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Atmospheric pressure

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pump ppt

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  • Atmospheric pressure

  • Vacuum

    The full or partial elimination of Atmospheric Pressure

    Atmospheric Pressure on the Moon = 0 = Full Vacuum1 Inch Hg Vacuum = 1.13 Ft of Water

  • Specific GravityThe ratio of the weight of anything to the weight of water.Example Specific Gravity of HCl = Weight of HCl (/) Weight of Water = 10.0 (/) 8.34 = 1.2

  • Pressure and Liquid Height Relationship (Head)1 PSI = 2.31 Ft of Water

  • Pressure, Liquid Height and Specific Gravity RelationshipPressure(PSI) = Head(FT) x Specific Gravity(SG) / 2.31Example - Water - 231Ft x 1.0 / 2.31 = 100 PSIExample - HCL - 231 Ft x 1.2 / 2.31 = 120 PSIExample - Gas - 231 Ft x .80 / 2.31 = 80 PSI

  • Vapor Pressure The pressure pushing against atmospheric pressure on liquids at elevated temperatures.

  • Suction HeadA Suction Head exists when the liquid is taken from an open to atmosphere tank where the liquid level is above the centerline of the pump suction, commonly known as a Flooded Suction.

  • Suction LiftA Suction Lift exists when the liquid is taken from an open to atmosphere tank where the liquid level is below the centerline of the pump suction.

  • Total Dynamic HeadTotal Dynamic Head (TDH) = Elevation(ft) + Friction(ft)

  • First, Let's Get A Big Picture PerspectiveOf Energy Flow in Pumping SystemsElectric utilityfeederTransformerMotor breaker/starterMotorAdjustablespeed drive(electrical)CouplingPumpFluidsystemUltimategoalAt each interface, there are inefficiencies. The goal should be to maximize the overall cost effectiveness of the pumping, or how much flow is delivered per unit of input energy.

  • Pumps and Pumping Systems

  • Centrifugal Pump

  • Centrifugal pumps

  • Pump Performance Curve

  • Hydraulic power, pump shaft power and electrical input power Hydraulic power Ph = Q (m3/s) x Total head, hd - hs (m) x (kg/m3) x g (m2/s) 1000 Where hd - discharge head, hs suction head, - density of the fluid, g acceleration due to gravity

    Pump shaft power Ps = Hydraulic power, Ph pump efficiency, Pump

    Electrical input power = Pump shaft power P Motor

  • Static Head

  • Dynamic (Friction) Head

  • System with high static head

  • System with low static head

  • Pump curve

  • Pump operating point

  • Typical pump characteristic curves

  • Selecting a pump Head Meters System Curve

  • Selecting a pump Head Meters 82% Pump Curve at Const. Speed System Curve Operating Point 500 m3/hr

  • Selecting a pump Head,m 82% Pump Curve at Const. Speed System Curve Operating Point 50030050

  • Selecting a pump Pump Efficiency 77% 82% Pump Curve at Const. Speed Partially closed valve Full open valve System Curves 500300 Head,m 5070

  • Selecting a pump Head Meters Pump Efficiency 77% 82% Pump Curve at Const. Speed Partially closed valve Full open valve System Curves Operating Points A B 500 m3/hr 300 m3/hr 50 m 70 m Static Head C 42 m

  • Efficiency Curves

  • If we select E, then the pump efficiency is 60%Hydraulic Power = Q (m3/s) x Total head, hd - hs (m) x (kg/m3) x g (m2/s) 1000 = (68/3600) x 47 x 1000 x 9.81 1000 = 8.7 kWShaft Power - 8.7 / 0.60 = 14.5 KwMotor Power - 14.8 / 0.9 = 16.1Kw (considering a motor efficiency of 90%)

  • If we select A, then the pump efficiency is 50% Hydraulic Power = Q (m3/s) x Total head, hd - hs (m) x (kg/m3) x g (m2/s) 1000 (68/3600) x 76 x 1000 x 9.81 1000 = 14 kWShaft Power - 14 / 0.50 = 28 KwMotor Power - 28 / 0.9 = 31 Kw (considering a motor efficiency of 90%)

  • Using oversized pump !As shown in the drawing, we should be using impeller "E" to do this, but we have an oversized pump so we are using the larger impeller "A" with the pump discharge valve throttled back to 68 cubic meters per hour, giving us an actual head of 76 meters.

    Hence, additional power drawn by A over E is 31 16.1 = 14.9 kW.Extra energy used - 8760 hrs/yr x 14.9 = 1,30,524 kw. = Rs. 5,22,096/annumIn this example, the extra cost of the electricity is more than the cost of purchasing a new pump.

