Question

An oil pump operating at steady state delivers oil at a rate of \(5.5 \mathrm{~kg} / \mathrm{s}\) and a velocity of \(6.8 \mathrm{~m} / \mathrm{s}\). The oil, which can be modeled as incompressible, has a density of \(1600 \mathrm{~kg} / \mathrm{m}^{3}\) and experiences a pressure rise from inlet to exit of \(28 \mathrm{Mpa}\). There is no significant elevation difference between inlet and exit, and the inlet kinetic energy is negligible. Heat transfer between the pump and its surroundings is negligible, and there is no significant change in temperature as the oil passes through the pump. If pumps are available in \(1 / 4\)-horsepower increments, determine the horsepower rating of the pump needed for this application.

Short answer

129.25 horsepower

Step-by-step solution

- Identify Given Data

Given data includes: \( \text{Mass flow rate} (\frac{dm}{dt}) = 5.5 \mathrm{~kg/s} \)\( \text{Exit velocity} (v) = 6.8 \mathrm{~m/s} \)\( \text{Density of oil} (\rho) = 1600 \mathrm{~kg/m^3} \)\( \text{Pressure rise} (\triangle p) = 28 \mathrm{~MPa} = 28 \times 10^6 \mathrm{~Pa} \)\( \text{No significant elevation difference, heat transfer, or temperature change} \)

- Calculate the Volume Flow Rate

Use the mass flow rate and the density to find the volume flow rate:\[ \text{Volume flow rate} (Q) = \frac{\text{Mass flow rate}}{\text{Density}} = \frac{5.5}{1600} = 0.0034375 \mathrm{~m^3/s} \]

- Determine the Kinetic Energy Change

Because the inlet kinetic energy is negligible, the kinetic energy change is based on the exit velocity:\[ KE = \frac{1}{2} \text{Mass flow rate} \times v^2 = \frac{1}{2} \times 5.5 \times 6.8^2 = 126.24 \mathrm{~J/s} \]

- Calculate the Work Done by the Pump

The work done by the pump (power) combines the pressure work and the kinetic energy change:\[ W = Q \times \triangle p + KE \]\[ W = 0.0034375 \times 28 \times 10^6 + 126.24 \]\[ W = 96250 + 126.24 = 96376.24 \mathrm{~W} \]

- Convert Power to Horsepower

Convert the power from watts to horsepower:\[ 1 \mathrm{~hp} = 745.7 \mathrm{~W} \]\[ \text{Horsepower} = \frac{96376.24}{745.7} = 129.26 \mathrm{~hp} \]

- Determine the Required Horsepower Rating

Given that pumps are available in 1/4 horsepower increments, round up to the nearest increment:\[ 129.25 \approx 129.25 \mathrm{~hp} \]

Key concepts

mass flow rate

The mass flow rate is a fundamental concept for understanding fluid dynamics in various applications, including the operation of an oil pump. It measures how much mass of a substance passes through a given point per unit time. This rate is typically measured in kilograms per second (kg/s). For instance, in our exercise, the oil pump delivers oil at a rate of 5.5 kg/s.
Mass flow rate is mathematically expressed as: \[ \frac{dm}{dt} \] \ where \( dm \) is the mass and \( dt \) is the time.
Understanding the mass flow rate helps in calculating other important variables, such as the volume flow rate and the power needed for the pump.

volume flow rate

Volume flow rate is another essential concept in fluid dynamics. It refers to the volume of fluid that flows through a cross-sectional area per unit of time. In the context of our oil pump problem, it can be determined using the mass flow rate and the oil's density.
It is given by: \[ Q = \frac{dm}{dt} \times \frac{1}{\rho} \] \ where: \( Q \) is the volume flow rate, \( \frac{dm}{dt} \) is the mass flow rate, and \( \rho \) is the density.
From our exercise, with a mass flow rate of 5.5 kg/s and a density of 1600 kg/m³, we derive the volume flow rate to be 0.0034375 m³/s.
This value is crucial for determining the work done by the pump and other subsequent calculations.

kinetic energy change

Kinetic energy change is a key factor in understanding the energy dynamics of fluid flow. It is the difference in kinetic energy between two points in the fluid system. In our example, the inlet kinetic energy is negligible, so the change in kinetic energy primarily comes from the exit velocity.
To calculate this, we use the equation: \[ KE = \frac{1}{2} \times \text{Mass flow rate} \times v^2 \] \ where \( v \) is the exit velocity.
Using the given values, the kinetic energy change comes out to be 126.24 J/s. This value helps in calculating the total power required by the pump to ensure the oil flows effectively.

pressure work

Pressure work involves the work needed to move a fluid from one pressure level to another. This concept is particularly important in pump operations where the fluid is forcibly moved against a pressure difference. In our oil pump problem, the pressure rise from the inlet to the exit is 28 MPa.
The formula to calculate the work done by pressure is: \[ W_{pressure} = Q \times \triangle p \] \ where: \( Q \) is the volume flow rate and \( \triangle p \) is the pressure difference.
Using the calculated volume flow rate of 0.0034375 m³/s and the pressure rise of 28 MPa (or 28 x 10⁶ Pa), we determine the pressure work to be 96,250 J/s. This considerable value is crucial for later steps to calculate the total power required.

power conversion

Lastly, power conversion is essential for translating the calculated work into practical units like horsepower. This helps in determining the Real-World Requirements for the pump. Power produced by the pump is calculated by combining the kinetic energy change and the pressure work.
The total work done (power) by the pump is expressed as: \[ W = W_{pressure} + KE \]
From our calculations, this value sums up to 96,376.24 W.
To convert this into horsepower, a common unit of power, we use the conversion factor:
<\[ 1 \text{~hp} = 745.7 \text{~W}\ \]
Thus, \[ \text{Horsepower} = \frac{W}{745.7} \]
This helps us determine that the pump needs approximately 129.25 hp. Considering pumps are available in 1/4 horsepower increments, the final requirement rounds up to a 129.25 hp pump. Thus, understanding these conversions ensures that the system is appropriately designed and functions efficiently.