10.3: Calculating Power Costs
- Page ID
- 7188
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)It is important for water managers to determine the potential costs in electricity for pumping water. Units used for measuring electrical usage are typically in kilowatt-hours (kW-Hr). In order to convert horsepower to kilowatts of power, the following conversion factor is used.
1 horsepower = 0.746 kilowatts of power
Once you know the hp that is needed you can then determine the amount of kW-Hr needed. Then, costs can be determined depending on what the local electric company charges per kW-Hr. Water utilities will calculate estimated budgets for pumping costs since these are typically the largest operating costs.
Exercises
Solve the following problems.
-
A well flows an estimated 3,200 gpm against a discharge head pressure of 95 psi. What is the corresponding hp and kW‐Hr if the pump has an efficiency of 70% and the motor 88%?
-
Based on the above question, how much would the electrical costs be if the rate is $0.12 per kW‐Hr and the pump runs for 10 hours a day?
-
A utility has 3 pumps that run at different flow rates and supply water to a 500,000-gallon storage tank. The TDH for the pumps is 210 ft. The utility needs to fill the tank daily and power costs are to be calculated at a rate of $0.135 per kW‐Hr. Complete the table below.
Pump |
Flow Rate |
Hp |
Efficiency |
Run Time |
Total Cost |
---|---|---|---|---|---|
1 |
500 gpm |
50 |
|||
2 |
1,000 gpm |
75 |
|||
4 |
2,000 gpm |
250 |
- A well draws water from an aquifer that has an average water level of 150 ft bgs and pumps to a tank 225 ft above it. Friction loss to the tank is approximately 22 psi. If the well pumps at a rate of 2,300 gpm and has a wire‐to‐water efficiency of 62% how much will it cost to run this well 14 hours per day. Assume the electrical rate is $0.13 per kW‐Hr.
-
A utility manager is trying to determine which hp motor to purchase for a pump station. A 400 hp motor with a wire‐to‐water efficiency of 65% can pump 3,000 gpm. Similarly, a 250 hp motor with a wire‐to‐water efficiency of 75% can pump 2,050 gpm. With an electrical rate of $0.155 per kW‐Hr, how much would it cost to run each motor to achieve a daily flow of 2 MG? Which one is less expensive to run?
-
Approximately 224 kW of power are needed to run a certain booster pump. If the booster has a wire‐to‐water efficiency of 67.5% and is pumping against 135 psi of head pressure, what is the corresponding flow in gpm?
-
It costs $88.77 in electricity to run a well for 7 hours a day. The well has a TDH of 100 psi and an overall efficiency of 58.3%. The cost per kW‐Hr is $0.17. What is the cost of the water per gallon?
-
Complete the table below based on the information provided.
Well |
Flow (gpm) |
Run Time (Hr/Day) |
Wire-to-Water Eff |
Head Pressure (psi) |
hp |
Cost/Year ($) @ $0.135/kW-Hr |
---|---|---|---|---|---|---|
A |
750 |
18 |
68% |
110 |
||
B |
1,800 |
13 |
61% |
85 |
||
C |
2,750 |
12 |
57% |
95 |