8.6: Power Requirements
- Page ID
- 44582
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)A pump transfers energy from an electric motor or engine to the water (Figure 8.15). Since a pump cannot be 100% efficient, pump efficiency (Ep) is used to account for the energy lost in pumping and is defined as:
\(E_p= \dfrac{\text{Output of energy or power}}{\text{Input of energy of power}}\)
It is also necessary to determine how large of an engine or motor is required to pump the water. Horsepower (hp) is the typical unit of power in the USCS system and is defined as:
1 hp = 33,000 ft-lb/min
Thus, to lift 33,000 pounds of water at the rate of 1 foot per minute, 1 horsepower would be required. A gallon of water weighs 8.33 pounds, so 1 horsepower would lift approximately 3,960 gallons of water at the rate of 1 foot per minute. The power required to pressurize and lift water (called water horsepower) may be expressed by:
Figure 8.15. Pumping plants including a well with a vertical turbine pump and a power supply: electric motor (left) and internal combustion engine (right).
\(whp = \dfrac{(Q \times TDH)}{3960}\)
where: whp = water horsepower,
Q = flow rate (gpm), and
TDH = total dynamic head (ft).
Water horsepower is the power that is actually added to the water. Since the pump has some inefficiency, the power input to the pump must be more than the water horsepower. The power input to the pump is called the brake horsepower (bhp) or pump horsepower and is determined by:
\(bhp=\dfrac{whp}{E_p} \)
Example 8.6 A pump operating at 80% efficiency lifts water form a reservoir a vertical distance of 100 feet and also develops a pressure of 50 psi. If the flow rate is 800 gpm, what is the water horsepower requirement? What is the brake horsepower requirement?
Given: Q = 800 gpm P = 50 psi
L = 100 ft Ep = 0.80
Find: wph, bhp
Solution
\(TDH = 2.31 \dfrac{\text{ft}}{psi}P+L \) (Equation 8.14)
\(TDH = 2.31 \dfrac{\text{ft}}{psi}(50 \text{ psi})+100 \text{ ft}=216 \text{ ft} \)
\(whp= \dfrac{Q \times TDH}{3960}\) (Equation 8.18)
\(whp= \dfrac{(800 \text{ gpm})(216 \text{ ft})}{3960}=44 \text{ hp}\)
\(bhp= \dfrac{whp}{E_p}\) (Equation 8.19)
\(bhp= \dfrac{44 \text{ hp}}{0.80}= 55 \text{ hp}\)