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dedylja [7]
3 years ago
5

A pipe of 10 cm inner diameter is used to send crude oil over distance of 400 meters. The entire pipe was laid horizontal. The v

iscosity of the oil is 10 cp. The density of oil is 800 kg/m3 . a. (20 pts) The desired volumetric flow rate is 0.1 m3 /min. What is the Reynolds number of this flow? Is the flow laminar or turbulent? What is the pressure difference needed to generate this flow rate? b. (15 pts) Three month later, the operator found that they had to triple the pressure difference to maintain the flow rate at 0.1 m3 /min. The operator thought that wax deposition from the oil had reduced the inner diameter of the tube. Based on this assumption, can you estimate the reduced inner diameter of the tube? What is the Reynolds number of the flow?

Engineering
1 answer:
Arisa [49]3 years ago
7 0

Answer:

See explaination

Explanation:

Looking at Reynolds number, we can go ahead and describe the Reynolds number as a dimensionless value that is used to determine whether the fluid is exhibiting laminar flow (R less than 2300) or turbulent flow (R greater than 4000). Laminar flow is when a fluid moves smoothly and is predictable.

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Compare and contrast ""centralized"" and ""decentralized"" routing algorithms. (What are the advantages and disadvantages of eac
pantera1 [17]

Answer:

Comparison between centralized and decentralized routing algorithms:

The major similarity between both centralized and decentralized routing algorithms is that they are both communication serving systems. They both utilize node system(e.g, computer) and are both a communication liking system( e.g, Cable).

Contrasting between centralized and decentralized routing algorithms:

There are few differences between these two type of communication link or path way system but to name a couple of them,

For centralized, there is a single client server distribution node which simply means that one or more client server system are connected to a central processing server.

This also means that if the central processing server or pathway fails, it leads to the failure of the entire system. That is, there is no sending, responding or general processing of any form of requests. Example of a system that uses the centralized routing algorithm is the google search engine.

WHILE:

for the decentralized routing algorithms, there are multiple client server distribution pathway and each server makes its own decision. Here,there is no single entity that receives and responds to the request therefore,failure of any form of central path way processing node does not lead to the failure of the whole system unlike for the centralized system.

Advantages of centralized routing algorithm:

It can be easily protected or secured due to the nature of the system. If the central node is been secured, it generally translate to the different client node being secured.

It is easy to disconnect a connected client node from the central pathway node or central server as the case may be.

Disadvantages of centralized routing algorithm:

The client nodes are totally dependent on the central node or server so if there is a failure in the central server, the client node is then totally shut down.

Advantages of decentralized routing algorithm:

There is a random distribution of data on all the processing node or server which automatically creates a form of balance within the system. This leads to minimal or no down time processing client request.

Disadvantages of decentralized routing algorithm:

Due to the nature or fact that there are multiple processing system for different client node, it is difficult to detect which client node or request processing server is faulty. This can lead to delay in the fixing of fault in the system should it arise.

Why do we prefer a ""decentralized"" algorithm for routing messages through the internet?

The major reason why decentralized algorithm routing is preferred is because of the level of security attached to it. Each processing servers are secured independently and there is privilege of utilizing independent networking system.

5 0
3 years ago
The lead of the threaded shaft of the C-clamp is 0.05 in., and the mean radius of the thread is r = 0.15 in. Assume mus = 0.18.
barxatty [35]

The omitted part of the question shown in bold format

The lead of the threaded shaft of the C-clamp is 0.05 in., and the mean radius of the thread is r = 0.15 in. Assume μs = 0.18 and μk = 0.16. What couple must be applied to the shaft to exert a 30-lb force on the clamped object?

Answer:

C = 0.967 in. lb

Explanation:

Given that:

The lead of the threaded shaft of the C-clamp is 0.05 in.

∴ the pitch of the screw = 0.05 in

the mean radius of the thread is r = 0.15 in.

Assuming:

(μs)= 0.18  which implies the  coefficient of the static friction

(μk) = 0.16  (coefficient of kinetic friction)

Force = 30-lb

What couple must be applied to the shaft to exert a 30-lb force on the clamped object?

To determine the Couple (C) that must be applied; we use the expression:

C = Fr  tan (\theta_k + \alpha)

where; F = force

r =  mean radius

\theta_k= angle of kinetic friction

\alpha = pitch angle

NOW, let's take then one after the other.

From the coefficient of the static friction and the kinetic friction; we can solve for their respective angles, so we have:

Angle of static friction (\theta_s)  = tan^{-1}(U_s)

(\theta_s) = tan^{-1}(0.18)

(\theta_s) = 10.204°

Angle of kinetic friction (\theta_k)  = tan^{-1}(U_k)

\theta_k = tan^{-1}(0.16)

\theta_k = 9.0903°

To determine the pitch angle(\alpha); we apply the expression:

(\alpha) = tan^{-1}(\frac{p}{2 \pi r} )

(\alpha) = tan^{-1}(\frac{0.05}{2 \pi 0.15} )

(\alpha) = tan^{-1}(0.0530516 )

(\alpha) = 3.0368°

Have gotten our parameters to solve for Couple (C); then we have:

C = Fr  tan (\theta_k + \alpha)

substituting our values; we have:

C = (30 × 0.15)  tan ( 9.0903 + 3.0368)

C = 4.5  × tan ( 12.1271)

C = 4.5  × 0.2148761968

C = 0.96669428854 in.lb

C = 0.967 in. lb

Therefore, 0.967 in. lb couple  must be applied to the shaft to exert a 30-lb force on the clamped object.

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