Chapter 8 Flow in Pipes. Piping Systems and Pump Selection

Transcription

1 Piping Systems and Pump Selection 8-6C For a piping system that involves two pipes o dierent diameters (but o identical length, material, and roughness connected in series, (a the low rate through both pipes is the same and (b the pressure drop across smaller diameter pipe is larger. 8-6C For a piping system that involves two pipes o dierent diameters (but o identical length, material, and roughness connected in parallel, (a the low rate through the larger diameter pipe is larger and (b the pressure drop through both pipes is the same. 8-6C The pressure drop through both pipes is the same since the pressure at a point has a single value, and the inlet and exits o these the pipes connected in parallel coincide. 8-65C Yes, when the head loss is negligible, the required pump head is equal to the elevation dierence between the ree suraces o the two reservoirs. 8-66C The pump installed in a piping system will operate at the point where the system curve and the characteristic curve intersect. This point o intersection is called the operating point. 8-67C The plot o the head loss versus the low rate is called the system curve. The experimentally determined pump head and pump eiciency versus the low rate curves are called characteristic curves. The pump installed in a piping system will operate at Head the point where the system curve and the characteristic curve intersect. This point o intersection is called the operating point. h pump η pump Operating point System demand curve Flow rate 8- PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution

2 8-68 The pumping power input to a piping system with two parallel pipes between two reservoirs is given. The low rates are to be determined. ssumptions The low is steady and incompressible. The entrance eects are negligible, and thus the low is ully developed. The elevations o the reservoirs remain constant. The minor losses and the head loss in pipes other than the parallel pipes are said to be negligible. 5 The lows through both pipes are turbulent (to be veriied. Properties The density and dynamic viscosity o water at 0 C are ρ 998 kg/m and µ kg/m s. Plastic pipes are smooth, and their roughness is zero, ε 0. nalysis This problem cannot be solved directly since the velocities (or low rates in the pipes are not known. Thereore, we would normally use a trial-and-error approach here. However, nowadays the equation solvers such as EES are widely available, and thus below we will simply set up the equations to be solved by an equation solver. The head supplied by the pump to the luid is determined rom ghpump, u (998 kg/m (9.8m/s hpump, u W & ρ & & elect,in 7000 W ( η 0.68 pump-motor We choose points and B at the ree suraces o the two reservoirs. Noting that the luid at both points is open to the atmosphere (and thus P P B P atm and that the luid velocities at both points are zero ( B 0, the energy equation or a control volume between these two points simpliies to or where P + α + z g + h pump, u h u (9 + h pump, ( h h h,, PB + α B ( ( B + z g We designate the -cm diameter pipe by and the 5-cm diameter pipe by. The average velocity, Reynolds number, riction actor, and the head loss in each pipe are expressed as & & & c, π / π (0.0m & & & π / π (0.05m c, B + h / / turbine, e (5 (6 + h Reservoir z m cm Pump h pump, u 5 m ( z B z 5 cm + h Reservoir B z B m Re Re ρ µ ρ µ Re Re (998 kg/m (0.0 m.00 0 kg/m s /.0 log ε.7 /.0 log ε.7 (998 kg/m (0.05 m.00 0 kg/m s.5 + Re.5 + Re (7 (8.5.0 log0 + Re.5.0 log0 + Re (9 (0 h 5 m, h, g 0.0 m (9.8m/s ( 8- PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution

3 h 5 m, h, g 0.05 m (9.8m/s & & + & ( ( This is a system o equations in unknowns, and their simultaneous solution by an equation solver gives & 0.08 m /s, & m /s, & 0.06 m /s, 5.0 m/s, 7. m/s, h h, h, 9.5 m, h pump,u 6.5 m Re 58,00, Re 69,700, 0.06, 0.09 Note that Re > 000 or both pipes, and thus the assumption o turbulent low is veriied. iscussion This problem can also be solved by using an iterative approach, but it will be very time consuming. Equation solvers such as EES are invaluable or this kind o problems. 8- PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution

4 8-69E The low rate through a piping system connecting two reservoirs is given. The elevation o the source is to be determined. ssumptions The low is steady and incompressible. The entrance eects are negligible, and thus the low is ully developed. The elevations o the reservoirs remain constant. There are no pumps or turbines in the piping system. Properties The density and dynamic viscosity o water at 70 F are ρ 6.0 lbm/t and µ.60 lbm/t h lbm/t s. The roughness o cast iron pipe is ε t. nalysis The piping system involves 0 t o -in diameter piping, a well-rounded entrance (K 0.0, standard langed elbows (K 0. each, a ully open gate valve (K 0., and a sharp-edged exit (K.0. We choose points and at the ree suraces o the two reservoirs. Noting that the luid at both points is open to the atmosphere (and thus P P P atm, the luid velocities at both points are zero ( 0, the ree surace o the lower reservoir is the reerence level (z 0, and that there is no pump or turbine (h pump,u h turbine 0, the energy equation or a control volume between these two points simpliies to P + α + z + hpump, u + α + z + hturbine, e + h z h P g g where h h, total h,major + h,minor + K g since the diameter o the piping system is constant. The average velocity in the pipe and the Reynolds number are & & 0/60 t /s 7.6 t/s π / π ( / t / c ρ (6. lbm/t (7.6 t/s(/ t Re 60,700 µ.07 0 lbm/t s in which is greater than 000. Thereore, the low is turbulent. The relative roughness o the pipe is 0 t /min t ε / / t The riction actor can be determined rom the Moody chart, but to avoid the reading error, we determine it rom the Colebrook equation using an equation solver (or an iterative scheme, ε / log +.0 log +.7 Re.7 60, 700 It gives The sum o the loss coeicients is K K, entrance + K,elbow + K,valve + K, exit Then the total head loss and the elevation o the source become 0 t (7.6 t/s h + K ( g / t (. t/s.t 0 t z z h.t Thereore, the ree surace o the irst reservoir must be. t above the ree surace o the lower reservoir to ensure water low between the two reservoirs at the speciied rate. iscussion Note that /.0 in this case, which is almost 0 olds o the total minor loss coeicient. Thereore, ignoring the sources o minor losses in this case would result in an error o about 0%. 8- PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution

