The strength and stability of tires may be enhanced by heating both sides of the rubber ( 0.14 W/m·K, 6.35 × 10^-8m^2/s) in a steam chamber for which T[infinity]= 200°C. In the heating process, a 20-mm-thick rubber wall (assumed to be untreaded) is taken from an initial temperature of 35°C to a midplane temperature of 170°C. If steam flow over the tire surfaces maintains a convection coefficient of 200 W/m^2·K. How long will it take to achieve the desired midplane temperature?

Answer:

Gear Y would turn Counter-Clockwise do to the opposite force created from gear X.

                          Hope this helped!  Have a great day!

Answer: downward velocity = 6.9×10^-4 cm/s

Explanation: Given that the

Diameter of the smoke = 0.05 mm = 0.05/1000 m = 5 × 10^-5 m

Where radius r = 2.5 × 10^-5 m

Density = 1200 kg/m^3

Area of a sphere = 4πr^2

A = 4 × π× (2.5 × 10^-5)^2

A = 7.8 × 10^-9 m^2

Volume V = 4/3πr^3

V = 4/3 × π × (2.5 × 10^-5)^3

V = 6.5 × 10^-14 m^3

Since density = mass/ volume

Make mass the subject of formula

Mass = density × volume

Mass = 1200 × 6.5 × 10^-14

Mass M = 7.9 × 10^-11 kg

Using the formula

V = sqrt( 2Mg/ pCA)

Where

g = 9.81 m/s^2

M = mass = 7.9 × 10^-11 kg

p = density = 1200 kg/m3

C = drag coefficient = 24

A = area = 7.8 × 10^-9m^2

V = terminal velocity

Substitute all the parameters into the formula

V = sqrt[( 2 × 7.9×10^-11 × 9.8)/(1200 × 24 × 7.8×10^-9)]

V = sqrt[ 1.54 × 10^-9/2.25×10-4]

V = 6.9×10^-6 m/s

V = 6.9 × 10^-4 cm/s

Answer:

The uniform pressure for the necessary axial force  is  W = 945 N

The uniform wear for the necessary axial force is  W = 970.15 N

Explanation:

Solution

Given that:

r₁ = 0.1 m

r₂ = 0.05m

μ = 0.35

p = 12 N or kW

N = 1500 rpm

W = 1000 N

The angular velocity is denoted as  ω= 2πN/60

Here,

ω = 2π *1500/60 = 157.07 rad/s

Now, the power transferred becomes

P = Tω this is the equation (1)

Thus

12kW = T * 157.07 rad/s

T = 76.4 N.m

Now, when we look at the uniform condition, we have what is called the torque that is frictional which acts at the frictional surface of the clutch dented as :

T = nμW R this is the equation (2)

The frictional surface of the mean radius is denoted by

R =2/3 [(r₁)³ - (r₂)³/(r₁)² - (r₂)²]

=[(0.1)³ - (0.05)³/[(0.1)² - (0.05)²]

R is =0.077 m

Now, we replace this values and put them into the equation (2)

It gives us this, 76. 4 N.m = n * 0.35* 1000 N * 0.077 m

n = 2.809 = 3

The number of pair surfaces is = 3

Secondly, we determine the uniform wear.

So, the mean radius is denoted as follows:

R = r₁ + r₂/ 2

=0.1 + 0.05/2

=0.075 m

Now, we replace the values and put it into the equation (2) formula

76. 4 N.m = n *0.35* 1000 N * 0.075 m

n= 2.91 = 3

Again, the number of pair surfaces = 3

However, for the uniform pressure with regards to the number of clutch plates is 3 we can derive the necessary axial force from the equation (2)

76. 4 N.m = 3 * 0.35 * W *0.077 m

W = 945 N

Also, for the uniform wear with regards to the number of clutch plates is 3 we can derive the necessary axial force from the equation (2)

76. 4 N.m = 3 * 0.35 * W *0.075 m

W = 970. 15 N

Answer:

The power produced by the turbine is 74655.936 kW.

