2.7. OVERALL ENERGY BALANCE

2.7A. Introduction

The second property to be considered in the overall balances on a control volume is energy. We shall apply the principle of the conservation of energy to a control volume fixed in space in much the same manner as the principle of conservation of mass was used to obtain the overall mass balance. The energy-conservation equation will then be combined with the first law of thermodynamics to obtain the final overall energy-balance equation.

We can write the first law of thermodynamics as

Equation 2.7-1


where E is the total energy per unit mass of fluid, Q is the heat absorbed per unit mass of fluid, and W is the work of all kinds done per unit mass of fluid upon the surroundings. In the calculations, each term in the equation must be expressed in the same type of units, such as J/kg (SI), btu/lbm, or ft · lbf/lbm (English).

Since mass carries with it associated energy due to its position, motion, or physical state, we will find that each of these types of energy will appear in the energy balance. In addition, we can also transport energy across the boundary of the system without transferring mass.

2.7B. Derivation of Overall Energy-Balance Equation

The entity balance for a conserved quantity such as energy is similar to Eq. (2.6-3) and is as follows for a control volume:

Equation 2.7-2


The energy E present within a system can be classified in three ways:

  1. Potential energy zg of a unit mass of fluid is the energy present because of the position of the mass in a gravitational field g, where z is the relative height in meters from a reference plane. The units for zg in the SI system are m · m/s2. Multiplying and dividing by kg mass, the units can be expressed as (kg · m/s2) · (m/kg), or J/kg. In English units the potential energy is zg/gc in ft · lbf/lbm.

  2. Kinetic energy ν2/2 of a unit mass of fluid is the energy present because of translational or rotational motion of the mass, where ν is the velocity in m/s relative to the boundary of the system at a given point. Again, in the SI system the units of ν2/2 are J/kg. In the English system the kinetic energy is ν2/2gc in ft · lbf/lbm.

  3. Internal energy U of a unit mass of a fluid is all of the other energy present, such as rotational and vibrational energy in chemical bonds. Again the units are in J/kg or ft · lbf/lbm.

The total energy of the fluid per unit mass is then

Equation 2.7-3


The rate of accumulation of energy within the control volume V in Fig. 2.6-4 is

Equation 2.7-4


Next we consider the rate of energy input and output associated with mass in the control volume. The mass added or removed from the system carries internal, kinetic, and potential energy. In addition, energy is transferred when mass flows into and out of the control volume. Net work is done by the fluid as it flows into and out of the control volume. This pressure–volume work per unit mass fluid is pV. The contribution of shear work is usually neglected. The pV term and U term are combined using the definition of enthalpy, H:

Equation 2.7-5


Hence, the total energy carried with a unit mass is (H + ν2/2 + zg).

For a small area dA on the control surface in Fig. 2.6-4, the rate of energy efflux is (H + ν2/2 + zg)(ρν)(dA cos α), where (dA cos α) is the area dA projected in a direction normal to the velocity vector ν and α is the angle between the velocity vector ν and the outward-directed unit normal vector n. We now integrate this quantity over the entire control surface to obtain

Equation 2.7-6


Now we have accounted for all energy associated with mass in the system and moving across the boundary in the entity balance, Eq. (2.7-2). Next we take into account heat and work energy which transfers across the boundary and is not associated with mass. The term q is the heat transferred per unit time across the boundary to the fluid because of a temperature gradient. Heat absorbed by the system is positive by convention.

The work , which is energy per unit time, can be divided into , purely mechanical shaft work identified with a rotating shaft crossing the control surface, and the pressure–volume work, which has been included in the enthalpy term H in Eq. (2.7-6). By convention, work done by the fluid upon the surroundings, that is, work out of the system, is positive.