5 8-70 water tank open to the atmosphere is initially illed with water. sharp-edged oriice at the bottom drains to the atmosphere. The initial velocity rom the tank and the time required to empty the tank are to be determined. ssumptions The low is uniorm and incompressible. The low is turbulent so that the tabulated value o the loss coeicient can be used. The eect o the kinetic energy correction actor is negligible, α. Properties The loss coeicient is K 0.5 or a sharp-edged entrance. nalysis (a We take point at the ree surace o the tank, and point at the exit o the oriice. We also take the reerence level at the centerline o the oriice (z 0, and take the positive direction o z to be upwards. Noting that the luid at both points is open to the atmosphere (and thus P P P atm and that the luid velocity at the ree surace is very low ( 0, the energy equation or a control volume between these two points (in terms o heads simpliies to P P + α + z + hpump, u + α + z + hturbine, e + h z α + g g g where the head loss is expressed as z h h K. Substituting and solving or gives g gz α + K gz ( α + K g g α + K where α. Noting that initially z m, the initial velocity is determined to be gz + K (9.8 m/s ( m 5.m/s The average discharge velocity through the oriice at any given time, in general, can be expressed as Water tank m m 0 cm gz + K where z is the water height relative to the center o the oriice at that time. (b We denote the diameter o the oriice by, and the diameter o the tank by 0. The low rate o water rom the tank can be obtained by multiplying the discharge velocity by the oriice area, & oriice π gz + K Then the amount o water that lows through the oriice during a dierential time interval dt is π gz d & dt dt ( + K which, rom conservation o mass, must be equal to the decrease in the volume o water in the tank, π0 d tank ( dz dz ( where dz is the change in the water level in the tank during dt. (Note that dz is a negative quantity since the positive direction o z is upwards. Thereore, we used dz to get a positive quantity or the amount o water discharged. Setting Eqs. ( and ( equal to each other and rearranging, π gz + K π dt 0 dz dt 0 + K gz dz dt 0 + K g z / dz 8-5 PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution

6 The last relation can be integrated easily since the variables are separated. etting t be the discharge time and integrating it rom t 0 when z z to t t when z 0 (completely drained tank gives 0 + t K / 0 + K z 0 + dt z dz t - 0 t g z z g + z Simpliying and substituting the values given, the draining time is determined to be K g z / t 0 z ( + K g ( m (0.m ( m( m/s 70 s.7 min iscussion The eect o the loss coeicient K on the draining time can be assessed by setting it equal to zero in the draining time relation. It gives t 0 z ( m ( m, zero loss 575 s 9.6 min g (0.m 9.8m/s Note that the loss coeicient causes the draining time o the tank to increase by (.7-9.6/ or 8%, which is quite signiicant. Thereore, the loss coeicient should always be considered in draining processes. 8-6 PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution

7 8-7 water tank open to the atmosphere is initially illed with water. sharp-edged oriice at the bottom drains to the atmosphere through a long pipe. The initial velocity rom the tank and the time required to empty the tank are to be determined. ssumptions The low is uniorm and incompressible. The draining pipe is horizontal. The low is turbulent so that the tabulated value o the loss coeicient can be used. The riction actor remains constant (in reality, it changes since the low velocity and thus the Reynolds number changes. 5 The eect o the kinetic energy correction actor is negligible, α. Properties The loss coeicient is K 0.5 or a sharp-edged entrance. The riction actor o the pipe is given to be nalysis (a We take point at the ree surace o the tank, and point at the exit o the pipe. We take the reerence level at the centerline o the pipe (z 0, and take the positive direction o z to be upwards. Noting that the luid at both points is open to the atmosphere (and thus P P P atm and that the luid velocity at the ree surace is very low ( 0, the energy equation or a control volume between these two points (in terms o heads simpliies to where P P + α + z + hpump, u + α + z + hturbine, e + h z α + g g g h h, total h,major + h,minor + K + K g g h since the diameter o the piping system is constant. Substituting and solving or gives z α + + K g g α gz + / + K where α. Noting that initially z m, the initial velocity is determined to be, i gz + / + K (9.8m/s ( m.5 m/s (00 m/(0.m The average discharge velocity at any given time, in general, can be expressed as gz + / + K Water tank m m 0 cm 00 m where z is the water height relative to the center o the oriice at that time. (b We denote the diameter o the pipe by, and the diameter o the tank by o. The low rate o water rom the tank can be obtained by multiplying the discharge velocity by the pipe cross-sectional area, & pipe π gz + / + K Then the amount o water that lows through the pipe during a dierential time interval dt is π gz d & dt dt ( + / + K which, rom conservation o mass, must be equal to the decrease in the volume o water in the tank, π0 d tan k ( dz dz ( where dz is the change in the water level in the tank during dt. (Note that dz is a negative quantity since the positive direction o z is upwards. Thereore, we used dz to get a positive quantity or the amount o water discharged. Setting Eqs. ( and ( equal to each other and rearranging, 8-7 PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution

8 gz π0 0 + / + K 0 dt dz dt dz + / + K gz π Chapter 8 Flow in Pipes + / + K The last relation can be integrated easily since the variables are separated. etting t be the discharge time and integrating it rom t 0 when z z to t t when z 0 (completely drained tank gives 0 t / + K / 0 + / + K z 0 + / + dt z dz t - 0 t g z z g g z Simpliying and substituting the values given, the draining time is determined to be g z dz K z t 0 z ( + / + K g ( m (0. m ( m[+ (0.05(00 m/(0.m + 0.5] 9.8 m/s s 8.9 min iscussion It can be shown by setting 0 that the draining time without the pipe is only.7 min. Thereore, the pipe in this case increases the draining time by more than olds. 8-8 PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution

9 8-7 water tank open to the atmosphere is initially illed with water. sharp-edged oriice at the bottom drains to the atmosphere through a long pipe equipped with a pump. For a speciied initial velocity, the required useul pumping power and the time required to empty the tank are to be determined. ssumptions The low is uniorm and incompressible. The draining pipe is horizontal. The low is turbulent so that the tabulated value o the loss coeicient can be used. The riction actor remains constant. 5 The eect o the kinetic energy correction actor is negligible, α. Properties The loss coeicient is K 0.5 or a sharp-edged entrance. The riction actor o the pipe is given to be The density o water at 0 C is ρ 996 kg/m. nalysis (a We take point at the ree surace o the tank, and point at the exit o the pipe. We take the reerence level at the centerline o the oriice (z 0, and take the positive direction o z to be upwards. Noting that the luid at both points is open to the atmosphere (and thus P P P atm and that the luid velocity at the ree surace is very low ( 0, the energy equation or a control volume between these two points (in terms o heads simpliies to P P + α + z + hpump, u + α + z + hturbine, e + h z + hpump, u α + g g g where α and h h, total h,major + h,minor + K + K g g h since the diameter o the piping system is constant. Substituting and noting that the initial discharge velocity is m/s, the required useul pumping head and power are determined to be ρ( π / (996 kg/m [ (0.m /]( m/s. kg/m m& ρ π c 00 m ( m/s pump, u + + K z + ( ( m.6 m g 0.m (9.8m/s h kn kw W & pump, u & P m& ghpump, u (. kg/s(9.8 m/s (.6 m.5 kw 000 kg m/s kn m/s Thereore, the pump must supply.5 kw o mechanical energy to water. Note that the shat power o the pump must be greater than this to account or the pump ineiciency. (b When the discharge velocity remains constant, the low rate o water becomes & c ( π / [ π (0.m /]( m/s 0.0 m The volume o water in the tank is tank 0 z ( π / z [ π ( m /]( m. m Then the discharge time becomes /s Water tank m 0 m 0 cm. m t 50 s 7.5 min & Pump 0.0 m /s iscussion Note that the pump reduces the discharging time rom 8.9 min to 7.5 min. The assumption o constant discharge velocity can be justiied on the basis o the pump head being much larger than the elevation head (thereore, the pump will dominate the discharging process. The answer obtained assumes that the elevation head remains constant at m (rather than decreasing to zero eventually, and thus it under predicts the actual discharge time. By an exact analysis, it can be shown that when the eect o the decrease in elevation is considered, the discharge time becomes 68 s 7.8 min. This is demonstrated below. The required pump head (o water is.6 m, which is more than 0. m o water column which corresponds to the atmospheric pressure at sea level. I the pump exit is at atm, then the absolute pressure at pump inlet must be negative ( -.6 m or. kpa, which is impossible. Thereore, the system 8-9 PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution 00 m m/s

10 cannot work i the pump is installed near the pipe exit, and cavitation will occur long beore the pipe exit where the pressure drops to. kpa and thus the pump must be installed close to the pipe entrance. detailed analysis is given below. emonstration or Prob. 8-7 (extra (the eect o drop in water level on discharge time Noting that the water height z in the tank is variable, the average discharge velocity through the pipe at any given time, in general, can be expressed as h pump, u + + K z g g( z + h pump, u + / + K where z is the water height relative to the center o the oriice at that time. We denote the diameter o the pipe by, and the diameter o the tank by 0. The low rate o water rom the tank can be obtained by multiplying the discharge velocity by the cross-sectional area o the pipe, & pipe π g( z + h pump, u + / + K Then the amount o water that lows through the oriice during a dierential time interval dt is π g( z + hpump, u d & dt dt ( + / + K which, rom conservation o mass, must be equal to the decrease in the volume o water in the tank, π0 d tan k ( dz dz ( where dz is the change in the water level in the tank during dt. (Note that dz is a negative quantity since the positive direction o z is upwards. Thereore, we used dz to get a positive quantity or the amount o water discharged. Setting Eqs. ( and ( equal to each other and rearranging, π g( z + hpump,u π dt + / + K 0 dz dt 0 + / + K g ( z + h pump,u dz The last relation can be integrated easily since the variables are separated. etting t be the discharge time and integrating it rom t 0 when z z to t t when z 0 (completely drained tank gives t + / + K 0 dt z g Perorming the integration gives t 0 - t 0 z + / + K g ( z + h pump 0 z 0 0 ( z + h pump, u / Substituting the values given, the draining time is determined to be dz ( z hpump( + / + K hpump( / K g g t ( m ( +.6 m[ / ] (.6 m[ / ] (0.m 9.8m/s 9.8m/s 68 s 7.8 min 8-0 PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution

11 emonstration or Prob. 8-7 (on cavitation We take the pump as the control volume, with point at the inlet and point at the exit. We assume the pump inlet and outlet diameters to be the same and the elevation dierence between the pump inlet and the exit to be negligible. Then we have z z and. The pump is located near the pipe exit, and thus the pump exit pressure is equal to the pressure at the pipe exit, which is the atmospheric pressure, P P atm. lso, the can take h 0 since the rictional eects and loses in the pump are accounted or by the pump eiciency. Then the energy equation or the pump (in terms o heads reduces to P + α + z + h g Solving or P and substituting, P P h,abs atm pump, u pump, u P + α + z g + h turbine, e + h P,abs + h pump, u Patm kn kpa (0. kpa (996 kg/m (9.8m/s (.6 m -0.7 kpa 000 kg m/s kn/m which is impossible (absolute pressure cannot be negative. The technical answer to the question is that cavitation will occur since the pressure drops below the vapor pressure o.6 kpa. The practical answer is that the question is invalid (void since the system will not work anyway. Thereore, we conclude that the pump must be located near the beginning, not the end o the pipe. Note that when doing a cavitation analysis, we must work with the absolute pressures. (I the system were installed as indicated, a water velocity o m/s could not be established regardless o how much pump power were applied. This is because the atmospheric air and water elevation heads alone are not suicient to drive such low, with the pump restoring pressure ater the low. To determine the urthest distance rom the tank the pump can be located without allowing cavitation, we assume the pump is located at a distance * rom the exit, and choose the pump and the discharge portion o the pipe (rom the pump to the exit as the system, and write the energy equation. The energy equation this time will be as above, except that h (the pipe losses must be considered and the pressure at (pipe inlet is the cavitation pressure, P.6 kpa: P + α + z g + h pump, u P + α + z g + h turbine, e + h P,abs atm + hpump, u + P * g * P,abs Patm or + hpump, u g Substituting the given values and solving or * gives * ( m/s (.6 0. kn/m 000 kg m/s ( (.6 m *.5 m 0.m (9.8 m/s (996 kg/m (9.8m/s kn Thereore, the pump must be at least.5 m rom the pipe exit to avoid cavitation at the pump inlet (this is where the lowest pressure occurs in the piping system, and where the cavitation is most likely to occur. Cavitation onset places an upper limit to the length o the pipe on the suction side. pipe slightly longer would become vapor bound, and the pump could not pull the suction necessary to sustain the low. Even i the pipe on the suction side were slightly shorter than m, cavitation can still occur in the pump since the liquid in the pump is usually accelerated at the expense o pressure, and cavitation in the pump could erode and destroy the pump. lso, over time, scale and other buildup inside the pipe can and will increase the pipe roughness, increasing the riction actor, and thereore the losses. Buildup also decreases the pipe diameter, which increases pressure drop. Thereore, low conditions and system perormance may change (generally decrease as the system ages. new system that marginally misses cavitation may degrade to where cavitation becomes a problem. Proper design avoids these problems, or where cavitation cannot be avoided or some reason, it can at least be anticipated. 8- PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution

12 8-7 Oil is lowing through a vertical glass unnel which is always maintained ull. The low rate o oil through the unnel and the unnel eectiveness are to be determined. ssumptions The low is steady and incompressible. The entrance eects are negligible, and thus the low is ully developed (to be veriied. The rictional loses in the cylindrical reservoir are negligible since its diameter is very large and thus the oil velocity is very low. Properties The density and viscosity o oil at 0 C are ρ 888. kg/m and µ 0.87 kg/m s. nalysis We take point at the ree surace o the oil in the cylindrical reservoir, and point at the exit o the unnel pipe which is also taken as the reerence level (z 0. The luid at both points is open to the atmosphere (and thus P P P atm and that the luid velocity at the ree surace is negligible ( 0. For the ideal case o rictionless low, the exit velocity is determined rom the Bernoulli equation to be P P + + z + + z g g Substituting,,max,max gz (9.8m/s (0.0 m.80m/s This is the low velocity or the rictionless case, and thus it is the maximum low velocity. Then the maximum low rate and the Reynolds number become & Re max,max,max ( π (.80m/s[ π (0.0m / / ].0 0 ρ (888. kg/m (.80m/s(0.0m 9.7 µ 0.87 kg/m s m /s gz Oil cm 5 cm 5 cm which is less than 00. Thereore, the low is laminar, as postulated. (Note that in the actual case the velocity and thus the Reynolds number will be even smaller, veriying the low is always laminar. The entry length in this case is h 0.05 Re (0.0m 0.05 m which is much less than the 0.5 m pipe length. Thereore, the entrance eects can be neglected as postulated. Noting that the low through the pipe is laminar and can be assumed to be ully developed, the low rate can be determined rom the appropriate relation with θ -90 since the low is downwards in the vertical direction, ( P sinθ π & 8µ where P P P ( P + ρ gh P h is the pressure dierence across the pipe inlet pipe exit atm cylinder atm cylinder pipe, h pipe, and sin θ sin ( Substituting, the low rate is determined to be ( h + cylinder hpipe π (888. kg/m (9.8m/s ( m π (0.0 m & 8µ 8(0.87 kg/m s(0.5 m m /s Then the unnel eectiveness becomes 6 &.09 0 m /s E & max.0 0 m /s or.86% iscussion Note that the low is driven by gravity alone, and the actual low rate is a small raction o the low rate that would have occurred i the low were rictionless. 8- PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution

13 8-7 Oil is lowing through a vertical glass unnel which is always maintained ull. The low rate o oil through the unnel and the unnel eectiveness are to be determined. ssumptions The low is steady and incompressible. The entrance eects are negligible, and thus the low is ully developed (to be veriied. The rictional loses in the cylindrical reservoir are negligible since its diameter is very large and thus the oil velocity is very low. Properties The density and viscosity o oil at 0 C are ρ 888. kg/m and µ 0.87 kg/m s. nalysis We take point at the ree surace o the oil in the cylindrical reservoir, and point at the exit o the unnel pipe, which is also taken as the reerence level (z 0. The luid at both points is open to the atmosphere (and thus P P P atm and that the luid velocity at the ree surace is negligible ( 0. For the ideal case o rictionless low, the exit velocity is determined rom the Bernoulli equation to be P P + + z + + z g g,max gz (a Case : Pipe length remains constant at 5 cm, but the pipe diameter is doubled to cm: Substitution gives,max gz (9.8m/s (0.0 m.80m/s This is the low velocity or the rictionless case, and thus it is the maximum low velocity. Then the maximum low rate and the Reynolds number become & max Re m /s ( π / (.80m/s[ π (0.0 m,max,max ρ (888. kg/m (.80m/s(0.0 m 59. µ 0.87 kg/m s / ] Oil cm 5 cm 5 cm which is less than 00. Thereore, the low is laminar. (Note that in the actual case the velocity and thus the Reynolds number will be even smaller, veriying the low is always laminar. The entry length is h 0.05 Re (0.0 m m which is considerably less than the 0.5 m pipe length. Thereore, the entrance eects can be neglected (with reservation. Noting that the low through the pipe is laminar and can be assumed to be ully developed, the low rate can be determined rom the appropriate relation with θ -90 since the low is downwards in the vertical direction, ( P sinθ π & 8µ where P P P ( P + ρ gh P h is the pressure dierence across the pipe inlet pipe exit atm cylinder atm cylinder pipe, h pipe, and sin θ sin ( Substituting, the low rate is determined to be ( h + cylinder hpipe π (888. kg/m (9.8m/s ( m π (0.0 m & 8µ 8(0.87 kg/m s(0.5 m m /s Then the unnel eectiveness becomes & E & max m /s 0.07 m /s or 7.% 8- PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution

14 (b Case : Pipe diameter remains constant at cm, but the pipe length is doubled to 50 cm: Substitution gives,max gz (9.8m/s (0.65 m.57m/s This is the low velocity or the rictionless case, and thus it is the maximum low velocity. Then the maximum low rate and the Reynolds number become & max ( π / (.57m/s[ π (0.0m,max Re m / s,max ρ (888. kg/m (.57m/s(0.0m 7.87 µ 0.87 kg/m s / ] Oil cm 5 cm 50 cm which is less than 00. Thereore, the low is laminar. (Note that in the actual case the velocity and thus the Reynolds number will be even smaller, veriying the low is always laminar. The entry length is h 0.05 Re (0.0m 0.09 m which is much less than the 0.50 m pipe length. Thereore, the entrance eects can be neglected. Noting that the low through the pipe is laminar and can be assumed to be ully developed, the low rate can be determined rom the appropriate relation with θ -90 since the low is downwards in the vertical direction, ( P sinθ π & 8µ where P P P ( P + ρ gh P h is the pressure dierence across the pipe inlet pipe exit atm cylinder atm cylinder pipe, h pipe, and sin θ sin ( Substituting, the low rate is determined to be ( h + cylinder hpipe π (888. kg/m (9.8m/s ( m π (0.0 m & 8µ 8(0.87 kg/m s(0.50 m. 0-6 m /s Then the unnel eectiveness becomes & E & max m /s 0.08 m /s or.8% iscussion Note that the unnel eectiveness increases as the pipe diameter is increased, and decreases as the pipe length is increased. This is because the rictional losses are proportional to the length but inversely proportional to the diameter o the low sections. 8- PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution

15 8-75 Water is drained rom a large reservoir through two pipes connected in series. The discharge rate o water rom the reservoir is to be determined. ssumptions The low is steady and incompressible. The pipes are horizontal. The entrance eects are negligible, and thus the low is ully developed. The low is turbulent so that the tabulated value o the loss coeicients can be used. 5 The pipes involve no components such as bends, valves, and other connectors. 6 The piping section involves no work devices such as pumps and turbines. 7 The reservoir is open to the atmosphere so that the pressure is atmospheric pressure at the ree surace. 8 The water level in the reservoir remains constant. 9 The eect o the kinetic energy correction actor is negligible, α. Properties The density and dynamic viscosity o water at 5 C are ρ 999. kg/m and µ kg/m s, respectively. The loss coeicient is K 0.5 or a sharp-edged entrance, and it is 0.6 or the sudden contraction, corresponding to d / / The pipes are made o plastic and thus they are smooth, ε 0. nalysis We take point at the ree surace o the reservoir, and point at the exit o the pipe, which is also taken to be the reerence level (z 0. Noting that the luid at both points is open to the atmosphere (and thus P P P atm, the luid level in the reservoir is constant ( 0, and that there are no work devices such as pumps and turbines, the energy equation or a control volume between these two points (in terms o heads simpliies to P P + α + z + hpump, u + α + z + hturbine, e + h z α + g g g where α. Substituting, 8 m + h ( (9.8m/s where h h, total h,major + h,minor + K g h Water tank 8 m 0 m 5 m Note that the diameters o the two pipes, and thus the low velocities through them are dierent. enoting the irst pipe by and the second pipe by, and using conservation o mass, the velocity in the irst pipe can be expressed in terms o as ( cm m & ρ ρ 0. 6 (0 cm &m ( Then the head loss can be expressed as h K,entrance K,contraction g g or 0 m 5 m h ( 0.0 m (9.8m/s 0.0 m (9.8m/s The low rate, the Reynolds number, and the riction actor are expressed as & ( π / & [ π (0.0 m Re Re ρ µ ρ µ Re Re (999. kg/m (0.0 m.8 0 kg/m s /.0 log ε.7 (999. kg/m (0.0 m.8 0 kg/m s.5 + Re / ] (5 (6 (.5.0 log0 + Re 8-5 PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution (7

16 /.0 log ε Re.5.0 log0 + Re Chapter 8 Flow in Pipes This is a system o 8 equations in 8 unknowns, and their simultaneous solution by an equation solver gives & m /s, m/s,.7 m/s, h h + h m, Re 66,500, Re 66,00, 0.096, 0.06 Note that Re > 000 or both pipes, and thus the assumption o turbulent low is valid. iscussion This problem can also be solved by using an iterative approach by assuming an exit velocity, but it will be very time consuming. Equation solvers such as EES are invaluable or this kind o problems. (8 8-6 PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution

17 8-76E The low rate through a piping system between a river and a storage tank is given. The power input to the pump is to be determined. ssumptions The low is steady and incompressible. The entrance eects are negligible, and thus the low is ully developed. The low is turbulent so that the tabulated value o the loss coeicients can be used (to be veriied. The elevation dierence between the ree suraces o the tank and the river remains constant. 5 The eect o the kinetic energy correction actor is negligible, α. Properties The density and dynamic viscosity o water at 70 F are ρ 6.0 lbm/t and µ.60 lbm/t h lbm/t s. The roughness o galvanized iron pipe is ε t. nalysis The piping system involves 5 t o 5-in diameter piping, an entrance with negligible loses, standard langed 90 smooth elbows (K 0. each, and a sharp-edged exit (K.0. We choose points and at the ree suraces o the river and the tank, respectively. We note that the luid at both points is open to the atmosphere (and thus P P P atm, and the luid velocity is 6 t/s at point and zero at point ( 6 t/s and 0. We take the ree surace o the river as the reerence level (z 0. Then the energy equation or a control volume between these two points simpliies to P + α + z + hpump, u + α + z + hturbine, e + h α + hpump, u z + h P g g where α and h h, total h,major + h,minor + K g since the diameter o the piping system is constant. The average velocity in the pipe and the Reynolds number are & c & π / /.0 t/s ρ (6. lbm/t (.0 t/s(5/ t Re 5,500 µ lbm/t s.5 t /s π (5 / t River 5 t 5 in.5 t /s g t Water tank which is greater than 000. Thereore, the low is turbulent. The relative roughness o the pipe is t ε / / t The riction actor can be determined rom the Moody chart, but to avoid the reading error, we determine it rom the Colebrook equation using an equation solver (or an iterative scheme, ε / log +.0 log +.7 Re.7 5, 500 It gives 0.0. The sum o the loss coeicients is K K, entrance + K,elbow + K, exit Then the total head loss becomes 5 t (.0 t/s h + K ( t g 5/ t (. t/s The useul pump head input and the required power input to the pump are hpump, u z + h g (6 t/s (. t/s 6.9 t 8-7 PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution

18 W& pump W & η pump, u pump & h η pump, u pump (.5 t /s(6.0 lbm/t (. t/s (6.9 t lb lbm t/s.87 kw Thereore,.87 kw o electric power must be supplied to the pump. Chapter 8 Flow in Pipes kw 77 lb t/s iscussion The riction actor could also be determined easily rom the explicit Haaland relation. It would give 0.0, which is identical to the calculated value. The riction coeicient would drop to 0.05 i smooth pipes were used. Note that / 6. in this case, which is about times the total minor loss coeicient o.9. Thereore, the rictional losses in the pipe dominate the minor losses, but the minor losses are still signiicant. 8-8 PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution

19 8-7 In Prob. 8-76E, the eect o the pipe diameter on pumping power or the same constant low rate is to be investigated by varying the pipe diameter rom in to 0 in in increments o in. g. 5 inch/ z rho6.0 numu/rho mu e0.70 Re*/nu pi*(^/ dot/ dot.5 6 eps reps/ /sqrt(-*log0(r/.7+.5/(re*sqrt( K.9 H(*(/+K*(^/(*g hpumpz+h-^/(*. Wpump(dot*rho*hpump/e/77, in W pump, kw, t/s Re E E E E+05.56E+05.60E+05.E+05.7E+05.0E+05.78E Wpump, kw , in 8-9 PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution

20 8-78 solar heated water tank is to be used or showers using gravity driven low. For a speciied low rate, the elevation o the water level in the tank relative to showerhead is to be determined. ssumptions The low is steady and incompressible. The entrance eects are negligible, and thus the low is ully developed. The low is turbulent so that the tabulated value o the loss coeicients can be used (to be veriied. The elevation dierence between the ree surace o water in the tank and the shower head remains constant. 5 There are no pumps or turbines in the piping system. 6 The losses at the entrance and at the showerhead are said to be negligible. 7 The water tank is open to the atmosphere. 8 The eect o the kinetic energy correction actor is negligible, α. Properties The density and dynamic viscosity o water at 0 C are ρ 99. kg/m and µ kg/m s, respectively. The loss coeicient is K 0.5 or a sharp-edged entrance. The roughness o galvanized iron pipe is ε m. nalysis The piping system involves 0 m o.5-cm diameter piping, an entrance with negligible loss, miter bends (90 without vanes (K. each, and a wide open globe valve (K 0. We choose point at the ree surace o water in the tank, and point at the shower exit, which is also taken to be the reerence level (z 0. The luid at both points is open to the atmosphere (and thus P P P atm, and 0. Then the energy equation or a control volume between these two points simpliies to P P + α + z + hpump, u + α + z + hturbine, e + h z α + h g g g where h h, total h,major + h,minor + K g since the diameter o the piping system is constant. The average velocity in the pipe and the Reynolds number are & & m /s.96m/s π / π (0.05 m / c ρ (99. kg/m (.96m/s(0.05 m.5 cm Re 90,70 µ kg/m s which is greater than 000. Thereore, the low is turbulent. The 0 m 0.7 /s relative roughness o the pipe is m ε / m Showers The riction actor can be determined rom the Moody chart, but to avoid the reading error, we determine it rom the Colebrook equation using an equation solver (or an iterative scheme, ε / log +.0 log +.7 Re.7 90, 70 It gives The sum o the loss coeicients is K K + K + K + K Water tank, entrance,elbow,valve, exit Note that we do not consider the exit loss unless the exit velocity is dissipated within the system considered (in this case it is not. Then the total head loss and the elevation o the source become 0 m (.96m/s h + K ( m g 0.05 m (9.8m/s (.96m/s z α + h ( m 5. m g (9.8m/s since α. Thereore, the ree surace o the tank must be 5. m above the shower exit to ensure water low at the speciied rate. z 8-50 PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution

21 8-79 The low rate through a piping system connecting two water reservoirs with the same water level is given. The absolute pressure in the pressurized reservoir is to be determined. ssumptions The low is steady and incompressible. The entrance eects are negligible, and thus the low is ully developed. The low is turbulent so that the tabulated value o the loss coeicients can be used (to be veriied. There are no pumps or turbines in the piping system. Properties The density and dynamic viscosity o water at 0 C are ρ kg/m and µ kg/m s. The loss coeicient is K 0.5 or a sharp-edged entrance, K or swing check valve, K 0. or the ully open gate valve, and K or the exit. The roughness o cast iron pipe is ε m. nalysis We choose points and at the ree suraces o the two reservoirs. We note that the luid velocities at both points are zero ( 0, the luid at point is open to the atmosphere (and thus P P atm, both points are at the same level (z z. Then the energy equation or a control volume between these two points simpliies to P + α g + z + h pump, u P + α g + z + h turbine, e + h P Patm + h P P atm + h where h h, total h,major + h,minor + K g ir P since the diameter o the piping system is constant. The average low velocity and the Reynolds number are & & 0.00 m /s.8 m/s c π / π (0.0 m / Water 0 m Water ρ (999.7 kg/m (.8 m/s(0.0 m Re 58,00 µ.07 0 kg/m s cm which is greater than 000. Thereore, the low is turbulent. The. /s relative roughness o the pipe is m ε / m The riction actor can be determined rom the Moody chart, but to avoid the reading error, we determine it rom the Colebrook equation using an equation solver (or an iterative scheme, ε / log +.0 log +.7 Re.7 58, 00 It gives 0.0. The sum o the loss coeicients is K K Then the total head loss becomes h + K g Substituting,, entrance + K,check valve + K, gate valve + K,exit (0.0 0 m 0.0 m (.8 m/s +.7 (9.8m/s 65.8 m kn kpa P Patm + ρ gh (88 kpa + (999.7 kg/m (9.8m/s (65.8 m 7 kpa 000 kg m/s kn/m iscussion The absolute pressure above the irst reservoir must be 7 kpa, which is quite high. Note that the minor losses in this case are negligible (about % o total losses. lso, the riction actor could be determined easily rom the explicit Haaland relation (it gives the same result, 0.0. The riction coeicient would drop to 0.00 i smooth pipes were used. 8-5 PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution

22 8-80 tanker is to be illed with uel oil rom an underground reservoir using a plastic hose. The required power input to the pump is to be determined. ssumptions The low is steady and incompressible. The entrance eects are negligible, and thus the low is ully developed. The low is turbulent so that the tabulated value o the loss coeicients can be used (to be veriied. Fuel oil level remains constant. 5 Reservoir is open to the atmosphere. Properties The density and dynamic viscosity o uel oil are given to be ρ 90 kg/m and µ 0.05 kg/m s. The loss coeicient is K 0. or a slightly-rounded entrance and K 0. or a 90 smooth bend (langed. The plastic pipe is smooth and thus ε 0. The kinetic energy correction actor at hose discharge is given to be α.05. nalysis We choose point at the ree surace o oil in the reservoir and point at the exit o the hose in the tanker. We note the luid at both points is open to the atmosphere (and thus P P P atm and the luid velocity at point is zero ( 0. We take the ree surace o the reservoir as the reerence level (z 0. Then the energy equation or a control volume between these two points simpliies to P P + α + z + hpump, u + α + z + hturbine, e + h hpump, u α + z + h g g g where h h, total h,major + h,minor + K g since the diameter o the piping system is constant. The low rate is determined rom the requirement that the tanker must be illed in 0 min, tanker 8 m & 0.0m /s t (0 60 s Then the average velocity in the pipe and the Reynolds number become & & 0.0m /s 5.09 m/s π / π (0.05 m / c Tanker 8 m ρ (90 kg/m (5.09 m/s(0.05 m Re 506 Pump µ 0.05 kg/m s which is greater than 000. Thereore, the low is turbulent. The riction actor can be determined rom the Moody chart, but to avoid the reading error, we determine it rom the Colebrook equation, ε / log +.0 log Re 506 It gives The sum o the loss coeicients is K K + K , entrance, bend Note that we do not consider the exit loss unless the exit velocity is dissipated within the system (in this case it is not. Then the total head loss, the useul pump head, and the required pumping power become 0 m (5.09 m/s h + K ( m g 0.05 m (9.8m/s (5.09 m/s h pump, u + z + h m m 6.9 m g (9.8m/s h pump, u (0.0m /s(90 kg/m (9.8m/s (6.9 m kn kw W& & pump.96 kw η pump kg m/s kn m/s iscussion Note that the minor losses in this case are negligible (0.7/ or about 5% o total losses. lso, the riction actor could be determined easily rom the Haaland relation (it gives m 5 cm 5 m Fuel oil 8-5 PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution

23 8-8 Two pipes o identical length and material are connected in parallel. The diameter o one o the pipes is twice the diameter o the other. The ratio o the low rates in the two pipes is to be determined. ssumptions The low is steady and incompressible. The riction actor is given to be the same or both pipes. The minor losses are negligible. nalysis When two pipes are parallel in a piping system, the head loss or each pipe must be same. When the minor losses are disregarded, the head loss or ully developed low in a pipe o length and diameter can be expressed as h Solving or the low rate gives & & & & g g c g π / g π gπ h g.5.5 & π k (k constant o proportionality 8 When the pipe length, riction actor, and the head loss is constant, which is the case here or parallel connection, the low rate becomes proportional to the.5 th power o diameter. Thereore, when the diameter is doubled, the low rate will increase by a actor o since I Then &.5 k & k k( k & & B.5 B.5 Thereore, the ratio o the low rates in the two pipes is B 8-5 PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution

24 8-8 Cast iron piping o a water distribution system involves a parallel section with identical diameters but dierent lengths. The low rate through one o the pipes is given, and the low rate through the other pipe is to be determined. ssumptions The low is steady and incompressible. The entrance eects are negligible, and thus the low is ully developed. The minor losses are negligible. The low is ully turbulent and thus the riction actor is independent o the Reynolds number (to be veriied. Properties The density and dynamic viscosity o water at 5 C are ρ 999. kg/m and µ kg/m s. The roughness o cast iron pipe is ε m. nalysis The average velocity in pipe is & & 0. m /s 0. m m/s /s 0 cm 000 m c π / π (0.0 m / When two pipes are parallel in a piping system, the head loss or each pipe must be same. When the minor losses are disregarded, the head loss or ully developed low in a pipe o length and diameter is B 0 cm 000 m h g Writing this or both pipes and setting them equal to each other, and noting that B (given and B (to be veriied gives g B B B B g B B (5.659 m/s 000 m 000 m.67 m/s Then the low rate in pipe B becomes & B [ π / ] [ π (0.0 m / ](.67 m/s 0. m c B B Proo that low is ully turbulent and thus riction actor is independent o Reynolds number: The velocity in pipe B is lower. Thereore, i the low is ully turbulent in pipe B, then it is also ully turbulent in pipe. The Reynolds number in pipe B is ρ (999. kg/m (.67 m/s(0.0 m 6 Re B B µ.8 0 kg/m s which is greater than 000. Thereore, the low is turbulent. The relative roughness o the pipe is m ε / m From Moody s chart, we observe that or a relative roughness o , the low is ully turbulent or Reynolds number greater than about 0 6. Thereore, the low in both pipes is ully turbulent, and thus the assumption that the riction actor is the same or both pipes is valid. iscussion Note that the low rate in pipe B is less than the low rate in pipe because o the larger losses due to the larger length. /s 8-5 PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution

25 8-8 Cast iron piping o a water distribution system involves a parallel section with identical diameters but dierent lengths and dierent valves. The low rate through one o the pipes is given, and the low rate through the other pipe is to be determined. ssumptions The low is steady and incompressible. The entrance eects are negligible, and thus the low is ully developed. The minor losses other than those or the valves are negligible. The low is ully turbulent and thus the riction actor is independent o the Reynolds number. Properties The density and dynamic viscosity o water at 5 C are ρ 999. kg/m and µ kg/m s. The roughness o cast iron pipe is ε m. nalysis For pipe, the average velocity and the Reynolds number are & & 0. m /s m/s π / π (0.0 m / c ρ (999. kg/m (5.659 m/s(0.0 m Re.9 0 µ.8 0 kg/m s The relative roughness o the pipe is ε / m 0.0 m cm B 0. m /s 000 m 000 m 0 cm The riction actor corresponding to this relative roughness and the Reynolds number can simply be determined rom the Moody chart. To avoid the reading error, we determine it rom the Colebrook equation ε / log +.0 log Re It gives Then the total head loss in pipe becomes 000 m (5.659 m/s h, + K ( m g 0.0 m (9.8m/s When two pipes are parallel in a piping system, the head loss or each pipe must be same. Thereore, the head loss or pipe B must also be 07.9 m. Then the average velocity in pipe B and the low rate become h B B, B + K B & B g 000 m B 07.9 m ( m (9.8m/s [ π / ] [ π (0.0 m / ](. m/s 0.9 c B B m /s. m/s iscussion Note that the low rate in pipe B decreases slightly (rom 0. to 0.9 m /s due to the larger minor loss in that pipe. lso, minor losses constitute just a ew percent o the total loss, and they can be neglected i great accuracy is not required PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution

26 8-8 Geothermal water is supplied to a city through stainless steel pipes at a speciied rate. The electric power consumption and its daily cost are to be determined, and it is to be assessed i the rictional heating during low can make up or the temperature drop caused by heat loss. ssumptions The low is steady and incompressible. The entrance eects are negligible, and thus the low is ully developed. The minor losses are negligible because o the large length-to-diameter ratio and the relatively small number o components that cause minor losses. The geothermal well and the city are at about the same elevation. 5 The properties o geothermal water are the same as resh water. 6 The luid pressures at the wellhead and the arrival point in the city are the same. Properties The properties o water at 0 C are ρ kg/m, µ kg/m s, and C p.9 kj/kg C. The roughness o stainless steel pipes is 0-6 m. nalysis (a We take point at the well-head o geothermal resource and point at the inal point o delivery at the city, and the entire piping system as the control volume. Both points are at the same elevation (z z and the same velocity ( since the pipe diameter is constant, and the same pressure (P P. Then the energy equation or this control volume simpliies to P P + α + z + hpump, u + α + z + hturbine, e + h g g That is, the pumping power is to be used to overcome the head losses due to riction in low. The average velocity and the Reynolds number are & &.5 m /s 5.05 m/s π / π (0.60 m / c ρ (950.6 kg/m (5.05 m/s(0.60 m Re.87 0 µ kg/m s which is greater than 000. Thereore, the low is turbulent. The relative roughness o the pipe is 6 h pump, u h 7 Water.5 m /s km 60 cm 0 m 6 ε / m The riction actor can be determined rom the Moody chart, but to avoid the reading error, we determine it rom the Colebrook equation using an equation solver (or an iterative scheme, 6 ε / log +.0 log Re It gives Then the pressure drop, the head loss, and the required power input become P P h P W& electric, in ρ,000 m (950.6 kg/m (5.05 m/s m kn kpa 000 kg m/s kn/m g W& η,000 m (5.05 m/s ( m 0.60 m (9.8m/s pump, u pump-motor P & η pump-motor (.5 m /s(8 kpa kw 0.7 kpa m 96 kw /s 8 kpa Thereore, the pumps will consume 96 kw o electric power to overcome riction and maintain low. The pumps must raise the pressure o the geothermal water by 8 kpa. Providing a pressure rise o this magnitude at one location may create excessive stress in piping at that location. Thereore, it is more desirable to raise the pressure by smaller amounts at a several locations along the low. This will keep the maximum pressure in the system and the stress in piping at a saer level. (b The daily cost o electric power consumption is determined by multiplying the amount o power used per day by the unit cost o electricity, mount W & elect, in t (96 kw( h/day 07,900 kwh/day 8-56 PROPRIETRY MTERI. 006 The McGraw-Hill Companies, Inc. imited distribution