Energy Equation and Its Applications

Abstract

Section 4.1 introduces the energy equation, which is of prime importance in fluid mechanics. This section includes discussion of different forms of energy and their units in USCS as well as SI systems. The mechanical energy equation and its different forms (energy per unit weight and energy per unit mass) commonly used in fluid mechanics are explained in Sect. 4.2. Section 4.3 covers the pump power equation and the formulas for obtaining pump power required in horsepower and kilowatts with various forms of inputs such as gallons per minute cubic feet per second (cfs). The practical applications of the pump power equation are clearly illustrated through several example problems with detailed solutions. The famous Bernoulli’s equation is considered in Sect. 4.4. The pump performance parameters and the performance curves are described in Sect. 4.5, which also illustrates the determination of the pump operating point. In addition, Sect. 4.5 also examines the details related to the operation of pumps including pumps operating in series and pumps operating in parallel. Section 4.6 takes a look at the “affinity laws.” Pump cavitation is examined in Sect. 4.7. The topic of net positive suction head, with applications in the configuration and layout of pumping systems, is examined in Sect. 4.8.

Appendices

Practice Problems

Practice Problem 4.1

Fuel oil with specific gravity 0.92 and kinematic viscosity 1.6 × 10⁻⁵ m²/s flows through a 65 mm DN and Sch.40 horizontal steel pipe (62.7 mm ID, roughness, ε = 0.05 mm) at a mass flow rate of 25000 kg/hr. Calculate the head (m) to be added by the pump to handle this flow across 100 m of the pipe.

Practice Problem 4.2

The pump used in Practice Problem 4.1 consumes 1.75 kW of power. The motor driving the pump has an efficiency of 80%. Calculate the efficiency at which the pump is operating.

Practice Problem 4.3

The total friction loss in the piping system shown in the figure is equivalent to 1.50 psi. The liquid being pumped has a specific gravity of 1.27. Both the reservoirs are open to the atmosphere. The efficiency of the pump is 73%, and the brake horsepower used is measured to be 60 hp. Determine the maximum flow rate (gpm) of the liquid that can be handled by the pumping system.

Practice Problem 4.4

Items A and B shown in the figure are pressure gages attached to the suction and discharge of the pump. The readings in the gages are P_A = 0.1 bar(g) and P_B = 2.7 bar(g). The pump is sourcing water from an open reservoir. The water flow rate is 25 L/s, and it discharges into an open tank, where the water surface is 15 m above the water level in the source tank. The Darcy friction factor for the flow is 0.022. Determine the equivalent length (m) of 100 mm DN (ID = 102.3 mm) piping that can be handled by this system.

Practice Problem 4.5

30,000 bpd (1 barrel = 0.1590 m³) of crude oil (SG = 0.85) flows through DN 500 mm (ID = 478 mm), Sch.40 pipeline. The pumping stations along this line are located 50 km apart. The friction factor for the pipeline is 0.02, and the efficiency for the pump is 75%. Calculate the pump power required in kW.

Practice Problem 4.6

Given the pump and system curves shown in the figure, determine the number of pumps and the configuration (series, parallel, etc.) required to operate at system conditions represented by point A.

Practice Problem 4.7

300 gpm of water is pumped using a centrifugal pump supplying a head of 200 ft. The horsepower consumed by the pump is 20 hp. The impeller speed is 1200 rpm. Determine the efficiency, capacity, head supplied, and power consumption of the pump when the speed is increased to 1500 rpm, assuming that the efficiency of the pump remains constant.

Practice Problem 4.8

A centrifugal pump is used in lifting water from a sump where the lowest possible water level is 12 ft below the centerline of the pump. The water in the sump is at a temperature of 80 °F and at atmospheric pressure. The friction losses in the suction line are equivalent to 5 ft of water. Calculate the net positive suction head available (NPSH_A) for the pump.

