2.8 OPEN-SYSTEM ENERGY BALANCES ON PROCESS EQUIPMENT

In this section, we will examine examples of how to apply the fi rst law to common types of process equipment. These systems will be analyzed at steady-state, when the proper- ties at any place in the system do not change with time. Most cases will consist of one stream in and one stream out, which will be labeled streams 1 and 2, respectively. For these cases the mass balance becomes:

m#

15m#

2

And the energy balance, Equation (2.19) becomes:

05m#

11h^ 112VS21gz211m#

21h^ 112VS21gz22 (2.50) Equation (2.50) can be rewritten in molar terms as:

05n#

1¢h1MWVS2

2 1MWgz≤11n#

2¢h1MWVS2

2 1MWgz≤2 (2.50 molar) where the molecular weight is used in the macroscopic kinetic and potential energy terms to convert from a mass basis to a molar basis.

2.8 Open-System Energy Balances on Process Equipment ◄ 95

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96 ► Chapter 2. The First Law of Thermodynamics

It is important to remember that the examples in this section are restricted to cases when steady-state can be applied. If we are interested in start-up or shutdown of these processes, or the case where there are fl uctuations in feed or operating conditions, we must use the unsteady form of the energy balance.

EXAMPLE 2.18 Diffuser Final Temperature Calculation

The intake to the engine of a jet airliner consists of a diffuser that must reduce the air velocity to zero so that it can enter the compressor. Consider a jet fl ying at a cruising speed of 350 m/s at an altitude of 10,000 m where the temperature is 10ºC. What is the temperature of the air upon exiting the diffuser and entering the compressor?

SOLUTION A schematic diagram of the system, including the information that we know, is shown in Figure E2.15.

Figure E2.18 Schematic of the dif- fuser in Example 2.18.

→

Diffuser T1 = 10° C

V1= 350 m/s V→2 ≈ 0 m/s

T2 = ?

This steady-state process occurs in an open system with one stream in and one stream out.

In this case, we can write the fi rst law using Equation (2.50):

05n#

1(h1MWVS2

2 1MWgz)12n#

2(h1MWVS2

2 1MWgz)21Q# 1W#

s (E2.18A) where the negligible terms have been set to zero. Note that the reference state for potential energy is set at 10,000 m. A mole balance gives:

n#

15n#

2

so that Equation (E2.18A) can be simplifi ed to:

eK51

21MW2VS125 1h22h125 3

T2

T1

cP,air dT (E2.18B)

Looking up the value for heat capacity for air in Appendix A.3, we get:

A53.355, B50.57531023, and D5 20.0163105 EXAMPLE 2.18

Diffuser Final Temperature Calcualtion

0 0 0 0 0

Nozzles and Diffusers

These process devices convert between internal energy and kinetic energy by changing the cross-sectional area through which a fl uid fl ows. In a nozzle the fl ow is constricted, increasing eK. A diffuser increases the cross-sectional area to decrease the bulk fl ow velocity. An example of a process calculation through a diffuser follows.

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2.8 Open-System Energy Balances on Process Equipment ◄ 97

Using the defi nition of heat capacity, we get the following integral expression:

3

T2

T1

cPdT5R3

T2

283

3A1BT1DT224dT5RcAT1B

2T22D Td

283 T2

(E2.18C) Using Equation (E2.18C), Equation (E2.18B) becomes:

1

21MW2VS125RBA1T222832 1B

21T22228322 2D¢1 T2 2 1

283≤ R We now have one equation with one unknown, T2, which can be solved implicitly to give:

T25344 3K4

The temperature of the air increases because the kinetic energy of the inlet stream is being converted to internal energy.

You wish to pump 0.001 m3/s of water from a well to your house on a mountain, 250 m above.

Calculate the minimum power needed by the pump, neglecting the friction between the fl ow- ing water and the pipe.

SOLUTION Can you draw a schematic of this process? We need to write the energy balance.

This system is at steady-state, with one stream in and one stream out. When working with macroscopic potential energy, it is often convenient to write the balance on a mass (rather than mole) basis. We will neglect the bulk kinetic energy of the water at the inlet and outlet and the heat loss through the pipe. Since there are no frictional losses, the exit temperature is the same as the inlet; therefore, their enthalpy is equal. Thus, the fi rst law simplifi es to:

05m1(h1 1

2 VS21gz)12 m2 (h1 1

2 VS 21gz)21Q# 1W#

s

or, rearranging,

W#

s5m#

21gz225V#

2

v^2gz2

EXAMPLE 2.19 Pump Power Calculation

0 0 0 0 0 0

. .

