Process Fluid Mechanics
- 19.1 Basic Relationships in Fluid Mechanics
- 19.2 Fluid Flow Equipment
- 19.3 Frictional Pipe Flow
- 19.4 Other Flow Situations
- 19.5 Performance of Fluid Flow Equipment
- References
- Short Answer Questions
- Problems
This excerpt provides a more logical pedagogy for the design of basic process equipment including pipes, pumps, and compressors.
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This chapter is from the book
This chapter is from the book
This chapter is from the book
The purpose of this chapter is to introduce the concepts needed to design piping systems, including pumps, compressors, turbines, valves, and other components, and to evaluate the performance of these systems once designed and implemented. The scope is limited to steady-state situations. Derivations are minimized, and the emphasis is on providing a set of useful, working equations that can be used to design and evaluate the performance of piping systems.
19.1 Basic Relationships in Fluid Mechanics
In expressing the basic relationships for fluid flow, a general control volume is used, as illustrated in Figure 19.1. This control volume can be the fluid inside the pipes and equipment connected by the pipes, with the possibility of multiple inputs and multiple outputs. For the simple case of one input and one output, the subscript 1 refers to the upstream side and the subscript 2 refers to the downstream side.
Figure 19.1 General Control Volume
19.1.1 Mass Balance
At steady state, mass is conserved, so the total mass flowrate (
, mass/time) in must equal the total mass flowrate out. For a device with m inputs and n outputs, the appropriate relationship is given by Equation (19.1). For a single input and single output, Equation (19.2) is used.
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In describing fluid flow, it is necessary to write the mass flowrate in terms of both volumetric flowrate (
, volume/time) and velocity (u, length/time). These relationships are
)
where ρ is the density (mass/volume) and A is the cross-sectional area for flow (length2). From Equation (19.3), for an incompressible fluid (constant density) at steady state, the volumetric flowrate is constant, and the velocity is constant for a constant cross-sectional area for flow. However, for a compressible fluid flowing with constant cross-sectional area, if the density changes, the volumetric flowrate and velocity both change in the opposite direction, since the mass flowrate is constant. Accordingly, if the density decreases, the volumetric flowrate and velocity both increase. For problems involving compressible flow, it is useful to define the superficial mass velocity, G (mass/area/time), as
)
The advantage of defining a superficial mass velocity is that it is constant for steady-state flow in a constant cross-sectional area, unlike density and velocity, and it shows that the product of density and velocity remains constant.
For a system with multiple inputs and/or multiple outputs at steady state, as is illustrated in Figure 19.2, the total mass flowrate into the system must equal the total mass flowrate out, Equation (19.1). However, the output mass flowrate in each section differs depending on the size, length, and elevation of the piping involved. These problems are discussed later.
Figure 19.2 System with Multiple Inputs and Outputs
Example 19.1
Two streams of crude oil (specific gravity of 0.887) mix as shown in Figure 19.1. The volumetric flowrate of Stream 1 is 0.006 m3/s, and its pipe diameter is 0.078 m. The volumetric flowrate of Stream 2 is 0.009 m3/s, and its pipe diameter is 0.10 m.
Determine the volumetric and mass flowrates of Stream 3.
Determine the velocities in Streams 1 and 2.
If the velocity is not to exceed 1 m/s in Stream 3, determine the minimum possible pipe diameter.
Determine the superficial mass velocity Stream 3 using the pipe diameter calculated in Part (c).
Figure E19.1 Physical Situation in Example 19.1
Solution
Since the density is constant, the volumetric flowrate of Stream 3 is the sum of the volumetric flowrates of Streams 1 and 2, 0.015 m3/s. To obtain the mass flowrate,
, so 
. Alternatively, the mass flowrate of Streams 1 and 2 could be calculated and added to get the same result.From Equation (19.3), at constant density
. Therefore,The diameter at which u3 = 1 m/s can be calculated from Equation (19.3) at constant density.
If the diameter were smaller, the cross-sectional area would be smaller, and from Equation (19.3), the velocity would be larger. Hence, the result in Equation (E19.1c) is the minimum possible diameter. As shown later, actual pipes are available only in discrete sizes, so it is necessary to use the next higher pipe diameter.
From Equation (19.4), using the rounded values,
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19.1.2 Mechanical Energy Balance
The mechanical energy balance represents the conversion between different forms of energy in piping systems. With the exception of temperature changes for a gas undergoing compression or expansion with no phase change, temperature is assumed to be constant. The mechanical energy balance is
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In Equation (19.5) and throughout this chapter, the difference, Δ, represents the value at Point 2 minus the value at Point 1, that is, out − in. The units in Equation (19.5) are energy/mass or length2/time2. In SI units, since 1 J =1 kg m2/s2, it is clear that 1 J/kg = 1 m2/s2. In American Engineering units, since 1 lbf = 32.2 ft lbm/sec2, this conversion factor (often called gc) must be used to reconcile the units. The notation < > represents the appropriate average quantity.
