Design of a New Geometrically Similar Pump After graduation, you go to work for a pump manufacturing company. One of your company’s best-selling products is a water pump, which we shall call pump A. Its impeller diameter is DA = 6.0 cm, and its performance data when operating at nA = 1725 rpm

Question

Design of a New Geometrically Similar Pump

After graduation, you go to work for a pump manufacturing company. One of your company’s best-selling products is a water pump, which we shall call pump A. Its impeller diameter is [latex] D_{ A }=6.0 cm [/latex], and its performance data when operating at [latex] \dot{n}_{ A }=1725 rpm \left(\omega_{ A }=180.6 rad / s ight) [/latex] are shown in Table 14–2. The marketing research department is recommending that the company design a new product, namely, a larger pump (which we shall call pump B) that will be used to pump liquid refrigerant R-134a at room temperature. The pump is to be designed such that its best efficiency point occurs as close as possible to a volume flow rate of [latex] \dot{V}_{B}=2400 cm ^{3} / s [/latex] and at a net head of [latex] H_{B}=450 [/latex] cm (of R-134a). The chief engineer (your boss) tells you to perform some preliminary analyses using pump scaling laws to determine if a geometrically scaled-up pump could be designed and built to meet the given requirements. (a) Plot the performance curves of pump A in both dimensional and dimensionless form, and identify the best efficiency point. (b) Calculate the required pump diameter [latex] D_{ B } [/latex], rotational speed [latex] \dot{n}_{ B } [/latex], and brake horsepower [latex] bhp _{ B } [/latex] for the new product.

TABLE 14–2

Manufacturer’s performance data for a water pump operating at 1725 rpm and room temperature (Example 14–11)*

[latex]\begin{array}{ccc}\dot{V}, cm ^{3} / s & H, cm & \eta_{ ext {pump }}, \% \\hline 100 & 180 & 32 \200 & 185 & 54 \300 & 175 & 70 \400 & 170 & 79 \500 & 150 & 81 \600 & 95 & 66 \700 & 54 & 38\end{array}[/latex]

* Net head is in centimeters of water.

Accepted Answer

(a) For a given table of pump performance data for a water pump, we are to plot both dimensional and dimensionless performance curves and identify the BEP. (b) We are to design a new geometrically similar pump for refrigerant R-134a that operates at its BEP at given design conditions.
Assumptions 1 The new pump can be manufactured so as to be geometrically similar to the existing pump. 2 Both liquids (water and refrigerant R-134a) are incompressible. 3 Both pumps operate under steady conditions.
Properties At room temperature (20°C), the density of water is [latex] ho_{ ext {water }} =998.0 kg / m ^{3} [/latex] and that of refrigerant R-134a is [latex] ho_{ R -134 a }=1226 kg / m ^{3} [/latex].
Analysis (a) First, we apply a second-order least-squares polynomial curve fit to the data of Table 14–2 to obtain smooth pump performance curves.
These are plotted in Fig. 14–76, along with a curve for brake horsepower, which is obtained from Eq. 14–5. A sample calculation, including unit conversions, is shown in Eq. 1 for the data at [latex] \dot{V}_{ A }=500 cm ^{3} / s [/latex], which is approximately the best efficiency point:

[latex]\begin{aligned}\operatorname{bhp}_{ A } &=\frac{ ho_{ ext {water }} g \dot{V}_{ A } H_{ A }}{\eta_{ ext {pump,A }}} \&=\frac{\left(998.0 kg / m ^{3} ight)\left(9.81 m / s ^{2} ight)\left(500 cm ^{3} / s ight)(150 cm )}{0.81}\left(\frac{1 m }{100 cm } ight)^{4}\left(\frac{ W \cdot s }{ kg \cdot m / s ^{2}} ight) = 9.07 W\end{aligned}[/latex] (1)

Pump efficiency: [latex] \eta_{ ext {pump }}=\frac{\dot{W}_{ ext {water horsepower }}}{\dot{W}_{ ext {shaft }}}=\frac{\dot{W}_{ ext {water horsepower }}}{ ext { bhp }}=\frac{ ho g \dot{V} H}{\omega T _{ ext {shaft }}} [/latex] (14–5)

Note that the actual value of [latex] b h p_{A} [/latex] plotted in Fig. 14–76 at [latex] \dot{V}_{ A }=500 cm ^{3} / s [/latex] differs slightly from that of Eq. 1 due to the fact that the least-squares curve fit smoothes out scatter in the original tabulated data.

Next we use Eqs. 14–30 to convert the dimensional data of Table 14–2 into nondimensional pump similarity parameters. Sample calculations are shown in Eqs. 2 through 4 at the same operating point as before (at the approximate location of the BEP). At [latex] \dot{V}_{ A }=500 cm ^{3} / s [/latex] the capacity coefficient is approximatel

[latex] C_{Q}=\frac{\dot{V}}{\omega D^{3}}=\frac{500 cm ^{3} / s }{(180.6 rad / s )(6.0 cm )^{3}}=0.0128 [/latex] (2)

Dimensionless pump parameters: [latex] \begin{aligned} &C_{H}= ext { Head coefficient }=\frac{g H}{\omega^{2} D^{2}} \ &C_{Q}= ext { Capacity coefficient }=\frac{\dot{V}}{\omega D^{3}} \ &C_{P}= ext { Power coefficient }=\frac{b h p}{ ho \omega^{3} D^{5}} \end{aligned} [/latex] (14–30)

The head coefficient at this flow rate is approximately

[latex] C_{H}=\frac{g H}{\omega^{2} D^{2}}=\frac{\left(9.81 m / s ^{2} ight)(1.50 m )}{(180.6 rad / s )^{2}(0.060 m )^{2}}=0.125 [/latex] (3)

