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Mobile App Chapter 8: Problem 139
Engine oil flows in a \(15-\mathrm{cm}\)-diameter horizontal tube with a velocity of \(1.3 \mathrm{~m} / \mathrm{s}\), experiencing a pressure drop of \(12 \mathrm{kPa}\). The pumping power requirement to overcome this pressure drop is (a) \(190 \mathrm{~W}\) (b) \(276 \mathrm{~W}\) (c) \(407 \mathrm{~W}\) (d) \(655 \mathrm{~W}\) (e) \(900 \mathrm{~W}\)
Short Answer
Expert verified
Answer: (b) 276 W
Step by step solution
01
Find the flow rate
To determine the pumping power required, we first need to find the flow rate (Q) of the engine oil in the tube. We can do this using the formula:
\[Q = A * v\]
where A is the cross-sectional area of the tube, and v is the fluid velocity.
We are given the diameter (D) of the tube as \(15 \thinspace cm\), so we can find the area as follows:
\[A = \pi (\frac{D}{2})^{2}\]
02
Convert the diameter to meters and find the cross-sectional area
Before substituting values, we need to convert the diameter from centimeters to meters for the purpose of calculations, since the velocity given is in meters per second.
\[D = 15 \thinspace \mathrm{cm} = 0.15 \thinspace \mathrm{m}\]
Now find the area:
\[A = \pi (\frac{0.15}{2})^{2}\]
03
Calculate the flow rate
Now we can substitute the values for A and v into the flow rate equation:
\[Q = A * v \Rightarrow Q = (\pi (\frac{0.15}{2})^{2}) * 1.3 \thinspace \mathrm{m/s}\]
04
Calculate the pumping power requirement
Now that we have found the flow rate, we can calculate the pumping power (P) required to overcome the pressure drop using the formula:
\[P = \Delta p * Q\]
where ∆p is the pressure drop. We are given that the pressure drop is \(12 \thinspace \mathrm{kPa}\). We need to convert this value to pascals (Pa):
\[\Delta p = 12 \thinspace \mathrm{kPa} \times 1000 \thinspace \mathrm{Pa/kPa} = 12000 \thinspace \mathrm{Pa}\]
Now calculate the pumping power requirement:
\[P = \Delta p * Q \Rightarrow P = (12000 \thinspace \mathrm{Pa}) * ((\pi (\frac{0.15}{2})^{2}) * 1.3 \thinspace \mathrm{m/s})\]
05
Compare the result with the given options
Now we can compare our result for the pumping power requirement with the given options to find the closest match:
(a) \(190 \thinspace \mathrm{W}\)
(b) \(276 \thinspace \mathrm{W}\)
(c) \(407 \thinspace \mathrm{W}\)
(d) \(655 \thinspace \mathrm{W}\)
(e) \(900 \thinspace \mathrm{W}\)
After calculations, we find the pumping power requirement closest to one of the given options.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Pressure Drop in Fluid Flow
Understanding the pressure drop in fluid flow is fundamental in fluid mechanics and crucial for calculations relating to fluid transport systems. Pressure drop refers to the decrease in pressure of a fluid as it moves through a conduit, such as a pipe or a tube. This decrease in pressure is caused by the frictional forces between the fluid and the walls of the conduit as well as any viscous effects within the fluid itself.
Different factors can affect the magnitude of the pressure drop, including the fluid's viscosity, the flow velocity, the pipe material, and the diameter of the pipe. Our exercise involves a known pressure drop of 12 kPa as the engine oil flows through a tube. To calculate the power needed to pump the fluid and overcome this pressure drop, we use the relationship involving the volumetric flow rate and the pressure drop, given by the formula:
\[P = \frac{Q}{1000} \times \frac{DP}{1000}\]where P is the pumping power in kW, Q is the volumetric flow rate in L/s, and DP is the pressure drop in kPa. This formula simplifies to:
\[P = \frac{Q \times DP}{1000000}\]Given a fluid's pressure drop and flow rate, engineers can use this formula to design pumps and systems that are both efficient and capable of maintaining the desired flow under real-world conditions.
