Chapter 11 Utilizing Experience-Based Principles to Confirm the Suitability of a Process Design

Search in book...
Toggle Font Controls
Create new playlist

Name your new playlist

Playlist description (optional)
Sign In

Email address

Password

Forgot Password?

or

Continue with Facebook

Continue with Google
Sign Up

Full Name

Email address

Confirm Email Address

Password

or

Continue with Facebook

Continue with Google

Chapter 11: Utilizing Experience-Based Principles to Confirm the Suitability of a Process Design

WHAT YOU WILL LEARN

There are experienced-based heuristics that can be used to estimate unknown parameters and validate calculated parameters used to design a chemical process.

Experienced chemical engineers possess the skills necessary to perform detailed and accurate calculations for the design, analysis, and operation of equipment and chemical processes. In addition, these engineers will have formulated a number of experienced-based shortcut calculation methods and guidelines useful for the following:

Checking new process designs
Providing equipment size and performance estimates
Helping troubleshoot problems with operating systems
Verifying that the results of computer calculations and simulations are reasonable
Providing reasonable initial values for input into a process simulator required to achieve program convergence
Obtaining approximate costs for process units
Developing preliminary process layouts

These shortcut methods are forms of heuristics that are helpful to the practicing engineer. All heuristics are, in the final analysis, fallible and sometimes difficult to justify. They are merely plausible aids or directions toward the solution of a problem [1]. Especially for the heuristics described in this chapter, the four characteristics of any heuristic should be kept in mind:

A heuristic does not guarantee a solution.
It may contradict other heuristics.
It can reduce the time to solve a problem.
Its acceptance depends on the immediate context instead of on an absolute standard.

The fact that one cannot precisely follow all heuristics all the time is to be expected, as it is with any set of technical heuristics. However, despite the limitations of heuristics, they are nevertheless valuable guides for the process engineer.

In Chapter 6, process units and stream conditions that were identified as areas of special concern were analyzed. These areas were highlighted in a series of informational tables. In this chapter, the analysis of chemical processes will be completed by checking the equipment parameters and stream conditions in the PFD for agreement with observations and experiences in similar applications.

The required information to start an analysis is provided in a series of informational tables containing shortcut calculation techniques. In this chapter, the use of these resources is demonstrated by checking the conditions given in the basic toluene hydrodealkylation PFD.

11.1 The Role of Experience in the Design Process

The following short narrative illustrates a situation that could be encountered early in your career as an engineer.

You are given an assignment that involves writing a report that is due in two weeks. You work diligently and feel confident you have come up with a respectable solution. You present the findings of the written report personally to your director (boss), who asks you to summarize only your final conclusions. Immediately after you provide this information, your boss declares that “your results must be wrong” and returns your report unopened and unread.

You return to your desk angry. Your comprehensive and well-written report was not even opened and read. Your boss did not tell you what was wrong, and you did not receive any “partial credit” for all your work. After a while, you cool off and review your report. You find that you had made a “simple” error, causing your answer to be off by an order of magnitude. You correct the error and turn in a revised report.

What remains is the nagging question, “How could your boss know you made an error without having reviewed your report or asking any questions?”

The answer to this nagging question is probably a direct result of your director’s experience with a similar problem or knowledge of some guideline that contradicted your answer. The ability of your boss to transfer personal experience to new situations is one reason why he or she was promoted to that position.

It is important to be able to apply knowledge gained through experience to future problems.

11.1.1 Introduction to Technical Heuristics and Shortcut Methods

A heuristic is a statement concerning equipment sizes, operating conditions, and equipment performance that reduces the need for calculations. A shortcut method replaces the need for extensive calculations in order to evaluate equipment sizes, operating conditions, and equipment performance. These are referred to as “back-of-the-envelope calculations.” In this text, both of these experience-based tools are referred to as guidelines or heuristics.

