Pumps and Compressors

The compressor emergency shutdown system (ESD) has tripped the natural gas compressor off-line three times in the past 24 hours. Each time the operator goes to reset the compressor interlock, she notices the graphic display panel on the interlock system says “Separator boot high level” as the reason for the trip. After this last trip, operations decides to keep the compressor shut down for a few hours until your arrival to diagnose the problem. Your first diagnostic test is to look at the indicated boot level in the sightglass (LG-93). There, you see a liquid level appears to be normal:

First, explain why this first diagnostic test was a good idea. Then, identify what would your next diagnostic test be.

Finally, comment on the decision by operations to leave the compressor shut down until your arrival. Do you think this was a good idea or a bad idea, from a diagnostic perspective? Why or why not?

The compressor automatically shut down last night, tripped by LSHH-231. The control system alarm log showed a high level alarm LIR-92 about 15 minutes prior to the shutdown:

Identify the likelihood of each specified fault in this process. Consider each fault one at a time (i.e. no coincidental faults), determining whether or not each fault could independently account for all measurements and symptoms in this process.

$$\begin{array} {|l|l|} \hline Fault & Possible & Impossible \\ \hline 2-inch~line~plugged~at bottom~of~separator~vessel & & \\ \hline LT-92~failed~with~high~output~signal & & \\ \hline Air~supply~to~solenoid~valve~shut~off & & \\ \hline Solenoid~vent~line~plugged & & \\ \hline PSV-11~stuck~open~& & \\ \hline LSHH-231~failed~with~high~output~signal & & \\ \hline \end{array}$$

This amount of vacuum (negative pressure) in this knock-out drum is controlled by varying the compressor’s bypass valve:

An operator tells you there is a problem with this system, though: the vacuum gauge near the pressure transmitter registers -6.9 PSI, even though the controller faceplate registers -8.0 PSI which is the same as the setpoint. The same operator notes that the control valve position is approximately 30

Another instrument technician happens to be with you, and recommends the operator place the pressure controller in manual mode to “stroke-test” the control valve. Explain why this test would be a waste of time, and propose a better test for helping to pinpoint the location of the fault.

{\bullet} A valuable principle to apply in a diagnostic scenario such as this is correspondence: identifying which field variables correspond with their respective controller faceplate displays, and which do not. Apply this comparative test to the scenario described, and use it to explain why the technician’s proposed test was probably not the best first step.
{\bullet} A problem-solving technique useful for analyzing control systems is to mark the PV and SP inputs of all controllers with “+” and “-” symbols, rather than merely label each controller as “direct” or “reverse” action. Apply this technique to the control strategy shown here, identifying which controller input(s) should be labeled “+” and which controller input(s) should be labeled “-”.
{\bullet} Predict the effects resulting from one of the transmitters in this system failing with either a high or a low signal.
{\bullet} For those who have studied level measurement, explain how the level transmitter (which is nothing more than a DP transmitter) senses liquid level inside the knock-out drum.

This compressor control system uses a pressure transmitter and controller to regulate the discharge pressure to a constant setpoint, allowing either a power controller (JIC) or a suction pressure controller (PIC) to override. The power controller overrides the discharge pressure controller under conditions of high load, throttling back the suction valve to limit power. The suction pressure controller overrides them all under conditions of high inlet vacuum, opening the suction valve in order to ensure the compressor’s gland seals are not ruined by excessive vacuum:

In the event of a high inlet vacuum condition simultaneous with a high load condition, the suction pressure controller will “win” by overriding the power controller. Alter this system so that the override priority is vice-versa: the power controller is able to override the suction pressure controller, yet the suction controller is still able to override the discharge controller.

The following loop diagram shows a compressor surge control system. When the flow controller (FIC 42) detects a condition of high differential pressure across the compressor and a simultaneous condition of low flow through the compressor, it responds by opening the surge control valve (FV 42), bypassing flow from the outlet of the compressor directly back to the input of the compressor:

If the screw on terminal JB1-4 were to come loose, breaking the connection between the two wires joined at that point, what would this surge control valve do, and what effect do you think that would have on the compressor?

