Troubleshooting Pump Performance

A manufacturer’s pump performance curves contain data that can help HVAC technicians analyze a pumping installation. Pump curves help identify the system’s operating point, find reasons for a system not performing, and even determine a pump’s impeller diameter.

After designing a pump, the manufacturer usually produces a number of units for performance testing. The tests are necessary to establish how the pump will actually perform. The data collected often includes water flow operating against various system resistances, brake horsepower required, efficiency, and the net positive suction head required for proper operation for the various diameter impellers allowable in the pump volute.

This data is analyzed and then plotted and published as the pump operating characteristics. The pump curve shows how the pump will perform with varying head or flow requirements (see Figure 1).

It is not unusual for a pump’s nameplate to be missing because the information is so important (it usually includes manufacturer’s name, pump model, size, impeller diameter, head and flow for the duty point), that it is often removed for safekeeping. Unfortunately, at times it is so well safeguarded that it cannot be retrieved. Or, perhaps it is just painted over. Either way, the information on it is not available.

To identify the pump and reestablish the nameplate data, the manufacturer must be determined. Most pumps are made of castings and have casting part numbers and markings on them that identify the pump manufacturer. Once the pump manufacturer is known, the type or model and size can be determined with the help of published literature or a telephone call.

Since larger pumps generally have a family of impeller sizes which can be used with a given pump body, at this point the impeller diameter is unknown. A simple procedure using a pressure gauge and the pump’s curves will identify the impeller size in the pump.

IDENTIFYING IMPELLER SIZE

Close the pump discharge valve and take the suction and discharge pressures. This is the “dead-head” condition. Reopen the discharge valve and reset it to the position it was in prior to closing, if it was used to balance the flow. The algebraic difference between the discharge pressure and the suction pressure is the pressure head being generated by the pump. Convert this to feet of water head and determine the correct impeller diameter from the no flow point on the pump curve.

As an example: P discharge = 20.5 psi P suction = 4 in. of Hg vacuum (a negative pressure), but the curves are dimensioned “Head, feet of water.” Therefore, the gauge pressures must be converted. To convert pressure in psi to head in feet of water, multiply psi by 2.31 and divide by the specific gravity of the fluid being pumped. The specific gravity of water is 1. One inch of Hg is equal to 0.491 psi. Therefore:   

P discharge = 20.5 psi x 2.31 ft. of water per psi /1 = 47.4 ft. of water

   P suction=4 in. Hg x 0.491 psi per in.

Hg=-1.96 psi x 2.31 ft. of water per psi/1 = -4.5 ft. of water

   Pump head=P discharge

   P suction = 47.4 -(-4.5) ft. of water

Algebraically subtracting a minus is a plus, so:

Pump head = 47.4 + 4.5 ft. of water = 51.9 ft. of water

Locating this head, 52 ft., at 0 gpm flow on the pump curve in Figure 1, shows the pump impeller diameter to be seven inches.

The system operating point can also be determined by using gauge readings. Take the suction and discharge pressures while the system is operating with the discharge valve in the normal open position. Again, convert these into feet of water and subtract (algebraically) the suction pressure from the discharge pressure. This is the head of the pump at the operating flow. Follow the head line from the zero flow axis out to where it intersects the previously identified impeller characteristic curve. The flow at that point is the system’s operating flow.

For example: after determining the pump’s impeller diameter to be seven inches, gauge readings of the pump taken while it operated were:

P discharge = 17.5 psi

P suction = 4 in. Hg (vacuum)

Convert the gauge readings for the fluid being pumped to feet of water:

P discharge = 17.5 psi x 2.31/1 = 40.5 ft. of water

P suction = 4 in. Hg x 0.491 psi per in

Hg = -1.96 psi x 2.31 ft. of water per psi/1 = -4.5 ft. of water

Pump head = 40.5 – (-4.5) ft. of water = 45.0 ft. of water

The pump head of 45 ft. intersects the seven-inch diameter characteristic curve at 55 gpm, which then is the system operating flow. Being able to fully identify a pump, determine the installed impeller size and the system operating point are of key to troubleshooting (see sidebar below In The Field).

Understanding the information provided on pump curves by the pump manufacturer and taking some simple gauge readings are fundamental to analyzing pumped system problems.

Larry Konopacz is manager of training and education for Bell & Gossett Little Red Schoolhouse. He is a LEED AP and a member of ASHRAE, the Hydraulic Institute, ASPE, and the USGBC. This feature is adapted from a Little Red Schoolhouse article. http://bellgossett.com 

IN THE FIELD

A two hp base-mounted pump trips its circuit breaker regularly when pumping water. The nameplate specifies an 8 ½ in. diameter impeller with a duty point of 51 gpm at 74 ft. of head. Gauge readings at shutoff are 12 psi suction pressure and 46 psi discharge pressure. When the three-way valve is fully open to the coil, the suction pressure is still 12 psi and the discharge pressure is 44 psi.

When the three-way valve is fully open to the bypass, the suction pressure is still 12 psi, but the discharge pressure is 40 psi.

What is the problem? What is the solution? The first step to solving the problem is to analyze the pump readings.

Shutoff head is (discharge pressure minus suction pressure) x 2.31 = (46 psi – 12 psi) x 2.31 = 78.5 ft. of water

The 78.5 ft. at shutoff and the pump curve (Figure 1) confirms the impeller diameter as 8 ½ in.

With the three-way valve fully open to the coil: Pump head = (44 psi – 12 psi) x 2.31 = 74 ft. of water

The intersection of 74 ft. of head and the 8 ½ in. impeller curve on Figure 1 indicates a flow of 51 gpm. The horsepower required is 1 ¾. With the three-way valve open to the bypass, the pump head is calculated as follows: Pump head = (40 psi – 12 psi) x 2.31 = 64.7 ft. of water. The intersection of 64.7 ft. of head and the 8 ½ in. impeller curve on Figure 1 indicates a flow of 80 gpm. The horse-power required is 2 . The problem is  too much flow because the resistance to flow in the bypass circuit is too low.

The solution: increase the resistance in the bypass circuit by 9.3 ft. of water (74-64.7) so the resistance through the bypass circuit is the same as the resistance through the coil. The flow and the horsepower will then be reduced to the same as that flowing through the coil, eliminating breaker trips.