(Continued)

Turbines and Pumps (or Compressors)

These processes involve the transfer of energy via shaft work. A turbine serves to generate power as a result of a fl uid passing through a set of rotating blades. They are commonly found in power plants and used to produce energy locally as part of chemical plants. This process was described in Section 2.1 and illustrated in Example 2.4. Pumps and compressors use shaft work to achieve a desired outcome. The term compressor is reserved for gases, since they are compressible.

Typically they are used to raise the pressure of a fl uid. However, they can also be used to increase its potential energy, as illustrated in Example 2.19.

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98 ► Chapter 2. The First Law of Thermodynamics

where V#

is the volumetric fl ow rate. Solving for shaft work gives:

W#

s5 B 10.0013m3/s4 2

10.0013m3/kg4 2R19.83m/s24 2 1250 3m4 252.5 3kW4

Note that the sign for work is positive. Why? The actual work needed would be greater due to frictional losses.

Heat Exchangers

You plan to use a heat exchanger to bring a stream of saturated liquid CO2 at 0ºC to a superheated vapor state at 10ºC. The fl ow rate of CO2 is 10 mol/min. The hot stream available to the heat exchanger is air at 50ºC. The air must leave no cooler than 20ºC. The enthalpy of vaporization for CO2 at 0ºC is given by:

Dh^vap, CO25236 3kJ/kg4 at 0°C What is the required fl ow rate of air?

SOLUTION First, let’s draw a diagram of the system including the information that we know, shown in Figure E2.20A.

Air Air

Heat exchanger

Vapor CO2 Sat. liq.

CO2

nCO2 = 10 mol Boundary 1 min

T1 = 0°C

T4 = 20°C T3 = 50°C

T2 = 10°C Q

Figure E2.20A Schematic of heat exchanger with boundary 1 depicted.

There are several possible choices for our system boundary. We will choose a boundary around the CO2 stream, labeled “boundary 1” in Figure E2.20A. In this case, the heat transferred from the air stream to evaporate and warm the CO2 stream is labeled Q#

. We next need to EXAMPLE 2.20

Heat Exchanger Flow Rate Calculation

These processes are designed to “heat up” or “cool down” fl uids through thermal con- tact with another fl uid at a different temperature. The radiator in your automobile is an example of a heat exchanger. In this application, energy is removed from the engine block to keep it from overheating during combustion. The most common design is when the two streams are separated from each other by a wall through which energy, but not mass, can pass. A calculation on a system employing this design is given in Example 2.20.

An alternative design allows the fl uids to be mixed directly. An example of such an open feedwater heater is given in Example 2.21.

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2.8 Open-System Energy Balances on Process Equipment ◄ 99

perform a fi rst-law balance around boundary 1. The appropriate energy balance is for an open system at steady-state with one stream in and one stream out is:

05n#

1(h1MWVS2

2 1MWgz)12n#

2(h1MWVS2

2 1MWgz)21Q# 1W#

s

where we have set the bulk kinetic and potential energies and shaft work to zero. A mole balance yields:

n#

15n#

25n#

CO2

so the fi rst-law balance on boundary 1 simplifi es to:

Q# 5n#

CO21h22h12

To determine the change in enthalpy, we must account for the latent heat (vaporization) and the sensible heat of the CO2 stream, that is:

1h22h12 5 Dhvap,CO21 3

T2

T1

cP, CO2dT

These can be found, in [J/mol], as follows. The latent heat is given by:

Dhvap,CO25 ¢236 ckJ

kgd≤ ¢44 c kg

kmold≤510,400 c J mold and the sensible heat is given by:

3

T2

T1

cP,CO2dT5RcA1T22T12 1B

21T222T122 2D¢1 T22 1

T1bR5353 c J mold where the numerical values for the heat capacity parameters, A, B, and D, are given in Appendix A.3. Thus, the energy transferred via heat to boundary 1 is:

Q#

5100,753 3J/min4

Now that we know the rate at which energy must be supplied to the CO2 stream, we can fi nd the fl ow required for the air. We do this by choosing a different system boundary in the heat exchanger, which is labeled boundary 2 in Figure E2.20B.

Air Air

Heat exchanger

Vapor CO2 Sat. liq.

CO2 nCO2 = 10 mol

Boundary 2 min

T1 = 0°C

T4 = 20°C T3 = 50°C

T2 = 10°C Q

Figure E2.20B Schematic of heat exchanger with boundary 2 depicted.