{\bullet} Identify whether FV-42 is fail-open (FO) or fail-closed (FC).
{\bullet} What do the short arrows represent (located next to the individual instrument bubbles) in this loop diagram?

Examine this natural gas compressor system with inlet separator vessel:

Examine this P&ID and answer the following questions:

{\bullet} Why is it important that the separator vessel be placed upstream of the compressor? Why not place it on the discharge side of the compressor instead?

{\bullet} What might cause the compressor to vibrate excessively, thus requiring a vibration monitoring system?

{\bullet} What effect will opening the recycle valve (XV-76) have on the effective compression ratio of this compressor as it is operating?

{\bullet} What measured variables in this system might indicate that the compressor is being overloaded (i.e. working too hard as it tries to compress more natural gas than it safely can)?

This control system measures and regulates the amount of differential pressure across a gas compressor, by opening a recirculation valve to let high-pressure discharge gas go back to the low-pressure “suction” of the compressor. This control system needs to be very fast-acting, and currently it is anything but that, as revealed by the open-loop trend shown in the upper-right of this illustration:

Identify what type of problem you think you are dealing with here, as the compressor’s differential pressure should not take several seconds to stabilize following a sudden move by the recirculation valve. Also suggest a next diagnostic test or measurement to take, explaining how the result(s) of that test help further identify the location and/or nature of the fault.

{\bullet} Based on the evidence presented, how do you know this problem is definitely not caused by poor PID controller tuning?

{\bullet} What other methods exist for controlling differential pressure across a large gas compressor, other than using a recirculation valve?

This control system measures and regulates the differential pressure across a large motor-driven gas compressor by “recycling” gas from the compressor’s discharge line back to its suction line. It uses an air-to-close control valve so that the valve will fail open in the event of air pressure or signal loss. The controller’s output indication, however, is reverse-responding so that 0

Unfortunately, this system has a problem. The pressure differential indicating controller (PDIC) shows the process variable (PV) being about 50

Identify which of these four areas of the system the problem may be located in, and then describe a good test you could do in or to this system to narrow the problem location even further:

{\bullet} Problem with the measurement side (transmitter, wiring, controller analog input)?

{\bullet} Problem with the controller’s control action (its “decision-making”)?

{\bullet} Problem with the final control element side (valve, I/P, booster, controller analog output)?

{\bullet} Problem with the compressor itself (or other portions of the process)?

{\bullet} Could a shut recycle line hand valve account for what we’re seeing here? Explain why or why not.

Suppose a compressor is operating with the following suction and discharge parameters:

Suction:
{\bullet} Pressure = 45 PSIG
{\bullet} Volumetric flow = 1300 CFM
{\bullet} Temperature = 74 deg F

Discharge:
{\bullet} Pressure = 281 PSIG
{\bullet} Volumetric flow = 317 CFM
{\bullet} Temperature = 186 deg F

From these figures, calculate the operating {\it compression ratio} of this gas compressor.

Electrically-powered air compressors are commonly used in many different industries for supplying clean, dry compressed air to machines, instrument systems, and pneumatic tools. A simple compressor system consists of a compressor which works much like a bicycle tire pump (drawing in air, then compressing it using pistons), an electric motor to turn the compressor mechanism via a V-belt, a “receiver tank” to receive the compressed air discharged by the compressor mechanism, and some miscellaneous components installed to control the pressure of the compressed air in the receiver tank and drain any condensed water vapor that enters the receiver:

Electromechanical relay circuitry located inside the electrical enclosure decides when to turn the compressor motor on and off based on the statuses of the high- and low-pressure control switches (PSH = high pressure switch ; PSL = low pressure switch).

Your task is two-fold. First, you must figure out how to wire a new low-low pressure alarm switch (PSLL, shown on the left-hand end of the receiver) so that an alarm buzzer will activate if ever the compressed air pressure falls too low. A newly-installed hand switch located on the front panel of the electrical enclosure must be wired with this PSLL switch in such a way that the buzzer cannot energize if the hand switch is in the “alarm disable” position. Second, you must figure out how to wire a new high-level switch (LSH, shown on the “boot” of the receiver tank) so that the condensate drain valve will energize automatically to open up and drain water out of the receiver boot when the level gets too high, and then automatically shut again when the water in the boot drops down to an acceptable level.

