Name: centrifugal pump flow rate calculation
Uploaded: 2025-02-13T15:43:22.000Z
Duration: 1 min 4 s
Channel: JoVE

Se si deve calcolare la portata di una pompa centrifuga che applica una pressione di 7500 joule al secondo per sollevare l'acqua da un serbatoio di livello inferiore a un serbatoio di livello superiore, è possibile utilizzare l'equazione dell'energia.
La differenza di livello tra i serbatoi è di 10 metri e la perdita di carico è di 5 metri.
Poiché le superfici di entrambi i serbatoi sono aperte all'atmosfera, la pressione e la velocità sia in ingresso che in uscita sono zero, semplificando ulteriormente l'equazione energetica.
In questo caso, la prevalenza della pompa è ottenuta dal quoziente della potenza assorbita netta dall'albero e dal prodotto del peso unitario dell'acqua e della portata.
Se la prevalenza della pompa viene sostituita con l'equazione energetica semplificata, è possibile ottenere la portata della pompa.
Inoltre, sostituendo i valori forniti nell'espressione per la portata, si ottiene il valore finale che indica la velocità con cui l'acqua viene pompata.

applicazione dell'equazione dell’energia

L'applicazione dell'equazione dell’energia alle pompe centrifughe, è un principio fondamentale nella fluidodinamica e nell'ingegneria. In questo scenario, l'equazione viene utilizzata per calcolare la portata di una pompa centrifuga, responsabile del trasferimento dell'acqua tra due serbatoi a diverse altitudini. La pompa applica un input di energia di 7500 joule al secondo e la differenza verticale tra i serbatoi inferiore e superiore, è di 10 metri. Inoltre, la perdita di carico dovuta all'attrito e ad altre resistenze, è di 5 metri. Questi parametri sono essenziali per risolvere la portata utilizzando l'equazione dell’energia.

Poiché entrambi i serbatoi sono a cielo aperto, la pressione, sia all'ingresso che all'uscita della pompa, è uguale alla pressione atmosferica. Ciò consente di annullare i termini di pressione nell'equazione dell’energia. Inoltre, poiché le velocità all'ingresso e all'uscita sono trascurabili, anche i termini di velocità vengono eliminati. L'equazione energetica per questo caso specifico, dipende solo dalla prevalenza gravitazionale e dalla prevalenza della pompa.

La prevalenza generata dalla pompa, è calcolata dividendo la potenza netta in ingresso all'albero, per il prodotto della portata e del peso specifico dell'acqua. Sostituendo questa prevalenza nell'equazione energetica e risolvendo per la portata della pompa, si determina quest’ultima.

Applicazione dell'Equazione dell’Energia

Se la portata di una pompa centrifuga che applica una pressione di 7500 joule al secondo per sollevare l'acqua da un serbatoio a un livello inferiore a un serbatoio a ...

Education

Didattica

JoVE Core

Civil Engineering

Ingegneria Civile

Unit 1

Unità 1

Finite Control Volume Analysis

Analisi a volume di controllo finito

KEY TERMS AND CONCEPTS

TERMINI E CONCETTI CHIAVE

Application of the Energy Equation

centrifugal pump flow rate calculation

Centrifugal pump flow rate can be calculated with the energy equation. In this problem, water moves between two reservoirs at different elevations, and the pump provides the energy needed for the transfer. The pump input is 7500 joules per second, the elevation difference is 10 meters, and the head loss from friction and other resistances is 5 meters.

Both reservoirs are open to the atmosphere, so the pressure at the inlet and outlet is the same. That means the pressure terms in the energy equation cancel. The flow at the inlet and outlet is also negligible, so the velocity terms drop out as well.

With pressure and velocity terms removed, the equation depends on the gravitational head and the pump head. The pump head is found by dividing the net shaft power input by the product of the flow rate and the specific weight of water. Substituting that head into the energy equation makes it possible to solve for the pump flow rate.