To obtain the overall energy balance, we substitute Eqs. (2.7-4) and (2.7-6) into the entity balance Eq. (2.7-2) and equate the resulting equation to q:

Equation 2.7-7


2.7C. Overall Energy Balance for Steady-State Flow System

A common special case of the overall or macroscopic energy balance is that of a steady-state system with one-dimensional flow across the boundaries, a single inlet, a single outlet, and negligible variation of height z, density ρ, and enthalpy H across either inlet or outlet area. This is shown in Fig. 2.7-1. Setting the accumulation term in Eq. (2.7-7) equal to zero and integrating,

Equation 2.7-8


Figure 2.7-1. Steady-state flow system for a fluid.


For steady state, m1 = ρ1ν1 avA1 = m2 = m. Dividing through by m so that the equation is on a unit mass basis,

Equation 2.7-9


The term (ν3)av/(2νav) can be replaced by , where α is the kinetic-energy velocity correction factor and is equal to . The term α has been evaluated for various flows in pipes and is for laminar flow and close to 1.0 for turbulent flow. (See Section 2.7D.) Hence, Eq. (2.7-9) becomes

Equation 2.7-10


Some useful conversion factors from Appendix A.1 are as follows:


2.7D. Kinetic-Energy Velocity Correction Factor α

1. Introduction

In obtaining Eq (2.7-8) it was necessary to integrate the kinetic-energy term,

Equation 2.7-11


which appeared in Eq. (2.7-7). To do this we first take ρ as a constant and cos α = 1.0. Then multiplying the numerator and denominator by νavA, where νav is the bulk or average velocity, and noting that m = ρνavA, Eq. (2.7-11) becomes

Equation 2.7-12


Dividing through by m so that Eq. (2.7-12) is on a unit mass basis,

Equation 2.7-13


where α is defined as

Equation 2.7-14


and (ν3)av is defined as follows:

Equation 2.7-15


The local velocity ν varies across the cross-sectional area of a pipe. To evaluate (ν3)av and, hence, the value of α, we must have an equation relating ν as a function of position in the cross-sectional area.

2. Laminar flow

In order to determine the value of α for laminar flow, we first combine Eqs. (2.6-18) and (2.6-20) for laminar flow to obtain ν as a function of position r:

Equation 2.7-16


Substituting Eq. (2.7-16) into (2.7-15) and noting that A = πR2 and dA = r dr dθ (see Example 2.6-3), Eq. (2.7-15) becomes

Equation 2.7-17


Integrating Eq. (2.7-17) and rearranging,

Equation 2.7-18


Substituting Eq. (2.7-18) into (2.7-14),

Equation 2.7-19


Hence, for laminar flow the value of α to use in the kinetic-energy term of Eq. (2.7-10) is 0.50.

3. Turbulent flow

For turbulent flow a relationship is needed between ν and position. This can be approximated by the following expression:

Equation 2.7-20


where r is the radial distance from the center. Eq. (2.7-20) is substituted into Eq. (2.7-15) and the resultant integrated to obtain the value of (ν3)av. Next, Eq. (2.7-20) is substituted into Eq. (2.6-17) and this equation integrated to obtain νav and (νav)3. Combining the results for (ν3)av and (νav)3 into Eq. (2.7-14), the value of a α is 0.945. (See Problem 2.7-1 for solution.) The value of α for turbulent flow varies from about 0.90 to 0.99. In most cases (except for precise work) the value of α is taken to be 1.0.

2.7E. Applications of Overall Energy-Balance Equation

The total energy balance, Eq. (2.7-10), in the form given is not often used when appreciable enthalpy changes occur or appreciable heat is added (or subtracted), since the kinetic- and potential-energy terms are usually small and can be neglected. As a result, when appreciable heat is added or subtracted or large enthalpy changes occur, the methods for doing heat balances described in Section 1.7 are generally used. Examples will be given to illustrate this and other cases.

EXAMPLE 2.7-1. Energy Balance on Steam Boiler

Water enters a boiler at 18.33°C and 137.9 kPa through a pipe at an average velocity of 1.52 m/s. Exit steam at a height of 15.2 m above the liquid inlet leaves at 137.9 kPa, 148.9°C, and 9.14 m/s in the outlet line. At steady state, how much heat must be added per kg mass of steam? The flow in the two pipes is turbulent.

