Energy

Energy Balances ChE 2800 - Spring 2002 1/13

1 Basic Concepts

Definitions:

• System: Any specified mass of material or piece(s) of equipment we wish to devote our

attention to.

– closed: No transfer of mass across the system boundaries.

– open: Transfer of mass across the system boundaries is permitted.

• Surroundings: Anything that is not part of the system.

• Properites: A characteristic of the material that can be calculated or measured.

– extensive: A property that is the sum of the values of each of the subsystems

comprising the whole system. A property dependent on system size.

∗ Example: mass or volume.

– intensive: A property whose value is not additive, and does not vary with system

size.

∗ Example: temperature, pressure, density

– specific: An intensive property that is obtained by dividing an extensive property

by the total amount of process material.

∗ Example: A liquid where V = 200 cm3 and the mass = 200 g, then the specific

volume V = 1 cm3/g.

• State: A set of properites of the material at a point in time. The state of a system

depends only on the system’s intensive properties.

1.1 Work

The term “Work” is used widely in casual conversation, but in this context we refer to a specific

definition. Work is a form of energy that represents a transfer of energy between the system


and surroundings. For a mechanical force:

W =∫ state 2

state 1F · dl (1)

where F is an external force in the direction of l acting on the system, or a system force acting

on the surroundings.

The amount of work done by a system on its surroundings (or by the surroundings on the

systems) is dependent upon the inital and final states of the system and the path taken to get

from state 1 to state 2. For this reason, we say that work is a path function.

• Path function: A function who’s final value is dependent on the path taken to get from

the initial to final state.

– Example: work, heat.

• State function: A function that is only dependent upon the initial and final states of

the system.

– Example: internal energy, enthalpy.

1.1.1 Illustration of work as a path function

An ideal gas at 300 K and 500 kPa is enclosed in a cylinder capped by a frictionless piston. The

gas slowly expands from 0.1 m3 to 0.2 m3. Calculate the work done by the gas on the piston if:

• Path 1: expansion occurs at constant pressure (p = 500 kPa)

• Path 2: expansion occurs at constant temperature (T = 300 K)

W =∫ V2

V1

F

A· Adl =

∫ V2

V1

p dV (2)

Path 1

W =∫ V2

V1

p dV = p (V2 − V1) (3)

= 5.00× 105 kPa (0.2 m3 − 0.1 m3) = 50kJ


Remember 1 J = 1 N·m.

Path 2

W =∫ V2

V1

p dV =∫ V2

V1

nRT

VdV (4)

= nRT ln(V2/V1)

n =PV

RT=

(5.0× 105Pa)(0.1m3)

(300K)(8.314 Pa m3/mol K)= 20.0 mol (5)

W = (20.0mol)(8.314 Pa m3/mol K)(300K) ln(0.2/0.1) = 34.57kJ (6)

1.2 Heat

• Heat (Q): The part of the total energy flow across a system boundary that is caused by

a temperature difference between the system and its surroundings. Heat, like work, is a

path function.

1.3 Kinetic Energy

• Kinetic energy (KE): The energy a system possesses because of a velocity difference

between it and it’s surroundings at rest.

KE =1

2mv2 (7)

where m is the mass of the object and v is the object’s velocity relative to the reference state.

1.3.1 Example: Calculation of Kinetic Energy

Water is pumped from a storage tank into a 4.0 cm ID tube at a rate of 0.001 m3/s. What is

the kinetic energy of the water?

• Step 1, determine the linear velocity of the water

v = V /A =0.001m3/s

π(0.02)2m2= 0.795 m/s (8)


• Step 2, find the mass flow rate of mater

ρ = m/V ; ρV = m; (1000 kg/m3)(0.001 m3/s) = 1.0 kg/s (9)

• Step 3, calculate the kinetic energy

KE =1

2mv2 =

1

2(1.0 kg/s)(0.795 m/s)2 = 0.316 J/s (10)

• check units

– 1 N = kg · m/s2

– 1 J = N · m

– Therefore: (kg · m/s2)(m/s) = N· m/s = J/s

1.4 Potential Energy

• Potential Energy (PE): The energy a system possesses because of forces exerted on its

mass by gravitational or electromagnetic fields with respect to a reference surface.

PE = mgh (11)

where h is the distance from the reference surface, m is the mass of the object and g is the

acceleration due to gravity (9.807 m/s2).

1.4.1 Example: Calculation of Potential Energy

Water is pumped from one reservoir to a second reservoir 10 m above the water level of the first.

What is the increase in the potential energy of the water per kg of water moved (the increase

in the specific PE of the water)?

PE = mgh = (9.80 m/s2)(10m) = 98.0 J/kg (12)


1.5 Internal Energy

• Internal Energy (U): A macroscopic measure of the molecular, atomic and subatomic

energies contained within a fluid. Cannot be measured, but is instead calculated from

other variables that can be measured, such as pressure, temperature and volume.

