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3 years ago

What are the four causes of electrical faults?

Engineering

1 answer:

Arada [10]3 years ago

6 0

Answer:

Electrical faults are also caused due to human errors such as selecting improper rating of equipment or devices, forgetting metallic or electrical conducting parts after servicing or maintenance, switching the circuit while it is under servicing, etc.

Explanation:

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The stagnation chamber of a wind tunnel is connected to a high-pressure airbottle farm which is outside the laboratory building.

Natasha2012 [34]

This question is not complete, the complete question is;

The stagnation chamber of a wind tunnel is connected to a high-pressure air bottle farm which is outside the laboratory building. The two are connected by a long pipe of 4-in inside diameter. If the static pressure ratio between the bottle farm and the stagnation chamber is 10, and the bottle-farm static pressure is 100 atm, how long can the pipe be without choking? Assume adiabatic, subsonic, one-dimensional flow with a friction coefficient of 0.005

Answer:

the length of the pipe is 11583 in or 965.25 ft

Explanation:

Given the data in the question;

Static pressure ratio; p1/p2 = 10

friction coefficient f = 0.005

diameter of pipe, D =4 inch

first we obtain the value from FANN0 FLOW TABLE for pressure ratio of ( p1/p2 = 10 )so

4fL $_{max}$ / D = 57.915

we substitute

(4×0.005×L $_{max}$ ) / 4 = 57.915

0.005L $_{max}$ = 57.915

L $_{max}$ = 57.915 / 0.005

L $_{max}$ = 11583 in

Therefore, the length of the pipe is 11583 in or 965.25 ft

6 0

3 years ago

Please help I need it by today!!!

AfilCa [17]

Answer:

if engineering disappeared for a day i would be at a loss. i wouldnt know what to do with myself considering engineering is my life. one way that engineers improve my life is they help me to understand enything end everything

Explanation:

7 0

3 years ago

A solid circular rod that is 600 mm long and 20 mm in diameter is subjected to an axial force of P = 50 kN The elongation of the

Kay [80]

Answer:

a) V = 0.354

b) G = 25.34 GPA

Explanation:

Solution:

We first determine Modulus of Elasticity and Modulus of rigidity

Elongation of rod ΔL = 1.4 mm

Normal stress, δ = P/A

Where P = Force acting on the cross-section

A = Area of the cross-section

Using Area, A = π/4 · d²

= π/4 · (0.0020)² = 3.14 × 10⁻⁴m²

δ = 50/3.14 × 10⁻⁴ = 159.155 MPA

E(long) = Δl/l = 1.4/600 = 2.33 × 10⁻³mm/mm

Modulus of Elasticity Е = δ/ε

= 159.155 × 10⁶/2.33 × 10⁻³ = 68.306 GPA

Also final diameter d(f) = 19.9837 mm

Initial diameter d(i) = 20 mm

Poisson said that V = Е(elasticity)/Е(long)

= - <u>( 19.9837 - 20 /20)</u>

2.33 × 10⁻³

= 0.354,

∴ v = 0.354

Also G = Е/2. (1+V)

= 68.306 × 10⁹/ 2.(1+ 0.354)

= 25.34 GPA

⇒ G = 25.34 GPA

5 0

3 years ago

A sheet of steel 3-mm thick has nitrogen atomospheres on both sides at 900 C and is permitted to achieve a steady-state di usion

kati45 [8]

Answer:

$X_B = 1.8 \times 10^{-3} m = 1.8 mm$

Explanation:

Given data:

Diffusion constant for nitrogen is $= 1.85\times 10^{-10} m^2/s$

Diffusion flux $= 1.0\times 10^{-7} kg/m^2-s$

concentration of nitrogen at high presuure = 2 kg/m^3

location on which nitrogen concentration is 0.5 kg/m^3 ......?

from fick's first law

$J = D \frac{C_A C_B}{X_A X_B}$

Take C_A as point on which nitrogen concentration is 2 kg/m^3

$x_B = X_A + D\frac{C_A -C_B}{J}$

Assume X_A is zero at the surface

$X_B = 0 + ( 12\times 10^{-11} ) \frac{2-0.5}{1\times 10^{-7}}$

$X_B = 1.8 \times 10^{-3} m = 1.8 mm$

4 0

3 years ago

Air enters a cmpressor at 20 deg C and 80 kPa and exits at 800 kPa and 200 deg C. The power input is 400 kW. Find the heat trans

aksik [14]

Answer:

The heat is transferred is at the rate of 752.33 kW

Solution:

As per the question:

Temperature at inlet, $T_{i} = 20^{\circ}C$ = 273 + 20 = 293 K

Temperature at the outlet, $T_{o} = 200{\circ}C$ = 273 + 200 = 473 K

Pressure at inlet, $P_{i} = 80 kPa = 80\times 10^{3} Pa$

Pressure at outlet, $P_{o} = 800 kPa = 800\times 10^{3} Pa$

Speed at the outlet, $v_{o} = 20 m/s$

Diameter of the tube, $D = 10 cm = 10\times 10^{- 2} m = 0.1 m$

Input power, $P_{i} = 400 kW = 400\times 10^{3} W$

Now,

To calculate the heat transfer, $Q$ , we make use of the steady flow eqn:

$h_{i} + \frac{v_{i}^{2}}{2} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH' + p_{s}$

where

$h_{i}$ = specific enthalpy at inlet

$h_{o}$ = specific enthalpy at outlet

$v_{i}$ = air speed at inlet

$p_{s}$ = specific power input

H and H' = Elevation of inlet and outlet

Now, if

$v_{i} = 0$ and H = H'

Then the above eqn reduces to:

$h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}$

$Q = h_{o} - h_{i} + \frac{v_{o}^{2}}{2} + p_{s}$ (1)

Also,

$p_{s} = \frac{P_{i}}{ mass, m}$

Area of cross-section, A = $\frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}$

Specific Volume at outlet, $V_{o} = A\times v_{o} = 7.85\times 10^{- 3}\times 20 = 0.157 m^{3}/s$

From the eqn:

$P_{o}V_{o} = mRT_{o}$

$m = \frac{800\times 10^{3}\times 0.157}{287\times 473} = 0.925 kg/s$

Now,

$p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg$

Also,

$\Delta h = h_{o} - h_{i} = c_{p}\Delta T =c_{p}(T_{o} - T_{i}) = 1.005(200 - 20) = 180.9 kJ/kg$

Now, using these values in eqn (1):

$Q = 180.9 + \frac{20^{2}}{2} + 432.432 = 813.33 kW$

Now, rate of heat transfer, q:

q = mQ = $0.925\times 813.33 = 752.33 kW$

4 0

3 years ago