Solution: The process flow diagram is shown in Fig. 2.7-2. Rearranging Eq. (2.7-10) and setting α = 1 for turbulent flow and WS = 0 (no external work),

Equation 2.7-21


Figure 2.7-2. Process flow diagram for Example 2.7-1.


To solve for the kinetic-energy terms,


Taking the datum height z1 at point 1, z2 = 15.2 m. Then,


From Appendix A.2, steam tables in SI units, H1 at 18.33°C = 76.97 kJ/kg, H2 of superheated steam at 148.9°C = 2771.4 kJ/kg, and


Substituting these values into Eq. (2.7-21),


Hence, the kinetic-energy and potential-energy terms totaling 189.75 J/kg are negligible compared to the enthalpy change of 2.694 × 106 J/kg. This 189.75 J/kg would raise the temperature of liquid water about 0.0453°C, a negligible amount.


EXAMPLE 2.7-2. Energy Balance on a Flow System with a Pump

Water at 85.0°C is being stored in a large, insulated tank at atmospheric pressure, as shown in Fig. 2.7-3. It is being pumped at steady state from this tank at point 1 by a pump at the rate of 0.567 m3/min. The motor driving the pump supplies energy at the rate of 7.45 kW. The water passes through a heat exchanger, where it gives up 1408 kW of heat. The cooled water is then delivered to a second large open tank at point 2, which is 20 m above the first tank. Calculate the final temperature of the water delivered to the second tank. Neglect any kinetic-energy changes, since the initial and final velocities in the tanks are essentially zero.

Figure 2.7-3. Process flow diagram for energy balance for Example 2.7-2.


Solution: From Appendix A.2, steam tables, H1 (85°C) = 355.90 × 103 J/kg and ρ1 = 1/0.0010325 = 968.5 kg/m3. Then, for steady state,


Also, z1 = 0 and z2 = 20 m. The work done by the fluid is WS, but in this case work is done on the fluid and WS is negative:


The heat added to the fluid is also negative since it gives up heat and is


Setting and substituting into Eq. (2.7-10),


Solving, H2 = 202.71 × 103 J/kg. From the steam tables this corresponds to t2 = 48.41°C. Note that in this example, WS and g(z2z1) are very small compared to Q.


EXAMPLE 2.7-3. Energy Balance in Flow Calorimeter

A flow calorimeter is being used to measure the enthalpy of steam. The calorimeter, which is a horizontal insulated pipe, consists of an electric heater immersed in a fluid flowing at steady state. Liquid water at 0°C at a rate of 0.3964 kg/min enters the calorimeter at point 1. The liquid is vaporized completely by the heater, where 19.63 kW is added, and steam leaves point 2 at 250°C and 150 kPa absolute. Calculate the exit enthalpy H2 of the steam if the liquid enthalpy at 0°C is set arbitrarily as 0. The kinetic-energy changes are small and can be neglected. (It will be assumed that pressure has a negligible effect on the enthalpy of the liquid.)

Solution: For this case, WS = 0 since there is no shaft work between points 1 and 2. Also, and g(z2z1) = 0. For steady state, m1 = m2 = 0.3964/60 = 6.607 × 103 kg/s. Since heat is added to the system,


The value of H1 = 0. Equation (2.7-10) becomes


The final equation for the calorimeter is

Equation 2.7-22


Substituting Q = 2971 kJ/kg and H1 = 0 into Eq. (2.7-22), H2 = 2971 kJ/kg at 250°C and 150 kPa, which is close to the value from the steam table of 2972.7 kJ/kg.


