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3 years ago

How 2 mak robut??...

Engineering

2 answers:

olga55 [171]3 years ago

4 0

Answer:

I don't understand the question, could you try to explain in detail what you're trying to do?

FrozenT [24]3 years ago

3 0

Answer:

The question doesn't make sense

Explanation:

If you want to ask something please do make it understandable.

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A hydraulic jump is formed in a 4m wide outlet just downstream of a control gate, which is located at the upstream end of the ou

cupoosta [38]

Answer:

Width w = 4m

Glow depth = y1 = 20m

Outlet discharge = 40m

V1= velocity of flow = 40/20*4 = 1/2 = 0.5m/s

Froud number = v1/√gy1

= 0.5/√9.81x20 = 0.0356

1. Y2/20 = 1/2[-1+√1+8*(0.0356)²]

Y2 = 0.05

2. Energy loss in the jump = (20-0.05)²/4x20x0.05

= 1985m

7 0

3 years ago

A long aluminum wire of diameter 3 mm is extruded at a temperature of 280°C. The wire is subjected to cross air flow at 20°C at

Musya8 [376]

Answer:

Explanation:

Given:

Diameter of aluminum wire, D = 3mm

Temperature of aluminum wire, $T_{s}=280^{o}C$

Temperature of air, $T_{\infinity}=20^{o}C$

Velocity of air flow $V=5.5m/s$

The film temperature is determined as:

$T_{f}=\frac{T_{s}-T_{\infinity}}{2}\\\\=\frac{280-20}{2}\\\\=150^{o}C$

from the table, properties of air at 1 atm pressure

At $T_{f}=150^{o}C$

Thermal conductivity, $K = 0.03443 W/m^oC$ ; kinematic viscosity $v=2.860 \times 10^{-5} m^2/s$ ; Prandtl number $Pr=0.70275$

The reynolds number for the flow is determined as:

$Re=\frac{VD}{v}\\\\=\frac{5.5 \times(3\times10^{-3})}{2.86\times10^{-5}}\\\\=576.92$

sice the obtained reynolds number is less than $2\times10^5$ , the flow is said to be laminar.

The nusselt number is determined from the relation given by:

$Nu_{cyl}= 0.3 + \frac{0.62Re^{0.5}Pr^{\frac{1}{3}}}{[1+(\frac{0.4}{Pr})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{Re}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}$

$Nu_{cyl}= 0.3 + \frac{0.62(576.92)^{0.5}(0.70275)^{\frac{1}{3}}}{[1+(\frac{0.4}{(0.70275)})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{576.92}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}\\\\=12.11$

The covective heat transfer coefficient is given by:

$Nu_{cyl}=\frac{hD}{k}$

Rewrite and solve for $h$

$h=\frac{Nu_{cyl}\timesk}{D}\\\\=\frac{12.11\times0.03443}{3\times10^{-3}}\\\\=138.98 W/m^{2}.K$

The rate of heat transfer from the wire to the air per meter length is determined from the equation is given by:

$Q=hA_{s}(T_{s}-T{\infin})\\\\=h\times(\pi\timesDL)\times(T_{s}-T{\infinity})\\\\=138.92\times(\pi\times3\times10^{-3}\times1)\times(280-20)\\\\=340.42W/m$