Aprils supervisor asks her to find a replacement pump that will increase the flow rate through the

Question:

April’s supervisor asks her to find a replacement pump that will increase the flow rate through the piping system of Prob. 14–43 by a factor of 2 or greater. April looks through some online brochures, and finds a pump with the performance data shown in Table P14–46. All dimensions and parameters remain the same as in Prob. 14–43—only the pump is changed. 

(a) Perform a least-squares curve fit (regression analysis) of Havailable versus V̇2,
and calculate the best-fit values of coefficients H0 and a that
translate the tabulated data of Table P14–46 into the parabolic expression Havailable = H0 – aV̇2. Plot the data points as
symbols and the curve fit as a line for comparison.

(b) Use the expression obtained in part (a) to estimate the operating volume flow rate of the new pump if it were to replace the existing pump, all else being equal. Compare to the result of Prob. 14–43 and discuss. Has April achieved her goal?

(c) Generate a plot of required net head and available net head as functions of volume flow rate, and indicate the operating point on the plot.


TABLE P14-46 V, Lpm  5 10 15 20 25 30 H, m 46.5 46 42 37 29 16.5 0


Data from Problem 14–43

A water pump is used to pump water from one large reservoir to another large reservoir that is at a higher elevation. The free surfaces of both reservoirs are exposed to atmospheric pressure, as sketched in Fig. P14–43. The dimensions and minor loss coefficients are provided in the figure. The pump’s performance is approximated by the expression Havailable = H0 – aV̇2, where shutoff head H0 = 24.4 m of water column, coefficient a = 0.0678 m/Lpm2, available pump head Havailable is in units of meters of water column, and capacity V̇ is in units of liters per minute (Lpm). Estimate the capacity delivered by the pump.


Data from FIGURE P14–43

22- 22-21 7.85 m (elevation difference) D = 2.03 cm (pipe diameter) KL. , entrance = 0.50 (pipe entrance) KL,

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