Explanation:

A turbine is a device that operates at steady-state. Let suppose that turbine does not have heat interactions with surroundings, as well as changes in potential and kinetic energies are neglictible. Power output can be determined by First Law of Thermodynamics:

[tex]-\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0[/tex]

[tex]\dot W_{out} = \dot m\cdot (h_{in}-h_{out})[/tex]

Let suppose that water enters as saturated vapor and exits as saturated liquid. Specific enthalpies are, respectively:

[tex]h_{in} = 2748.1\,\frac{kJ}{kg}[/tex]

[tex]h_{out} = 225.94\,\frac{kJ}{kg}[/tex]

The power produce by the turbine is:

[tex]\dot W_{in} = \left(29.6\,\frac{kg}{s} \right)\cdot \left(2748.1\,\frac{kJ}{kg} - 225.94\,\frac{kJ}{kg} \right)[/tex]

[tex]\dot W_{in} = 74655.936\,kW[/tex]

Answer:

a. 20.265 MN

b. 0.555 MNm

c. 403.44 KW

Explanation:

Given:-

- The width ( w ) = 1000 mm

- Original thickness ( to ) = 10 mm

- Final thickness ( t ) = 5 mm

- The radius of the rollers ( R ) = 600 mm

- The peripheral speed of the roller ( v ) = 0.12

- Deformation efficiency ( ε ) = 55%

Find:-

a) the roller force ( F )

b) the roller torque ( T )

c) the performance on the pair of rollers. ( P )

Solution:-

- The process of flat rolling entails a pair of compressive forces ( F ) exerted by the rollers on the steel sheet that permanently deforms.

- The permanent deformation of sheet metal is seen as reduced thickness.

- We will assume that the compressive force ( F ) acts normal to the point of contact between rollers and metal sheet.

- The roll force ( F ) is defined as:

                                 [tex]F =L*w*Y_a_v_g[/tex]

Where,

                     L: The projected length of strip under compression

                     Y_avg: The yielding stress of the material = 370 MPa

- The projected length of strip under compression is approximated by the following relation:

                               [tex]L = \sqrt{R*( t_o - t_f )} \\\\L = \sqrt{0.6*( 0.01 - 0.005 )} \\\\L = 0.05477 m[/tex]

- The Roll force ( F ) can be determined as follows:

                            [tex]F = (0.05477)*(1 )*(370*10^6 )\\\\F = 20.265 MN[/tex]

- The roll torque ( T ) is given by the following relation as follows:

                               [tex]T = \frac{L}{2} * F\\\\T = \frac{0.05477}{2} * 20.265\\\\T = 0.555 MNm[/tex]

- The rotational speed of the rollers ( N ) is determined by the following procedure:

                               [tex]f = \frac{v}{2\pi* R} = \frac{0.12}{2*\pi 0.6} = 0.03181818 \frac{rev}{s} \\\\N = f*60 = 1.9090 rpm[/tex]

- The power consumed by the pair of rollers ( P ) is given by:

                              [tex]P = \frac{2\pi * F * L * N}{e*60,000} KW \\\\P = \frac{2\pi * ( 20.265*10^6) * (0.05477) * (1.90909 ) }{60,000*0.55} KW\\\\P = 403.44 KW[/tex]

I'd say number 4, number 3 looks like an exhaust valve

Answer:

screw thrust = ML[tex]T^{-2}[/tex] 

Explanation:

thrust of a screw propeller is given by the equation = p[tex]V^{2}[/tex][tex]D^{2}[/tex] x [tex]\frac{ND}{V}[/tex]Re

where,

D is diameter

V is the fluid velocity

p is the fluid density

N is the angular speed of the screw in revolution per second

Re is the Reynolds number which is equal to  puD/μ

where p is the fluid density

u is the fluid velocity, and

μ is the fluid viscosity = kg/m.s = M[tex]L^{-1}[/tex][tex]T^{-1}[/tex]

Reynolds number is dimensionless so it cancels out

The dimensions of the variables are shown below in MLT

diameter is m = L

speed is in m/s = L[tex]T^{-1}[/tex]

fluid density is in kg/[tex]m^{3}[/tex] = M[tex]L^{-3}[/tex]

N is in rad/s = L[tex]L^{-1}[/tex][tex]T^{-1}[/tex] =

If we substitute these dimensions in their respective places in the equation, we get

thrust = M[tex]L^{-3}[/tex][tex](LT^{-1}) ^{2}[/tex][tex]L^{2}[/tex][tex]\frac{T^{-1} L}{LT^{-1} }[/tex]

= M[tex]L^{-3}[/tex][tex]L^{2}[/tex][tex]T^{-2}[/tex]

screw thrust = ML[tex]T^{-2}[/tex] 

This is the dimension for a force which indicates that thrust is a type of force