  • Flow vs SpeedIf the speed of the impeller is increased from N1 to N2 rpm,the flow rate will increase from Q1 to Q2 as per the given formula:

  • The affinity law for a centrifugal pump with the impeller diameter held constant and the speed changed:

    Flow:Q1 / Q2 = N1 / N2Example: 100 / Q2 = 1750/3500 Q2 = 200 m3/hr

  • Head Vs speedThe head developed(H) will be proportional to the square of the quantity discharged, so that

  • Head:

    H1/H2 = (N12) / (N22) Example: 100 /H2 = 1750 2 / 3500 H2 = 400 m

  • Power Vs SpeedThe power consumed(W) will be the product of H and Q, and, therefore

  • Effect of speed variation

  • The affinity law for a centrifugal pump with the speed held constant and the impeller diameter changed

  • Reducing impeller diameterChanging the impeller diameter gives a proportional change in peripheral velocityDiameter changes are generally limited to reducing the diameter to about 75% of the maximum, i.e. a head reduction to about 50% Beyond this, efficiency and NPSH are badly affected However speed change can be used over a wider range without seriously reducing efficiency For example reducing the speed by 50% typically results in a reduction of efficiency by 1 or 2 percentage points. It should be noted that if the change in diameter is more than about 5%, the accuracy of the squared and cubic relationships can fall off and for precise calculations, the pump manufacturers performance curves should be referred to

  • Impeller Diameter Reduction on Centrifugal Pump Performance

  • Pump suction performance (NPSH)Net Positive Suction Head Available (NPSHA) NPSH Required (NPSHR) Cavitation NPSHR increases as the flow through the pump increases as flow increases in the suction pipework, friction losses also increase, giving a lower NPSHA at the pump suction, both of which give a greater chance that cavitation will occur

  • Pump control by varying speed:Pure friction headReducing speed in the friction loss system moves the intersection point on the system curve along a line of constant efficiency The affinity laws are obeyed

  • Pump control by varying speed:Static + friction headOperating point for the pump moves relative to the lines of constant pump efficiency when the speed is changed The reduction in flow is no longer proportional to speed A small turn down in speed could give a big reduction in flow rate and pump efficiency At the lowest speed illustrated, (1184 rpm), the pump does not generate sufficient head to pump any liquid into the system

  • Pumps in parallel switched to meet demand

  • Pumps in parallel with system curve

  • Energy conservation measuresConduct water balance minimise water consumptionAvoid idle cooling water circulation in DG sets, compressors, refrigeration systems In multiple pump operations, judiciously mix the operation of pumps and avoid throttlingHave booster pump for few areas of higher headReplace old pumps by energy efficient pumpsIn the case of over designed pump, provide variable speed drive, trim / replace impeller or replace with correct sized pump Remove few stages in multi-stage pump with over designed head

  • Energy Savings OpportunitiesGive efficiency of the pump due consideration while selecting a pump.Select pumps to match head flow requirements.Select a motor to match the load with high efficiency.Optimize the piping design.Monitor all important system parameters like: motor kW, pump head, flow temperature.Use pumps in series and parallel so that mismatch in system design or variations in operating conditions can be handled properly.

  • Energy Savings Opportunities (Contd.)Use variable speed drives for variations of flow due to process requirement.If the head flow is higher than needed by 5 to 15%, (i) The existing impeller should be trimmed to a smaller diameter, (ii) or a new impeller with a smaller diameter is to be put.In multistage pumps, add or remove stages to the existing pump, allowing an increase / decrease in delivered head of flow, if required.

    Use the white board and do the following

    Assume that we need to pump 68 m3/hr. to a 47 meter head with a pump that is 60% efficient at that point.Liquid Power - 68 x 47 / 360 = 8.9 KwWhere 360 is a constantShaft Power - 8.9 / 0.60 = 14.8 KwWhere 0.6 is the efficiency at that pointMotor Power - 14.8 / 0.9 = 16.4 KwWhere 0.9 is the motor efficiencyAs shown in the drawing, we should be using impeller "E" to do this, but we have an oversized pump so we are using the larger impeller "A" with the pump discharge valve throttled back to 68 cubic meters per hour, giving us an actual head of 76 meters. Now our Kilowatts look like this:68 x 76 / 360 = 14.3 Kw being produced by the pump, and 14.3 / 0.50 = 28.6 Kw required to do this. Subtracting the amount of kilowatts we should have been using gives us: 28.6 - 14.8 = 13.8 extra kilowatts being used to pump against the throttled discharge valve. Extra energy used - 8760 hrs/yr x 13.8 = 120,880 kw. = $ 10,000/annumIn this example the extra cost of the electricity could almost equal the cost of purchasing two or three pumps.

    As a first step, a water balance has to be made out quantifying requirements for various uses. Once this is done analysis needs to be made to reduce or recycle water. Reducing water consumption is the simplest way to save energy.Also look for standby equipment and equipment which are not in operation, but utilising water. For example a DG set which operates only during emergency might have water getting circulated for cooling. Such uses can be curtailed.Just to meet one or two applications where a higher pressure is required, the entire pumping system may be operating at a higher head. Reduce the pressure and provide booster pumps for such applications.

    First you should be sure of the flow and head requirements and then select the pump to operate in this point at maximum efficiency.The motor has to be selected so that the loading is maintained high. Energy efficient motor can be considered during new installations.Provide appropriate meters and log data so that analysis becomes easy.Where large variations are expected, multiple pumps have to be judiciously selected and operated for maximum efficiency.Variable speed drives can be considered only where continuous fluctuations are expected.Each pump casing has provision for replacement with lower or higher size impellers up to a certain limit and varies with manufacturer. Flow changes can be met by changing the impeller or trimming the existing impeller.In multistage pumps such as boiler feed pumps, too high a pressure than requirement can be dealt by removing a few stages of the pump.