The first term in Equation (19.5) is the enthalpy of the system. On the basis of the constant temperature assumption, only pressure is involved. For incompressible fluids, such as liquids, density is constant, and the term reduces to
)
For compressible fluids, the integral must be evaluated using an equation of state.
The second term in Equation (19.5) is the kinetic energy term. For turbulent flow, a reasonable assumption is that
)
For laminar flow,
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For simplicity, < u2 > is hereafter represented as < u>2, which is shortened to u2.
The third term in Equation (19.5) is the potential energy term. Based on the general control volume, Δz is positive if Point 2 is at a higher elevation than Point 1.
The fourth term in Equation (19.5) is often called the energy “loss” due to friction. Of course, energy is not lost—it is just expended to overcome friction, and it manifests as a change in temperature. The procedures for calculating frictional losses are discussed later.
The last term in Equation (19.5) represents the shaft work, that is, the work done on the system (fluid) by a pump or compressor or the work done by the system on a turbine. These devices are not 100% efficient. For example, more work must be applied to the pump than is transferred to the fluid, and less work is generated by the turbine than is expended by the fluid. In this book, work is defined as positive if done on the system (pump, compressor) and negative if done by the fluid (turbine). This convention is consistent with the flow of energy in or out of the system; however, many textbooks use the reverse sign convention. Equipment such as pumps, compressors, and turbines are described in terms of their power, where power is the rate of doing work. Therefore, a device power (
, energy/time) is defined as the product of the mass flowrate (mass/time) and the shaft work (energy/mass):
)
When efficiencies are included, the last term in Equation (19.5) becomes
)
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Example 19.2
Water in an open (source or supply) tank is pumped to a second (destination) tank at a rate of 5 lb/sec with the water level in the destination tank 25 ft above the water level in the source tank, and it is assumed that the water level does not change with the flow of water. The destination tank is under a constant 30 psig pressure. The pump efficiency is 75%. Neglect friction.
Determine the required horsepower of the pump.
Determine the pressure increase provided by the pump assuming the suction and discharge lines have the same diameter.
Solution
Turbulent flow in the pipes is assumed. The mechanical energy balance is
Figure 19.2 is an illustration of the system.
Figure E19.2 Physical Situation for Example 19.2
The control volume is the water in the tanks, the pipes, and the pump, and the locations of Points 1 and 2 are illustrated. The integral in the first term of Equation (19.2) is simplified to the first term in Equation (E19.2a), since the density of water is a constant. In general, the fluid velocity in tanks is assumed to be zero because tank diameters are much larger than pipe diameters, so the kinetic energy term for the liquid surface in the tank is essentially zero. Any fluid in contact with the atmosphere is at atmospheric pressure, so P1 = 1 atm = 0 psig. The friction term is assumed to be zero in this problem, as stated. So, Equation (E19.2a) reduces to
and
Converting to horsepower yields
To determine the pressure rise in the pump, the control volume is now taken as the fluid in the pump. So, the mechanical energy balance is written between Points 3 and 4. The mechanical energy balance reduces to
The kinetic energy term is zero because the suction and discharge pipes have the same diameter. Frictional losses are assumed to be zero in this example. The potential energy term is also assumed to be zero across the pump; however, since the discharge line of a pump may be higher than the suction line, in a more detailed analysis, that potential energy difference might be included. Solving
.
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Example 19.3
A nozzle is a device that converts pressure into kinetic energy by forcing a fluid through a small-diameter opening. Turbines work in this way because the fluid (usually a gas) with a high kinetic energy impinges on turbine blades, causing spinning, and allowing the energy to be converted to electric power.
Consider a nozzle that forces 2 gal/min of water at 50 psia in a tube of 1-in inside diameter through a 0.1-in nozzle from which it discharges to atmosphere. Calculate the discharge velocity.
Solution
The system is illustrated in Figure 19.3. It is assumed that the velocity at a small distance from the end of the nozzle is identical to the velocity in the nozzle, but the contact with the atmosphere makes the pressure atmospheric.
Figure E19.3 Illustration of Nozzle for Example 19.3
For the case when frictional losses may be neglected, the mechanical energy balance reduces to
)
which yields
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so u2 = 72.4 ft/sec. For a real system, there would be some frictional losses and the actual discharge velocity would be lower than calculated here.
This problem was solved under the assumption of turbulent flow. The criterion for turbulent flow is introduced later; however, for this system, the Reynolds number is about 2 × 105, which is well into the turbulent flow region.
19.1.3 Force Balance
The force balance is essentially a statement of Newton’s law. A common form for flow in pipes is
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where F are the forces on the system. The underlined parameters indicate vectors, since there are three spatial components of a force balance. For steady-state flow and the typical forces involved in fluid flow, Equation (19.12) reduces to
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where Fp is the pressure force on the system, Fd is the drag force on the system, Fg is the gravitational force on the system, and R is the restoring force on the system, that is, the force necessary to keep the system stationary. The term on the left side of Equation (19.12) is acceleration, confirming that Equation (19.12) is a statement of Newton’s law. The most common application of Equation (19.13) is to determine the restoring forces on an elbow. These problems are not discussed here.