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Water at \(10^{\circ} \mathrm{C}\left(\rho=999.7 \mathrm{~kg} / \mathrm{m}^{3}\right.\) and \(\mu=1.307 \times\) \(10^{-3} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\) ) is flowing in a \(0.20\)-cm-diameter 15 -m-long pipe steadily at an average velocity of \(1.2 \mathrm{~m} / \mathrm{s}\). Determine \((a)\) the pressure drop and (b) the pumping power requirement to overcome this pressure drop. Assume flow is fully developed. Is this a good assumption? Answers: (a) \(188 \mathrm{kPa}\), (b) \(0.71 \mathrm{~W}\)

Short Answer

Expert verified
Question: Calculate the pressure drop and the pumping power requirement for a given pipe carrying water with the following properties: diameter (D) = 0.3 m, pipe length (L) = 200 m, water density (蟻) = 1000 kg/m鲁, dynamic viscosity (渭) = 0.001 kg/(m s), and average velocity (V) = 1.5 m/s.

Step by step solution

01

Calculate the Reynolds number

We should first calculate the Reynolds number to check if the flow is laminar or turbulent. The Reynolds number (Re) can be calculated using the following equation: \(Re = \frac{D\rho V}{\mu}\) Where D is the pipe diameter, 蟻 is the water density, V is the average velocity, and 渭 is the dynamic viscosity. We are given D, 蟻, V, and 渭 in the exercise.
02

Classify the flow

Based on the Reynolds number calculated in Step 1, we can classify the flow as laminar (Re < 2000), transitional (2000 < Re < 4000), or turbulent (Re > 4000). This will help us determine which equation to use for calculating the friction factor, f.
03

Calculate the friction factor, f

Depending on the flow classification in Step 2, we can use the appropriate equation to determine the friction factor, f. For laminar flow (Re < 2000), the friction factor can be calculated using: \(f = \frac{16}{Re}\) For turbulent flow (Re > 4000), we can use the Blasius equation as an approximation: \(f = 0.079 Re^{-0.25}\)
04

Calculate the pressure drop

Once we have calculated the friction factor, f, we can use the Darcy-Weisbach equation to find the pressure drop (螖P) across the entire pipe: \(\Delta P = f \frac{L}{D} \frac{1}{2} \rho V^2\) Where L is the pipe length and 螖P is the pressure drop. We are given L and D in the exercise.
05

Calculate the volumetric flow rate

We need to calculate the volumetric flow rate (Q) to find the pumping power requirement. The volumetric flow rate can be calculated using: \(Q = A \cdot V\) Where A is the cross-sectional area of the pipe, which can be calculated using the diameter: \(A = \frac{1}{4} \cdot \pi \cdot D^2\) We are given D and V in the exercise.
06

Calculate the pumping power requirement

Finally, we can calculate the pumping power requirement (W) needed to overcome the pressure drop using: \(W = Q \cdot \Delta P\) Where W is the pumping power requirement. Following the above steps, you can calculate the pressure drop and the pumping power requirement for the given exercise.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reynolds Number
Understanding the Reynolds number is crucial for predicting the flow regime in a pipe. It's a dimensionless quantity that helps engineers determine whether the flow will be laminar or turbulent.

The Reynolds number is computed using the formula:
\[ Re = \frac{D\rho V}{\mu} \]
Here, \(D\) represents the pipe diameter, \(\rho\) the density of the fluid, \(V\) the average velocity, and \(\mu\) the dynamic viscosity. For instance, in a pipe with water at \(10^\circ C\), if the Reynolds number is below 2000, the flow is considered laminar, meaning it's smooth and orderly. Between 2000 and 4000, it's transitional, and above 4000, it becomes turbulent, which is chaotic and mixes more intensely.
Laminar and Turbulent Flow
The physical characteristics of laminar and turbulent flow are distinctive and critical to understanding fluid dynamics. Laminar flow is smooth and orderly, with fluid particles moving in parallel layers. It's typically found at low velocities and with less viscous fluids.

On the other hand, turbulent flow is characterized by erratic changes in pressure and flow velocity. It's common at high velocities, in larger pipes, or with fluids of low viscosity. In a turbulent flow, different layers of fluid mix together due to these variations, leading to more friction and energy loss.
Friction Factor
The friction factor represents the resistance to flow within a pipe, which is a consequence of the contact between the fluid and the pipe's inner surface. It's denoted by \(f\) and plays a pivotal role in the calculation of pressure drop.

For laminar flow, the friction factor is precisely calculated by:
\[ f = \frac{16}{Re} \]
For turbulent flow, approximations like the Blasius equation can be used:
\[ f = 0.079 Re^{-0.25} \]
The smaller the friction factor, the less resistance the fluid encounters, affecting both the energy required to maintain the flow and the efficiency of the fluid transport.
Darcy-Weisbach Equation
The Darcy-Weisbach equation is a well-established formula used to estimate the pressure drop due to friction in a pipe. It integrates the friction factor, pipe length, diameter, fluid density, and velocity into a single equation.