Solutions to Practice Problems

Practice Problem 4.1

• (Solution)

Calculate the density of fuel oil by using the definition of specific gravity (Eq. 1.5).

$$ {\displaystyle \begin{array}{l}{\rho}_{\mathrm{oil}}=\left({\mathrm{SG}}_{\mathrm{oil}}\right)\left({\rho}_{\mathrm{water},\mathrm{std}}\right)=(0.92)\left(1000\frac{\mathrm{kg}}{{\mathrm{m}}^3}\right)\\ {}=920\ \mathrm{kg}/{\mathrm{m}}^3\end{array}} $$

Convert the pipe ID to m.

$$ D=\frac{62.7\ \mathrm{m}\mathrm{m}}{\frac{1000\ \mathrm{m}\mathrm{m}}{\mathrm{m}}}=0.0627\ \mathrm{m} $$

Calculate the average velocity by using Eq. 3.5.

$$ \mathrm{v}=\frac{\dot{m}}{\rho A}=\frac{\left(\frac{25000\ \mathrm{kg}/\mathrm{hr}}{3600\ \mathrm{s}/\mathrm{hr}}\right)}{\left(920\frac{\mathrm{kg}}{{\mathrm{m}}^3}\right)\left(\frac{\pi }{4}\right){\left(0.0627\ \mathrm{m}\right)}^2}=2.44\ \mathrm{m}/\mathrm{s} $$

Calculate the Reynolds number for the flow by using Eq. 3.7.

$$ \operatorname{Re}=\frac{D\mathrm{v}}{\nu }=\frac{\left(0.0627\ \mathrm{m}\right)\left(2.44\ \frac{\mathrm{m}}{\mathrm{s}}\right)}{\left(1.6\times {10}^{-5}\ \frac{{\mathrm{m}}^2}{\mathrm{s}}\right)}=9.56\times {10}^3 $$

Calculate the relative roughness of the pipe surface. Using Eq. 3.17, the relative roughness is

$$ r=\frac{\ \varepsilon }{D}=\frac{0.05\ \mathrm{mm}}{62.7\ \mathrm{mm}}=0.0008 $$

Locate the intersection of the ordinate at Re = 9.56 × 10³ and the curve for r = 0.0008, as shown in the accompanying figure. At the point of intersection, f = 0.034

The head to be added by the pump is equivalent to the head loss in the pipe. Calculate the head loss in the pipe using the Darcy equation (Eq. 3.12).

$$ {\displaystyle \begin{array}{l}{h}_{\mathrm{Pump}}={h}_{\mathrm{f}}=f\left(\frac{L}{D}\right)\left(\frac{{\mathrm{v}}^2}{2g}\right)\\ {}=(0.034)\left(\frac{100\ \mathrm{m}\ }{0.0627\ \mathrm{m}}\right)\left[\frac{{\left(2.44\ \frac{\mathrm{m}}{\mathrm{s}}\right)}^2}{(2)\left(9.81\ \frac{\mathrm{m}}{{\mathrm{s}}^2}\right)}\right]\\ {}=16.45\ \mathrm{m}\ \mathrm{of}\ \mathrm{oil}\end{array}} $$

Practice Problem 4.2

• (Solution)

Calculate the volume flow rate of oil by using the data in Practice Problem 4.1. and Eq. 3.5.

$$ Q=\frac{\dot{m}}{\rho_{\mathrm{oil}}}=\frac{\left(\frac{25000\ \mathrm{kg}/\mathrm{hr}}{3600\ \mathrm{s}/\mathrm{hr}}\right)}{920\frac{\mathrm{kg}}{{\mathrm{m}}^3}}=0.0075\ {\mathrm{m}}^3/\mathrm{s} $$

Calculate the specific weight of oil by using Eq. 1.5.

$$ {\gamma}_{\mathrm{oil}}=\left({\mathrm{SG}}_{\mathrm{oil}}\right)\left({\gamma}_{\mathrm{water},\mathrm{std}}\right)=(0.92)\left(9.81\frac{\mathrm{kN}}{{\mathrm{m}}^3}\right)=9.025\ \mathrm{kN}/{\mathrm{m}}^3 $$

From Practice Problem 4.1, h_Pump = 16.45 m of oil.

