From h = 8LQ /(π g 5 ), we get Q =π g 5 h /(8L) = , rom which Q = 0.365/ (A). Also, R = 4Q/(πν) = 1.5x10 5 Q (B). An iterative solution proceeds as ollows: Assume = 0.03, (A) gives Q = 0.365/ 0.03 =.107 cs, R = 3.x10 5 Moody: = 0.019, (A) gives Q = 0.365/ =.65 cs, R = 4.03x10 5 Moody: = 0.019, convergence achieved, thus Q =.65 cs. Solution to Problem [3]: From h = 8LQ /(π g 5 ), we get 5 / = 8LQ /(π gh ) =33.98, rom which =.0 1/5 (A). The relative roughness is e/ = / (B), and the Reynolds number is R = 4Q/(πν) =.55x10 5 / (C). An iterative procedure is implemented as ollows: Assume = 0.03, (A) = 1.00 t, (B) e/ = , (C) R =.55x10 5 Moody: = 0.00, (A) = 0.93 t, (B) e/ = , (C) R =.76x10 Moody: = 0.01, (A) = 0.93 t, (B) e/ , (C) R =.74x10 5 Moody: = 0.01, convergence achieved, thus = 0.93 t. Solution using unctions ARCY, HA and SJ instead o the Moody diagram Use o the Moody diagram requires us to read the values o rom the diagram or known values o e/ and R. Functions ARCY (in HP calculators only), HA (or ha) and SJ (or sj) can be used to calculate the values o instead o reading them out o the Moody diagram. Solution o sample problems using the ARCY unction The ARCY unction being available only in the HP calculators, these solutions cannot be implemented in the TI calculator. Solutions using unctions HA and SJ or both calculators will be presented in subsequent sections. The ollowing solutions used unction ARCY to calculate the riction actor : Solution to Problem [1]: Given = 0.3 t, Q = 0.0 cs, g = 3. t/s, L = 1000 t, e = 0.00 in = t, and ν = 1.13x10-5 t /s, ind h. For e/ = , R = 4Q/(πν) = 7.5x10 4, the ARCY unction shows = , thereore, h = 8LQ /(π g 5 ) = 8.04 t. 5

6 Solution to Problem []: Given = 0.7 t, h = 15 t, g = 3. t/s, L = 750 t, e = in = t, and ν = 1.x10-5 t /s, ind Q. e/ = From h = 8LQ /(π g 5 ), we get Q =π g 5 h /(8L) = , rom which Q = 0.365/ (A). Also, R = 4Q/(πν) = 1.5x10 5 Q (B). An iterative solution proceeds as ollows: Assume = 0.03, (A) gives Q = 0.365/ 0.03 =.107 cs, R = 3.x10 5 ARCY: = , (A) gives Q = 0.365/ =.67 cs, R = 4.05x10 5 ARCY: = , (A) gives Q = 0.365/ =.69 cs, R = 4.09x10 5 ARCY: = , convergence achieved, thus Q =.69 cs. Solution to Problem [3]: Given Q = 3 cs, h = 10 t, g = 3. t/s, L = 1500 t, e = 0.01 in = t, and ν = 1.5x10-5 t /s, ind Q. From h = 8LQ /(π g 5 ), we get 5 / = 8LQ /(π gh ) =33.98, rom which =.0 1/5 (A). The relative roughness is e/ = / (B), and the Reynolds number is R = 4Q/(πν) =.55x10 5 / (C). An iterative procedure is implemented as ollows: Assume = 0.03, (A) = 1.00 t, (B) e/ = , (C) R =.55x10 5 ARCY: = 0.000, (A) = 0.94 t, (B) e/ = , (C) R =.35x10 5 ARCY: = 0.004, (A) = 0.97 t, (B) e/ , (C) R =.36x10 5 ARCY: = 0.004, convergence achieved, thus = 0.97 t 0.93 t. Solution o sample problems using the HA or ha unction Function HA or ha implement Haaland s equation (7) to calculate the riction actor. The solutions to the three sample problems using this equation or, instead o the Moody diagram, is shown next: Solution to Problem [1]: Given = 0.3 t, Q = 0.0 cs, g = 3. t/s, L = 1000 t, e = 0.00 in = t, and ν = 1.13x10-5 t /s, ind h. For e/ = , R = 4Q/(πν) = 7.5x10 4, unction HA shows 0.01, thereore, h = 8LQ /(π g 5 ) = 8.73 t. Solution to Problem []: Given = 0.7 t, h = 15 t, g = 3. t/s, L = 750 t, e = in = t, and ν = 1.x10-5 t /s, ind Q. e/ = From h = 8LQ /(π g 5 ), we get Q =π g 5 h /(8L) = , rom which Q = 0.365/ (A). Also, R = 4Q/(πν) = 1.5x10 5 Q (B). An iterative solution proceeds as ollows: Assume = 0.03, (A) gives Q = 0.365/ 0.03 =.107 cs, R = 3.x10 5 HA: = , (A) gives Q = 0.365/ =.67 cs, R = 3.96x10 5 HA: = , (A) gives Q = 0.365/ =.69 cs, R = 4.0x10 5 HA: = , convergence achieved, thus Q =.69 cs. 6

