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Likurg_2 [28]
4 years ago
5

Someone claims that in fully developed turbulent flow in a tube, the shear stress is a maximum at the tube surface. Is this clai

m correct? a. Yes b. No
Engineering
1 answer:
Alika [10]4 years ago
5 0

Answer:

Yes this claim is correct.

Explanation:

The shear stress at any point is proportional to the velocity gradient at any that point. Since the fluid that is in contact with the pipe wall shall have zero velocity due to no flow boundary condition and if we move small distance away from the wall the velocity will have a non zero value thus a maximum gradient will exist at the surface of the pipe hence correspondingly the shear stresses will also be maximum.

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Describe the risks associated with their working environment (such as the tools, materials and equipment that they use, spillage
Llana [10]

The risks which are associated with such working environment and not reporting accidental breakages of tools is pollution and damage to the body system of individuals.

<h3>What is Risk?</h3>

This is defined as the possibility of something causing a bad event to occur and is common in workplace when tools are left carelessly which can cause injuries etc.

In the case of a working environment with oil spills, it can lead to slipping of the individuals present in the area and not reporting accidental breakages of tools or equipment could also lead to pollution and different forms of accident which thereby making it the most appropriate choice in this context.

Read more about Risk here brainly.com/question/1224221

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3 0
2 years ago
A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu
Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,  

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

  • Negligible heat loss through the insulation
  • Steady state system
  • One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}

During steady state

\frac{dT}{dt} = 0 which gives k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0

From which we have \frac{d^{2}T }{dx^{2} }  = -\frac{q'_{G}}{k}

Considering the boundary condition at x =0 where there is no heat loss

 \frac{dT}{dt} = 0 also at the other end of the plane wall we have

-k\frac{dT }{dx } = hc (T - T∞) at point x = L

Integrating the equation we have

\frac{dT }{dx }  = \frac{q'_{G}}{k} x+ C_{1} from which C₁ is evaluated from the first boundary condition thus

0 = \frac{q'_{G}}{k} (0)+ C_{1}  from which C₁ = 0

From the second integration we have

T  = -\frac{q'_{G}}{2k} x^{2} + C_{2}

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k}  + C_{2}-T∞) → C₂ = q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞

T(x) = \frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞ and T(x) = T∞ + \frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )

∴ Tmax → when x = 0 = T∞ + \frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))

Substituting the values we get

T = 167 ° C

4 0
4 years ago
I need help with my autos
oksano4ka [1.4K]

Answer:

what is wrong with it and what is the question

Explanation:

6 0
3 years ago
An analog baseband audio signal with a bandwidth of 4kHz is transmitted through a transmission channel with additive white noise
siniylev [52]

Answer:

2k20

Explanation:

4k ✈

8 0
3 years ago
When your complex reaction time is compromised by alcohol, an impaired person's ability to respond to emergency or unanticipated
babunello [35]

Answer:

decreased

Explanation:

when impaired you react slower then you would sober.

3 0
3 years ago
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