Question

The velocity profile in fully developed laminar flow of water at \(40^{\circ} \mathrm{F}\) in a 140 -ft-long horizontal circular pipe, in \(\mathrm{ft} / \mathrm{s}\), is given by \(u(r)=0.8\left(1-625 r^{2}\right)\) where \(r\) is the radial distance from the centerline of the pipe in \(\mathrm{ft}\). Determine \((a)\) the volume flow rate of water through the pipe, \((b)\) the pressure drop across the pipe, and \((c)\) the useful pumping power required to overcome this pressure drop.

Short answer

Based on the given solution, determine the following: 1. The volumetric flow rate (Q) of water through the pipe. 2. The pressure drop (∆P) across the pipe. 3. The useful pumping power (P) required to overcome the pressure drop. Answer: 1. The volumetric flow rate (Q) of water through the pipe is approximately 0.0032 ft³/s. 2. The pressure drop (∆P) across the pipe is about 22.5 psi. 3. The useful pumping power (P) required to overcome the pressure drop is around 0.072 hp.

Step-by-step solution

Determine the cross-sectional area of the pipe

The given velocity profile is \(u(r) = 0.8(1-625 r^2)\), where \(r\) denotes radial distance from the centerline of the pipe in ft. To find the cross-sectional area \(A\) of the pipe, we square the radius \(R\) and multiply by \(\pi\). $$A = \pi R^2$$ As \(u(R) = 0\) (velocity is zero at the pipe wall), solving for \(R\) with the given velocity profile: $$0 = 0.8(1 - 625 R^2)$$ $$R^2 = \frac{1}{625}$$ $$R = \frac{1}{25}$$ Now we can calculate the cross-sectional area \(A\): $$A = \pi \left(\frac{1}{25}\right)^2$$

Calculate the volumetric flow rate \(Q\)

To find the volumetric flow rate, we must integrate the velocity profile over the cross-sectional area of the pipe. $$Q = \int_A u(r) dA$$ $$Q = 2\pi\int_0^R u(r)r dr$$ Substituting the given velocity profile \(u(r) = 0.8(1 - 625 r^2)\) and integrating: $$Q = 2\pi\int_0^{\frac{1}{25}} 0.8\left(1 - 625 r^2\right)r dr$$ $$Q = \frac{8}{15} \pi R^3$$ Now, substitute the value of \(R\) to find the volumetric flow rate: $$Q \approx 0.0032 \ \mathrm{ft^3/s}$$

Calculate the pressure drop \(\Delta P\)

Using the Hagen-Poiseuille equation, the pressure drop \(\Delta P\) in fully developed laminar flow can be determined as: $$\Delta P = \frac{32 \mu L Q}{(\pi R^4)}$$ Given the properties of water at \(40^{\circ}\mathrm{F}\), we can find the dynamic viscosity \(\mu = 2.2557 \times 10^{-5}\ \mathrm{lb\cdot s/ft^2}\), and use the given length of the pipe \(L = 140\ \mathrm{ft}\). Substituting the given values: $$\Delta P \approx 22.5\ \mathrm{psi}$$

Calculate useful pumping power \(P\)

The useful pumping power required can be calculated as: $$P = \Delta P \cdot Q$$ Substituting the values of \(\Delta P\) and \(Q\), we obtained earlier: $$P \approx 0.072\ \mathrm{hp}$$ In summary, the volume flow rate of water through the pipe is approximately \(0.0032\ \mathrm{ft^3/s}\), the pressure drop across the pipe is about \(22.5\ \mathrm{psi}\), and the useful pumping power required to overcome this pressure drop is around \(0.072\ \mathrm{hp}\).

Key concepts

Understanding Volume Flow Rate

Volume flow rate, denoted by the symbol \(Q\), is a critical parameter in fluid dynamics that represents the volume of fluid that passes through a given surface per unit time. It is often measured in cubic feet per second (\(\text{ft}^3\)/\(\text{s}\)) or liters per second (\(\text{L}\)/\(\text{s}\)). To establish this in the context of laminar flow, consider water moving smoothly with steady velocity through a pipe, such as the scenario in our exercise. The aim is to determine how much water is making its way through the pipe's cross-section at any given moment.

To calculate \(Q\) for laminar flow, one must integrate the velocity profile across the pipe's cross section. In the given exercise, the integration factor takes into account the varying speeds of the fluid at different radial distances from the centerline, which is due to the viscosity of the fluid causing it to stick to the pipe walls. The resulting formula, \(Q = 2\pi\int_0^R u(r)r \(dr\)\), reflects the physical reality that fluid at the center of a pipe travels faster than that near the edges.

It's essential for students to recognize that accurate calculation of the volume flow rate is fundamental to the understanding of systems involving fluid transportation. Precise flow rates dictate the efficiency and design parameters of such systems, affecting everything from household plumbing to large-scale industrial fluid transport.

Pressure Drop Calculation in Pipe Flow

Pressure drop is a concept that describes the decrease in pressure of a fluid as it moves through a conduit, like a pipe. This decrease is caused by frictional forces between the fluid particles and between the fluid and the pipe walls. For laminar flow, we use the Hagen-Poiseuille equation to quantify the pressure loss over a certain distance.

In mathematical terms, the pressure drop \(\Delta P\) can be expressed as \(\Delta P = \frac{32 \mu L Q}{(\pi R^4)}\), where \(\mu\) is the fluid's dynamic viscosity, \(L\) is the length of the pipe, \(Q\) is the volume flow rate, and \(R\) is the inside radius of the pipe. The viscosity \(\mu\) is a measure of a fluid's resistance to shear or flow, and it plays a significant role in determining \(\Delta P\).

For the student, it is critical to understand that any change in the fluid's velocity, pipe's dimensions, or the viscosity due to temperature can significantly alter the pressure drop. This concept is crucial in designing systems that need to maintain a specific pressure to function correctly. For instance, in our exercise, the calculated pressure drop across the 140-ft-long pipe can influence the selection of a pumping system needed to maintain adequate flow rate for the water at \(40^\circ\mathrm{F}\).

Calculating Pumping Power for Fluid Transport

Pumping power is the energy required per unit time to move a fluid through a pipe or other type of conduit against the resistance posed by pressure drop. Expressed in horsepower (hp) or watts (W), it represents the work a pump must perform to ensure the desired flow rate is achieved. The formula to find the pumping power \(P\) is \(P = \Delta P \cdot Q\), where \(\Delta P\) is the pressure drop across the pipe, and \(Q\) is the volumetric flow rate.

The importance of this calculation lies in the design and energy consumption analysis of pumping systems. Students must note that the power demands escalate with greater volume flow rates and pressure drops. Therefore, selecting a pump with sufficient power while optimizing its efficiency is critical for reducing operational costs and ensuring reliability.

In our exercise example, once the pressure drop and the volume flow rate are known, the calculation of pumping power gives us an idea of the required energy input. For systems transferring liquids over vast distances or with high friction, this insight is particularly valuable, allowing for precise cost and resource planning in real-world applications such as municipal water supply or irrigation systems.