Design and analyze cam profiles. Using MATLAB code,

A light pressure vessel is made of 2024-T3 aluminum alloy tubing with suitable end closures. This cylinder has a 90mm OD, a 1.65

Andru [333]

Given:

Outer Diameter, OD = 90 mm

thickness, t = 0.1656 mm

Poisson's ratio, $\mu$ = 0.334

Strength = 320 MPa

Pressure, P =3.5 MPa

Formula Used:

1). $Axial Stress_{max}$ = $P[\frac{r_{o}^{2} + r_{i}^{2}}{r_{o}^{2} - r_{i}^{2}}]$

2). factor of safety, m = $\frac{strength}{stress_{max}}$

Explanation:

Now, for Inner Diameter of cylinder, ID = OD - 2t

ID = 90 - 2(1.65) = 86.7 mm

Outer radius, $r_{o}$ = 45mm

Inner radius, $r_{i}$ = 43.35 mm

Now by using the given formula (1)

$Axial Stress_{max}$ = $3.5[\frac{45^{2} + 43.35^{2}}{45^{2} - 43.35^{2}}]$

$Axial Stress_{max}$ = $3.5\times 26.78$ =93.74 MPa

Now, Using formula (2)

factor of safety, m = \frac{320}{93.74} = 3.414

7 0

4 years ago

A centrifugal pump is required to pump water to an open water link situated 4 km away from the location of the pump through a pi

11111nata11111 [884]

Answer:

P= 5.5 bar

Explanation:

Given that

L= 4000 m

d= 0.2 m

Friction factor(F) = 0.01

speed V= 2 m/s

Head = 5 m

Head loss due to friction

$h_f=\dfrac{FLV^2}{2gd}$

$h_f=\dfrac{0.01\times 4000\times 2^2}{2\times 9.81\times 0.2}$

$h_f=40.77m$

So the total head(H) = 5 + 40.77 + 10.3 =56.07

Where 10.3 m is the atmospheric head.

We know that

P=ρ g H

So total Pressure

P= 1000 x 9.81 x 56.07 Pa

$P=5.5\times 10^5\ Pa$

P= 5.5 bar

5 0

4 years ago

Water flows around a 6-ft diameter bridge pier with a velocity of 12 ft/s. Estimate the force (per unit length) that the water e

jolli1 [7]

Answer: hello the diagram related to your question is missing please the third image is the missing part of the question

Fx = 977.76 Ib/ft

Explanation:

<u>Estimate the force that water exerts on the pier </u>

V = 12 ft/s

D( diameter ) = 6 ft

first express the force on the first half of the cylinder as

Fx1 = - $-2\int\limits^\pi _\frac{\pi }{2} {Ps*cos\beta *a} \, d\beta$ ---------------- ( 1 )

where ; Fy = 0

Ps = Po + 1/2 Pv^2 ( 1 - 4 sin^2β ) ------------- ( 2 )

Input equation (2) into equation ( 1 ) (note : assuming Po = 0 )

attached below is the remaining part of the solution

3 0

3 years ago

An experiment to determine the convection coefficient associated with airflow over the surface of a thick stainless steel castin

Maksim231197 [3]

Answer:

h = 375 KW/m^2K

Explanation:

Given:

Thermo-couple distances: L_1 = 10 mm , L_2 = 20 mm

steel thermal conductivity k = 15 W / mK

Thermo-couple temperature measurements: T_1 = 50 C , T_2 = 40 C

Air Temp T_∞ = 100 C

Assuming there are no other energy sources, energy balance equation is:

E_in = E_out

q"_cond = q"_conv

Since, its a case 1-D steady state conduction, the total heat transfer rate can be found from Fourier's Law for surfaces 1 and 2

q"_cond = k * (T_1 - T_2) / (L_2 - L_1) = 15 * (50 - 40) / (0.02 - 0.01)

=15KW/m^2

Assuming SS is solid, temperature at the surface exposed to air will be 60 C since its gradient is linear in the case of conduction, and there are two temperatures given in the problem. Convection coefficient can be found from Newton's Law of cooling:

q"_conv = h * ( T_∞ - T_s ) ----> h = q"_conv / ( T_∞ - T_s )

h = 15000 W / (100 - 60 ) C = 375 KW/m^2K

4 0

3 years ago

Fuel filters are being replaced on a HPCR diesel

saw5 [17]

Answer:) The correct answer is B. at the end of the fuel rail.

2) The one who is correct is the Technician A.

Explanation:

7 0

4 years ago