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

Se a vazão de uma bomba centrífuga que aplica uma pressão de 7500 joules por segundo para elevar a água de um reservatório em um nível inferior para um reservatório em um nível superior tiver que ser calculada, a equação de energia pode ser usada.
A diferença de nível entre os reservatórios é de 10 metros e a perda de carga é de 5 metros.
Como as superfícies de ambos os reservatórios estão abertas para a atmosfera, a pressão e a velocidade na entrada e na saída são zero, simplificando ainda mais a equação de energia.
Aqui, a altura manométrica da bomba é obtida pelo quociente entre a entrada de potência líquida do eixo e o produto do peso unitário da água e da vazão.
Se a cabeça da bomba for substituída pela equação de energia simplificada, a vazão da bomba pode ser obtida.
Além disso, substituir os valores fornecidos na expressão pela vazão fornece o valor final que indica a taxa na qual a água é bombeada.

aplicação da equação de energia

A aplicação da equação de energia a bombas centrífugas é um princípio fundamental em dinâmica de fluidos e engenharia. Neste cenário, a equação de energia é usada para calcular a vazão de uma bomba centrífuga responsável por transferir água entre dois reservatórios em diferentes elevações. A bomba aplica uma entrada de energia de 7500 joules por segundo, e a diferença vertical entre os reservatórios inferior e superior é de 10 metros. Além disso, a perda de carga devido ao atrito e outras resistências é de 5 metros. Esses parâmetros são essenciais para resolver a vazão usando a equação de energia.

Como ambos os reservatórios estão abertos para a atmosfera, a pressão na entrada e na saída da bomba é igual à pressão atmosférica. Isso permite que os termos de pressão na equação de energia se cancelem. Além disso, como as velocidades na entrada e na saída são desprezíveis, os termos de velocidade na equação também são eliminados. A equação de energia, nesse caso específico, depende apenas da carga gravitacional e da carga da bomba.

A altura manométrica gerada pela bomba é determinada pela divisão da entrada de potência líquida do eixo pelo produto da vazão e do peso específico da água. Substituindo essa altura manométrica na equação de energia e resolvendo-a, a vazão da bomba pode ser determinada.

Aplicação da equação de energia

Se a vazão de uma bomba centrífuga que aplica uma pressão de 7500 joules por segundo para elevar a água de um reservatório em um nível mais baixo para um reservatório em ...

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Civil Engineering

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Finite Control Volume Analysis

Análise de Volume de Controle Finito

KEY TERMS AND CONCEPTS

TERMOS E CONCEITOS-CHAVE

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

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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...

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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.

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Centrifugal Pump Flow Rate Calculation

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