2.7F. Overall Mechanical-Energy Balance

A more useful type of energy balance for flowing fluids, especially liquids, is a modification of the total energy balance to deal with mechanical energy. Engineers are often concerned with this special type of energy, called mechanical energy, which includes the work term, kinetic energy, potential energy, and the flow work part of the enthalpy term. Mechanical energy is a form of energy that is either work or a form that can be directly converted into work. The other terms in the energy-balance equation (2.7-10), heat terms and internal energy, do not permit simple conversion into work because of the second law of thermodynamics and the efficiency of conversion, which depends on the temperatures. Mechanical-energy terms have no such limitation and can be converted almost completely into work. Energy converted to heat or internal energy is lost work or a loss in mechanical energy caused by frictional resistance to flow.

It is convenient to write an energy balance in terms of this loss, Σ F, which is the sum of all frictional losses per unit mass. For the case of steady-state flow, when a unit mass of fluid passes from inlet to outlet, the batch work done by the fluid, W', is expressed as

Equation 2.7-23


This work W' differs from the W of Eq. (2.7-1), which also includes kinetic- and potential-energy effects. Writing the first law of thermodynamics for this case, where ΔE becomes ΔU,

Equation 2.7-24


The equation defining enthalpy, Eq. (2.7-5), can be written as

Equation 2.7-25


Substituting Eq. (2.7-23) into (2.7-24) and then combining the resultant with Eq. (2.7-25), we obtain

Equation 2.7-26


Finally, we substitute Eq. (2.7-26) into (2.7-10) and 1/ρ for V, to obtain the overall mechanical-energy-balance equation:

Equation 2.7-27


For English units the kinetic- and potential-energy terms of Eq. (2.7-27) are divided by gc.

The value of the integral in Eq. (2.7-27) depends on the equation of state of the fluid and the path of the process. If the fluid is an incompressible liquid, the integral becomes (p2p1)/ρ and Eq. (2.7-27) becomes

Equation 2.7-28


EXAMPLE 2.7-4. Mechanical-Energy Balance on Pumping System

Water with a density of 998 kg/m3 is flowing at a steady mass flow rate through a uniform-diameter pipe. The entrance pressure of the fluid is 68.9 kN/m2 abs in the pipe, which connects to a pump that actually supplies 155.4 J/kg of fluid flowing in the pipe. The exit pipe from the pump is the same diameter as the inlet pipe. The exit section of the pipe is 3.05 m higher than the entrance, and the exit pressure is 137.8 kN/m2 abs. The Reynolds number in the pipe is above 4000 in the system. Calculate the frictional loss Σ F in the pipe system.

Solution: First a flow diagram of the system is drawn (Fig. 2.7-4), with 155.4 J/kg mechanical energy added to the fluid. Hence, WS = −155.4, since the work done by the fluid is positive.

Figure 2.7-4. Process flow diagram for Example 2.7-4.


Setting the datum height z1 = 0, z2 = 3.05 m. Since the pipe is of constant diameter, ν1 = ν2. Also, for turbulent flow α = 1.0 and


Since the liquid can be considered incompressible, Eq. (2.7-28) is used:


Using Eq. (2.7-28) and solving for Σ F, the frictional losses,

Equation 2.7-29


Substituting the known values, and solving for the frictional losses,



EXAMPLE 2.7-5. Pump Horsepower in Flow System

A pump draws 69.1 gal/min of a liquid solution having a density of 114.8 lbm/ft3 from an open storage feed tank of large cross-sectional area through a 3.068-in.-ID suction line. The pump discharges its flow through a 2.067-in.-ID line to an open overhead tank. The end of the discharge line is 50 ft above the level of the liquid in the feed tank. The friction losses in the piping system are Σ F = 10.0 ft-lb force/lb mass. What pressure must the pump develop and what is the horsepower of the pump if its efficiency is 65% (η = 0.65)? The flow is turbulent.

Solution: First, a flow diagram of the system is drawn (Fig. 2.7-5). Equation (2.7-28) will be used. The term WS in Eq. (2.7-28) becomes

Equation 2.7-30


Figure 2.7-5. Process flow diagram for Example 2.7-5.


where −WS = mechanical energy actually delivered to the fluid by the pump or net mechanical work, η = fractional efficiency, and Wp is the energy or shaft work delivered to the pump.

