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2 years ago

Which of the following activities can help expand engineers' creative thinking capabilities?

Engineering

2 answers:

Papessa [141]2 years ago

4 0

Answer: its not team building excercize

Explanation:

Assoli18 [71]2 years ago

3 0

Answer:

team building exercises

Explanation:

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The wall shear stress in a fully developed flow portion of a 12-in.-diameter pipe carrying water is 2.00 lb/ft^2. Determine the

soldier1979 [14.2K]

Answer:

a) -8 lb / ft^3

b) -70.4 lb / ft^3

c) 54.4 lb / ft^3

Explanation:

Given:

- Diameter of pipe D = 12 in

- Shear stress t = 2.0 lb/ft^2

- y = 62.4 lb / ft^3

Find pressure gradient dP / dx when:

a) x is in horizontal flow direction

b) Vertical flow up

c) vertical flow down

Solution:

- dP / dx as function of shear stress and radial distance r:

(dP - y*L*sin(Q))/ L = 2*t / r

dP / L - y*sin(Q) = 2*t / r

Where dP / L = - dP/dx,

dP / dx = -2*t / r - y*sin(Q)

Where r = D /2 ,

dP / dx = -4*t / D - y*sin(Q)

a) Horizontal Pipe Q = 0

Hence, dP / dx = -4*2 / 1 - 62.4*sin(0)

dP / dx = -8 + 0

dP/dx = -8 lb / ft^3

b) Vertical pipe flow up Q = pi/2

Hence, dP / dx = -4*2 / 1 - 62.4*sin(pi/2)

dP / dx = 8 - 62.4

dP/dx = -70.4 lb / ft^3

c) Vertical flow down Q = -pi/2

Hence, dP / dx = -4*2 / 1 - 62.4*sin(-pi/2)

dP / dx = -8 + 62.4

dP/dx = 54.4 lb / ft^3

7 0

2 years ago

What are the main causes of injuries when using forklifts?

astra-53 [7]

Answer:

overturns

Explanation:

6 0

3 years ago

Read 2 more answers

A smoking lounge is to accommodate 13 heavy smokers. The minimum fresh air requirement for smoking lounges is specified to be 30

n200080 [17]

Answer:

The minimum required flow rate of fresh air that needs to be supplied to the lounge is 390L/s.

The diameter of the duct mus be of 24.9cm at least if thte air velocity is not to exceed 8m/s.

Explanation:

First, in order to find the minimum required flow rate of fresh air, we need to take the ASHRAE standard and multiply it by the number of heavy smokers, so we get:

$V'=30\frac{L}{s-persons}*13persons=390\frac{L}{s}$

This value can be converted to $m^{3}/s$

$390\frac{L}{s}*\frac{1000cm^{3}}{1L}*\frac{1m^{3}}{(100cm)^{3}}=0.39m^{3}/s$

Once we got this value we use the volumetric flow formula, which tells us that:

$V'=Av$

where A is the area of the duct and v is the velocity of the air flow.

We also know that:

$A=\frac{\pi d^{2}}{4}$

so we can substitute that into our equation, so we get:

$V'=\frac{\pi d^{2}}{4}v$

which can be solved for the diameter, which yields:

$d=\sqrt{\frac{4V'}{\pi v}}$

and we can next substitute the values the problem provides us with:

$d=\sqrt{\frac{4()0.39m^{3}/s}{\pi (8m/s)}}=0.249m$

so the duct must have a diameter of at least 24.9cm for the velocity not to exceed 8m/s.

4 0

2 years ago

Air at 20 kPa and 5 °C enters a 1.5 cm diameter tube at a uniform velocity of 1.5 m/s. The tube walls are maintained at a unifor

ankoles [38]

Answer:

The distance from the entrance at which the flow becomes fully developed (entrance lenght) is:

$L_{E}=1.752 m$

Explanation:

First, we need to know if the flow is laminar or turbulent using the equation for the Reynolds number in a circular tube, which is:

$Re=\frac{VD}{v}$ (Equation 1)

We know that for

$Re\leq 2300$ , the flow is laminar

$2300\leq Re\leq 1x10^{5}$ , the flow is turbulent

Then, tanking into account that for air at 20 kPa and 5°C, kinematic viscosity $(v)$ is $1.252x10^{-5} \frac{m^{2}}{s}$ (taken from Table A-9, Cengel's book), we use the equation 1 ,

$Re=\frac{(1.5 m/s)(0.015m)}{1.252x10^{-5}m^{2}/s}=1797.12$

And, we can conclude that the flow is laminar. Then, we can use the relationship between the entrance length $(L_{E})$ , which is the distance from the entrance at wich the flow becomes fully developed, and diameter for a laminar flow in a circular tube, which is: