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The following data pertain to the Waikiki Sands Hotel for the month of March.

 Flexible Budget March  Actual Results March  (in thousands)*  (in thousands)*  Banquets and Catering $650$658 Restaurants 1,8001,794 Kitchen staff wages (85)(86) Food (690)(690) Paper products (125)(122) Variable overhead (75)(78) Fixed overhead (90)(93)\begin{array}{lcc} & \textbf{ Flexible Budget March } & \textbf{ Actual Results March } \\ & \textbf{ (in thousands)* } & \textbf{ (in thousands)* } \\ \text{ Banquets and Catering } & \$ 650 & \$ 658 \\ \text{ Restaurants } & 1,800 & 1,794 \\ \text{ Kitchen staff wages } & (85) & (86) \\ \text{ Food } & (690) & (690) \\ \text{ Paper products } & (125) & (122) \\ \text{ Variable overhead } & (75) & (78) \\ \text{ Fixed overhead } & (90) & (93) \end{array}

^*Numbers without parentheses denote profit; numbers with parentheses denote expenses.

Prepare a March performance report. The report should have six numerical columns with headings analogous. Your performance report should cover only the Food and Beverage Department and the Kitchen. Draw arrows to show the relationships between the numbers in the report. For the year-to-date columns in your report, use the previous data given. You will need to update those figures using the March data given above.

Question

Consider the piping system of the figure, with all the dimensions, parameters, minor loss coefficients, etc., of the earlier problem. The pump's performance follows a parabolic curve fit, Hwailate =H0aV˙2H_{\text {wailate }}=H_0-a \dot{V}^2, where H0=19.8 mH_0=19.8 \mathrm{~m} is the pump's shutoff head, and a=0.00426 m/(Lpm)2a=0.00426 \mathrm{~m} /(\mathrm{Lpm})^2 is a coefficient of the curve fit. Find the operating volume flow rate V˙\dot{V} in Lpm (liters per minute), and compare it with that of the earlier problem. Discuss.

Solution

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The pump’s performance curve is given by:

Havailable=H0aV˙2\begin{equation*} H_{\text{available}}=H_0-a \dot V^2 \end{equation*}

where H0=19.8H_0=19.8 m is the pump’s shutoff head and a=0.00426a= 0.00426 m/(Lpm)2^2 is a coefficient of the curve fit.

The coefficient aa can be expressed as:

a=0.00426mLpm2(11000160)2(m3/s)2Lpm2=1.533107m(m3/s)2\begin{equation*} a=0.00426\:\frac{\text{m}}{\text{Lpm}^2\cdot\bigg(\frac{1}{1000}\cdot\frac{1}{60}\bigg)^2\:\frac{(\text{m}^3/\text{s})^2}{\text{Lpm}^2}}=1.533\cdot10^7\:\frac{\text{m}}{(\text{m}^3/\text{s})^2} \end{equation*}

Therefore, the pump's performance curve is:

Havailable=19.61.533107V˙2\begin{equation*} H_{\text{available}}=19.6-1.533\cdot10^7\dot V^2 \end{equation*}

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