M E 320 Professor John M. Cimbala Lecture 27

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1 M E 30 Professor John M. Cimbala Lecture 7 Toay, we will: Do some examples of complex piping networks (multiple pipes with branches, etc.) Briefly mention flow meters an velocity measurement Complex piping networks Summary: For each section of pipe, nee to write a separate equation for Re, f, h L, etc. For pipe sections in series, V 1 = V = 3. For splitting pipe sections (1 splitting into an 3), V 1 = V + 3 Write a separate energy equation from inlet to outlet for each branch of the network. z 1 1 Branch 1 D 1 Tee Branch 3 Valve D 3 Elbow Branch 3 z 3 D D z When all equations are written, solve simultaneously for all unknowns. If you o everything right, there shoul be the same number of equations as unknowns. 4. Examples

Example: Taking a Shower an Flushing a Toilet (E.g. 8-9, Çengel an Cimbala) Given: This is a very practical everyay example of a parallel piping network! You are taking a shower. The piping is 1.

2 Example: Taking a Shower an Flushing a Toilet (E.g. 8-9, Çengel an Cimbala) Given: This is a very practical everyay example of a parallel piping network! You are taking a shower. The piping is 1.50-cm coper pipes with threae connectors as sketche. The gage pressure at the inlet of the system is 00 kpa, an the shower is on. The hot water is from a separate supply only the col water system is shown here. To o: (a) Calculate the volume flow rate through the shower hea when there is no water flowing through the toilet. (b) Calculate the volume flow rate through the shower hea when someone flushes the toilet, an water flows into the toilet reservoir. Solution: (copie from the textbook) This is a simplifying assumption that may or may not be vali. We shoul check the valiity later. We consier only the col water line. The hot water line is separate, an is not connecte to the toilet, so the volume flow rate of hot water through the shower remains constant. The col water, however, is affecte by flushing the toilet.

Alternatively, if V 1 = V, then these two terms cancel each other out.

3 Part (a) is not a parallel system since no flow through the toilet. We neglect the velocity heas in the energy equation. Alternatively, if V 1 = V, then these two terms cancel each other out. P = P atm, an therefore P 1 P = P 1,gage. Only one Re an one Colebrook equation for Part (a)

4 Answer to part (a) no toilet flushing This is where EES comes in hany solving all these simultaneous equations! (1 equations an 1 unknowns) Now we nee three Colebrook equations one for each branch!

5 Answer to part (b) with toilet flushing

6 EES Solution Toilet Flushing Example Problem Part (a) EES Equation winow: Part (a) EES Solution:

7 Part (b) EES Equation Winow: Part (b) EES Solution:

8 Example: Parallel Pipe Network an Pump Bypass System Backgroun: In some applications (e.g., nuclear reactor cooling), it is critical that the volume flow rate of a flui remains constant, regarless of hea changes in the system ownstream (within specifie limits, of course). One metho of ensuring a constant volume flow rate is to install an oversize pump to rive the flow. A bypass line is then installe in parallel with the pump so that some of the flui (bypass volume flow rate b ) recirculates through the bypass line as shown. Base on feeback from a ownstream volume flowmeter, the bypass valve is then ajuste by a computer to control both b an the volume flow rate through the pump p such that the volume flow rate ownstream of the system remains constant. Bypass line (subscript b) b Computer-controlle valve D b D p Pump p 1 Pump line (subscript p) Given: In this particular case, water (ρ = 998 kg/m 3, µ = kg/m s) nees to be supplie at a constant ownstream volume flow rate of = 0.0 m 3 /s. The pump s performance curve is given by the pump manufacturer as hpump,u,supply = a( b c p ) where a = 100 m, b = 1.0, an c = 1.0 s /m 6. The units of h pump,u,supply are [m] an the units of p are [m 3 /s]. The pump line has a iameter D p = 0.50 m an length L p = 3.0 m, while the bypass line has a iameter D b = 0.50 m an length L b = 5.0 m. All pipes have roughness ε = 0.00 m. Minor losses in the system inclue two flange 90 o bens (K L = 0.0 each) an a gate valve (0. < K L < ) in the bypass line, an two flange tees (line flow K L = 0.0, an branch flow K L = 1.00). To Do: Calculate an plot how volume flow rates V p, V b, an vary with the minor loss coefficient of the valve as it goes from fully open (K L,valve = 0.0) to fully close (K L,valve ). Solution: First, as always, we nee to pick a control volume. In this case, we nee two control volumes since there are two branches in the parallel pipe system. After careful thought (an experience), we ecie that the most appropriate control volumes go between points (1) an () as labele on the above sketch. We re-raw the flow system