The guidelines provided in this chapter are limited to materials specifically covered in this text (including problems at the end of the chapters). All such material is likely to be familiar to final-year B.S. chemical engineering students and new graduates as a result of their education. Upon entering the work force, engineers will develop guidelines that apply specifically to their area of responsibility.

Guidelines and heuristics must be applied with an understanding of their limitations. In most cases, a novice chemical engineer should have sufficient background to apply the rules provided in this text.

The narrative started earlier is now revisited. The assignment remains the same; however, the approach to solving the problem changes.

Before submitting your report, you apply a heuristic that highlights an inconsistency in your initial results. You then review your calculations, find the error, and make corrections before submitting your report. Consider two possible responses to this report:

Your boss accepts the report and notes that the report appears to be excellent and he or she looks forward to reading it.
Your boss expresses concern and returns the report as before. In this case, you have a reasoned response available. You show that your solution is consistent with the heuristic you used to check your work. With this supporting evidence your boss would have to rethink his or her response and provide you with an explanation regarding his or her concern.

In either case, your work will have made a good impression.

Guidelines and heuristics are frequently used to make quick estimates during meetings and conferences and are valuable in refreshing one’s memory with important information.

11.1.2 Maximizing the Benefits Obtained from Experience

No printed article, lecture, or text is a substitute for the perceptions resulting from experience. An engineer must be capable of transferring knowledge gained from one or more experiences to resolve future problems successfully.

To benefit fully from experience, it is important to make a conscious effort to use each new experience to build a foundation upon which to increase your ability to handle and to solve new problems.

An experienced engineer retains a body of information, made up largely of heuristics and shortcut calculation methods, that is available to help solve new problems.

The process by which an engineer uses information and creates new heuristics consists of three steps. These three steps are predict, authenticate, and reevaluate, and they form the basis of the PAR process. The elements of this process are presented in Table 11.1, which illustrates the steps used in the PAR process.

Table 11.1 PAR Process to Maximize Benefits of Experience: Predict, Authenticate, Reevaluate

Predict: This is a precondition of the PAR process. It represents your best prediction of the solution. It often involves making assumptions and applying heuristics based on experience. Calculations should be limited to back-of-the-envelope or shortcut techniques.
Authenticate/Analyze: In this step, you seek out equations and relationships, do research relative to the problem, and perform the calculations that lead to a solution. The ability to carry out this activity provides a necessary but not sufficient condition to be an engineer. When possible, information from actual operations is included in order to achieve the best possible solution.
Reevaluate/Rethink: The best possible solution from Step 2 is compared with the predicted solution in Step 1. When the prediction is not acceptable, it is necessary to correct the reasoning that led to the poor prediction. It becomes necessary to remove, revise, and replace assumptions made in Step 1. This is the critical step in learning from experience.

Example 11.1

Evaluate the film heat transfer coefficient for water at 93°C (200°F) flowing at 3.05 m/s (10 ft/s) inside a 38 mm (1.5 in) diameter tube. From previous experience, you know that the film heat transfer coefficient for water, at 21°C (70°F) and 1.83 m/s (6 ft/s), in these tubes is 5250 W/m²°C. Follow the PAR process to establish the heat transfer coefficient at the new conditions.

Step 1—Predict: Assume that the velocity and temperature have no effect.

Predicted heat transfer coefficient = 5250 W/m²°C

Step 2—Authenticate/Analyze: Using the properties given below, it is found that the Reynolds number for the water in the tubes is

Re = vρD_pipe/μ = (1.83)(997.4)(1.5)(0.0254)/(9.8 × 10⁻⁴) = 71 × 10³ → Turbulent Flow

Use the Sieder-Tate equation [2] to check the prediction:

h D / k = (0.023) {(D u ρ / μ)}^{0.8} {(C_{p} μ / k)}^{1 / 3} (E11.1a)

