A 0.25-hp prototype water impulse turbine extracts 13 ft of freely available head with a freely available discharge of 252 US gpm. The diameter of the prototype turbine runner blades is 9.5 in with a runner rotational speed of 800 rpm. A smaller model water turbine of the prototype water impulse is

Question

A 0.25-hp prototype water impulse turbine extracts 13 ft of freely available head with a freely available discharge of 252 US gpm. The diameter of the prototype turbine runner blades is 9.5 in with a runner rotational speed of 800 rpm. A smaller model water turbine of the prototype water impulse is designed in order to study the flow characteristics of the flow through a turbine. The model scale is 0.5. (a) Design a homologous model water impulse turbine. (b) Determine if the resulting designed model is a true model or a distorted model.

Accepted Answer

(a) In order to design a homologous model water impulse turbine, the affinity laws (similarity rules) for the efficiency of homologous turbines (Equations 11.178 through 11.181)

[latex] \underbrace{\left[\left(\eta _{turbine} ight)_{p} ight] }_{\left(\frac{C_{P}}{C_{Q}C_{H}} ight)_{p} } = \underbrace{\left[\left(\eta _{turbine} ight)_{m} ight] }_{\left(\frac{C_{p} }{C_{Q}C_{H}} ight)_{m} } [/latex] (11.178)

[latex] \underbrace{\left[\left(\frac{g h_{turbine}}{ \omega^2 D^{2}} ight)_{p} ight] }_{(C_{H})_{p}} = \underbrace{\left[\left(\frac{g h_{turbine}}{ \omega^2 D^{2}} ight)_{m} ight] }_{(C_{H})_{m}} [/latex] (11.179)

[latex] \underbrace{\left[\left(\frac{Q}{ \omega D^{3}} ight)_{p} ight] }_{(C_{Q})_{p}} = \underbrace{\left[\left(\frac{Q}{ \omega D^{3}} ight)_{m} ight] }_{(C_{Q})_{m}} [/latex] (11.180)

[latex] \underbrace{\left[\left(\frac{\omega T}{ ho \omega ^{3}D^{5}} ight)_{p} ight] }_{(C_{p})_{p}} = \underbrace{\left[\left(\frac{\omega T}{ ho \omega ^{3}D^{5}} ight)_{m} ight] }_{(C_{p})_{m}} [/latex] (11.181)

are applied as follows:

But, first, the turbine design coefficients, the hydraulic power input to the turbine, the turbine efficiency, and the Reynolds number are computed for the larger prototype water impulse turbine as follows:

[latex] slug: = 1 lb \frac{sec^{2}}{ft} [/latex] [latex] ho _{p} : = 1.936 \frac{slug}{ft^{3}} [/latex] [latex] g: = 32.174 \frac{ft}{sec^{2}} [/latex]

[latex] \gamma _{p}: = ho _{p} . g = 62.289 \frac{lb}{ft^{3}} [/latex] [latex] h_{p}: = 550 \frac{ftlb}{sec} [/latex] [latex] P_{toutp}: = 0.25 hp = 137.5 \frac{ft.lb}{s} [/latex]

[latex] h_{turbinep}: = 13 ft [/latex] [latex] USgal: = 0.133681 ft^{3} [/latex] [latex] Q_{p}: = 252 \frac{USgal}{min} = 0.561 \frac{ft^{3}}{sec} [/latex]

[latex] D_{p}: = 9.5 in = 0.792 ft [/latex] [latex] \omega _{p}: = 800 rpm = 83.776 \frac{1}{s} [/latex]

[latex] C_{Hp}: = \frac{g.h_{turbinep} }{\omega ^{2}_{p}.D^{2}_{p}} = 0.095 [/latex] [latex] C_{Qp}: = \frac{Q_{p}}{\omega _{p}.D^{3}_{p}} = 0.014 [/latex]

[latex] T_{p}: = \frac{P_{toutp}}{\omega _{p}} = 1.641 ft.lb [/latex] [latex] C_{p_{p}}: = \frac{\omega _{p}.T_{p}}{ ho_{p}.\omega _{p}^{3}.D_{p}^{5} } = 3.884 imes 10^{-4} [/latex]