The following schematic diagram shows the basic motor control circuit for this air compressor, with the new switches, buzzer, and drain valve shown unwired:

Complete this control circuit by sketching connecting wires between the new switches, buzzer, and drain valve solenoid. Remember that the way all switches are drawn in schematic diagrams is in their “normal” states as defined by the manufacturer: the state of minimum stimulus (when the switch is un-actuated). For pressure switches, this “normal” state occurs during a low pressure condition; for liquid level switches, this “normal” state occurs during a low-level (dry) condition. Note that each of the new process switches has SPDT contacts, allowing you to wire each one as normally-open (NO) or as normally-closed (NC) as you see fit.

Describe all that is represented by this P&ID:

A centrifugal pump works by spinning a disk with radial vanes called an “impeller,” which flings fluid outward from the center of the disk to the edge of the disk. This kinetic energy imparted to the fluid translates to potential energy in the form of pressure when the fluid molecules strike the inner wall of the pump casing:

The performance of a centrifugal pump is often expressed in a special graph known as a {\it pump curve}. A typical centrifugal pump curve appears here:

Examine this pump curve, and explain in your own words what it tells us about the performance behavior of this pump.

{\bullet} One way to describe the operation of a centrifugal pump is to say it generates discharge pressure by converting kinetic energy into potential energy. Elaborate on this statement, explaining exactly where and how kinetic energy gets converted to potential energy. Hint: this might be easier to answer if you consider the “limiting case” of maximum discharge pressure described by the pump curve, where flow is zero and pressure is maximum.
{\bullet} Appealing to the conversion of energy between kinetic and potential forms, explain why discharge pressure for a centrifugal pump falls off as flow rate increases.

Suppose operators submitted a “trouble-call” to your instrument shop, claiming sump V-15 had an excessive liquid level inside of it (as indicated by LIR-134), and that the pump was not pumping that level down as it should:

Identify at least three possible faults, each one independently capable of accounting for the high sump level indication. Also, identify any diagnostic tests you could perform on this system to pinpoint the nature and location of the fault.

{\bullet} Suppose this trouble-call came to you during a very cold winter day, when the outside temperature was well below freezing. How might this alter the list of potential faults?
{\bullet} Explain the purpose for having check valves on the discharge lines of the two submersible sump pumps.
{\bullet} Identify some of the different pressure-measurement accessory devices visible in this P&ID.

In this oily water sump process, two submersible pumps move water out of the sump based on liquid level measurement inside the sump:

Examine this P&ID and answer the following questions:

{\bullet} Why are “check” valves installed on the discharge lines of pumps P-407 and P-408?

{\bullet} Supposing pump P-408 is the only one running, qualitatively determine the effects of pinching off the block valve immediately upstream of pressure gauge PG-419 on the following variables (e.g. increase, decrease, or remain the same):
item{\rightarrow} Pressure at the pump’s discharge port
item{\rightarrow} Pressure registered by PG-419
item{\rightarrow} Electrical current to the driving motor
item{\rightarrow} Flow rate of water through filter S-401
item{\rightarrow} Pressure registered by PG-419
item{\rightarrow} Oily water level inside the sump

A level control system uses a variable-frequency motor drive (VFD) to control the speed of a pump drawing liquid out of the vessel. The greater the liquid level, the faster the pump spins, drawing liquid out at a faster rate. A low-level cutoff switch is also part of this control system, forcing the pump to a full stop to protect it from running dry if ever a low-level condition is sensed by the switch:

Unfortunately, this system seems to have a problem. The pump refuses to start even though the liquid level is greater than the controller’s setpoint (as indicated by both the controller and the sightglass). It was running just fine yesterday, and no technician has touched any of the components since then.

A fellow instrument technician helping you troubleshoot this problem decides to perform a simple test: he uses his multimeter (configured to measure DC current) as a “jumper” wire to momentarily short together terminals 5 and 7 on terminal strip TB13. Still, the motor remains off and does not start up as it should.