7 Solution to Problem [3]: Given Q = 3 cs, h = 10 t, g = 3. t/s, L = 1500 t, e = 0.01 in = t, and ν = 1.5x10-5 t /s, ind Q. From h = 8LQ /(π g 5 ), we get 5 / = 8LQ /(π gh ) =33.98, rom which =.0 1/5 (A). The relative roughness is e/ = / (B), and the Reynolds number is R = 4Q/(πν) =.55x10 5 / (C). An iterative procedure is implemented as ollows: Assume = 0.03, (A) = 1.00 t, (B) e/ = , (C) R =.55x10 5 HA: = , (A) = 0.88 t, (B) e/ = , (C) R =.89x10 5 HA: = 0.003, (A) = 0.93 t, (B) e/ , (C) R =.74x10 5 HA: = 0.00, convergence achieved, thus = 0.93 t. Solution o sample problems using the SJ or sj unction Function SJ or sj implements Swamme-Jain s equation (8) to calculate the riction actor. The solutions to the three sample problems using this equation or, instead o the Moody diagram, is shown next: Solution to Problem [1]: Given = 0.3 t, Q = 0.0 cs, g = 3. t/s, L = 1000 t, e = 0.00 in = t, and ν = 1.13x10-5 t /s, ind h. For e/ = , R = 4Q/(πν) = 7.5x10 4, the Moody diagram shows 0.016, thereore, h = 8LQ /(π g 5 ) = 8.89 t. Solution to Problem []: Given = 0.7 t, h = 15 t, g = 3. t/s, L = 750 t, e = in = t, and ν = 1.x10-5 t /s, ind Q. e/ = From h = 8LQ /(π g 5 ), we get Q =π g 5 h /(8L) = , rom which Q = 0.365/ (A). Also, R = 4Q/(πν) = 1.5x10 5 Q (B). An iterative solution proceeds as ollows: Assume = 0.03, (A) gives Q = 0.365/ 0.03 =.107 cs, R = 3.x10 5 SJ: = , (A) gives Q = 0.365/ =.66 cs, R = 4.04x10 5 SJ: = , (A) gives Q = 0.365/ =.68 cs, R = 4.07x10 5 SJ: = , convergence achieved, thus Q =.68 cs. Solution to Problem [3]: Given Q = 3 cs, h = 10 t, g = 3. t/s, L = 1500 t, e = 0.01 in = t, and ν = 1.5x10-5 t /s, ind Q. From h = 8LQ /(π g 5 ), we get 5 / = 8LQ /(π gh ) =33.98, rom which =.0 1/5 (A). The relative roughness is e/ = / (B), and the Reynolds number is R = 4Q/(πν) =.55x10 5 / (C). An iterative procedure is implemented as ollows: Assume = 0.03, (A) = 1.00 t, (B) e/ = , (C) R =.55x10 5 SJ: = 0.000, (A) = 0.9 t, (B) e/ = , (C) R =.77x10 5 SJ: = 0.004, (A) = 0.93 t, (B) e/ = , (C) R =.74x10 5 SJ: = 0.004, convergence achieved, thus = 0.93 t. 7