$h D / k = (0.023) {(D u ρ / μ)}^{0.8} {(C_{p} μ / k)}^{1 / 3} (E11.1a)$

Property	21°C (70°F)	93°C (200°F)	Ratio of (New/Old)
ρ (kg/m³)	997.4	963.2	0.966
k (W/m°C)	0.604	0.678	1.12
C_p (kJ/kg°C)	4.19	4.20	1.00
μ (kg/m/s)	9.8 × 10⁻⁴	3.06 × 10⁻⁴	0.312

Take the ratio of Equation (E11.1a) for the two conditions given above, and rearrange and substitute numerical values. Using ′ to identify the new condition at 93°C,

\begin{matrix} h^{'} / h = {(D / D)}^{0.2} {(u^{'} / u)}^{0.8} {(ρ^{'} / ρ)}^{0.8} {(μ / μ^{'})}^{0.47} {({C^{'}}_{p} / C_{p})}^{0.33} {(k^{'} / k)}^{0.67} \\ = (1) (1.50) (0.973) (1.73) (1.00) (1.08) = 2.725 \end{matrix} (E11.1b)

$\begin{matrix} h^{'} / h = {(D / D)}^{0.2} {(u^{'} / u)}^{0.8} {(ρ^{'} / ρ)}^{0.8} {(μ / μ^{'})}^{0.47} {({C^{'}}_{p} / C_{p})}^{0.33} {(k^{'} / k)}^{0.67} \\ = (1) (1.50) (0.973) (1.73) (1.00) (1.08) = 2.725 \end{matrix} (E11.1b)$

h^{'} = (2.725) (5250) W / m^{2} ° C = 14, 300 W / m^{2} ° C (E11.1c)

$h^{'} = (2.725) (5250) W / m^{2} ° C = 14, 300 W / m^{2} ° C (E11.1c)$

The initial assumption that the velocity and temperature do not have a significant effect is incorrect. Equation (E11.1c) reveals a velocity effect of a factor of 1.5 and a viscosity effect of a factor of 1.73. All other factors are close to 1.0.

Step 3—Reevaluate/Rethink: The original assumptions that velocity and temperature had no effect on the heat transfer coefficient have been rejected. Improved assumptions for future predictions are as follows:

The temperature effect on viscosity must be evaluated.
The effects of temperature on Cp, ρ, and k are negligible.
Pipe diameter has a small effect on h (all other things being equal).
Results are limited to the range where the Sieder-Tate equation is valid.

With these assumptions, the values for water at 21°C are substituted into Equation (E11.1b). This creates a useful heuristic for evaluating the heat transfer coefficients for water flowing inside tubes at turbulent flow conditions.

h^{'} (W / m^{2 °} C) = 125 u'^{0.8} / μ'^{0.47} for u' (m / s), μ' (kg/m / s)

$h^{'} (W / m^{2 °} C) = 125 u'^{0.8} / μ'^{0.47} for u' (m / s), μ' (kg/m / s)$

Although it takes longer to obtain a solution when you start to apply the PAR process, the development of the heuristic and the addition of a more in-depth understanding of the factors that are important offer substantial long-term advantages.

There are hundreds of heuristics covering all areas of chemical engineering—some general, and others specific to a given application, process, or material. The next section presents a number of these rules that can be used to make predictions to start the PAR analysis.

11.2 Presentation of Tables of Technical Heuristics and Guidelines

A number of these guidelines are provided in this section. The information given is limited to operations most frequently encountered in this text. Most of the information was extracted from a collection presented in Couper et al [3]. In addition, this excellent reference also includes additional guidelines for the following equipment:

Conveyors for particulate solids
Cooling towers
Crystallization from solution
Disintegration
Drying of solids
Evaporators
Size separation of particles

The heuristics or rules are contained in a number of tables and apply to operating conditions that are most often encountered. The information provided is used in Example 11.2 and should be used to work problems at the end of the chapter and to check information on any PFD.