• U is a state function. From the Gibbs phase rule we know for a one componet, one phase

system that specifying two intensive variables will definite U .

• U is commonly defined in terms of T and V

U = f(T, V ) (13)

• Taking the derivative yields

dU =

(∂U

∂T

)

V

dT +

(∂U

∂V

)

T

dV (14)

• Note:

Cv =

(∂U

∂T

)

V

;

(∂U

∂V

)

T

≈ 0 (15)

• Therefore, changes in U can be calculated by integration of equation 14.

U2 − U1 =∫ T2

T1

Cv dT (16)

1.6 Enthalpy

• Enthalpy (H): Enthalpy, like U , is a state function and is defined:

H = U + pV (17)

• for a one component, one phase system:

H = f(T, p) (18)

• Taking the derivative yields:

dH =

(∂H

∂T

)

p

dT +

(∂H

∂p

)

T

dp (19)


• Note:

Cp =

(∂H

∂T

)

p

;

(∂H

∂p

)

T

≈ 0 (20)

• Therefore, changes in H can be calculated by integration of equation 19.

H2 − H1 =∫ T2

T1

Cp dT (21)

Again, we note that the absolute values of both U and H cannot be calculated directly. Only

differences in U or H relative to a reference state may be calculated.


2 Heat Capacities

Definitions:

• Sensible heat: The heat that must be transfered to raise or lower the temperature of a

substance(s).

Q = ∆U (closed system) (22)

Q = ∆H (open system) (23)

• Latent heat: The specific enthalpy change associated with a phase transition (e.g. liquid-

vapor) of a substance at constant T and P .

– Heat of vaporization (∆Hv): The specific enthalpy difference between the liquid

and vapor forms of a substance at a given T and P .

– Heat of fusion (∆Hm): The specific enthalpy difference between the solid and

liquid forms of a substance at a given T and P .

– NOTE: Both ∆Hv and ∆Hm are strong functions of T , but only weak functions of

P .

• Knowing the sensible heat and the latent heat, it is possible to determine the change in

internal energy (∆U , closed system), or the enthalpy (∆H, open system).

• The determination of ∆U for a system that does not undergo a phase transition is straight-

forward.

U2 − U1 =∫ T2

T1

Cv dT (24)

• This equation is valid under the following conditions:

– Exact for ideal gases

– A good approximation for liquids and solids

– Only valid for a non-ideal gas if V = constant.


• We can write a similar expression for ∆H:

H2 − H1 =∫ T2

T1

Cp dT (25)

• This equation is valid under the following conditions:

– Exact for ideal gases

– Only valid for a non-ideal gas if P = constant.

– For liquids or solids:

H2 − H1 = V ∆P +∫ T2

T1

Cp dT (26)

However, the V ∆P term in the above equation is usually negligible compared to the

second term.

• Cp and Cv are related to each other in the following way:

– For liquids and solids Cp ≈ Cv

– Ideal gases Cp = Cv + R

2.1 Example of a ∆U Calculation from Cv.

1 mol of liquid cyclohexane is stored in a covered beaker at 1 atm and 20 ◦C. Calculate the

energy input required to heat the cyclohexane to 60 ◦C.

• For a closed system, the energy balance becomes:

Q = n∆U (27)

• ∆U may be calculated from integration of 14. Cp ≈ Cv for liquid cyclohexane, taken from

Table B.2 [kJ/(mol ◦C)]is:

Cv = 94.140× 10−3 + 49.62× 10−5T − 31.90× 10−8T 2 + 80.63× 10−12T 3 (28)


• Integration of 28 yields:

∆U = 94.140× 10−3(T2 − T1) (29)

+49.62× 10−5

2(T 2

2 − T 21 )

−31.90× 10−8

3(T 3

2 − T 31 )

+80.63× 10−12

4(T 4

2 − T 41 )

∆U = 4.582 kJ/mol (30)

• Since Q = n∆U , Q =(1.0 mol)(4.582 kJ/mol) = 4.582 kJ

If the fluid undergoes a change of phase (solid to liquid, liquid to vapor, etc), a more complex

procedure to calculate the amount of heat required.

2.2 Hypothetical Process Paths

Suppose we wish to calculate the heat required to turn solid phenol at 25◦C and 1 atm to vapor

at 300 ◦C at 3 atm.

∆H = H(vapor, 300◦C, 1 atm)− H(solid, 25◦C, 3 atm) (31)

• Problem - we don’t know H for each state.

• Fortunately H is a state function, so it doesn’t matter what path we take to get from

point A to point B.

• ∆H can be evaluated by constructing a convenient alternate path.

The change in the enthalpy for the process is simply the sum of the enthalpy changes for each

leg of the hypothetical process path.

∆H = ∆H1 + ∆H2 + ∆H3 + ∆H4 + ∆H5 + ∆H6 (32)


25 C, 1 atm, solid

42.5 C, solid 42.5 C, liquid

181.4, vapor

300 C, vapor, 1 atm 300 C, vapor, 3 atm

dH2

dH6

dH1

dH3

dH4

dH5

181.4 C, liquid

Figure 1: Hypothetical process path for the vaporization of Phenol. Red line denotes actual

path, black line denotes the hypothetical process path.

2.3 Example of ∆H Calculation

100 mol of liquid hexane at 25 ◦ C and 7 bar is vaporized and heated to 300 ◦ C at constant

pressure.

Step 1, collect data

• From table B.1, ∆Hv = 28.85 kJ/mol at 69 ◦C.

• Find Tboil(7 bar).

log p∗ = (6.88555− 1175.817

T + 224.867) (33)

• Solve for T , Tboil = 144 ◦C.

Since Tboil 6= Tn−boil, we need to construct a hypothetical process path to get from the initial to

final states. The change in enthalpy for this process is given by:

∆H = ∆H1 + ∆H2 + ∆H3 (34)

Step 2, determine the ∆H values for each leg of our hypothetical process.

∆H1 = V ∆P +∫ 69

25216.3× 10−3dT (35)


69 C, 1 bar, vap

300 C, 7 bar, vap

dH1

dH2

dH3

69 C, 1 bar, liq

25 C, 7 bar, liq

Figure 2: Hypothetical process path for the vaporization of Hexane.

• ρ = 0.659 g/cm3, V = 1/ρ = 1.517× 10−3m3/kg.

• 100 mol*86.17 g/mol = 8617 g or 8.617× 10−3 kg

∆H1 = (8.617)× (1.517× 10−3)(1.01325− 7.0)

×(1.01325× 105)

+100(0.2163(69− 25)) (36)

∆H1 = −0.0785 kJ/h + 951.7 kJ/h (37)

∆H2 = (100)(28.85) = 2885 kJ/h (38)

∆H3 =∫ 300

69137.44× 10−3 + 40.85× 10−5T

+23.92× 10−8T 2 + 57.66× 10−12T 3 dT (39)

∆H3 = 4714.85 kJ/h (40)

Sum values of ∆H determined above:

∆H = −0.0785 + 951.7 + 2885 + 4714.8 = 8551 kJ/h (41)


2.4 Estimation of Latent Heats

• Trouton’s Rule:

∆Hv ≈ 0.088Tb (non− polar fluids)

≈ 0.109Tb (water, low MW alcohols) (42)

– where ∆Hv = [kJ/mol] and Tb =[K].

• For greater accuracy, one should use Chen’s equation

∆Hv =Tb[0.0331(Tb/Tc)− 0.0327 + 0.0297 log Pc]

1.07− (Tb/Tc)(43)

– where ∆Hv = [kJ/mol], Tb, Tc =[K] and Pc = [atm].

• If ∆Hv is known for one temperature, Watson’s Correlation can be used to predict

∆Hv at a different temperature

∆Hv(T2) = ∆Hv(T1)(

Tc − T2

Tc − T1

)0.38)

(44)

• The following relationships can be used to predict heats ofr fusion (∆Hm) when experi-

mental data is not availible

∆Hm ≈ 0.0092Tm (metallic elements)

≈ 0.0025Tm (inorganic compounds)

≈ 0.050Tm (organic compounds) (45)

3 Energy Balances on Closed Systems

Remember, a closed system is one where there is no transfer of material across the system

boundary. In this case, the general balance equation beomes:

accumulation = input − output (46)

or

final system energy − inital system energy =

net energy transfered to system (47)


• initial system energy

Ui + KEi + PEi (48)

• final system energy

Uf + KEf + PEf (49)

• energy transfered

Q + W (50)

• Combining the above equations we can write the energy balance for a closed system

∆U + ∆KE + ∆PE = Q + W (51)

• NOTES:

– U depends almost entirely on chemical composition, temperature, and phase (gas,

solid, liquid) of the system materials.

– For a closed system with no ∆T , phase changes or chemical reactions, ∆U ≈ 0

– If the system is not accelerating or decelerating (v = 0), ∆KE = 0.

– If the system is not rising or falling, ∆PE = 0.

– If the system and its surroundings are at the same temperature, or the system is

perfectly insultated, Q = 0. This kind of process is refered to as adiabatic.

– If there are no moving parts, electrical currents or radiation at the system boundary,

W = 0.

4 Energy Balances for Open Systems

In an open system, mass is allowed to cross the system boundaries. The general balance equation

in this case becomes simply:

input = output

and

∆H + ∆KE + ∆PE = Q + W (52)

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