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3 years ago

What is need for using fins?

1 answer:

7 0

Answer: It is a term of heat transfer process in which fins are surface that are the extension of the object to work for the heat exchangers to increase the heat exchanging rate.

Explanation: Fins are considered to help the heat exchanger surface to lead the process of heat transfer by increasing the are of the surface which is exposed to the surroundings. Fins work really well with materials having high thermal conductivity and will be more effective. They are preferred because they increase the rate of exchange of heat by increment in the convection.

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If a particle moving in a circular path of radius 5 m has a velocity function v = 4t2 m/s, what is the magnitude of its total ac

rjkz [21]

Answer:

8.62m/s²

Explanation:

the particle is experiencing both translational and circular motion

v=4t²

$a_{t}$ = $\frac{dv}{dt}$ =8t

at t=1s, $\frac{dv}{dt}$ =8(1)=8m/s²

$a_{c}$ = $\frac{v^{2} }{r}$

at t=1, v= 4(1)² = 4m/s

$a_{c}$ =4²/5

$a_{c}$ =3.2m/s²

∴ magnitude of total acceleration, a

a= $\sqrt{a_{t} ^{2} + a_{c} {2} }$

a= $\sqrt{8^{2} +3.2^{2} }$

a= $\sqrt{64+10.24}$

a= $\sqrt{74.24}$

a=8.62m/s²

5 0

3 years ago

Read 2 more answers

Yield and tensile strengths and modulus of elasticity . with increasing temperature. (increase/decrease/independent)

meriva

Answer:

Yield strength, tensile strength decreases with increasing temperature and modulus of elasticity decreases with increasing in temperature.

Explanation:

The modulus of elasticity of a material is theoretically a function of the shape of curve plotted between the potential energy stored in the material as it is loaded versus the inter atomic distance in the material. The temperature distrots the molecular structure of the metal and hence it has an effect on the modulus of elasticity of a material.

Mathematically we can write,

$E(t)=E_o[1-a\frac{T}{T_m}]$

where,

E(t) is the modulus of elasticity at any temperature 'T'

$E_o$ is the modulus of elasticity at absolute zero.

$T_{m}$ is the mean melting point of the material

Hence we can see that with increasing temperature modulus of elasticity decreases.

In the case of yield strength and the tensile strength as we know that heating causes softening of a material thus we can physically conclude that in general the strength of the material decreases at elevated temperatures.

6 0

3 years ago

Please help me on this it’s due now

lbvjy [14]

Answer:

road closures on deliver routes.. i think..

5 0

2 years ago

Select four items that an industrial engineer must obtain in order to practice in the field.

alex41 [277]

Answer:

Professional engineering license

Bachelor's degree

Computer science classes

job recommendations

3 0

2 years ago

Read 2 more answers

Air at 293k and 1atm flow over a flat plate at 5m/s. The plate is 5m wide and 6m long. (a) Determine the boundary layer thicknes

loris [4]

Answer:

a). 8.67 x $10^{-3}$ m

b).0.3011 m

c).0.0719 m

d).0.2137 N

e).1.792 N

Explanation:

Given :

Temperature of air, T = 293 K

Air Velocity, U = 5 m/s

Length of the plate is L = 6 m

Width of the plate is b = 5 m

Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X $10^{-5}$ Pa-s

We know density of air is ρ = 1.21 kg / $m^{3}$

Now we can find the Reyonld no at x = 1 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 1}{1.822\times 10^{-5} }$

Re = 332052.6

Therefore the flow is laminar.

Hence boundary layer thickness is

δ = $\frac{5.x}{\sqrt{Re}}$

= $\frac{5\times 1}{\sqrt{332052.6}}$

= 8.67 x $10^{-3}$ m

a). Boundary layer thickness at x = 1 is δ = 8.67 X $10^{-3}$ m

b). Given Re = 100000

Therefore the critical distance from the leading edge can be found by,

Re = $\frac{\rho .U.x}{\mu }$

100000 = $\frac{1.21\times5\times x}{1.822 \times10^{-5}}$

x = 0.3011 m

c). Given x = 3 m from the leading edge

The Reyonld no at x = 3 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 3}{1.822\times 10^{-5} }$

Re = 996158.06

Therefore the flow is turbulent.

Therefore for a turbulent flow, boundary layer thickness is

δ = $\frac{0.38\times x}{Re^{\frac{1}{5}}}$

= $\frac{0.38\times 3}{996158.06^{\frac{1}{5}}}$

= 0.0719 m

d). Distance from the leading edge upto which the flow will be laminar,

Re = $\frac{\rho \times U\times x}{\mu }$

5 X $10^{5}$ = $\frac{1.21 \times 5\times x}{1.822\times 10^{-5}}}$

x = 1.505 m

We know that the force acting on the plate is

$F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

and $C_{D}$ at x= 1.505 for a laminar flow is = $\frac{1.328}{\sqrt{Re}}$

= $\frac{1.328}{\sqrt{5\times10 ^{5}}}$

= 1.878 x $10^{-3}$

Therefore, $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

= $\frac{1}{2}\times 1.878\times 10^{-3}\times 1.21\times (5\times 1.505)\times 5^{2}$

= 0.2137 N

e). The flow is turbulent at the end of the plate.

Re = $\frac{\rho \times U\times x}{\mu }$

= $\frac{1.21 \times 5\times 6}{1.822\times 10^{-5} }$

= 1992316

Therefore $C_{D}$ = $\frac{0.072}{Re^{\frac{1}{5}}}$

= $\frac{0.072}{1992316^{\frac{1}{5}}}$

= 3.95 x $10^{-3}$

Therefore $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho\times A\times U^{2}$

= $\frac{1}{2}\times 3.95\times 10^{-3}\times 1.21\times (5\times 6)\times 5^{2}$

= 1.792 N

3 0

3 years ago