Example 11.2

Refer to the information given in Chapter 1 for the toluene hydrodealkylation process, namely, Figure 1.7 and Tables 1.5 and 1.7. Using the information provided in the tables in this chapter, estimate the size of the equipment and other operating parameters for the following units:

V-102
E-105
P-101
C-101
T-101
H-101

Compare your findings with the information given in Chapter 1.

V-102 High-Pressure Phase Separator

From Table 11.6, the following heuristics are used:

Rule 3 → Vertical vessel

Rule 4 → L/D between 2.5 and 5 with optimum at 3.0

Rule 5 → Liquid holdup time is 5 min based on 1/2 volume of vessel

Rule 9 → Gas velocity u is given by

u = k \sqrt{\frac{ρ_{l}}{ρ_{v}} - 1} m / s

$u = k \sqrt{\frac{ρ_{l}}{ρ_{v}} - 1} m / s$

where k = 0.0305 for vessels without mesh entrainers

Rule 12 → Good performance obtained at 30%–100% of u from Rule 9; typical value is 75%

From Table 1.5,

Vapor flow = Stream 8 = 9200 kg/h, P = 23.9 bar, T = 38°C

Liquid flow = Streams 17 + 18 = 11,570 kg/h, P = 2.8 bar, T = 38°C

ρ_v = 8 kg/m³ and ρ_l = 850 kg/m³ (estimated from Table 1.7)

From Rule 9, u = 0.0305[850/8 – 1]^0.5 = 0.313 m/s

Use u_act = (0.75)(0.313) = 0.23 m/s

Now mass flowrate of vapor = uρ_vπD²/4 = 9200/3600 = 2.56 kg/s

Solving for D, D = 1.33 m

From Rule 5, the volume of liquid = 0.5 LπD²/4 = 0.726L m³

5 min of liquid flow = (5)(60)(11,570)/850/3600 = 1.13 m³

Equating the two results above, L = 1.56 m

From Rule 4, L/D should be in the range 2.5 to 5. For this case L/D = 1.56/1.33 = 1.17

Because this is out of range, change to L = 2.5D = 3.3 m

Heuristics from Table 11.6 suggest that V-102 should be a vertical vessel with

D = 1.33 m, L = 3.3 m

From Table 1.7, the actual V-102 is a vertical vessel with D = 1.1 m, L = 3.5 m

It should be concluded that the design of V-102 given in Chapter 1 is consistent with the heuristics given in Table 11.6. The small differences in L and D are to be expected in a comparison such as this one.

E-105 Product Cooler

From Table 11.11 use the following heuristics:

Rule 1 → Set F = 0.9

Rule 6 → Min. ΔT = 10°C

Rule 7 → Water enters at 30°C and leaves at 40°C

Rule 8 → U = 850 W/m²°C

It is observed immediately from Table 1.5 and Figure 1.5 that Rule 6 has been violated because DT_min = 8°C.

For the moment, ignore this and return to the heuristic analysis:

ΔT_lm = [(105 − 40) − (38 − 30)]/ln[(105 − 40)/(38 − 30)] = 27.2°C

Q = 1085 MJ/h = 301 kW (from Table 1.7)

A = Q/UΔT_lmF = (301,000)/(850)/(27.2)/(0.90) = 14.46 m²

From Rule 9, Table 11.11, this heat exchanger should be a double-pipe or multiple-pipe design.

Comparing this analysis with the information in Table 1.7 shows

Heuristic: Double-pipe design, area = 14.5 m²

Table 1.7: Multiple-pipe design, area = 12 m²

Again, the heuristic analysis is close to the actual design. The fact that the minimum approach temperature of 10°C has been violated should not cause too much concern, because the actual minimum approach is only 8°C and the heat exchanger is quite small, suggesting that a little extra area (due to a smaller overall temperature driving force) is not very costly.