0 0 0 0 0

(Continued)

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100 ► Chapter 2. The First Law of Thermodynamics

A balance similar to that above yields:

2Q# 5n#

air1h42h32 (E2.20A)

Note that we must be careful about signs! We have included a negative sign on Q#

since the heat that enters boundary 1 must leave boundary 2. Rearranging Equation (E2.20A) gives:

n#

air5 2 Q#

1h42h32 5 2 Q#

3

T4

T3

cP,air dT

5 Q#

RBA1T42T321B

21T422T3222D¢1 T42 1

T3≤ R

Looking up values for the heat capacity parameters in Appendix A.3, we get:

n#

air5 1100,753 3J/min4 2

877 3J/mol4 5123 3mol/min4

Alternatively, this problem could have been solved with a system boundary around the entire heat exchanger. In that case, a fi rst-law balance would give:

05n#

CO21h22h12 1n#

air1h42h32 which could then be solved for n#

air.

Superheated water vapor at a pressure of 200 bar, a temperature of 500ºC, and a fl ow rate of 10 kg/s is to be brought to a saturated vapor state at 100 bar in an open feedwater heater.

This process is accomplished by mixing this stream with a stream of liquid water at 20ºC and 100 bar. What fl ow rate is needed for the liquid stream?

SOLUTION The fi rst step is to draw a diagram of the system with the known information, as shown in Figure E2.21.

This example has two inlet streams in, so Equation (2.50) does not apply. If we assume that the rate of heat transfer and the bulk kinetic energy of the streams are negligible and the bulk potential energy and shaft work are set to zero, an energy balance reduces to:

05m#

1h^11m#

2h^22m#

3h^3 (E2.21A)

Similarly, a mass balance at steady-state gives:

05m#

11m#

22m#

3 (E2.21B)

Rearranging Equation (E2.21B) and substituting into (E2.21A) gives:

05m#

1h^11m#

2h^221m#

11m#

22h^3 (E2.21C)

We can look up values for the enthalpies from the steam tables (Appendix B). For state 1, the superheated steam is at 500ºC and 200 bar 15 20 MPa2, so:

h^153238.2 3kJ/kg4 EXAMPLE 2.21

Open Feedwater Heater Calculation

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For state 2, we use subcooled liquid at 20ºC and 100 bar:

h^2593.3 3kJ/kg4 and the saturated vapor at 100 bar (10 MPa) for state 3 is:

h^352724.7 3kJ/kg4

Finally, rearranging Equation (E2.21C) and plugging in values gives:

m#

25m#

11h^12h^32

1h^32h^22 51.95 ckg s d Open

feedwater heater Superheated

vapor Saturated vapor

Subcooled liquid

kg s

T2 = 20°C P2 = 100 bar

T1 = 500°C P1 = 200 bar

P3 = 100 bar

m1 = 10

Figure E2.21 Schematic of the open feedwater heater in Example 2.21.

2.8 Open-System Energy Balances on Process Equipment ◄ 101

Water at 350ºC fl ows into a porous plug from a 10-MPa line. It exits at 1 bar. What is the exit temperature?

SOLUTION First, let’s draw a diagram of the system, as shown in see Figure E2.22.

A steady-state energy balance with one stream in and one stream out is appropriate for this system. We will assume that the bulk kinetic energy of the stream is negligible and that the porous plug is suffi ciently small as not to allow a signifi cant rate of heat transfer. Rewriting Equation (2.50) on a mass basis, we get:

05m#

11h^ 1 VS2

2 1gz212m#

21h^ 1 VS2

2 1gz22 EXAMPLE 2.22

Throttling Device Calculation

1Q# 1W#

s

0 0 0 0 0 0

(Continued)

Throttling Devices

These components are used to reduce the pressure of fl owing streams. The pressure reduction can be accomplished by simply placing a restriction in the fl ow line, such as a partially opened valve or a porous plug. Since these devices occupy a relatively small volume, the residence time of the passing fl uid is small. Hence, there is little energy loss by the transfer of heat. Consequently, we can neglect heat transfer. Since there is also no shaft work, the energy balance reduces to a very simple equation, as the next example illustrates.

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102 ► Chapter 2. The First Law of Thermodynamics

Porous plug T1 = 350°C

P1 = 10 MPa P2 = 1 bar

Figure E2.22 Schematic of the throttling device in Example 2.22.

Hence, the energy balance reduces this system to an isenthalpic process:11 h^15h^2

Looking up the value for the inlet stream from Appendix B.4, we fi nd:

h^152923.4 3kJ/kg4

Since the enthalpy of stream 2 equals that of stream 1, we have two intensive properties to constrain the exit state: h^2 and P2. To fi nd the temperature of stream 2, we must use linear interpolation. Inspection of the steam tables shows that T2 is somewhere between 200 and 250ºC. The following are taken from the superheated steam table at 100 kPa.

P5100 kPa

T[ºC] h^3kJ/kg4

200 2875.3

250 2974.3

Interpolation gives:

T2520013DT4 B h^22h^at 200

h^at 2502h^at 200R520013504 c2923.422875.3

2974.322875.3d 52243°C4