Finally, the power coefficient at [latex] \dot{V}_{ A }=500 cm ^{3} / s [/latex] is approximately

[latex] C_{P}=\frac{ ext { bhp }}{ ho \omega^{3} D^{5}}=\frac{9.07 W }{\left(998 kg / m ^{3} ight)(180.6 rad / s )^{3}(0.060 m )^{5}}\left(\frac{ kg \cdot m / s ^{2}}{ W \cdot s } ight)=0.00198 [/latex] (4)

These calculations are repeated (with the aid of a spreadsheet) at values of [latex] \dot{V}_{A} [/latex] between 100 and 700 cm³/s. The curve-fitted data are used so that the normalized pump performance curves are smooth; they are plotted in Fig. 14–77. Note that [latex] \eta _{ ext {pump }} [/latex] is plotted as a fraction rather than as a percentage. In addition, in order to fit all three curves on one plot with a single ordinate, and with the abscissa centered nearly around unity, we have multiplied [latex] C_{Q} [/latex] by 100, [latex] C_{H} [/latex] by 10, and [latex] C_{P} [/latex] by 100. You will find that these scaling factors work well for a wide range of pumps, from very small to very large. A vertical line at the BEP is also sketched in Fig. 14–77 from the smoothed data. The curve-fitted data yield the following nondimensional pump performance parameters at the BEP:

[latex] C_{Q}^{*}=0.0112 \quad C_{H}^{*}=0.133 \quad C_{P}^{*}=0.00184 \quad \eta_{ ext {pump }}^{*}=0.812 [/latex] (5)

(b) We design the new pump such that its best efficiency point is homologous with the BEP of the original pump, but with a different fluid, a different pump diameter, and a different rotational speed. Using the values identified in Eq. 5, we use Eqs. 14–30 to obtain the operating conditions of the new pump. Namely, since both [latex] \dot{V}_{ B } ext { and } H_{ B } [/latex] are known (design conditions), we solve simultaneously for [latex] D_{ B } ext { and } \omega_{ B } [/latex]. After some algebra in which we eliminate [latex] \omega_{B} [/latex], we calculate the design diameter for pump B,

[latex] D_{ B }=\left(\frac{\dot{V}_{ B }^{2} C_{H}^{*}}{\left(C_{Q}^{*} ight)^{2} g H_{ B }} ight)^{1 / 4}=\left(\frac{\left(0.0024 m ^{3} / s ight)^{2}(0.133)}{(0.0112)^{2}\left(9.81 m / s ^{2} ight)(4.50 m )} ight)^{1 / 4}= 0 . 1 0 8 ~ m [/latex] (6)

In other words, pump A needs to be scaled up by a factor of [latex] D_{ B } / D_{ A } [/latex] = 10.8 cm/6.0 cm = 1.80. With the value of [latex] D_{ B } [/latex] known, we return to Eqs. 14–30 to solve for [latex] \omega_{B} [/latex], the design rotational speed for pump B,

[latex] \omega_{ B }=\frac{\dot{V}_{ B }}{\left(C_{Q}^{*} ight) D_{ B }^{3}}=\frac{0.0024 m ^{3} / s }{(0.0112)(0.108 m )^{3}}=168 rad / s \quad ightarrow \quad \dot{n}_{ B }=1610 rpm [/latex] (7)

Finally, the required brake horsepower for pump B is calculated from Eqs. 14–30,

[latex]\begin{aligned}b h p_{ B } &=\left(C_{P}^{*} ight) ho_{ B } \omega_{ B }^{3} D_{ B }^{5} \&=(0.00184)\left(1226 kg / m ^{3} ight)(168 rad / s )^{3}(0.108 m )^{5}\left(\frac{ W \cdot s }{ kg \cdot m ^{2} / s } ight)= 1 6 0 W\end{aligned}[/latex] (8)

An alternative approach is to use the affinity laws directly, eliminating some intermediate steps. We solve Eqs. 14–38a and b for [latex] D_{ B } [/latex] by eliminating the ratio [latex] \omega_{B} / \omega_{A} [/latex]. We then plug in the known value of [latex] D_{ A } [/latex] and the curve-fitted values of [latex] \dot{V}_{ A } ext { and } H_{ A } [/latex] at the BEP (Fig. 14–78). The result agrees with those calculated before. In a similar manner we can calculate [latex] \omega_{ B } ext { and } b h p_{ B } [/latex].

Affinity laws: [latex] \frac{\dot{V}_{ B }}{\dot{V}_{ A }}=\frac{\omega_{ B }}{\omega_{ A }}\left(\frac{D_{ B }}{D_{ A }} ight)^{3} [/latex] (14–38a)

Discussion Although the desired value of [latex] \omega_{B} [/latex] has been calculated precisely, a practical issue is that it is difficult (if not impossible) to find an electric motor that rotates at exactly the desired rpm. Standard single-phase, 60-Hz, 120-V AC electric motors typically run at 1725 or 3450 rpm. Thus, we may not be able to meet the rpm requirement with a direct-drive pump. Of course, if the pump is belt-driven or if there is a gear box or a frequency controller, we can easily adjust the configuration to yield the desired rotation rate. Another option is that since [latex] \omega_{B} [/latex] is only slightly smaller than [latex] \omega_{A} [/latex], we drive the new pump at standard motor speed (1725 rpm), providing a somewhat stronger pump than necessary. The disadvantage of this option is that the new pump would then operate at a point not exactly at the BEP.

Question 14.11: Design of a New Geometrically Similar Pump After graduation,...

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