P-101

From Table 11.9, use the following heuristics:

Rule 1 → Power(kW) = (1.67)[Flow(m³/min)]ΔP(bar)/ε

Rules 4–7 → Type of pump based on head

From Figure 1.5 and Tables 1.5 and 1.7,

Flowrate (Stream 2) = 13,300 kg/h

Density of fluid = 870 kg/m³

ΔP = 25.8 − 1.2 = 24.6 bar = 288 m of liquid (head = ΔP/ρ g)

Volumetric flowrate = (13,300)/(60)/(870) = 0.255 m³/min

Fluid pumping power = (1.67)(0.255)(24.6) = 10.5 kW

From Rules 4–7, pump choices are multistage centrifugal, rotary, and reciprocating. Choose reciprocating to be consistent with Table 1.7. Typical ε = 0.75.

Power (shaft power) = 10.5/0.75 = 14.0 kW → compares with 14.2 kW from Table 1.7

C-101

From Table 11.10, use the following heuristics:

Rule 2 → W_{rev adiab} = $\dot{m}$ $\dot{m}$ z₁RT₁[(P₂/P₁)^a – 1]/a

From Table 1.7, flow = 6770 kg/h, T₁ = 38°C = 311 K, mw = 8.45, P₁ = 23.9 bar, P₂ = 25.5

k = 1.41 (assume) and a = 0.2908

$\dot{m}$ $\dot{m}$ = (6770)/(3600)/(8.45) = 0.223 kmol/s

W_{rev adiab} = (223)(1.0)(8.314)(311){ (25.5/23.9)^0.2908 – 1)/0.2908 = 37.7 kW using a compressor efficiency of 75%

W_actual = (37.7)/(0.75) = 50.3 kW → This checks with the shaft power requirement given in Table 1.7.

T-101

From Table 11.13, use the following heuristics:

Rule 5 → Optimum reflux in the range of 1.2–1.5R_min

Rule 6 → Optimum number of stages approximately 2N_min

Rule 7 → N_min = ln{ [x/(1 – x)]_ovhd/[x/(1 – x)]_bot}/ln α

Rule 8 → R_min = {F/D}/(α – 1)

Rule 9 → Use a safety factor of 10% on number of trays

Rule 14 → L_max = 53 m and L/D < 30

From Table 11.14, use the following heuristics:

Rule 2 → F_s = uρ_v^0.5 = 1.2 → 1.5 m/s(kg/m³)^0.5

Rule 3 → ΔP_tray = 0.007 bar

Rule 4 → ε_tray = 60% – 90%

x_ovhd = 0.9962, x_ovhd = 0.0308, α_ovhd = 2.44, α_bot = 2.13, α_{geom ave} = (α_ovhdα_bot)^0.5 = 2.28

N_min = ln{[0.9962/(1 – 0.9962)]/[0.0308/(1 – 0.0308)]} /ln (2.28) = 10.9

R_min = {142.2/105.6}/(2.28 – 1) = 1.05

Range of R = (1.2 → 1.5)R_min = 1.26 → 1.58

N_theoretical ≈ (2)(10.9) = 21.8

ε_tray = 0.6

N_actual ≈ (21.6/0.6)(1.1) = 40 trays

ρ_v = 6.1 kg/m³

u = (1.2 → 1.5)/6.1^0.5 = 0.49 → 0.60 m/s

Vapor flowrate (Stream 13) = 22,700 kg/h

Vol. flowrate, v = 1.03 m³/s

D_tower = [4v/πu]^0.5 = [(4)(1.03)/(3.142)/(0.49 → 0.60)]^0.5 = 1.64 − 1.48 m

ΔP_tower = (N_actual)(ΔP_tray) = (40)(0.007) = 0.28 bar

A comparison of the actual equipment design and the predictions of the heuristic methods are given below.