8 The ollowing table summarizes the solutions o the three sample problems using (a) the Moody diagram, (b) the ARCY unction in the HP calculators, (c) the HA or ha unction, and (d) the SJ or sj unction. Table 3. Solutions to sample problems through dierent methods o obtaining Moody ARCY HA SJ Problem [1] = 0.0 = = 0.01 = h = 9.05 t h = 8.04 t h = 8.73 t h = 8.89 t Problem [] = = = = Q =.65 cs Q =.69 cs Q =.69 cs Q =.68 cs Problem [3] = 0.01 = = 0.00 = = 0.93 t = 0.97 t = t = 0.93 t 3. Combining the arcy-weisbach equation with various equations or arcy-weisbach and Coolebrook-White equation We can combine the arcy-weisbach equation in terms o the velocity V (1) and the Coolebrook-White equation (5) as ollows. First, rom (1) we get and gh 1 V =, (9) L 1 L = V. (10) gh The last term within the logarithmic unction in (5) is written as ollows in terms o the deinition o R (3) and (10): ν L ν L =.51 =.51 V =.51 R R V gh gh. (11) Replacing 1/ in (5) within equation (9), with the result o (11) included, results in the equation [see Eq. 8.56a, page 93, in Finnemore and Franzini]: V gh e ν L =.88 log (1) L gh Equation (1) can be re-written in terms o the discharge by using the continuity equation () [see Eq. 8.56b, page 93, in Finnemore and Franzini]: 8

9 Q =. g 5 L h log 0.7 e ν L gh. WCWQ(13) Equation (13) is explicit in Q, thus, it is appropriate or a direct solution o problems o type, the discharge problem. Solutions o type 1 (head loss) and type 3 (sizing) problems using equation (13) require the use o numerical solutions. Equation (13) is reerred to by the name WCWQ, i.e., arcy-weisbach + Coolebrook-White in terms o Q. arcy-weisbach and Haaland equations Combining the arcy-weisbach equation in terms o the discharge Q, equation (1), and the Haaland equation (6), and combining numeric terms results in: Q =.0 g 5 L h log 0.34 e 1.11 ν WHAQ(14) Q This equation is implicit in Q and, but explicit in h. Thus, this equation is ideal or solving type 1 problems. Type and 3 problems, however, will require a numerical solution. Equation (14) is reerred to by the name WHAQ, i.e., arcy-weisbach + HAaland in terms o Q. arcy-weisbach and Swamee-Jain equations Combining the arcy-weisbach equation in terms o the discharge Q, equation (1), and the Swamee-Jain equation (8) requires taking the square root o the riction actor. In such operation we keep the negative value o the square root as shown in the ollowing equation: Q =. 5 g h L log 0.7 e ν Q 0.9. WSJQ(15) The reason or using the negative value in equation (15) is that the logarithmic unction has an argument that is smaller than 1, thus producing negative logarithms. Since the discharge in (15) must be a positive quantity, the use o the negative sign in that equation is needed. As with equation (14), equation (15) is implicit in Q and, but explicit in h. Thus, this equation is ideal or solving type 1 problems. Type and 3 problems, however, will require a numerical solution. Equation (15) is reerred to by the name WSJQ, i.e., arcy-weisbach + Swamee-Jain in terms o Q. Solution o sample problems using equations (13) through (15) The solution o sample problems [1] through [3] using equations (13) through (15) can be implemented using the numerical solvers in the HP and TI calculators. In order to 9

