Question 11.19: A 3-hp prototype water pump is designed to deliver a total h...

(a) In order to design a homologous model water pump, the affinity laws (similarity rules) for the efficiency of homologous pumps (Equations 11.169  through 11.172 )

\underbrace{\left[\left(\eta _{pump}\right)_{p} \right] }_{\left(\frac{C_{Q}C_{H}}{C_{P}} \right)_{p} } = \underbrace{\left[\left(\eta _{pump}\right)_{m} \right] }_{\left(\frac{C_{Q}C_{H}}{C_{P}} \right)_{m} }       (11.169)

\underbrace{\left[\left(\frac{g h_{pump}}{\omega^2 D^{2}} \right)_{p} \right] }_{(C_{H})_{p}} = \underbrace{\left[\left(\frac{g h_{pump}}{ \omega^2 D^{2}} \right)_{m} \right] }_{(C_{H})_{m}}                 (11.170)

\underbrace{\left[\left(\frac{Q}{\omega D^{3}} \right)_{p} \right] }_{(C_{Q})_{p}} = \underbrace{\left[\left(\frac{Q}{ \omega D^{3}} \right)_{m} \right] }_{(C_{Q})_{m}}                       (11.171)

\underbrace{\left[\left(\frac{\omega T}{\rho \omega ^{3}D^{5}} \right)_{p} \right] }_{(C_{p})_{p}} = \underbrace{\left[\left(\frac{\omega T}{\rho \omega ^{3}D^{5}} \right)_{m} \right] }_{(C_{p})_{m}}              (11.172)

are applied as follows:
But, first, the pump design coefficients, the hydraulic power output from the pump, the pump efficiency, and the Reynolds number are computed for the larger prototype water pump as follows:

slug: = 1 lb \frac{sec^{2}}{ft}                               \rho _{p} : = 1.936 \frac{slug}{ft^{3}}                               g: = 32.174 \frac{ft}{sec^{2}}

\gamma _{p}: = \rho _{p} . g = 62.289 \frac{lb}{ft^{3}}                               h_{p}: = 550 \frac{ft.lb}{sec}                               P_{pinp}: = 3.hp = 1.65 \times 10^{3} \frac{ft.lb}{s}

h_{pumpp}: = 140 ft                               USgal: = 0.133681 ft^{3}                               Q_{p}: = 42 \frac{USgal}{min} = 0.094 \frac{ft^{3}}{sec}

D_{p}: = 8.5 in = 0.708 ft                               \omega _{p}: = 1750 rpm = 138.26 \frac{1}{s}

C_{Hp}: = \frac{g.h_{pumpp} }{\omega ^{2}_{p}.D^{2}_{p}} = 0.267                               C_{Qp}: = \frac{Q_{p}}{\omega _{p}.D^{3}_{p}} = 1.437 \times 10^{-3}

T_{p}: = \frac{P_{pinp}}{\omega _{p}} = 9.004 ftlb                               C_{p_{p}}: = \frac{\omega _{p}.T_{p}}{\rho_{p}.\omega _{p}^{3}.D_{p}^{5} } = 7.776 \times 10^{-4}

\eta _{pumpp}: = \frac{C_{Qp}. C_H{p}}{C_{p_{p}}} = 0.495                               P_{poutp}: =\gamma _{p}. Q_{p}.h_{pumpp} = 816.03 \frac{ft.lb}{s}

\eta _{pumpp}: = \frac{ P_{poutp}}{ P_{pinp}} = 0.495                               \mu _{p} : = 21 \times 10^{-6} \frac{lb . sec}{ft^{2}}

V_{p}: = \omega _{p}.D_{p} = 129.809 \frac{ft}{s}                               R_{p}: = \frac{\rho _{p} .V_{p} .D_{p}}{\mu _{p}} = 8.477 \times 10^{6}