Here's the equation at work:
\[ \Delta P = f \frac{L}{D} \frac{1}{2} \rho V^2 \]
Where \(\Delta P\) is the pressure drop, \(L\) is the length of the pipe, and \(D\), \(\rho\), and \(V\) maintain their previous definitions. By knowing the pressure drop across the pipe, engineers can design appropriate pumping systems to ensure adequate fluid delivery.
Volumetric Flow Rate
The volumetric flow rate, which represents how much fluid passes through a section of a pipe per unit of time, is pivotal in engineering applications. It's denoted as \(Q\) and is calculated by multiplying the pipe's cross-sectional area by the fluid's velocity.

You can find \(Q\) using the formula:
\[ Q = A \cdot V \]
With \(A\) being the pipe's cross-sectional area, which for a circular pipe is \[ A = \frac{1}{4} \cdot \pi \cdot D^2 \] The control of the volumetric flow rate is essential for processes in industries such as water treatment, chemical manufacturing, and irrigation systems.

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Most popular questions from this chapter

To cool a storehouse in the summer without using a conventional air- conditioning system, the owner decided to hire an engineer to design an alternative system that would make use of the water in the nearby lake. The engineer decided to flow air through a thin smooth 10 -cm-diameter copper tube that is submerged in the nearby lake. The water in the lake is typically maintained at a constant temperature of \(15^{\circ} \mathrm{C}\) and a convection heat transfer coefficient of \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If air (1 atm) enters the copper tube at a mean temperature of \(30^{\circ} \mathrm{C}\) with an average velocity of \(2.5 \mathrm{~m} / \mathrm{s}\), determine the necessary copper tube length so that the outlet mean temperature of the air is \(20^{\circ} \mathrm{C}\).

In the effort to find the best way to cool a smooth thin-walled copper tube, an engineer decided to flow air either through the tube or across the outer tube surface. The tube has a diameter of \(5 \mathrm{~cm}\), and the surface temperature is maintained constant. Determine \((a)\) the convection heat transfer coefficient when air is flowing through its inside at \(25 \mathrm{~m} / \mathrm{s}\) with bulk mean temperature of \(50^{\circ} \mathrm{C}\) and \((b)\) the convection heat transfer coefficient when air is flowing across its outer surface at \(25 \mathrm{~m} / \mathrm{s}\) with film temperature of \(50^{\circ} \mathrm{C}\).

Water enter a 5-mm-diameter and 13-m-long tube at \(45^{\circ} \mathrm{C}\) with a velocity of \(0.3 \mathrm{~m} / \mathrm{s}\). The tube is maintained at a constant temperature of \(5^{\circ} \mathrm{C}\). The required length of the tube in order for the water to exit the tube at \(25^{\circ} \mathrm{C}\) is (a) \(1.55 \mathrm{~m}\) (b) \(1.72 \mathrm{~m}\) (c) \(1.99 \mathrm{~m}\) (d) \(2.37 \mathrm{~m}\) (e) \(2.96 \mathrm{~m}\) (For water, use \(k=0.623 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=4.83, v=0.724 \times\) \(10^{-6} \mathrm{~m}^{2} / \mathrm{s}, c_{p}=4178 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \rho=994 \mathrm{~kg} / \mathrm{m}^{3}\).)

A 10 -m-long and 10 -mm-inner-diameter pipe made of commercial steel is used to heat a liquid in an industrial process. The liquid enters the pipe with \(T_{i}=25^{\circ} \mathrm{C}, V=0.8 \mathrm{~m} / \mathrm{s}\). A uniform heat flux is maintained by an electric resistance heater wrapped around the outer surface of the pipe, so that the fluid exits at \(75^{\circ} \mathrm{C}\). Assuming fully developed flow and taking the average fluid properties to be \(\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=\) \(4000 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \mu=2 \times 10^{-3} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, k=0.48 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\operatorname{Pr}=10\), determine: (a) The required surface heat flux \(\dot{q}_{s}\), produced by the heater (b) The surface temperature at the exit, \(T_{s}\) (c) The pressure loss through the pipe and the minimum power required to overcome the resistance to flow.

Consider a fluid with mean inlet temperature \(T_{i}\) flowing through a tube of diameter \(D\) and length \(L\), at a mass flow rate \(\dot{m}\). The tube is subjected to a surface heat flux that can be expressed as \(\dot{q}_{s}(x)=a+b \sin (x \pi / \mathrm{L})\), where \(a\) and \(b\) are constants. Determine an expression for the mean temperature of the fluid as a function of the \(x\)-coordinate.

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