Posts Tagged ‘heat exchanger’

Manipulating Water Temperature to Control Cavitation

Monday, April 9th, 2018

    As we learned previously, cavitation bubbles form at a centrifugal pump’s inlet when the thermodynamic properties of water, namely temperature and pressure, are right.   Today we’ll see how just manipulating water temperature can control cavitation.

 Manipulating Water Temperature to Control Cavitation

Manipulating Water Temperature to Control Cavitation

   

    Some centrifugal pumps draw water from an external heat source such as a heat exchanger in order to provide heat to buildings, generate power, and perform manufacturing processes.   On some exchangers heat is applied at a fixed rate and can’t be varied.   On others heat can be varied by using a heat exchanger fitted with a temperature control.   This makes it easy to reduce or lower water temperature introduced at the pump’s inlet.   If the temperature is kept low enough relative to the pressure at the inlet, cavitation bubbles won’t form.

    Let’s say water enters the pump’s inlet from a heat exchanger at 59ºF and internal pump pressure is 0.25 pounds per square inch (PSI).   With these parameters in place water boils and cavitation bubbles will form in the pump inlet.   But if the heat exchanger is adjusted so that temperature is lowered by a mere two degrees to 57ºF, cavitation ceases.   This is in accordance with the boiling points of water, listed for various pressures and temperatures, as published in engineering thermodynamic texts.

    If it’s not possible to lower water temperature at the pump inlet, an alternate method to control cavitation is to raise water pressure, which can be accomplished in different ways.   We’ll review those options next time.

Copyright 2018 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog

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A Pulley Speed Ratio Formula Application

Friday, April 21st, 2017

    Last time we saw how pulley diameter governs speed in engineering scenarios which make use of a belt and pulley system.   Today we’ll see how this phenomenon is defined mathematically through application of the Pulley Speed Ratio Formula, which enables precise pulley diameters to be calculated to achieve specific rotational speeds.   Today we’ll apply this Formula to a scenario involving a building’s ventilating system.

    The Pulley Speed Ratio Formula is,

                                                D1 × N1 =  D2 × N2                             (1)

where, D1 is the diameter of the driving pulley and D2 the diameter of the driven pulley.

A Pulley Speed Ratio Formula Application

A Pulley Speed Ratio Formula Application

   

The pulleys’ rotational speeds are represented by N1 and N2,  and are measured in revolutions per minute (RPM).

    Now, let’s apply Equation (1) to an example in which a blower must deliver a specific air flow to a building’s ventilating system.   This is accomplished by manipulating the ratios between the driven pulley’s diameter, D2, with respect to the driving pulley’s diameter, D1.   If you’ll recall from our discussion last time, when both the driving and driven  pulleys have the same diameter, the entire assembly moves at the same speed, and this would be bad for our scenario.

    An electric motor and blower impeller moving at the same speed is problematic because electric motors are designed to spin at much faster speeds than typical blower impellers in order to produce desired air flow.   If their pulleys’ diameters were the same size, it would result in an improperly working ventilating system in which air passes through the furnace heat exchanger and air conditioner cooling coils far too quickly to do an efficient job of heating or cooling.

    To bear this out, let’s suppose we have an electric motor turning at a fixed speed of 3600 RPM and a belt-driven blower with an impeller that must turn at 1500 RPM to deliver the required air flow according to the blower manufacturer’s data sheet.   The motor shaft is fitted with a pulley 3 inches in diameter.   What pulley diameter do we need for the blower to turn at the manufacturer’s required 1500 RPM?

    In this example known variables are D1 = 3 inches, N1 = 3600 RPM, and N2 = 1500 RPM.   The diameter D2 is unknown.   Inserting the known values into equation (1), we can solve for D2,

                               (3 inches) × (3600 RPM) = D2 × (1500 RPM)         (2)

Simplified, this becomes,

                                            D2 = 7.2 inches                                      (3)

    Next time we’ll see how friction affects our scenario.

   

Copyright 2017 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog

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