Identify the likelihood of each specified fault for this control system. Consider each fault one at a time (i.e. no coincidental faults), determining whether or not each fault could independently account for {\it all} measurements and symptoms in this system.

$$\begin{array} {|l|l|} \hline Fault & Possible & Impossible \\ \hline No~AC~power~to~VFD & & \\ \hline Controller~has~dead~4-20~mA~output & & \\ \hline Level~transmitter~out~of~calibration & & \\ \hline Level~switch~contacts~failed~shorted & & \\ \hline Level~switch~contacts~failed~open & & \\ \hline 250~ohm~resistor~failed~open & & \\ \hline Cable~between~TB12~and~TB13~failed~open & & \\ \hline Cable~between~TB13~and~LSL~failed~open & & \\ \hline \end{array}$$

How much work is done pumping 3,000 gallons of water from reservoir “A” to reservoir “B” over the hill?

If the pump’s power output is 250 horsepower, how long will it take to pump all 3000 gallons to reservoir “B”?

{\bullet} Calculate the amount of pressure at the discharge port of the pump as it lifts water up to reservoir “B”

One laborer working on the top of a building uses a manual hoist to lift 10 gallons of water 30 feet up from ground level, while a second laborer uses an electric pump to do the same:

First, calculate the amount of work needed to lift 10 gallons up to the same roof. Then, calculate the time required for the pump to do this job, assuming a rating of 1.5 horsepower.

A surface-mounted water pump pulls water out of a well by creating a vacuum, though it might be more technically accurate to say that the pump works by reducing pressure in the inlet pipe to a level less than atmospheric pressure, allowing atmospheric pressure to then push water from the well up the pump’s inlet pipe:

Based on this description of pump operation, what is the theoretical maximum height that any pump can lift water out of a well, assuming the well is located at sea level?

Water wells located at altitudes other than sea level will have different theoretical maximum lifting heights (i.e. the farthest distance a surface-mounted pump may suck water out of the well). Research the average barometric pressure in Denver, Colorado (the “mile-high” city) and determine how far up a surface pump may draw water from a well in Denver.

Domestic water wells may be hundreds of feet deep. How can water be pumped out of wells this deep, given the height limitation of vacuum pumping?

{\bullet} If the liquid in question was something other than water, would the maximum “lift” depth be different? Why or why not?

Examine this P&ID:

Each pump is of the “reciprocating” type, a form of positive displacement machine. In essence, each rotation of the motor shaft causes the pump to move a measured quantity of liquid from its inlet to its outlet.

What will happen to the liquid level inside the vessel over time if one pump is moving more liquid flow?

Would you characterize this process as inherently self-regulating or inherently integrating?

Suppose two identical pumps exhibit the exact same pump curve shown below:

If these two pumps are connected in series with a suction pressure of 45 PSI ($P_1$ = 45 PSI), calculate pressures $P_2$ and $P_3$ as well as total discharge flow rate when the flow rate through each pump is 150 GPM:

If these two pumps are connected in parallel with a suction pressure of 45 PSI ($P_1$ = 45 PSI), calculate pressures $P_2$ and $P_3$ as well as total discharge flow rate when the flow rate through each pump is 150 GPM:

{\bullet} Explain how series and parallel pumps act much the same as series and parallel electrical sources.
{\bullet} Where might an engineer choose to use series pumps versus parallel pumps in a piping system?

Given the following pump curve for a water pump driven at a constant speed by an AC induction motor, determine the maximum flow rate of water it can deliver to different heights above the pump’s discharge port:

{\bullet} Maximum flow at 10 feet above pump level =

{\bullet} Maximum flow at 50 feet above pump level =

{\bullet} Maximum flow at 100 feet above pump level =

{\bullet} Maximum flow at 150 feet above pump level =

Sketch the pump curve for a positive-displacement pump turned at a constant speed by an electric motor:

{\bullet} What will change about the pump curve graph if the driving motor’s speed changes?