Solve Eq. 4.11 for the pump efficiency, and substitute the known values

$$ {\displaystyle \begin{array}{l}{\eta}_{\mathrm{Pump}}=\frac{Q\gamma {h}_{\mathrm{Pump}}}{P_{\mathrm{kW}}{\eta}_{\mathrm{Motor}}}=\frac{\left(0.0075\frac{{\mathrm{m}}^3}{\mathrm{s}}\right)\left(9.025\frac{\mathrm{kN}}{{\mathrm{m}}^3}\right)\left(16.45\ \mathrm{m}\right)}{\left(1.75\ \mathrm{kW}\right)(0.80)}\\ {}=0.7953\ \left(79\%\right)\end{array}} $$

Practice Problem 4.3

• (Solution)

Calculate the specific weight of the liquid using Eq. 1.5.

$$ {\displaystyle \begin{array}{l}{\gamma}_{\mathrm{liquid}}=\left(\mathrm{SG}\right)\left({\gamma}_{\mathrm{water},\mathrm{std}}\right)=(1.27)\left(62.4\frac{\mathrm{lbf}}{{\mathrm{ft}}^3}\right)\\ {}=79.25\ \mathrm{lbf}/{\mathrm{ft}}^3\end{array}} $$

From Eq.4.4, the head added by the pump is

$$ {h}_{\mathrm{Pump}}=\frac{P_2-{P}_1}{\gamma }+\frac{{\mathrm{v}}_2^2-{\mathrm{v}}_1^2}{2g}+\left({z}_2-{z}_1\right)+{h}_{\mathrm{f}} $$

Select the reference points 1 and 2 as shown in the figure.

Since the liquid surfaces are open to the atmosphere,

$$ {P}_1\ \left(\mathrm{gage}\right)=0\ \mathrm{psig}\ \mathrm{and}\ {P}_2\ \left(\mathrm{gage}\right)=0\ \mathrm{psig}. $$

Since the cross sections of the tanks are large and since the level of the liquid in each tank is constant, the velocity at the reference points 1 and 2 are zero, that is, v₁ = 0, and v₂ = 0.

The difference in elevation between points 1 and 2 is z₂ – z₁ = 190 ft.

Convert the friction loss from pressure units to equivalent head of the liquid by using Eq. 2.1.

$$ {h}_{\mathrm{f}}=\frac{\Delta {P}_{\mathrm{f}}}{\gamma }=\frac{1.50\frac{\mathrm{lbf}}{{\mathrm{in}}^2}\times \frac{144\ {\mathrm{in}}^2}{{\mathrm{f}\mathrm{t}}^2}}{79.25\frac{\mathrm{lbf}}{{\mathrm{f}\mathrm{t}}^3}}=2.726\ \mathrm{ft}\ \mathrm{liquid} $$

Substitute the known values into the equation for the head added by the pump.

$$ {\displaystyle \begin{array}{l}{h}_{\mathrm{Pump}}=\frac{P_2-{P}_1}{\gamma }+\frac{{\mathrm{v}}_2^2-{\mathrm{v}}_1^2}{2g}+\left({z}_2-{z}_1\right)+{h}_{\mathrm{f}}\\ {}=0+0+190\ \mathrm{ft}+2.726\ \mathrm{ft}\\ {}=192.73\ \mathrm{ft}\end{array}} $$

Solve Eq. 4.7 for the volume flow rate, Q_gpm, and substitute the known values.

$$ {\displaystyle \begin{array}{l}{Q}_{\mathrm{gpm}}=\frac{\left(\mathrm{BHP}\right)\left({\eta}_{\mathrm{Pump}}\right)\left(3960\frac{\mathrm{ft}\hbox{-} \mathrm{gpm}}{\mathrm{hp}}\right)}{\left(\mathrm{SG}\right)\left({h}_{\mathrm{Pump}}\right)}\\ {}=\frac{\left(60\ \mathrm{hp}\right)(0.73)\left(3960\frac{\mathrm{ft}\hbox{-} \mathrm{gpm}}{\mathrm{hp}}\right)}{(1.27)\left(192.73\ \mathrm{ft}\right)}\\ {}=708.62\ \mathrm{gpm}\end{array}} $$