10 activate such solvers we need to store the equations into variables. For example, in the HP calculators, within sub-directory PIPES, we can deine variables WCWQ, WHAQ, and WSJQ, which store the ollowing expressions: WCWQ: Q=-.* (g*^5*h/l)*log(0.7*ee/+1.775*nu/* (L/(g**h))) WHAQ: Q=-.0* (g*^5*h/l)*log(0.34*(ee/)^ *nu*/q) WSJQ: Q=-.* (g*^5*h/l)*log(0.7*ee/+4.6*(nu*/q)^0.9) Similar expressions can be stored in variables within TI-89 or TI-9 calculators and solved using the numerical solvers available in those calculators. Notice that in the calculators h represents the head loss h, ee represents the absolute roughness e, and Nu represents the kinematic viscosity ν. Solution with HP calculators The numerical solver in the HP calculators is obtained by using [ ][SOLVE][ OK ] in the HP 48 G series, or [ ][NUM.SLV][ OK ] in the HP 49 G, HP 49G+, and HP 48 GII calculators. Unless an equation is already stored in variable EQ, you will be prompted to enter an equation in the Eq: ield. Type equation one o the equations above (must be between quotes), or load an existing equation, then press [ENTER]. The resulting input orm will include input ields or the variables h, ee,, Q, Nu, L, and g. To solve or any o the unknowns, irst, enter the values o the six known variables, pressing [ OK ] ater each value entered. Then, using the arrow keys, select the ield o the unknown variable, and press [SOLVE]. I the value returned is too large to see directly in the input orm, press the [EIT] sot menu key to see the ull value in the stack. Solution with TI calculators The numerical solver in the TI calculators is obtained by pressing the [APPS] key, and selecting the option Numeric Solver. Type an equation in the eqn: ield, and press [ENTER], or load an existing equation. The numeric solver screen will now show the equation and a list o variable names (h, ee, d, q, nu, l, g) ollowed by equal signs. The last item in the list represents the bounds or the solution, with deault value bound = {- 1.E14,1.E14}. Using the arrow keys move rom ield to ield and enter the values o the six known variables. Change the bounds in the last item in the list i need be (e.g., you may require your solution to be a positive number, say, in the interval [0.0,100.0], thus, you could use bound = {0.0,100.0}). Then, move the cursor to the unknown variable, enter an initial guess or the result, and, while keeping the cursor in that position, press F-Solve. The result will be shown at the cursor position. [NOTE: i, in the process o inding a solution with the TI calculator you get a domain error message, simply change the initial guess o the solution to a larger or smaller value until a solution is produced]. 10

11 Summary o solutions The ollowing table summarizes the solution o the three sample problems using numerical solutions o the equations WCWQ, WHAQ, and WSJQ in the HP calculators: Table 3. Solutions to sample problems with equations (13) through (15) WCWQ(13) WHAQ(14) WSJQ(15) Problem [1] h = 8.86 t h = 8.73 t h = 8.90 t Problem [] Q =.69 cs Q =.70 cs Q =.68 cs Problem [3] = 0.98 t = 0.97 t = 0.99 t 4 - Numerical solution o riction problems using unctions ARCY, HA and SJ We can use the calculators numerical solver to solve riction loss problems when the arcy-weisbach equation is written in terms o unctions ARCY (HP calculators only), HA, and SJ. For example, in the HP calculators, the arcy-weisbach equation given in (1) in terms o velocity V can be entered in any o the ollowing orms i using unctions ARCY, HA, or SJ or the riction actor: h = ARCY(ee/,V*/Nu)*L/*V^/(*g) h = HA(ee/,V*/Nu)*L/*V^/(*g) h = SJ(ee/,V*/Nu)*L/*V^/(*g) WV(16) WHV(17) WSV(18) The names o the variables listed next to the equation numbers above relect the equation being implemented (W means arcy-weisbach), the riction actor used (e.g., means given by the ARCY unction), and V means Velocity. In the TI calculators, we can deine the ollowing variables representing the arcy- Weisbach equation with riction actors given by the Haaland s and Swamee-Jain s equations, respectively: h = ha(ee/d,v*d/nu)*l/d*v^/(*g) h = sj(ee/d,v*d/nu)*l/d*v^/(*g) dwhv(19) dwhv(0) Notice that variable Nu or nu in these equations represents the kinematic viscosity ν. [NOTE: Neither the HP nor the TI calculators include the Greek letter ν in their collection o characters.] Next, we present some examples o numerical solutions o equations (16) through (0): Example 1 Consider the ollowing data: Q = 0.05 m 3 /s, L = 1 km, = 0.0 m, ee = 0.1 mm, g = m/s, and ν = m /s. The velocity is calculated with equation (), V = 4Q/(π ) = /(π 0.0 ) = 1.59 m/s. With the values o V, L,, e, g, and ν, given above, we ind that h = 1.04 m (HP calculator, ARCY) or h = 1.10 m (TI calculator, sj). 11