An alternative to using a pressure-relief valve to control pressure in a hydraulic system is to use a variable-displacement pump with hydraulic feedback:

As hydraulic pressure increases, the pump mechanism automatically adjusts to give less volume displacement per rotation. Explain how this works to regulate pressure, and also why it saves energy compared to the more traditional design of a constant-displacement pump combined with a pressure-relief valve.

Hydraulic (liquid) power systems require pressure regulation just like pneumatic (air) power systems. However, pressure control must be done differently in a hydraulic system. In a pneumatic system, the electric motor driving the air compressor is simply turned on and off to maintain air system pressure between two setpoints. In a hydraulic system, the electric motor driving the positive-displacement pump continually runs, with a pressure relief valve regulating line pressure:

If not for the pressure-relief valve, the hydraulic pump would “lock up” and refuse to turn whenever the control valve was placed in the “stop” position (as shown in the diagram). With the pressure-relief valve in place, the pump will continue to spin and hydraulic pressure will be maintained.

Explain why a positive-displacement hydraulic pump will “lock up” if its outlet line is blocked, and explain the operating principle of the pressure-relief valve.

{\bullet} Identify what would have to be altered in this fluid power system to reverse the direction of the motor.
{\bullet} Would this system function adequately if the pressure relief valve were relocated to a location “downstream” of the spool valve?
{\bullet} If the filter were to entirely plug and prevent flow through it, would the hydraulic pump “lock up” in the same way it would having its discharge port blocked?
{\bullet} The Law of Energy Conservation states that energy cannot be created or destroyed, but must be accounted for in every system. When the spool valve is left in the “off” position and the motor does not move, where does all the energy go that is input by the pump into the fluid system? For example, if the hydraulic pump is being spun by a 1-horsepower motor, what happens to all that power if it is not directed to the motor to do mechanical work?

A centrifugal pump works by spinning a disk with radial vanes called an “impeller,” which flings fluid outward from the center of the disk to the edge of the disk. This kinetic energy imparted to the fluid translates to potential energy in the form of pressure when the fluid molecules strike the inner wall of the pump casing:

The energy conveyed by the liquid exiting the discharge port of this pump comes in two forms: pressure head and velocity head. Ignoring differences in elevation (height), we may apply Bernoulli’s equation to describe this fluid energy:

$$\hbox{Fluid Energy at discharge port} = {\rho v^2 \over 2} + P$$

Where,

Fluid Energy = expressed in units of pounds per square foot, or PSF

$P$ = Gauge pressure (pounds per square foot, or PSF)

$\rho$ = Mass density of fluid (slugs per cubic foot)

$v$ = Velocity of fluid (feet per second)

When the discharge port is completely blocked by an obstruction such as a closed valve or a blind, there is no velocity at the port ($v = 0$) and therefore the total energy is in the potential form of pressure ($P$). When the discharge port is completely unobstructed, there will be no pressure at the port ($P = 0$) and therefore the total energy is in kinetic form (${\rho v^2 \over 2}$). During normal operation when the discharge experiences some degree of resistance, the discharge fluid stream will possess some velocity as well as some pressure.

Assuming that the fluid molecules’ maximum velocity is equal to the speed of the impeller’s rim, calculate the discharge pressure under these conditions for a pump having an 8 inch diameter impeller spinning at 1760 RPM and a discharge port of 2 inches diameter, with water as the fluid (mass density $\rho$ = 1.951 slugs per cubic foot) and assuming atmospheric pressure at the suction port:

{\bullet} Discharge flow = 0 GPM ; $P$ =

{\bullet} Discharge flow = 100 GPM ; $P$ =

{\bullet} Discharge flow = 200 GPM ; $P$ =

{\bullet} Discharge flow = 300 GPM ; $P$ =

{\bullet} Discharge flow = 400 GPM ; $P$ =

{\bullet} Discharge flow = 500 GPM ; $P$ =

Next, calculate the maximum flow rate out of the pump with a completely open discharge port ($P = 0$).

Calculate the volumetric flow rate (in units of cubic feet per minute) for water flowing out of the 10-inch diameter discharge pipe of a centrifugal pump at a velocity of 25 feet per second. Then, convert that flow rate into units of gallons per minute.