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dedylja [7]
4 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]4 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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1. (15) A truck scale is made of a platform and four compression force sensors, one at each corner of the platform. The sensor i
Elanso [62]

Answer:

a). 139498.24 kg

b). 281.85 ohm

c). 10.2 ohm

Explanation:

Given :

Diameter, d = 22 m

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Gauge resistance, R = 340 Ω

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$P = 0.03 \times 30 \times 10^9\times \frac{\pi}{4}\times (0.022)^2$

 = 342119.44 N

For the four sensors,

Maximum weight = 4 x P

                            =  4 x 342119.44

                            = 1368477.76 N

Therefore, weight in kg is $m=\frac{W}{g}=\frac{1368477.76}{9.81}$

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b). Change in resistance

k=\frac{\Delta R/R}{\Delta L/L}

$\Delta R = k. \epsilon R$    , since $\epsilon= \Delta L/ L$

$\Delta R = 6.9 \times 0.03 \times 340$

$\Delta R = 70.38 $ Ω

For 4 resistance of the sensors,

$\Delta R = 70.38 \times 4 = 281.52$ Ω

c). $k=\frac{\Delta R/R}{\epsilon}$

If linear strain,

$\frac{\Delta R}{R} \approx \frac{\Delta L}{L}$  , where k = 1

$\Delta R = \frac{\Delta L}{L} \times R$

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When a material is found to be in the tertiary phase of creep, the following procedure should be implemented that is the component should be replaced immediately. Therefore, Option C is correct.

<h3>What do you mean by a tertiary degree of creep?</h3>

Tertiary Creep has an extended creep rate and terminates when the material breaks or ruptures. It is related to each necking and formation of grain boundary voids. The wide variety of possible stress-temperature- time combos is infinite.

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