Practice Problem 4.4

• (Solution)

Convert the differential pressure across the pump into equivalent head added by the pump (in meters) by using Eq. 3.8.

$$ {\displaystyle \begin{array}{l}{h}_{\mathrm{pump}}=\frac{\Delta {P}_{\mathrm{pump}}}{\gamma }=\frac{\left(2.7\ \mathrm{bar}-0.1\ \mathrm{bar}\right)\left(\frac{100\ \mathrm{kPa}}{1\ \mathrm{bar}}\right)}{9.81\frac{\mathrm{kN}}{{\mathrm{m}}^3}}\\ {}=26.50\ \mathrm{m}\ \mathrm{of}\ \mathrm{water}\end{array}} $$

• Note: kPa ≡ kN/m²

Apply the mechanical energy equation (Eq. 4.4) between the water surfaces in the source tank (reference point 1) and in the discharge tank (reference point 2). Since the water surfaces are open to the atmosphere, P₁ (gage) = 0 psig and P₂ (gage) = 0 psig. Since the cross sections of the tanks are large and since the level of the liquid in each tank is constant, the velocity at the reference points 1 and 2 are zero, that is, v₁ = 0, and v₂ = 0. The difference in elevation between points 1 and 2 is z₂ – z₁ = 15 m. Simplify Eq. 4.4 using the preceding substitutions.

$$ {\displaystyle \begin{array}{l}{h}_{\mathrm{Pump}}=\frac{P_2-{P}_1}{\gamma }+\frac{{\mathrm{v}}_2^2-{\mathrm{v}}_1^2}{2g}+\left({z}_2-{z}_1\right)+{h}_{\mathrm{f}}\\ {}=0+0+15\ \mathrm{m}+{h}_{\mathrm{f}}\end{array}} $$

As calculated earlier, the head added by the pump is equivalent to 26.50 m of water. Therefore, the friction head loss is

$$ {h}_{\mathrm{f}}={h}_{\mathrm{pump}}-15\ \mathrm{m}=26.50\ \mathrm{m}-15\ \mathrm{m}=11.50\ \mathrm{m} $$

Calculate the velocity in the pipe by using the continuity equation (Eq. 3.4).

$$ \mathrm{v}=\frac{Q}{A}=\frac{Q}{\frac{\pi {D}^2}{4}}=\frac{\left(25\ \frac{\mathrm{L}}{\mathrm{s}}\right)\left(\frac{1\ {\mathrm{m}}^3}{1000\ \mathrm{L}}\right)}{\frac{\uppi}{4}{\left(0.1023\ \mathrm{m}\right)}^2}=3.04\ \mathrm{m}/\sec $$

Calculate the equivalent length of the piping system that can be handled by the pump by using the Darcy equation (Eq. 3.12).

$$ {h}_{\mathrm{f}}=f\left(\frac{L_{\mathrm{eq}}}{D}\right)\left(\frac{{\mathrm{v}}^2}{2g}\right) $$

$$ {\displaystyle \begin{array}{l}{L}_{\mathrm{eq}}=\frac{2{gDh}_{\mathrm{f}}}{f{\mathrm{v}}^2}=\frac{2\times 9.81\frac{\mathrm{m}}{{\mathrm{s}}^2}\times 0.1023\ \mathrm{m}\times 11.50\ \mathrm{m}}{(0.022){\left(3.04\frac{\mathrm{m}}{\mathrm{s}}\right)}^2}\\ {}=113.53\ \mathrm{m}\end{array}} $$

Practice Problem 4.5

• (Solution)

Convert the volume flow rate to m³/s.