	From Tables 1.5 and 1.7 and Figure 1.5	From Heuristics
Tower diameter	1.5 m	1.48 → 1.64 m
Reflux ratio, R	1.75	1.26 → 1.58
Number of trays	42	40
Pressure drop, DP_tower	0.30 bar	0.28 bar

H-101

From Table 11.11, use the following heuristics:

Rule 13 → Equal heat transfer in radiant and convective sections

Radiant rate = 37.6 kW/m², convective rate = 12.5 kW/m²

Duty = 27,040 MJ/h = 7511 kW

Area radiant section = (0.5)(7511)/(37.6) = 99.9 m² (106.8 m² in Table 1.7)

Area convective section = (0.5)(7511)/(12.5) = 300.4 m² (320.2 m² in Table 1.7)

From the earlier worked examples, it is clear that the sizing of the equipment in Table 1.7 agrees well with the predictions of the heuristics presented in this chapter. Exact agreement is not to be expected. Instead, the heuristics should be used to check calculations performed using more rigorous methods and to flag any inconsistencies.

11.3 Summary

In this chapter, a number of heuristics have been introduced that allow the reasonableness of the results of engineering calculations to be checked. These heuristics or guidelines cannot be used to determine absolutely whether a particular answer is correct or incorrect. However, they are useful guides that allow the engineer to flag possible errors and help focus attention on areas of the process that may require special attention. Several heuristics, provided in the tables at the end of this chapter, were used to check the designs provided in Table 1.5 for the toluene hydrodealkylation process.

List of Informational Tables

Table	Description
11.2(a)	Physical Property Heuristics
11.2(b)	Typical Physical Property Variations with Temperature and Pressure
11.3	Capacities of Process Units in Common Usage
11.4	Effect of Typical Materials of Construction on Product Color, Corrosion, Abrasion, and Catalytic Effects
11.5	Heuristics for Drivers and Power Recovery Equipment
11.6	Heuristics for Process Vessels (Drums)
11.7	Heuristics for Vessels (Pressure and Storage)
11.8	Heuristics for Piping
11.9	Heuristics for Pumps
11.10	Heuristics for Compressors, Fans, Blowers, and Vacuum Pumps
11.11	Heuristics for Heat Exchangers
11.12	Heuristics for Thermal Insulation
11.13	Heuristics for Towers (Distillation and Gas Absorption)
11.14	Heuristics for Tray Towers (Distillation and Gas Absorption)
11.15	Heuristics for Packed Towers (Distillation and Gas Absorption)
11.16	Heuristics for Liquid-Liquid Extraction
11.17	Heuristics for Reactors
11.18	Heuristics for Refrigeration and Utility Specifications

Table 11.2(a) Physical Property Heuristics

	Units	Liquids	Liquids	Gases	Gases	Gases
		Water	Organic Material	Steam	Air	Organic Material
Heat capacity	kJ/kg°C	4.2	1.0–2.5	2.0	1.0	2.0–4.0
Density	kg/m³	1000	700–1500		1.29@STP
Latent heat	kJ/kg	1200–2100	200–1000
Thermal conductivity	W/m°C	0.55–0.70	0.10–0.20	0.025–0.07	0.025–0.05	0.02–0.06
Viscosity	kg/m s	0°C 1.8 × 10^–3	Wide Range	10–30 × 10^–6	20–50 × 10^–6	10–30 × 10^–6
		50°C 5.7 × 10^–4
		100°C 2.8 × 10^–4
		200°C 1.4 × 10^–4
Prandtl no.		1–15	10–1000	1.0	0.7	0.7–0.8

Table 11.2(b) Typical Physical Property Variations with Temperature and Pressure

	Liquids	Liquids	Gases	Gases
Property	Temperature	Pressure	Temperature	Pressure
Density	ρ_ı ∝ (T_c – T)^0.3	Negligible	ρ_g = (MW)P/zRT	ρ_g = (MW)P/zRT
Viscosity	μ_ı = Ae^B/T	Negligible	$μ_{g} \propto \frac{T^{1.5}}{(T + 1.47 T_{b})}$ $μ_{g} \propto \frac{T^{1.5}}{(T + 1.47 T_{b})}$	Significant only for P > 10 bar
Vapor pressure	P* = ae^{b/(T + c)}	—	—	—
T is temperature (K), T_c is the critical temperature (K), T_b is the normal boiling point (K), MW is molecular weight, P is pressure, Z is compressibility, R is the gas constant, and P* is the vapor pressure.