From Appendix A.5, the cross-sectional area of the 3.068-in. pipe is 0.05134 ft2 and of the 2.067-in. pipe, 0.0233 ft2. The flow rate is


ν1 = 0, since the tank is very large. Then The pressure p1 = 1 atm and p2 = 1 atm. Also, α = 1.0 since the flow is turbulent. Hence,


Using the datum of z1 = 0, we have


Using Eq. (2.4-28), solving for WS, and substituting the known values,


Using Eq. (2.7-30) and solving for Wp,



To calculate the pressure the pump must develop, Eq. (2.7-28) must be written over the pump itself between points 3 and 4 as shown on the diagram:


Since the difference in level between z3 and z4 of the pump itself is negligible, it will be neglected. Rewriting Eq. (2.7-28) between points 3 and 4 and substituting known values (Σ F = 0, since this is for the piping system),

Equation 2.7-31



2.7G. Bernoulli Equation for Mechanical-Energy Balance

In the special case where no mechanical energy is added (WS = 0) and for no friction (Σ F= 0), then Eq. (2.7-28) becomes the Bernoulli equation, Eq. (2.7-32), for turbulent flow, which is of sufficient importance to deserve further discussion:

Equation 2.7-32


This equation covers many situations of practical importance and is often used in conjunction with the mass-balance equation (2.6-2) for steady state:

Equation 2.6-2


Several examples of its use will be given.

EXAMPLE 2.7-6. Rate of Flow from Pressure Measurements

A liquid with a constant density ρ kg/m3 is flowing at an unknown velocity ν1 m/s through a horizontal pipe of cross-sectional area A1 m2 at a pressure p1 N/m2, and then it passes to a section of the pipe in which the area is reduced gradually to A2 m2 and the pressure is p2. Assuming no friction losses, calculate the velocities ν1 and ν2 if the pressure difference (p1p2) is measured.

Solution: In Fig. 2.7-6, the flow diagram is shown with pressure taps to measure p1 and p2. From the mass-balance continuity equation (2.6-2), for constant ρ where ρ1 = ρ2 = ρ,

Equation 2.7-33


Figure 2.7-6. Process flow diagram for Example 2.7-6.


For the items in the Bernoulli equation (2.7-32), for a horizontal pipe,


Then Eq. (2.7-32) becomes, after substituting Eq. (2.7-33) for ν2,

Equation 2.7-34


Rearranging,

Equation 2.7-35


Equation 2.7-36


Performing the same derivation but in terms of ν2,

Equation 2.7-37



EXAMPLE 2.7-7. Rate of Flow from a Nozzle in a Tank

A nozzle of cross-sectional area A2 is discharging to the atmosphere and is located in the side of a large tank, in which the open surface of the liquid in the tank is H m above the center line of the nozzle. Calculate the velocity ν2 in the nozzle and the volumetric rate of discharge if no friction losses are assumed.

Solution: The process flow is shown in Fig. 2.7-7, with point 1 taken in the liquid at the entrance to the nozzle and point 2 at the exit of the nozzle.

Figure 2.7-7. Nozzle flow diagram for Example 2.7-7.


Since A1 is very large compared to A2, ν1 ≅ 0. The pressure p1 is greater than 1 atm (101.3 kN/m2) by the head of fluid of H m. The pressure p2, which is at the nozzle exit, is at 1 atm. Using point 2 as a datum, z2 = 0 and z1 = 0 m. Rearranging Eq. (2.7-32),

Equation 2.7-38


Substituting the known values,

Equation 2.7-39


Solving for ν2,

Equation 2.7-40


Since p1p3 = Hρg and p3 = p2 (both at 1 atm),

Equation 2.7-41


where H is the head of liquid with density ρ. Then Eq. (2.4-40) becomes

Equation 2.7-42


The volumetric flow rate is

Equation 2.7-43


To illustrate the fact that different points can be used in the balance, points 3 and 2 will be used. Writing Eq. (2.7-32),

Equation 2.7-44


Equation 2.7-45



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