[latex] \eta _{turbinep}: = \frac{C_{p_{p}}}{ C_{Q_p}. C_{H_p}} = 0.302 [/latex] [latex] P_{tinp}: =\gamma _{p}. Q_{p}.h_{turbinep} = 454.645 \frac{ft.lb}{s} [/latex]

[latex] \eta _{turbinep}: = \frac{ P_{toutp}}{ P_{tinp}} = 0.302 [/latex] [latex] \mu _{p} : = 21 imes 10^{-6} \frac{lb . sec}{ft^{2}} [/latex]

[latex] V_{p}: = \omega _{p}.D_{p} = 66.323 \frac{ft}{s} [/latex] [latex] R_{p}: = \frac{ ho _{p} .V_{p} .D_{p}}{\mu _{p}} = 4.84 imes 10^{6} [/latex]

Then, in order to design a homologous smaller model water turbine, the affinity laws (similarity rules) for the efficiency of homologous turbines (Equations 11.178 through 11.181) are applied as follows:

[latex] \lambda : = 0.5 [/latex] [latex] ho _{m} : = ho _{p} = 1.936 \frac{slug}{ft^{3}} [/latex] [latex] \gamma _{m}: = \gamma _{p}= 62.289 \frac{lb}{ft^{3}} [/latex]

Guess value: [latex] D_{m}: = 6 in [/latex] [latex] h_{turbinem}: = 10 ft [/latex] [latex] \omega _{m}: = 100 rpm = 10.472^{\frac{-1}{s}} [/latex]

[latex] Q_{m}: = 20 \frac{USgal}{min} = 0.045 \frac{ft^{3}}{sec} [/latex] [latex] T_{m}: = 2.ft.lb [/latex] [latex] \eta _{pumpm}: =1 [/latex]

[latex] C_{Hm}: = 1 [/latex] [latex] C_{Qm}: = 1 [/latex] [latex] C_{pm}: = 1 [/latex]

Given

[latex] \lambda = \frac{D_{m}}{D_{p}} [/latex] [latex] \frac{g.h_{turbinep} }{\omega ^{2}_{p}.D^{2}_{p}} = \frac{g.h_{turbinem} }{\omega ^{2}_{m}.D^{2}_{m}} [/latex] [latex] \frac{Q_{p}}{\omega _{p}.D^{3}_{p}} = \frac{Q_{m}}{\omega _{m}.D^{3}_{m}} [/latex]

[latex] \frac{\omega _{p}.T_{p}}{ ho_{p}.\omega _{p}^{3}.D_{p}^{5} } = \frac{\omega _{m}.T_{m} }{ ho_{m}.\omega _{m}^{3}.D_{m}^{5} } [/latex] [latex] C_{Hm} = \frac{g.h_{turbinem} }{\omega ^{2}_{m}.D^{2}_{m}} [/latex] [latex] C_{Qm} = \frac{Q_{m}}{\omega _{m}.D^{3}_{m}} [/latex]

[latex] C_{p_{m}} = \frac{\omega _{m}.T_{m}}{ ho_{m}.\omega _{m}^{3}.D_{m}^{5} } [/latex] [latex] \eta _{turbinem} = \frac{C_{p_{p}}}{C_{Qp}. C_{Hp}} [/latex] [latex] \eta _{turbinep} = \eta _{turbinem} [/latex]
[latex] \left ( \begin{matrix} D_{m} \ h_{turbinem} \ \omega _{m} \ Q_{m} \ T_{m} \ \eta _{turbinem} \ C_{Hm} \ C_{Qm} \ C_{pm} \end{matrix} ight ) : = Find (D_{m}, h_{turbinem}, \omega _{m}, Q_{m}, T_{m}, \eta _{turbinem}, C_{Hm}, C_{Qm}, C_{pm}) [/latex]

[latex] D_{m}= 4.75 in [/latex] [latex] h_{turbinem} = 0.241 ft [/latex]

[latex] \omega _{m}= 22.807 \frac{1}{s} [/latex] [latex] \omega _{m}= 217.793 rpm [/latex]

[latex] Q_{m}= 0.019 \frac{ft^{3}}{sec} [/latex] [latex] Q_{m}= 8.576 \frac{USgal}{min} [/latex]

