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Example 4.4 If the pump in Example 4.3 is operated at 1,170 rpm, calculate the resulting H, Q, Pbrake, and n.

Solution:

Hgl=J_Hgl^ EjbL = ^ Hb = 10(^)= 17.88 m Ans b a2—2I a2—21 ab—2 (Ü„—2" ( 875

Example 4.5 If a homologous 30-cm pump is to be used for the problem in Example 4.4, calculate the resulting H, Q, Pbrake, and n for the same rpm.

Solution: The diameter of the pump represented by Figure 4.8 is 375 mm.

H^ I =J_Hg EbL = Eag ^ Hb = 17.88(-3?25) =11.44 m Ans ooD2[oO D2(oD2b a2D2a (37.5

brake brake brakeb brakeb

Lpo D[pm Dpm Db poo Da b

37.55

4.5 pump specific speed

Raising the flow coefficient Cq = Q/(oD ) to the power ;-, the head coefficient CH = gH/(o D2) to the power 3, and forming the ratio of the former to that of the latter, D will be eliminated. Calling this ratio as Ns produces the expression

If the dimensions of Ns are substituted, it will be found dimensionless and because it is dimensionless, it can be used as a general characterization for a whole variety of pumps without reference to their sizes. Thus, a certain range of the value of Ns would be a particular type of pump such as axial (no size considered), and another range would be another particular type of pump such as radial (no size considered). Ns is called specific speed.

By characterizing all the pumps generally like this, Ns is of great applicability in selecting the proper type of pump, whether radial or axial or any other type. For example, refer to Figure 4.9. The radial-vane pumps are in the range of Ns = 9.6 to 19.2; the Francis-vane pumps are in the range of 28.9 to 76.9; and so on. Therefore, if Q, a, and H are known, Ns can be computed using Equation (4.43); thus, depending upon the value obtained, the type of pump can be specified.

Just how is the chart of specific speeds obtained? Remember that one of the characteristics curves of a pump is the plot of the efficiency. Referring to Figure 4.8, along any curve characterized by a parameter such as rpm, there are an infinite number of efficiency values. Of these infinite number of values, there is only one maximum. As mentioned previously, this maximum is the best efficiency point.

If values of a, Q, and H are taken from the characteristics curves at the best operating efficiencies and substituted into Equation (4.43), values of specific speeds are obtained at these efficiencies. For example, from Figure 4.8 at a Q of 0.16 m /s, the best efficiency is approximately 66% corresponding to a total developed head

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