To use wiring diagrams, an understanding of the symbols, abbreviations, and connector coding used in the diagrams

Calculate the wire pressure for a round copper bar with an original cross-sectional area of 12.56 mm2 to a 30% reduction of area

dybincka [34]

Answer:153.76 MPa

Explanation:

$Initial Area\left ( A_0\right )=12.56 mm^2$

$Final Area\left ( A_f\right )=0.7\times 12.56 mm^2=8.792 mm^2$

$Die angle=30^{\circ}$

$\alpha =\frac{30}{2}=15^{\circ}$

$\mu =0.08$

$Yield stress\left ( \sigma _y \right )=350 MPa$

$B=\mu cot\left ( \aplha\right )=0.2985$

$\sigma _{pressure}=\sigma _y\left [\frac{1+B}{B}\right ]\left [ 1-\frac{A_f}{A_0}\right ]^B$

$\sigma _{pressure}=350\left [\frac{1+0.2985}{0.2985}\right ]\left [ 1-\frac{8.792}{12.56}\right ]^{0.2985}$

$\sigma _{pressure}=153.76 MPa$

8 0

3 years ago

Water flows through a horizontal plastic pipe with a diameter of 0.15 m at a velocity of 15 cm/s. Determine the pressure drop pe

Sonja [21]

Answer:0.1898 Pa/m

Explanation:

Given data

Diameter of Pipe $\left ( D\right )=0.15m$

Velocity of water in pipe $\left ( V\right )=15cm/s$

We know viscosity of water is $\left (\mu\right )=8.90\times10^{-4}pa-s$

Pressure drop is given by hagen poiseuille equation

$\Delta P=\frac{128\mu \L Q}{\pi D^4}$

We have asked pressure Drop per unit length i.e.

$\frac{\Delta P}{L} =\frac{128\mu \ Q}{\pi D^4}$

Substituting Values

$\frac{\Delta P}{L}=\frac{128\times8.90\times10^{-4}\times\pi \times\left ( 0.15^{3}\right )}{\pi\times 4 \times\left ( 0.15^{2}\right )}$

$\frac{\Delta P}{L}$ =0.1898 Pa/m

4 0

3 years ago

The metal control joints used to relieve stresses caused by expansion and contraction in large ceiling or wall expenses in inter

timurjin [86]

Answer:

It’s called Expansion Joints

6 0

3 years ago

A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16

abruzzese [7]

Answer:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The net work per cycle is 845.88 kJ/kg

The power developed in horsepower ≈ 45374 hP

Explanation:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The dimension of the cylinder bore diameter = 3.7 in. = 0.09398 m

Stroke length = 3.4 in. = 0.08636 m.

The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³

The clearance volume = 16% of cylinder volume = 0.16×5.99×10⁻⁴ m³

The clearance volume, v₂ = 9.59 × 10⁻⁵ m³

p₁ = 14.5 lbf/in.² = 99973.981 Pa

T₁ = 60 F = 288.706 K

$\dfrac{T_{2}}{T_{1}} = \left (\dfrac{v_{1}}{v_{2}} \right )^{K-1}$

Otto cycle T-S diagram

T₂ = 288.706* $6.25^{0.393}$ = 592.984 K

The maximum temperature = T₃ = 5200 R = 2888.89 K

$\dfrac{T_{3}}{T_{4}} = \left (\dfrac{v_{4}}{v_{3}} \right )^{K-1}$

T₄ = 2888.89 / $6.25^{0.393}$ = 1406.5 K

Work done, W = $c_v$ ×(T₃ - T₂) - $c_v$ ×(T₄ - T₁)

0.718×(2888.89 - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg

The power developed in an Otto cycle = W×Cycle per second

= 845.88 × 2400 / 60 = 33,835.377 kW = 45373.99 ≈ 45374 hP.

8 0

4 years ago

A parallel plates capacitor is filled with a dielectric of relative permittivity ε = 12 and a conductivity σ = 10^-10 S/m. The c

monitta

Answer:

t = 1.06 sec

Explanation:

Once disconnected from the battery, the capacitor discharges through the internal resistance of the dielectric, which can be expressed as follows:

R = (1/σ)*d/A, where d is is the separation between plates, and A is the area of one of the plates.

The capacitance C , for a parallel plates capacitor filled with a dielectric of a relative permittivity ε, can be expressed in this way:

C = ε₀*ε*A/d = 8.85*10⁻¹² *12*A/d

The voltage in the capacitor (which is proportional to the residual charge as it discharges through the resistance of the dielectric) follows an exponential decay, as follows:

V = V₀*e(-t/RC)

The product RC (which is called the time constant of the circuit) can be calculated as follows:

R*C = (1/10⁻¹⁰)*d/A*8.85*10⁻¹² *12*A/d

Simplifying common terms, we finally have:

R*C = 8.85*10⁻¹² *12 / (1/10⁻¹⁰) sec = 1.06 sec

If we want to know the time at which the voltage will decay to 3.67 V, we can write the following expression:

V= V₀*e(-t/RC) ⇒ e(-t/RC) = 3.67/10 ⇒ -t/RC = ln(3.67/10)= -1

⇒ t = RC = 1.06 sec.

3 0

3 years ago