9 incluing the two control volumes. In this parallel pipe system it is useful to imagine that the flui in the bypass line (CV b ) is colore re, while that in the pump line (CV p ) is colore blue. This is an artificial separation of the flui into the two branches since the flui mixes because of turbulence. However, it is useful as an ai to rawing the control volumes. Note that CV p has its inlet at (1) an its outlet at (), while CV b has its inlet at () an its outlet at (1). CV b b p 1 CV p We apply conservation of mass at either tee: V p = V b + Note: This equation couples the two control volumes together. Other than this, we treat the two control volume separately in the analysis below. We apply the hea form of the energy equation for CV p from inlet (1) to outlet (), assuming that the flow at both (1) an () is fully evelope turbulent pipe flow so that both velocity an kinetic energy correction factor are the same at (1) as at (): P1 V1 α ρ g + + z 1 1 g z 1 = z P V + hpump,u = + α z ρg g + + hturbine,e + h L, p V 1 = V an α 1 = α We call this h pump,u,system since it is the require pump hea for the given piping system. P P1 Or, hpump,u,system = + hl, p. ρg We next apply the hea form of the energy equation for CV b from inlet () to outlet (1), recognizing again that V 1 = V an α 1 = α, an that there is no pump in this CV: P V α ρ g + z g + + hpump,u P1 V1 = + α z ρg 1 1 g + + hturbine,e + h L, b P P1 Or, hlb, =. ρg Combining the above two results, we get hpump,u,system = hl, b + hl, p. Note: P P 1 is the same regarless of whether we are consiering the pump line or the branch line.

10 We note that since the velocities (an therefore the Reynols numbers) in the pump line an the bypass line are not the same, we therefore nee two Colebrook or Churchill equations to solve the problem, one for the pump line an one for the bypass line. We write all the equations, an solve them simultaneously [I use EES]. Here is what I type into the main Equations Winow of EES:

I create a parametric table, an selecte four of the variables (Table-New Parametric Table): I specifie the values of K L,valve, ranging from the minimum of 0.

Finally, I plotte all three of the epenent volume flow rates as functions of K L,valve : Verify: remains constant regarless of the valve setting. V = V p b = constant = 0.

11 I create a parametric table, an selecte four of the variables (Table-New Parametric Table): I specifie the values of K L,valve, ranging from the minimum of 0. (fully open gate valve) to a very large number (valve nearly close). The rest of the values are automatically calculate an fille in when you click on the green arrow at the upper left. Finally, I plotte all three of the epenent volume flow rates as functions of K L,valve : Verify: remains constant regarless of the valve setting. V = V p b = constant = 0.0 m 3 /s at any value of K L valve. When the valve is fully open, we get the maximum flow rate through both the pump an bypass lines. When the valve is nearly close, there is harly any flow through the bypass line nearly all the flow is through the pump line.

M E 320 Professor John M. Cimbala Lecture 24

M E 320 Professor John M. Cimbala Lecture 24 M E 30 Professor John M. Cimbala Lecture 4 Today, we will: Discuss pump performance curves Discuss how to match a pump and a piping system, and do some example problems. Pump Performance a. Pump performance