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3 years ago

What’s a pigtail when wiring

Engineering

1 answer:

Levart [38]3 years ago

7 0

Answer:

A pigtail when wiring is technique that is used in connecting a lot of wires together.

Explanation:

A pigtail is a wire that is short in length. It has two ends. One end has a connector while the other end has other wires connected to it.

Pigtail when wiring is the connection of more than one wire in a circuit to another device. Pigtail when wiring helps to extend the length of the wire in a circuit if the wire used it short).

Pigtail when wiring is a technique what helps to keep the circuit organised because it prevents the wires from getting tangled.

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Oil with a density of 800 kg/m3 is pumped from a pressure of 0.6 bar to a pressure of 1.4 bar, and the outlet is 3 m above the i

Naddik [55]

Answer:

23.3808 kW

20.7088 kW

Explanation:

ρ = Density of oil = 800 kg/m³

P₁ = Initial Pressure = 0.6 bar

P₂ = Final Pressure = 1.4 bar

Q = Volumetric flow rate = 0.2 m³/s

A₁ = Area of inlet = 0.06 m²

A₂ = Area of outlet = 0.03 m²

Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s

Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s

Height between inlet and outlet = z₂ - z₁ = 3m

Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation

$\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m$

Work done by pump

$W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 14.896\\\Rightarrow W_{p}=23380.8\ W$

∴ Power input to the pump 23.3808 kW

Now neglecting kinetic energy

$h=\frac{P_2-P_1}{\rho g}+z_2-z_1\\\Righarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+3\\\Righarrow h=13.19\ m\\$

Work done by pump

$W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 13.193\\\Rightarrow W_{p}=20708.8\ W$

∴ Power input to the pump 20.7088 kW

6 0

3 years ago

What are some advantages of generating electrical energy from tides instead of from fossil fuels

harina [27]

Answer:

Efficent at low speeds.

Explanation:

Since water is 1,000 times more dense than air, electricity can generated from tides much more efficiently at slower speeds than wind turbines can.

7 0

3 years ago

A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.6 mm; the spe

ankoles [38]

Answer:

F = 8849 N

Explanation:

Given:

Load at a given point = F = 4250 N

Support span = L = 44 mm

Radius = R = 5.6 mm

length thickness of tested material = 12 mm

First compute the flexural strength for circular cross section using the formula below:

σ $_{fs} = F_{f} L / \pi R^{3}$

σ = FL / π R³

Putting the given values in the above formula:

σ = 4250 ( 44 x 10⁻³ ) / π ( 5.6 x 10⁻³ ) ³

= 4250 ( 44 x 10⁻³ ) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 4250 (44 x 1 /1000 )) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 4250 ( 11 / 250 ) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 187 / 3.141593 ( 5.6 x 1 / 1000 ) ³

= 187 / 3.141593 (0.0056)³

= 338943767.745358

= 338.943768 x 10⁶

σ = 338 x 10⁶ N/m²

Now we compute the load i.e. F from the following formula:

$F_{f}$ = 2 σ $_{fs}$ d³/3 L

F = 2σd³/3L

= 2(338 x 10⁶)(12 x 10⁻³)³ / 3(44 x 10⁻³)

= 2 ( 338 x 1000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 676000000 ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 676000000 ( 12 x 1/1000 )³ / 3 ( 44 x 10⁻³)

= 676000000 ( 3 / 250 )³ / 3 ( 44 x 10⁻³)

= 676000000 ( 27 / 15625000 ) / 3 ( 44 x 10⁻³)

= 146016 / 125 / 3 ( 44 x 1 / 1000 )

= ( 146016 / 125 ) / (3 ( 11 / 250 ))

= 97344 / 11

F = 8849 N

4 0

3 years ago

It is desired to produce and aligned carbon fiber-epoxy matrix composite having a longitudinal tensile strength of 630 MPa. Calc

ratelena [41]

Answer:

The answer is below

Explanation:

Given that:

Diameter (D) = 0.03 mm = 0.00003 m, length (L) = 2.4 mm = 0.0024 m, longitudinal tensile strength $(\sigma_{cd})=630\ MPa = 630*10^6\ Pa$ , Fracture strength

$(\sigma_f)=5100\ MPa=5100*10^6\ Pa,fiber-matrix\ stres(\sigma_m)=17.5\ MPa=17.5*10^6\ Pa,matrix\ strength=\tau_c=17\ MPa=17 *10^6\ Pa$

a) The critical length ( $L_c$ ) is given by:

$L_c=\sigma_f*(\frac{D}{2*\tau_c} )=5100*10^6*\frac{0.00003}{2*17*10^6}=0.0045\ m=4.5\ mm$

The critical length (4.5 mm) is greater than the given length, hence th composite can be produced.

b) The volume fraction (Vf) is gotten from the formula:

$\sigma_{cd}=\frac{L*\tau_c}{D}*V_f+\sigma_m(1-V_f)\\\\V_f=\frac{\sigma_{cd}-\sigma_{m}}{\frac{L*\tau_c}{D}-\sigma_{m}} \\\\Substituting:\\\\V_f=\frac{630*10^6-17.5*10^6}{\frac{0.0024*17*10^6}{0.00003} -17.5*10^6} \\\\V_f=0.456$

6 0

2 years ago

Technician A says that a defective crankshaft position sensor can cause a no spark condition technician B says that a faulty ign

Aliun [14]

Both tech distributor ignition

5 0

3 years ago