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3 years ago

All of the following are categories for clutch covers except

Engineering

1 answer:

Lostsunrise [7]3 years ago

7 0

Whichever one is not one of these.

Molded disc. This type of disc produces less shock when the clutch engages. ...

Metal disc. Metal material is durable and therefore offers longer life cycle. ...

Single-plate type. This refers to clutches that use a single clutch disc between the flywheel and pressure plate. ...

Dual-plate type.

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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

Why the velocity potential Φ(x,y,z,t) exists only for irrotational flow

Black_prince [1.1K]

Answer:

$\omega_y,\omega_x,\omega_Z$ all are zero.

Explanation:

We know that if flow is possible then it will satisfy the below equation

$\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}+\dfrac{\partial w}{\partial z}=0$

Where u is the velocity of flow in the x-direction ,v is the velocity of flow in the y-direction and w is the velocity of flow in z-direction.

And velocity potential function $\phi$ given as follows

$u=-\frac{\partial \phi }{\partial x},v=-\frac{\partial \phi }{\partial y},w=-\frac{\partial \phi }{\partial z}$

Rotationality of fluid is given by $\omega$

$\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=\omega_Z$

$\frac{\partial v}{\partial z}-\frac{\partial w}{\partial y}=\omega_x$

$\frac{\partial w}{\partial x}-\frac{\partial u}{\partial z}=\omega_y$

So now putting value in the above equations ,we will find

$\omega =\frac{\partial \phi }{\partial x},u=\frac{\partial \phi }{\partial x},$

$\omega_y=\dfrac{\partial^2 \phi }{\partial z\partial x}-\dfrac{\partial^2 \phi }{\partial z\partial x}$

So $\omega_y$ =0

Like this all $\omega_y,\omega_x,\omega_Z$ all are zero.

That is why velocity potential flow is irroational flow.

5 0

4 years ago

A product whose total work content time = 50 min is assembled on a manual production line at a rate of 24 units per hour. From p

valentinak56 [21]

Answer:

a)Cycle time = 2.37 min

b)Numbers of workers =21

c)Stations on the line =24

Explanation:

Given that

Total work content time(TWC) = 50 min

Production rate Rp= 24 units/hr

manning level will be close =1.5

Line balancing efficiency =0.94

Cycle time

$T_c=\dfrac{60E}{R_P}$

$T_c=\dfrac{60\times 0.95}{24}$

Cycle time = 2.37 min

Numbers of workers ,W

$W=\dfrac{TWC}{T_c}$

$W=\dfrac{50}{2.37}$

W= 21

Numbers of workers =21

Stations on the line(n)

Lets find service time Ts

Ts = Cycle time - Time for repositioning

Ts = Tc- Tr

Ts= 2.37 - 9/ 60 min

Ts= 2.22 min

We know that efficiency

$\eta=\dfrac{TWC}{n.T_s}$

$0.94=\dfrac{50}{n\times 2.22}$

n=23.94 ⇒n=24

n=24

Stations on the line =24

7 0

3 years ago

A capillary tube is immersed vertically in a water container. Knowing that water starts to evaporate when the pressure drops bel

REY [17]

Answer: 10.12m, 20micro meter

Explanation:

8 0

3 years ago

A soil specimen was tested to have a moisture content of 32%, a void ratio of 0.95, and a specific gravity of soil solids of 2.7

belka [17]

Answer:

a. 0.9263

b. 0.4872

c. 13.83kN/m $^{3}$

Explanation:

moisture content (ω) = 0.32

void ratio (e) = 0.95

specific gravity ( $G_{s}$ ) = 2.75

the degree of satruation (S) = $\frac{w . G_{s} }{e}$ =0.32×2.75/0.95 = 0.9263

b. porosity (n) = $\frac{e}{e + 1}$ = 0.95/(0.95 + 1)= 0.4872

c. dry unit weight (γ $_{d}$ ) = $\frac{G_{s} . V_{w} }{1 + e}$

taking specific unit weight of water (V $_{w}$ )= 9.81kN/m $^{3}$

γ $_{d}$ = 2.75 × 1000/(1 + 0.95) = 13.83kN/m $^{3}$

5 0

3 years ago