12 Example Suppose that we use the same data as in Example 1, above, but we now need to ind the diameter required to maintain a velocity o V = 0.75 m/s i the head loss in a km o pipeline is to be 5.5 m. Thus, we keep all the values used above, except or V = 0.75, h = 5.5, and solve or. The result is = m (HP calculator, ARCY) or = m (TI calculator, sj). Example 3 Using the same data as in Example 1, we now determine the velocity in a 0.5-mdiameter, 500-m-long pipe, i the require riction loss is.5 m, i.e., use = 0.5, L = 500, h =.5. The result is V = 1.16 m/s in either the HP calculator (ARCY) or TI calculator (sj). Example 4 Reservoir-pipe system Consider the case o a reservoir whose ree surace is located and an elevation z 1 = 60 m, draining through a pipe open to the atmosphere whose outlet is located at an elevation z = 40 m. The system is depicted in the ollowing igure. Point 1 in the energy equation is at the reservoir ree surace where p 1 = 0 and V 1 = 0. Point, on the other hand, is at the pipe outlet where p = 0 and V = V, the pipe velocity. In order to make the problem as general as possible, we ll write z 1 = z + H, i.e., H = z 1 z = 60 m 40 m = 0 m, or this case. Writing out the energy equation between points 1 and, thus, we ind: p1 V1 p V z = z h, γ g γ g which simpliies to V L V z + H + + = z + + +, γ g γ g g V H = 1+ g e V, ν L. (1) In this equation, we ll use H = 0 m, L = 100 m, e = mm, = 0.5 m, g = m/s, and ν = m /s. The problem requires us to ind the low velocity V. Equation (8) can be solved using the numerical solvers in either the HP or the TI calculators, entering the equation as: 1

13 in the HP calculators, or as in the TI calculators. H = V^/(*g)*(1+ARCY(ee/,V*/nu)*L/) () h = v^/(*g)*(1+sj(ee/d,v*d/nu)*l/d) (3) Using an approach similar to that o the previous examples, we ind that V = m/s (HP calculator) or V = m/s (TI calculator). Solving riction loss problems in terms o low discharge Instead o using the low velocity in the arcy-weisbach equation, we can write out the equations in terms o the discharge as shown in equation (1). The corresponding equations to be entered in the calculators are: h = ARCY(ee/,4*Q/(π*nu*))*8*L*Q^/(π^*g*^5) h = HA(ee/,4*Q/(π*nu*))*8*L*Q^/(π^*g*^5) h = SJ(ee/,4*Q/(π*nu*))*8*L*Q^/(π^*g*^5) WQ(4) WHQ(5) WSQ(6) in the HP calculators, or, in the TI calculators: h = ha(ee/d,4*q/(π*nu*d))*8*l*q^/(π^*g*d^5) h = sj(ee/d,4*q/(π*nu*d))*8*l*q^/(π^*g*d^5) dwhq(7) dwhq(8) Solutions to sample problems [1] through [3] The ollowing table summarizes the solution o the three sample problems using numerical solutions o the equations WQ, WHQ, and WSQ in the HP calculators: Table 3. Solutions to sample problems with equations (13) through (15) WQ(4) WHQ(5) WSQ(6) Problem [1] h = 8.85 t h = 8.73 t h = 8.88 t Problem [] Q =.69 cs Q =.70 cs Q =.68 cs Problem [3] = 0.98 t = 0.97 t = 0.99 t Additional solutions in terms o discharge are shown next. Example 5 Consider the data: Q = 0.05 m 3 /s, L = 1 km, = 0.0 m, ee = 0.1 mm, g = m/s, and ν = m /s. With the equations (14) and (15) given in terms o discharge, there is no need to calculate the velocity. Instead, we program the equations in the corresponding calculators and solve directly or h. The result is h = 1.06 m (HP calculator) or h = 1.1 m (TI calculator). 13

