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3 years ago

40 POINTS IF ANSERED WITHIN 10 MINS

Engineering

2 answers:

jek_recluse [69]3 years ago

3 0

A It can show changes that occur in many different parts of Earth at the same time is the answer (I think)…

Nady [450]3 years ago

3 0

Right ones are. It can show changes that occur in many different parts of the Earth at the same time, and It can be used to show how the parts of the cycle relate to one another.

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A strip of chicken skin was excised for mechanical testing in tension. The initial dimensions of the rectangular specimen were 3

Stells [14]

Answer:

Table and chart are attached below

Explanation:

4 0

4 years ago

Air, at a free-stream temperature of 27.0°C and a pressure of 1.00 atm, flows over the top surface of a flat plate in parallel f

Morgarella [4.7K]

Answer:

Explanation:

Given that:

V = 12.5m/s

L= 2.70m

b= 0.65m

$T_{ \infty} = 27^0C= 273+27 = 300K$

$T_s= 127^0C = (127+273)= 400K$

P = 1atm

Film temperature

$T_f = \frac{T_s + T_{\infty}}{2} \\\\=\frac{400+300}{2} \\\\=350K$

dynamic viscosity =

$\mu =20.9096\times 10^{-6} m^2/sec$

density = 0.9946kg/m³

Pr = 0.708564

K= 229.7984 * 10⁻³w/mk

Reynolds number,

$Re = \frac{SUD}{\mu} =\frac{\ SUl}{\mu}$

$=\frac{0.9946 \times 12.5\times 2.7}{20.9096\times 10^-^6} \\\\Re=1605375.043$

we have,

$Nu=\frac{hL}{k} =0.037Re^{4/5}Pr^{1/3}\\\\\frac{h\times2.7}{29.79\times 10^-63} =0.037(1605375.043)^{4/5}(0.7085)^{1/3}\\\\h=33.53w/m^2k$

we have,

heat transfer rate from top plate

$\theta _1 =hA(T_s-T_{\infty})\\\\A=Lb\\\\=2.7*0.655\\\\ \theta_1=33.53*2.7*0.65(127/27)\\\\ \theta_1=5884.51w$

7 0

3 years ago

A cyclone is operated in a closed circuit with a ball mill. The cyclone is feed from a rod mill with a slurry that has a density

pav-90 [236]

Here is the flow sheet. Hope this helps have a great day!!

3 0

3 years ago

Two vertical parallel plates are spaced 0.01 ft apart. If the pressure decreases at a rate of 60 psf/ft in the vertical z-direct

Whitepunk [10]

Answer:

umax = 0.1259ft/s

Explanation:

Given:

•Distance between plates, B = 0.01ft

•Pressure difference decrease, $\frac{dp}{dz}=60ps/ft$

•Fluid viscosity, u = 10^-³lbf-s/ft²

Specific gravity, S = 0.80

Max velocity in the z-direction will be:

$u_max= [\frac{B^2y}{8u}]\frac{dh}{ds}$

$But h = \frac{P}{y}+z$

Substituting for h in the first equation, we have:

$\frac{d}{dz}[\frac{p}{y}+z]$

$\frac{dh}{dz}=\frac{1}{y}\frac{dp}{ds}+\frac{dz}{dz}$

$= \frac{1}{0.8*62.4}(-60)+1$

= -0.20192

Substituting dh/dz value in the first equation (umax), we have:

$umax = \frac{0.01^2(0.8*62.4)}{8*10^-^3}(-0.20192)$

umax = 0.1259ft/s

4 0

3 years ago

A 1-ft rod with a diameter of 0.5 in. is subjected to a tensile force of 1,300 lb and has an elongation of 0.009 in. The modulus

iragen [17]

Answer:

E = 8.83 kips

Explanation:

First, we determine the stress on the rod:

$\sigma = \frac{F}{A}\\\\$

where,

σ = stress = ?

F = Force Applied = 1300 lb

A = Cross-sectional Area of rod = 0.5 $\pi \frac{d^2}{4} = \pi \frac{(0.5\ in)^2}{4} = 0.1963\ in^2$

Therefore,

$\sigma = \frac{1300\ lb}{0.1963\ in^2} \\\\\sigma = 6.62\ kips$

Now, we determine the strain:

$strain = \epsilon = \frac{elongation}{original\ length} \\\\\epsilon = \frac{0.009\ in}{12\ in}\\\\\epsilon = 7.5\ x\ 10^{-4}$

Now, the modulus of elasticity (E) is given as:

$E = \frac{\sigma}{\epsilon}\\\\E = \frac{6.62\ kips}{7.5\ x\ 10^{-4}}$

7 0

3 years ago