Table 11.3 Capacities of Process Units in Common Usage^a

Process Unit	Capacity Unit	Max. Value	Min. Value	Comment
Horizontal vessel	Pressure (bar) Temper. (°C) Height (m) Diameter (m) L/D	400 400^b 10 2 5	Vacuum –200 2 0.3 2	L/D typically 2–5, see Table 11.6
Vertical vessel	Pressure (bar) Temper. (°C) Height (m) Diameter (m) L/D	400 400^b 10 2 5	400 –200 2 0.3 2	L/D typically 2–5, see Table 11.6
Towers	Pressure (bar) Temper. (°C) Height (m) Diameter (m) L/D	400 400^b 50 4 30	Vacuum –200 2 0.3 2	Normal limits Diameter L/D 0.5 3.0–40^c 1.0 2.5–30^c 2.0 1.6–23^c 4.0 1.8–13^c
Pumps Reciprocating Rotary and positive Displacement Centrifugal	Power^d (kW) Pressure (bar) Power^d (kW) Pressure (bar) Power^d (kW) Pressure (bar)	250 1000 150 300 250 300	< 0.1 < 0.1 < 0.1
Compressors Axial, centrifugal + recipr. Rotary	Power^d (kW) Power^d (kW)	8000 1000	50 50
Drives for compressors Electric Steam turbine Gas turbine Internal combustion eng.	Power^e (kW) Power^e (kW) Power^e (kW) Power^e (kW)	15,000 15,000 15,000 15,000	< 1 100 10 10
Process heaters	Duty (MJ/h)	500,000	10,000	Duties different for reactive heaters/furnaces
Heat exchangers	Area (m²) Tube dia. (m) Length (m) Pressure (bar) Temp. (°C)	1000 0.0254 6.5 150 400^b	10 0.019 2.5 Vacuum –200	For area < 10 m² use double pipe exchanger For 150 < P < 400 bar need special design
^aMost of the limits for equipment sizes shown here correspond to the limits used in the costing program (CAPCOST) introduced in Chapter 7. ^bMaximum temperature and pressure are related to the materials of construction and may differ from values shown here. ^cFor 20 < L/D < 30 special design may be required. Diameters up to 9 m possible but greater than 4 m must usually be fabricated on site. ^dPower values refer to fluid/pumping power. ^ePower values refer to shaft power.

Process Unit

Capacity Unit

Max. Value

Min. Value

Comment

Horizontal vessel

Pressure (bar)

Temper. (°C)

Height (m)

Diameter (m)

L/D

400

400^b

Vacuum

–200

0.3

L/D typically 2–5, see Table 11.6

Vertical vessel

Pressure (bar)

Temper. (°C)

Height (m)

Diameter (m)

L/D

400

400^b

400

–200

0.3

L/D typically 2–5, see Table 11.6

Towers

Pressure (bar)

Temper. (°C)

Height (m)

Diameter (m)

L/D

400

400^b

Vacuum

–200

0.3

Normal limits

Diameter L/D

0.5 3.0–40^c

1.0 2.5–30^c

2.0 1.6–23^c

4.0 1.8–13^c

Pumps

Reciprocating

Rotary and positive

Displacement

Centrifugal

Power^d (kW)

Pressure (bar)

Power^d (kW)

Pressure (bar)

Power^d (kW)

Pressure (bar)

250

1000

150

300

250

300

< 0.1

Compressors

Axial, centrifugal + recipr.

Rotary

Power^d (kW)

8000

1000

Drives for compressors

Electric

Steam turbine

Gas turbine

Internal combustion eng.