[latex] T_{m} = 3.801 imes 10^{-3} ft.lb [/latex] [latex] \eta _{turbinem} = 0.302 [/latex]

[latex] C_{Hm} = 0.095 [/latex] [latex] C_{Qm} = 0.014 [/latex] [latex] C_{pm} = 3.884 imes 10^{-4} [/latex]

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

[latex] P_{tinm}: = \gamma _{m} . Q_{m}. h_{turbinem} = 0.287 \frac{ft.lb}{s} [/latex] [latex] P_{tinm} = 5.212 imes 10^{-4} hp [/latex]

[latex] P_{toutm}=\omega _{m} .T_{m} =0.08 \frac{ft.lb}{s} [/latex] [latex] P_{toutm} = 1.576 imes 10^{-4} hp [/latex]

[latex] \eta_{turbinem}: = \frac{ P_{toutm}}{P_{tinm}} = 0.302 [/latex] [latex] \mu _{m}: = 21 imes 10^{-6} \frac{lb.sec}{ft^{2}} [/latex]

[latex] V_{m}: = \omega _{m}.D_{m} = 9.028 \frac{ft}{s} [/latex] [latex] R_{m}: = \frac{ ho _{m} .V_{m} .D_{m}}{\mu _{m}} = 3.294 imes 10^{5} [/latex]

(b) In order to determine if the resulting designed model is a true model or a distorted model, one must first determine whether the application of the affinity laws (similarity rules) for the efficiency of homologous turbines (Equations 11.178 through 11.181) has yielded a practically feasible designed model. If the resulting designed smaller model water turbine is practically feasible (design parameters: [latex] D_{m}= 4.75 in [/latex], [latex] h_{turbinem} = 0.241 ft [/latex], [latex] \omega _{m}= 217.793 rpm [/latex], [latex] Q_{m}= 0.019 cfs = 8.576 US gpm [/latex], [latex] T_{m} = 3.801 imes 10^{-3} ft-lb [/latex], [latex] p_{toutm} = 1.576 imes 10^{-4} hp [/latex]), then it is considered to be a true model (model and prototype are homologous turbines). However, if the resulting designed smaller model water turbine 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 turbines). Upon examination of the resulting model design parameters (design parameters: [latex] D_{m}= 4.75 in [/latex], [latex] h_{turbinem} = 0.241 ft [/latex], [latex] \omega _{m}= 217.793 \ rpm [/latex], [latex] Q_{m}= 0.019 \ cfs = 8.576 \ US \ gpm [/latex], [latex] T_{m} = 3.801 imes 10^{-3} ft-lb [/latex], [latex] p_{toutm} = 1.576 imes 10^{-4} hp [/latex]) and a comparison with some turbine manufacture design data parameters, it appears that such a model water turbine is indeed practically feasible. Furthermore, the [latex] R_{p} = 4.84 imes 10^{6} [/latex], while the [latex] R_{m} = 3.294 imes 10^{5} [/latex], which indicates that turbulent flow is maintained between the prototype turbine and the model turbine. 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 [latex] h_{turbinem} [/latex] and/or a limitation on the available [latex] Q_{m} [/latex] or [latex] \omega _{m} [/latex] , etc. that it is capable of providing (“design conditions”). Furthermore, for a smaller model turbine, 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 turbines (Equations 11.178 through 11.181) will yield a distorted model (model and prototype are nonhomologous turbines). 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 11.182 [latex] \frac{1 - \eta _{p} }{1 - \eta _{m} } = \left(\frac{D_{m}}{D_{p}} ight)^{1/5} [/latex]) as follows:

Guess value: [latex] \eta _{turbinem}: = 0.1 [/latex]

Given
[latex] \frac{1 - \eta _{turbinep} }{1 - \eta _{turbinem} } = \left(\frac{D_{m}}{D_{p}} ight)^{\frac{1}{5} } [/latex]
[latex] \eta _{turbinem}: = Find (\eta _{turbinem} ) = 0.199 [/latex]

which indicates that the estimated turbine efficiency of the designed smaller model water turbine is only 0.199, rather than the designed turbine efficiency of the model turbine of 0.302; larger turbines are typically more efficient than the smaller turbines.

Question 11.20: A 0.25-hp prototype water impulse turbine extracts 13 ft of ...

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