14 Example 6 Suppose that we use the same data as in Example 1, above, but we now need to ind the diameter required to maintain a discharge o Q = 0.10 m/s i the head loss in a km o pipeline is to be 5.5 m. Thus, we keep all the values used above, except or Q = 0.10, h = 5.5, and solve or. The result is = m (HP calculator) or = m (TI calculator). Example 7 Using the same data as in Example 1, we now determine the discharge in a 0.5-mdiameter, 500-m-long pipe, i the require riction loss is.5 m, i.e., use = 0.5, L = 500, h =.5. The result is Q = m 3 /s in either the HP calculator or TI calculator. Example 8 Reservoir-pipe system Consider the low described by Equation (8), presented above. This equation was given in terms o the low velocity. I we re-write it now in terms o the discharge, we ind the ollowing equation: H 8Q = π g 4 1+ e 4Q, πν L. (9) In this equation, we ll use H = 0 m, L = 100 m, e = mm, = 0.5 m, g = m/s, and ν = m /s. The problem requires us now to ind the discharge Q. Equation (16) can be solved using the numerical solvers in either the HP or the TI calculators, entering the equations as: H = 8*Q^/(π^*g*^4)*(1+ARCY(ee/,4*Q/(π*nu*))*L/) (30) in the HP calculators, or as in the TI calculators. h = 8*q^/(π^*g*d^4)*(1+sj(ee/d,4*q/(π*nu*d))*l/d) (31) Using an approach similar to that o the previous examples, we ind that Q =.10 m 3 /s (HP calculator) or Q =.10 m 3 /s (TI calculator). Example 9 Two reservoirs connected by a single pipe case 1 Two reservoirs (A) and (B) are connected by a pipe with L = 1500 t, = 0.50 t, and e = t. The level o reservoir (A) is maintained at an elevation H = 0 t above that o reservoir (B). Take the kinematic viscosity o water to be ν = 1.3x10-5 t /s. Assuming that all minor losses are negligible, except or the discharge loss at the entrance to reservoir (B), calculate the discharge Q. The system is depicted in the ollowing igure. 14

15 Setting up the energy equation between the ree suraces o reservoirs (A) and (B), with p A = p B = 0, V A = V B = 0, and z A = z B + H, and including riction losses and discharge losses, the resulting equation turns out to be the same as equation (9). Entering equation (30) into the HP calculators or equation (31) into the TI calculators, with the given data, we ind Q = 1.0 cs. Example 10 Two reservoirs connected by a single pipe case Suppose that or the system o Example 9 we have Q = 3 cs and we are asked to ind. Using the equation or Example 9 with the same data except or Q, and solving or results in = 0.75 t. Example 11 Two reservoirs connected by a single pipe case 3 Suppose that or the system o Example 9 we have Q = 3 cs and = 0.6 t, and we are asked to determine the length L o the pipe while all other data remain the same as in Example 9. Using the equation or Example 9 with the same data except or Q and, and solving or L results in L = t. Example 1 Two reservoirs connected by a single pipe case 4 Suppose that or the system o Example 9 we have Q =.5 cs, = 0.6 t, and L = 100 t, and we are asked to determine the elevation H o reservoir (A) over that o reservoir (B) while all other data remain the same as in Example 9. Using the equation or Example 9 with the same data except or Q,, and L, and solving or H results in H = t. 5 Empirical equations or single-pipe low For water low in pipes there are a couple o empirical equations that are oten used to calculate low velocities in pipelines. These are the Hazen-Williams equation and the Manning s equation. The Hazen-Williams equation This equation is valid or water in pipes whose diameters are larger than inches and or low velocities less than 10 ps. The Hazen-Williams equation is given, in BG and SI units, as ollows: BG units: V = C HW R h 0.63 S 0.54, with V(ps), R h (t), S (t/t) (3) SI units: V = C HW R h 0.63 S 0.54, with V(m/s), R h (m), S (m/m) (3) 15

