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inna [77]
3 years ago
7

How can any student outside apply for studying engineering at Cambridge University​

Engineering
1 answer:
telo118 [61]3 years ago
3 0
Admission to the Engineering course at Cambridge is highly competitive, both in terms of the numbers and quality of applicants. In considering applicants, Colleges look for evidence both of academic ability and of motivation towards Engineering. There are no absolute standards required of A Level achievement, but it should be noted that the average entrant to the Department has three A* grades. You need to get top marks in Maths and Physics.All Colleges strongly prefer applicants for Engineering to be taking a third subject that is relevant to Engineering.
Hope that helps and good luck if you are applying. Can you please mark this as brainliest and press the thank you button and if you have any further questions please let me know!!
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Someone claims that in fully developed turbulent flow in a tube, the shear stress is a maximum at the tube surface. Is this clai
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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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Mercury flows inside a copper tube 9m long with a 5.1сm inside diameter at an average velocity of 7.0 m/s. The inside surface te
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Answer:

rate of heat transfer = 9085708.80 W

Explanation:

Given:

Inside diameter, D = 5.1 cm

                               = 5.1 x 10^{-2} m

Average velocity, V = 7 m/s

Mean temperature, T = (66+38) /2

                                    = 52°C

Therefore kinematic viscosity at 52°C is ν = 0.104 X 10^{-6} m^{2} / s

Prandtl no., Pr = 0.021

We know Renold No. is

Re = \frac{V\times D}{\nu }

Re = \frac{7\times 5.1\times 10^{-2}}{0.104\times 10^{-6}}

     = 3.432 X 10^{6}

Therefore the flow is turbulent.

Since the flow is turbulent and the ratio of L/D is greater than 60 we can use Dittua-Boelter equation.

Nu = 0.023 Re^{0.8}.Pr^{0.3}

     = 0.023 x (3.432 \times10^{6})^{0.8} x (0.021)^{0.3}

     = 1221.52

Since Nu = \frac{h.D}{k}

          h = \frac{k\times Nu}{D}

             = \frac{9.4\times 1221.52}{5.1\times 10^{-2}}

             = 225143.3

Therefore rate of heat transfer, q = h.A(T-T_{\infty }

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             = 225143.3 X 2π X \frac{5.1\times10^{-2}}{2}\times 9\times 28

              = 9085708.80 W

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