$$ {\displaystyle \begin{array}{l}Q=\left(30000\frac{\mathrm{barrels}}{\mathrm{day}}\right)\left(\frac{0.1590\ {\mathrm{m}}^3}{1\ \mathrm{barrel}}\right)\left(\frac{1\ \mathrm{day}}{24\ \mathrm{hr}\mathrm{s}}\right)\left(\frac{1\ \mathrm{hr}}{3600\ \mathrm{s}}\right)\\ {}=0.0552\ {\mathrm{m}}^3/\mathrm{s}\end{array}} $$

Calculate the velocity in the pipeline by using the continuity equation (Eq. 3.4).

$$ \mathrm{v}=\frac{Q}{A}=\frac{Q}{\frac{\pi {D}^2}{4}}=\frac{0.0552\frac{{\mathrm{m}}^3}{\mathrm{s}}\ }{\frac{\pi }{4}{\left(0.478\ \mathrm{m}\right)}^2}=0.3076\ \mathrm{m}/\mathrm{s} $$

Calculate the friction head loss in the pipeline using the Darcy equation (Eq. 3.12).

$$ {\displaystyle \begin{array}{l}{h}_{\mathrm{f}}=f\left(\frac{L}{D}\right)\left(\frac{{\mathrm{v}}^2}{2g}\right)=(0.02)\left(\frac{50000\ \mathrm{m}}{0.478\ \mathrm{m}}\right)\left(\frac{{\left(0.3076\frac{\mathrm{m}}{\mathrm{s}}\right)}^2}{2\times 9.81\frac{\mathrm{m}}{{\mathrm{s}}^2}}\right)\\ {}=10.09\ \mathrm{m}\end{array}} $$

The head to be supplied by the pump is equivalent to the friction head loss in the pipeline.

$$ {h}_{\mathrm{Pump}}={h}_{\mathrm{f}}=10.09\ \mathrm{m}\ \mathrm{of}\ \mathrm{oil} $$

Calculate the pump power required using Eq. 4.10.

$$ {\displaystyle \begin{array}{l}\mathrm{Power}\ \left(\mathrm{kW}\right)=\frac{Q{\gamma}_{\mathrm{oil}}{h}_{\mathrm{Pump}}}{\eta_{\mathrm{Pump}}}\\ {}=\frac{\left(0.0552\frac{{\mathrm{m}}^3}{\mathrm{s}}\right)\left(0.85\times 9.81\frac{\mathrm{kN}}{{\mathrm{m}}^3}\right)\left(10.09\ \mathrm{m}\right)}{0.75}\\ {}=6.19\ \mathrm{kW}\end{array}} $$

• Note: kN.m/s ≡ kJ/s ≡ kW since N.m ≡ J, and J/s ≡ W

Practice Problem 4.6

• (Solution)

Determine the capacity at operating point A as shown in the figure. Therefore, at operating point A, the required flow rate is 750 gpm, and the head added is 200 ft. From the pump curve, at a capacity of 400 gpm, the head added is 100 ft. To meet the flow requirement of 750 gpm, two pumps in parallel will be required. To meet the head requirement of 200 ft, two additional pumps in series will be required in conjunction with the first set of parallelly operating pumps as shown in the figure.

Each pump will handle 400 gpm, providing a head of 100 ft. The combined flow from the second set of pumps can be throttled to 750 gpm as shown in the figure.

Practice Problem 4.7

• (Solution)

Calculate the hydraulic horsepower required by using Eq. 4.6.

$$ {\displaystyle \begin{array}{l}\mathrm{HHP}=\frac{Q_{\mathrm{gpm}}\left(\mathrm{SG}\right){h}_{\mathrm{Pump}}}{3960\frac{\mathrm{ft}-\mathrm{gpm}}{\mathrm{hp}}}=\frac{\left(300\ \mathrm{gpm}\right)(1.0)\left(200\ \mathrm{ft}\right)}{3960\frac{\mathrm{ft}-\mathrm{gpm}}{\mathrm{hp}}}\\ {}=15.15\ \mathrm{hp}\end{array}} $$

Calculate the efficiency of the pump by using Eq. 4.7.

$$ {\eta}_{\mathrm{Pump}}=\frac{\mathrm{HHP}}{\mathrm{BHP}}=\frac{15.15\ \mathrm{hp}}{20\ \mathrm{hp}}=0.76\ \left(76\%\right) $$

The efficiency of the pump remains at 76% when the speed is increased to 1500 rpm.

Determine the capacity, head supplied, and power consumption at 1500 rpm by using the affinity laws.

The capacity at 1500 rpm can be determined from Eq. 4.18.

$$ {\displaystyle \begin{array}{l}{Q}_2={Q}_1\left(\frac{N_2}{N_1}\right)=\left(300\ \mathrm{gpm}\right)\left(\frac{1500\ \mathrm{rpm}}{1200\ \mathrm{rpm}}\right)\\ {}=375\ \mathrm{gpm}\end{array}} $$

The head supplied at 1500 rpm can be determined from Eq. 4.19.

$$ {\displaystyle \begin{array}{l}{h}_2={h}_1{\left(\frac{N_2}{N_1}\right)}^2=\left(200\ \mathrm{ft}\right){\left(\frac{1500\ \mathrm{rpm}}{1200\ \mathrm{rpm}}\right)}^2\\ {}=312.5\ \mathrm{ft}\end{array}} $$

The power consumed at 1500 rpm can be determined from Eq. 4.20.

$$ {\displaystyle \begin{array}{l}{P}_2={P}_1{\left(\frac{N_2}{N_1}\right)}^3=\left(20\ \mathrm{hp}\right){\left(\frac{1500\ \mathrm{rpm}}{1200\ \mathrm{rpm}}\right)}^3\\ {}=39.1\ \mathrm{hp}\end{array}} $$

Practice Problem 4.8

• (Solution)

Convert the absolute pressure of water, which is atmospheric pressure, to equivalent column of water by using Eq. 2.1 or alternatively use the known pressure equivalents.

$$ {h}_{\mathrm{abs}}=\frac{P}{\gamma }=\frac{\left(14.7\frac{\mathrm{lbf}}{{\mathrm{in}}^2}\right)\left(\frac{144\ {\mathrm{in}}^2}{{\mathrm{ft}}^2}\right)}{62.4\frac{\mathrm{lbf}}{{\mathrm{ft}}^3}}=33.92\ \mathrm{ft}\ \mathrm{of}\ \mathrm{water} $$

Pressure Equivalents: 1 atm = 14.7 psi = 101 kPa = 33.92 ft of water

Find the saturation pressure (vapor pressure) of water at 80 °F from steam tables, and convert it into equivalent height of water column.

$$ {h}_{\mathrm{sat}}=\frac{P_{\mathrm{sat}}\ \mathrm{at}\ 80{}^{\circ}\mathrm{F}}{\gamma }=\frac{\left(0.51\frac{\mathrm{lbf}}{{\mathrm{in}}^2}\right)\left(\frac{144\ {\mathrm{in}}^2}{{\mathrm{ft}}^2}\right)}{62.4\frac{\mathrm{lbf}}{{\mathrm{ft}}^3}}=1.18\ \mathrm{ft}\ \mathrm{of}\ \mathrm{water} $$

Substitute the known values into Eq. 4.22. Since the water level in the sump is below the centerline of the pump, the elevation will be negative.

$$ {\displaystyle \begin{array}{l}{\mathrm{NPSH}}_{\mathrm{A}}={h}_{\mathrm{abs}}+{h}_{\mathrm{elev}}-{h}_{\mathrm{f}}-{h}_{\mathrm{sat}}\\ {}=33.92\ \mathrm{ft}-12\ \mathrm{ft}-5\ \mathrm{ft}-1.18\ \mathrm{ft}\\ {}=15.74\ \mathrm{ft}\end{array}} $$