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DanielleElmas [232]

3 years ago

9

Technician A says that 18 gauge AWG wire can carry more current flow that 12 gauge AWG wire. Technician B says that metric wire

is sized by its cross-sectional area. Who is correct? Select one: a. Both Technicians A and B O b. Neither Technician A nor B C. Technician A d. Technicain B

1 answer:

denpristay [2]3 years ago

6 0

Answer:

Technician B

Explanation:

Both AWG and metric are sized by cross-sectional area.

Technician A is wrong: 12 gauge wire is larger diameter rated for 20 amps in free air. 18 awg is smaller diameter and typically used for speaker wiring, Class II or low voltage and sub-circuits within appliances.

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Steam at 4 MPa and 350°C is expanded in an adiabatic turbine to 125kPa. What is the isentropic efficiency (percent) of this turb

guajiro [1.7K]

Answer:

$\eta_{turbine} = 0.603 = 60.3\%$

Explanation:

First, we will find actual properties at given inlet and outlet states by the use of steam tables:

AT INLET:

At 4MPa and 350°C, from the superheated table:

h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

AT OUTLET:

At P₂ = 125 KPa and steam is saturated in vapor state:

h₂ = $h_{g\ at\ 125KPa}$ = 2684.9 KJ/kg

Now, for the isentropic enthalpy, we have:

P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than $s_g$ and greater than $s_f$ at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

$x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88$

Now, we will find $h_{2s}$ (enthalpy at the outlet for the isentropic process):

$h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg$

Now, the isentropic efficiency of the turbine can be given as follows:

$\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%$

3 0

3 years ago

Suppose the working pressure for a boiler is 10 psig, then what is the corresponding absolute pressure?

yanalaym [24]

Answer:

The corresponding absolute pressure of the boiler is 24.696 pounds per square inch.

Explanation:

From Fluid Mechanics, we remember that absolute pressure ( $p_{abs}$ ), measured in pounds per square inch, is the sum of the atmospheric pressure and the working pressure (gauge pressure). That is:

$p_{abs} = p_{atm}+p_{g}$ (1)

Where:

$p_{atm}$ - Atmospheric pressure, measured in pounds per square inch.

$p_{g}$ - Working pressured of the boiler (gauge pressure), measured in pounds per square inch.

If we suppose that $p_{atm} = 14.696\,psi$ and $p_{g} = 10\,psi$ , then the absolute pressure is:

$p_{abs} = 14.696\,psi+10\,psi$

$p_{abs} = 24.696\,psi$

The corresponding absolute pressure of the boiler is 24.696 pounds per square inch.

8 0

2 years ago

engineering uses data from pareto charts to analyse motor faults caused during their production. Explain one advantage of using

lions [1.4K]

The advantage of a pareto chart is to make sure they have all of their tools

3 0

3 years ago

An aggregate blend is composed of 55% aggregate A (Sp. Gr. 2.631), 25% aggregate B (Sp. Gr. 2.331) and 20% sand (Sp. Gr. 2.609).

Phoenix [80]

Answer:

The right choice would be Option b (2.545).

Explanation:

The given values are:

The aggregate blend will be:

$W_A$ = 55%

$G_ A$ = 2.631

$W_ B$ = 25%

$G_B$ = 2.331

$W_C$ = 20%

$G_C$ = 2.609

Now,

On applying the formula, we get

⇒ $G_{BA}=\frac{W_A+W_B+W_C}{\frac{W_A}{G_A} +\frac{W_B}{G_B} +\frac{W_C}{G_C}}$

On substituting the values, we get

⇒ $=\frac{55+25+20}{\frac{55}{2.631} +\frac{25}{2.331} +\frac{20}{2.609}}$

⇒ $= \frac{100}{\frac{55}{2.631} +\frac{25}{2.331} +\frac{20}{2.609}}$

⇒ $=2.545$

8 0

3 years ago

In an orthogonal cutting operation, the tool has a rake angle = 12°. The chip thickness before the cut = 0.32 mm and the cut yie

Snezhnost [94]

Answer:

The shear plane angle and shear strain are 28.21° and 2.155 respectively.

Explanation:

(a)

Orthogonal cutting is the cutting process in which cutting direction or cutting velocity is perpendicular to the cutting edge of the part surface.

Given:

Rake angle is 12°.

Chip thickness before cut is 0.32 mm.

Chip thickness is 0.65 mm.

Calculation:

Step1

Chip reduction ratio is calculated as follows:

$r=\frac{t}{t_{c}}$

$r=\frac{0.32}{0.65}$

r = 0.4923

Step2

Shear angle is calculated as follows:

$tan\phi=\frac{rcos\alpha}{1-rsin\alpha}$

Here, $\phi$ is shear plane angle, r is chip reduction ratio and $\alpha$ is rake angle.

Substitute all the values in the above equation as follows:

$tan\phi=\frac{rcos\alpha}{1-rsin\alpha}$

$tan\phi=\frac{0.4923cos12^{\circ}}{1-0.4923sin12^{\circ}}$

$tan\phi=\frac{0.48155}{0.8976}$

$\phi=28.21^{\circ}$

Thus, the shear plane angle is 28.21°.

(b)

Step3

Shears train is calculated as follows:

$\gamma=cot\phi+tan(\phi-\alpha)$

$\gamma=cot28.21^{\circ}+tan(28.21^{\circ}-12^{\circ})$ $\gamma = 2.155$ .

Thus, the shear strain rate is 2.155.

6 0

3 years ago