Centrifugal Pump Flow Rate Calculation

conservation of mass in finite cotrol volume

conservation-of-mass-in-finite-cotrol-volume

Continuity Equation for a Fixed Control Volume

The continuity equation describes conservation of mass for a fixed control volume. In fluid mechanics, a system is a collection of the same contents over time, and its mass stays constant. That idea is the starting point for deriving the equation. To analyze flow, a fixed, non-deforming control volume is chosen so it matches the system at one instant. The Reynolds transport theorem then links the system to the control volume. It shows that the rate of change of system mass has two parts: mass...

conservation of mass in fixed, nondeforming control volume

conservation-of-mass-in-fixed-nondeforming-control-volume

Continuity and Velocity in Pipe Flow

Conservation of mass explains how fluid moves through a fixed control volume such as a pipe or duct. In a fixed, nondeforming control volume, the total mass stays constant unless mass enters or leaves through the control surfaces. For steady, incompressible flow, the mass entering the system equals the mass leaving it. A sewer pipe is a useful example of this idea. It can be treated as a fixed control volume with inlet and outlet control surfaces around the flow region. Because the fluid is...

conservation of mass in moving, nondeforming control volume

conservation-of-mass-in-moving-nondeforming-control-volume

Stormwater Detention Basin Mass Balance

Stormwater detention basins use mass balance to manage runoff during heavy rainfall. These basins are especially important in urban areas, where impervious surfaces can increase flooding risk. Engineers use the conservation of mass to track how water enters, is stored in, and leaves the basin. In a detention basin, the control volume is the water inside the basin. Its boundaries are the basin walls, the inflow pipes, and the outflow structures. During a rain event, water comes in through the...

linear momentum in control volume

Momentum Balance for Fluid Control Volumes

Linear momentum balance for a fluid control volume links Newton’s second law to moving fluid. The law says that the rate of change of linear momentum equals the sum of the external forces acting on the system. In a fluid system, this idea helps explain how motion and force are related. A control volume can match the fluid system at a specific moment. When that happens, the forces on the control volume and the system are the same. Reynolds transport theorem explains this connection by splitting...

application of the linear momentum equation

application-of-the-linear-momentum-equation

Forces in a 180° Pipe Bend

Forces in a 180-degree pipe bend can be found by applying the linear momentum equation to flowing water. The pipe has a constant cross-sectional area of 0.01 square meters, and the water moves at 15 meters per second. The pressure at the inlet is 0.2 megapascals, and the pressure at the outlet is 0.16 megapascals. The goal is to find the force components in the x and y directions that hold the bend in place. In this case, the flow is only in the y direction at both the entrance and the exit.

moment-of-momentum equation

Wind Turbine Torque and Angular Momentum

The moment-of-momentum equation helps explain how wind turbine blades create torque. Torque is the turning effect of a force around an axis. In rotating systems, this equation links the motion of the wind with the force acting on it. The equation begins with Newton's second law for a fluid particle. That law says the rate of change of linear momentum equals the external force on the particle. To find torque, the position of the particle relative to the rotation axis is combined with that force.

conservation of energy in control volume

Turbine Work from Energy Changes

A turbine can turn fluid energy into shaft work when flow passes through a control volume. The fluid enters the turbine at one point and leaves at another. Under steady-flow conditions, there is no energy buildup inside the control volume, so the stored energy does not change with time. This steady-flow setup makes the energy equation simpler. The turbine is treated as doing mechanical work, and heat transfer is taken as negligible, so Q̇ is approximately zero. The flow at the inlet and outlet...

application of the energy equation

Centrifugal pump flow rate can be calculated with the energy equation. In this problem, water moves between two reservoirs at different elevations, and the pump provides the energy needed for the transfer. The pump input is 7500 joules per second, the elevation difference is 10 meters, and the head loss from friction and other resistances is 5 meters. Both reservoirs are open to the atmosphere, so the pressure at the inlet and outlet is the same. That means the pressure terms in the energy...

design example: forces in sluice gate

Sluice Gate Water Pressure and Reaction Force

Sluice gates control water flow in channels, reservoirs, and irrigation systems. They work as vertical barriers with an opening that can be raised or lowered. Changing the opening height changes the water velocity and pressure below the gate. The forces on a sluice gate depend on the water depths upstream and downstream, written as z1 and z2. When the gate is closed, the upstream water depth is greater. That creates a strong pressure difference across the gate and a net force toward the...

key terms and concepts

Use the energy equation to find the flow rate of a centrifugal pump lifting water between reservoirs. Includes elevation change, head loss, and pump head.

Centrifugal Pump Flow Rate Calculation | JoVE K12

JoVE Core Civil Engineering provides a foundational exploration of essential construction materials in civil engineering, focusing on bricks, stones, steel, wood, and concrete through easy-to-understand animated videos. This collection covers the properties, manufacturing processes, and practical applications of these materials. Additionally, the accompanied science-in-action videos demonstrate the research experiments performed in laboratories.

civil engineering

Centrifugal Pump Flow Rate Calculation

Description

Transcript

Tags