Then, in order to design a homologous smaller model water pump, the affinity laws (similarity rules) for the efficiency of homologous pumps (Equations 11.169 \underbrace{\left[\left(\eta _{pump}\right)_{p} \right] }_{\left(\frac{C_{Q}C_{H}}{C_{P}} \right)_{p} } = \underbrace{\left[\left(\eta _{pump}\right)_{m} \right] }_{\left(\frac{C_{Q}C_{H}}{C_{P}} \right)_{m} } through 11.172 \underbrace{\left[\left(\frac{\omega T}{\rho \omega ^{3}D^{5}} \right)_{p} \right] }_{(C_{p})_{p}} = \underbrace{\left[\left(\frac{\omega T}{\rho \omega ^{3}D^{5}} \right)_{m} \right] }_{(C_{p})_{m}} ) are applied as follows:

\lambda : = 0.4                               \rho _{m} : = \rho _{p} = 1.936 \frac{slug}{ft^{3}}                               \gamma _{m}: = \gamma _{p}= 62.289 \frac{lb}{ft^{3}}

Guess value:                              D_{m}: = 6 in                               h_{pumpm}: = 50 ft                               \omega _{m}: = 10,000 rpm = 1.047 \times 10^{3} \frac{1}{s}

Q_{m}: = 20 \frac{USgal}{min} = 0.045 \frac{ft^{3}}{sec}                               T_{m}: = 2ftlb                               \eta _{pumpm}: =1

C_{Hm}: = 1                               C_{Qm}: = 1                               C_{pm}: = 1

Given

\lambda = \frac{D_{m}}{D_{p}}                               \frac{g.h_{pumpp} }{\omega ^{2}_{p}.D^{2}_{p}} = \frac{g.h_{pumpm} }{\omega ^{2}_{m}.D^{2}_{m}}                               \frac{Q_{p}}{\omega _{p}.D^{3}_{p}} = \frac{Q_{m}}{\omega _{m}.D^{3}_{m}}

\frac{\omega _{p}.T_{p}}{\rho_{p}.\omega _{p}^{3}.D_{p}^{5} } = \frac{\omega _{m}.T_{m} }{\rho_{m}.\omega _{m}^{3}.D_{m}^{5} }                               C_{Hm} = \frac{g.h_{pumpm} }{\omega ^{2}_{m}.D^{2}_{m}}                             C_{Qm} = \frac{Q_{m}}{\omega _{m}.D^{3}_{m}}

C_{p_{m}} = \frac{\omega _{m}.T_{m}}{\rho_{m}.\omega _{m}^{3}.D_{m}^{5} }                               \eta _{pumpm} = \frac{C_{Qm}. C_H{m}}{C_{p_{m}} }                             \eta _{pumpp} = \eta _{pumpm}  

D_{m}= 3.4 in                           h_{pumpm} = 81.322 ft                           \omega _{m}= 349.178 \frac{1}{s}                           \omega _{m}= 3.334 \times 10^{3} rpm

Q_{m}= 0.011 \frac{ft^{3}}{sec}                           Q_{m}= 5.122 \frac{USgal}{min}                            T_{m} = 0.335 ft.lb                           \eta _{pumpm} = 0.495

C_{Hm} = 0.267                           C_{Qm} = 1.437 \times 10^{-3}                           C_{pm} = 7.766 \times 10^{-4}

And, finally, the mechanical power input to the pump, the hydraulic power output from the pump, the pump efficiency, and the Reynolds number are computed for the smaller model water pump as follows:

P_{poutm}: = \gamma _{m} . Q_{m}. h_{pumpm} = 57.803 \frac{ft.lb}{s}                           P_{poutm} = 0.105 hp

P_{pinm}: =\omega _{m} .T_{m} = 116.876 \frac{ft.lb}{s}                           P_{pinm} = 0.213 hp

\eta_{pumpm}: = \frac{ P_{poutm}}{P_{pinm}} = 0.495                           \mu _{m}: = 21 \times 10^{-6} \frac{lb.sec}{ft^{2}}

V_{m}: = \omega _{m}.D_{m} = 98.934 \frac{ft}{s}                           R_{m}: = \frac{\rho _{m} .V_{m} .D_{m}}{\mu _{m}} = 2.584 \times 10^{6}