Different factors can affect the magnitude of the pressure drop, including the fluid's viscosity, the flow velocity, the pipe material, and the diameter of the pipe. Our exercise involves a known pressure drop of 12 kPa as the engine oil flows through a tube. To calculate the power needed to pump the fluid and overcome this pressure drop, we use the relationship involving the volumetric flow rate and the pressure drop, given by the formula:
\[P = \frac{Q}{1000} \times \frac{DP}{1000}\]where P is the pumping power in kW, Q is the volumetric flow rate in L/s, and DP is the pressure drop in kPa. This formula simplifies to:
\[P = \frac{Q \times DP}{1000000}\]Given a fluid's pressure drop and flow rate, engineers can use this formula to design pumps and systems that are both efficient and capable of maintaining the desired flow under real-world conditions.
Fluid Velocity Calculation
The fluid velocity calculation is a key step in determining the flow rate for any fluid transport system and is directly linked with kinetic energy of the flowing fluid. Fluid velocity (\(v\)) is the speed at which a fluid flows in a given direction and is measured in meters per second (m/s). It's a vector quantity which tells us both the magnitude and the direction of flow.
To find the fluid velocity, we often rely on the continuity equation when dealing with incompressible fluids, stating that the product of cross-sectional area and velocity is constant throughout a streamline. From our problem, the velocity is already given as 1.3 m/s, and there’s no need to calculate it further. However, in practice, not all situations provide the velocity directly, and it might have to be determined from other parameters such as flow rate or pressure gradients, using Bernoulli's equation or other fluid dynamics principles.
Understanding the velocity profile of a fluid flow is also important for various engineering applications, especially in designing systems to minimize pressure losses, optimize flow rates, or control turbulent flows that can introduce efficiency losses or even cause damage to the system.
To find the fluid velocity, we often rely on the continuity equation when dealing with incompressible fluids, stating that the product of cross-sectional area and velocity is constant throughout a streamline. From our problem, the velocity is already given as 1.3 m/s, and there’s no need to calculate it further. However, in practice, not all situations provide the velocity directly, and it might have to be determined from other parameters such as flow rate or pressure gradients, using Bernoulli's equation or other fluid dynamics principles.
Understanding the velocity profile of a fluid flow is also important for various engineering applications, especially in designing systems to minimize pressure losses, optimize flow rates, or control turbulent flows that can introduce efficiency losses or even cause damage to the system.
Fluid Flow Rate
The fluid flow rate is a critical aspect in the study of fluid dynamics and is commonly used to measure the volume of fluid passing through a section of a system over a specific period. It is usually denoted as Q and expressed in cubic meters per second (\(\thinspace m^3/s\)) or in liters per second (\(\thinspace L/s\)).
To calculate the flow rate in our exercise, we first find the cross-sectional area of the pipe (A) where the fluid is flowing. Since the pipe has a circular cross-section, we can calculate the area using the radius (r) or diameter (D) of the pipe with the formula:
\[A = \thinspace \text{π}(\text{D}/2)^2\]Then, the volumetric flow rate (Q) is obtained by the product of the cross-sectional area (A) and the fluid velocity (v):
\[Q = A \times v\]Knowing the flow rate is not just about quantity; it helps predict the transfer rates of mass and heat, it is vital in sizing equipment such as pumps and pipes, and it helps establish whether the fluid flow is laminar or turbulent – each of which behaves differently and has unique implications for system design and energy consumption. Calculating an accurate flow rate is a key step in ensuring that systems operate within their intended specifications, promoting efficiency and longevity.
To calculate the flow rate in our exercise, we first find the cross-sectional area of the pipe (A) where the fluid is flowing. Since the pipe has a circular cross-section, we can calculate the area using the radius (r) or diameter (D) of the pipe with the formula:
\[A = \thinspace \text{π}(\text{D}/2)^2\]Then, the volumetric flow rate (Q) is obtained by the product of the cross-sectional area (A) and the fluid velocity (v):
\[Q = A \times v\]Knowing the flow rate is not just about quantity; it helps predict the transfer rates of mass and heat, it is vital in sizing equipment such as pumps and pipes, and it helps establish whether the fluid flow is laminar or turbulent – each of which behaves differently and has unique implications for system design and energy consumption. Calculating an accurate flow rate is a key step in ensuring that systems operate within their intended specifications, promoting efficiency and longevity.