Power^e (kW)

15,000

< 1

100

Process heaters

Duty (MJ/h)

500,000

10,000

Duties different for reactive heaters/furnaces

Heat exchangers

Area (m²)

Tube dia. (m)

Length (m)

Pressure (bar)

Temp. (°C)

1000

0.0254

6.5

150

400^b

0.019

2.5

Vacuum

–200

For area < 10 m² use

double pipe exchanger

For 150 < P < 400 bar

need special design

^aMost of the limits for equipment sizes shown here correspond to the limits used in the costing program (CAPCOST) introduced in Chapter 7.

^bMaximum temperature and pressure are related to the materials of construction and may differ from values shown here.

^cFor 20 < L/D < 30 special design may be required. Diameters up to 9 m possible but greater than 4 m must usually be fabricated on site.

^dPower values refer to fluid/pumping power.

^ePower values refer to shaft power.

Table 11.4 Effect of Typical Materials of Construction on Product Color, Corrosion,^a Abrasion, and Catalytic Effects (see also Table 7.9)

Metals
*Material*	*Advantages*	*Disadvantages*
Carbon steel	Low cost, readily available, resists abrasion, standard fabrication, resists alkali	Poor resistance to acids and strong alkali, often causes discoloration and contamination
Stainless steel	Resists most acids, reduces discolora-tion, available with a variety of alloys, abrasion less than mild steel	Not resistant to chlorides, more expensive, fabrication more difficult, alloy materials may have catalytic effects
Monel-Nickel	Little discoloration, contamination, resistant to chlorides	Not resistant to oxidizing environments, expensive
Hasteloy	Improved over Monel-Nickel	More expensive than Monel-Nickel
Other exotic	Improves specific properties	Can be very high cost metals
Nonmetals
*Material*	*Advantages*	*Disadvantages*
Glass	Useful in laboratory and batch systems, low diffusion at walls	Fragile, not resistant to high alkali, poor heat transfer, poor abrasion resistance
Plastics	Good at low temperature, large variety to select from with various characteristics, easy to fabricate, seldom discolors, low cost	Poor at high temperature, low strength, not resistant to high-alkali conditions, low heat transfer. Minor catalytic effects possible
Ceramics	Withstands high temperatures, variety of formulations available, modest cost	Poor abrasion properties, high diffusion at walls (in particular hydrogen), low heat transfer, may encourage catalytic reactions
Source: ^aIn addition, see Chapter 7, Table 7.9 for preliminary selection of materials of construction.

Table 11.5 Heuristics for Drivers and Power Recovery Equipment

Efficiency is greater for larger machines. Electric motors are 85%–95%; steam turbines are 42%–78%; gas engines and turbines are 28%–38% efficient (see Figure 8.7).
For less than 74.6 kW (100 hp), electric motors are used almost exclusively. They are made for services up to 14,900 kW (20,000 hp).
Steam turbines are competitive higher than 76.6 kW (100 hp). They are speed controllable. They are frequently used as spares in case of power failure.
Combustion engines and turbines are restricted to mobile and remote locations.
Gas expanders for power recovery may be justified at capacities of several hundred horsepower; otherwise any pressure reduction in process is done with throttling valves.
The following useful definitions are given:

$\begin{array}{l} Shaft power = \frac{theoretical power to pump fluid (liquid or gas)}{efficiency of pump or compressor, ε_{s h}} \\ Drive power = \frac{shaft power}{efficiency of drive, ε_{d r}} \\ Overall efficiency = ε_{o v} = ε_{s h} ε_{d r} \end{array}$ $\begin{array}{l} Shaft power = \frac{theoretical power to pump fluid (liquid or gas)}{efficiency of pump or compressor, ε_{s h}} \\ Drive power = \frac{shaft power}{efficiency of drive, ε_{d r}} \\ Overall efficiency = ε_{o v} = ε_{s h} ε_{d r} \end{array}$

ε_dr values are given in this table and Figure 8.7.

ε_sh values are given in Tables 11.9 and 11.10. Usually ε_sh are given on PFD.