(b) In order to determine if the resulting designed model is a true model or a distorted model, one must first determine if the application of the affinity laws (similarity rules) for the efficiency of homologous pumps (Equations 11.169 \underbrace{\left[\left(\eta _{pump}\right)_{p} \right] }_{\left(\frac{C_{Q}C_{H}}{C_{P}} \right)_{p} } = \underbrace{\left[\left(\eta _{pump}\right)_{m} \right] }_{\left(\frac{C_{Q}C_{H}}{C_{P}} \right)_{m} } through 11.172 \underbrace{\left[\left(\frac{\omega T}{\rho \omega ^{3}D^{5}} \right)_{p} \right] }_{(C_{p})_{p}} = \underbrace{\left[\left(\frac{\omega T}{\rho \omega ^{3}D^{5}} \right)_{m} \right] }_{(C_{p})_{m}} ) have yielded a practically feasible designed model or not. If the resulting designed smaller model water pump is practically feasible (design parameters: D_{m}= 3.4 in , h_{pumpm} = 81.322 ft , \omega _{m}= 3334 rpm , Q_{m}= 0.011 cfs = 5.122 US gpm, T_{m} = 0.335 ft-lb , P_{pinm} = 0.213 hp ), then it is considered to be a true model (model and prototype are homologous pumps). However, if the resulting designed smaller model water pump is not practically feasible (at least one of the design parameters is not practically feasible), then it is considered to be a distorted model (model and prototype are nonhomologous pumps). Upon examination of the resulting model design parameters (design parameters: D_{m}= 3.4 in , h_{pumpm} = 81.322 ft , \omega _{m}= 3334 rpm , Q_{m}= 0.011 cfs = 5.122 US gpm, T_{m} = 0.335 ft-lb , P_{pinm} = 0.213 hp ) and a comparison with some pump manufacture design data parameters, it appears that such a model water pump is indeed practically feasible. Furthermore, the R_{p} = 8.477 \times 10^{6} , while the R_{m} = 2.584 \times 10^{6} , which indicates that turbulent flow is maintained between the prototype pump and the model pump. However, in most laboratory situations, there will be some constraints imposed upon the model design; these are called “design conditions.” For instance, a certain laboratory may impose a limitation on the available h_{pumpm} and/or a limitation on the available, Q_{m} or \omega _{m} , etc. that it is capable of providing (“design conditions”). Furthermore, for a smaller model pump, it may not be practically feasible to attain high speeds, and/or cavitation may occur at higher speeds. Thus, when there are certain “design conditions” imposed due to practical limitations/constraints, the application of the affinity laws (similarity rules) for the efficiency of homologous pumps (Equations 11.169 \underbrace{\left[\left(\eta _{pump}\right)_{p} \right] }_{\left(\frac{C_{Q}C_{H}}{C_{P}} \right)_{p} } = \underbrace{\left[\left(\eta _{pump}\right)_{m} \right] }_{\left(\frac{C_{Q}C_{H}}{C_{P}} \right)_{m} } through 11.172 \underbrace{\left[\left(\frac{\omega T}{\rho \omega ^{3}D^{5}} \right)_{p} \right] }_{(C_{p})_{p}} = \underbrace{\left[\left(\frac{\omega T}{\rho \omega ^{3}D^{5}} \right)_{m} \right] }_{(C_{p})_{m}} ) will yield a distorted model (model and prototype are nonhomologous pumps). If a “distorted model” results,  appropriate correction measures are taken in order to model the scaling effects (i.e., account for a “distorted model”) between the model and its prototype, using the empirical equation developed by Moody (Equation 11.175 \frac{1 – \eta _{p} }{1 – \eta _{m} } = \left(\frac{D_{m}}{D_{p}} \right)^{1/5} )as follows:

Guess value:                           \eta _{pumpm}: = 0.1

Given
\frac{1 – \eta _{pumpp} }{1 – \eta _{pumpm} } = \left(\frac{D_{m}}{D_{p}} \right)^{\frac{1}{5} }
\eta _{pumpm}: = Find (\eta _{pumpm} ) = 0.393

which indicates that the estimated pump efficiency of the designed smaller model water pump is only 0.393, rather than the designed pump efficiency of the model pump of 0.495; larger pumps are typically more efficient than smaller pumps.