16 Where V is the low velocity, R h is the hydraulic radius (or a pipe lowing ull, R h = /4), and S is the energy gradient, S = h /L. The Hazen-Williams coeicient (C HW ) depends on the pipe material. Some typical values are: C HW = 140, C HW = 110, C HW = 90, C HW = 80, smooth straight pipe riveted steel or vitriied pipe old pipes tuberculated pipes The Manning s equation The Manning s equation was developed or open-channel low applications, but can also be used in pipelines. The equation, in BG and SI units, is given as ollows: BG units: SI units: V V R S / 3 1/ = h, with V(ps), R h (t), S (t/t) (3) nm 1 R S / 3 1/ = h, with V(m/s), R h (m), S (m/m) (3) nm The Manning s resistance coeicient (n m ) depends on the pipe material. Some typical values are: n m = 0.008, brass or plastic pipe n m = 0.01, concrete pipes n m = 0.014, drainage tile, vitriied sewer pipe n m = old pipes n m = 0.035, tuberculated cast-iron pipes Non-rigorous head-loss equations Unlike the arcy-weisbach s equation with a riction actor that depends on the relative roughness (e/) and the Reynolds number (R), the Hazen-Williams and Manning s equations use coeicients that are constant. Since the rigorous derivation o the arcy- Weisbach equation demonstrated that such should not be the case or pipe low, the Hazen-Williams and Manning s equations produce non-rigorous, yet practical, head-loss equations. Using either the Hazen-Williams equations the head loss can be written in the general orm: h =K Q n. (33) Sometimes, to avoid the iterative process typically involved in solving the arcy- Weisbach equation, it can be assumed that the low is in the ully-rough regime and that a constant value o applies. Thus, equation (33) would also apply or this non-rigorous arcy-weisbach equation. 16

17 The values o K and n or the arcy-weisbach, Hazen-Williams, and Manning s equations (the two latter in the BG system only) are given by: 8 L arcy-weisbach: K =, n = (34) π 5 g 4.77L Hazen-Williams (BG): K =, n = (35) CHW 4.66 nm L Manning s (BG): K =, n = (36) 16 / 3 Example 13 Two reservoirs connected by a single pipe case 1 Two reservoirs (A) and (B) are connected by a pipe with L = 1500 t, and = 0.50 t. The level o reservoir (A) is maintained at an elevation H = 0 t above that o reservoir (B). Use a Hazen-Williams coeicient C HW = 140 and a Manning s coeicient n m = Assuming that all minor losses are negligible, except or the discharge loss at the entrance to reservoir (B), calculate the discharge Q. The system is depicted in the ollowing igure: Setting up the energy equation between the ree suraces o reservoirs (A) and (B), with p A = p B = 0, V A = V B = 0, and z A = z B + H, and including riction losses and discharge losses, the resulting equation turns out to be V H h = 0. (37) g Replacing the ollowing expression or the velocity head V g = 1 4Q g π = 8Q π g 4 = K V Q, (38) and equation (33) or the riction loss h, equation (37) becomes: H = KQ n + K V Q (37) where K is given by either equation (35) or (36), and K V 8 =. (38) π 4 g 17

18 For this problem we ind K v = and the ollowing values or K and n: 4.77L Hazen-Williams, K HW = C HW 4.66 n Manning s, K m = m L 16/ 3 = 1.98, n HW = = 18.04, n m = Thus, the system equation (37), or the Hazen-Williams ormula is: 0 = 1.98Q Q. (39) A numerical solution with a calculator solver produces the result Q = 0.94 cs. I using the Manning s equation, equation (37) becomes whose solution is Q = 1.04 cs. 0 = 18.04Q Q, (40) 18