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4 years ago

A single phase, 50-KVA, 2400-240-volt, 60 Hz distribution transformer has the following parameters: •

Engineering

1 answer:

never [62]4 years ago

8 0

Answer:

B) voltage at the sending end of the feeder = 2483.66 v

Explanation:

attached below is the the equivalent circuits and the remaining solution for option A

B) voltage = 2400 v

I = $\frac{50*10^3}{2400}$ = 20.83 A

calculate voltage at sending end ( Vs )

Vs = 2400 + 20.83 ∠ -cos^-1 (0.8) ( 0.75*2 + 0.5 + j 2 + j2 )

hence Vs = 2483.66 ∠ 0.961

therefore voltage at the sending end = 2483.66 v

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A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. Determine the outside diameter of the

Olegator [25]

Answer:

119.35 mm

Explanation:

Given:

Inside diameter, d = 100 mm

Tensile load, P = 400 kN

Stress = 120 MPa

let the outside diameter be 'D'

Now,

Stress is given as:

stress = Load × Area

also,

Area of hollow pipe = $\frac{\pi}{4}(D^2-d^2)$

Area of hollow pipe = $\frac{\pi}{4}(D^2-100^2)$

thus,

400 × 10³ N = 120 × $\frac{\pi}{4}(D^2-100^2)$

D² = tex]\frac{400\times10^3+30\pi\times10^4}{30\pi}[/tex]

D = 119.35 mm

7 0

3 years ago

Read 2 more answers

The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elas

Sedaia [141]

Answer:

a) 254.6 GPa

b) 140.86 GPa

Explanation:

a) Considering the expression of rule of mixtures for upper-bound and calculating the modulus of elasticity for upper bound;

Ec(u) = EmVm + EpVp

To calculate the volume fraction of matrix, 0.63 is substituted for Vp in the equation below,

Vm + Vp = 1

Vm = 1 - 0.63

Vm = 0.37

In the first equation,

Where

Em = 68 GPa, Ep = 380 GPa, Vm = 0.37 and Vp = 0.63,

The modulus of elasticity upper-bound is,

Ec(u) = EmVm + EpVp

Ec(u) = (68 x 0.37) + (380 x 0.63)

Ec(u) = 254.6 GPa.

b) Considering the express of rule of mixtures for lower bound;

Ec(l) = (EmEp)/(VmEp + VpEm)

Substituting values into the equation,

Ec(l) = (68 x 380)/(0.37 x 380) + (0.63 x 68)

Ec(l) = 25840/183.44

Ec(l) = 140.86 GPa

6 0

4 years ago

Consider a turbojet powered airplane flying at a standard altitude of 30,000ft at a velocity of 500 mph. The turbojet engine its

grandymaker [24]

Answer:

T = 20.42 N

Explanation:

given data

standard altitude = 30,000 ft

velocity Ca = 500 mph = 0.4 m/s

inlet areas Aa = 7 ft² = 0.65 m²

exit areas Aj = 4.5 ft² = 0.42 m²

velocity at exit Cj = 1600 ft/s = 487.68 m/s

pressure exit $\rho$ j = 640 lb/ft² = 0.3 bar

solution

we get here thrust of the turbojet that is express as

thrust of the turbojet T = Mg × Cj - Ma × Ca + ( $\rho$ j Aj - $\rho$ a Ag ) .............1

here Ma = Mg

Ma = $\rho$ a × Ca Aa = 0.042 kg/s

put value in equation 1 we get

T = 0.042 × (487.68 -0.14) + ( 0.3 × - 0.3 × 0.65 )

T = 20.42 N

5 0

3 years ago

What are some qualities of a good engineer? Explain why each characteristic is important. (2-3 sentences minimum)

natali 33 [55]

A good engineer is smart.

8 0

3 years ago

For a bolted assembly with six bolts, the stiffness of each bolt is kb=Mlbf/in and the stiffness of the members is km=12Mlbf/in.

makkiz [27]

Answer:

nP ≈ 4.9

nL = 1.50

Explanation:

GIVEN DATA

external load applied (p) = 85 kips

bolt stiffness ( Kb ) = 3(10^6) Ibf / in

Member stiffness (Km) = 12(10^6) Ibf / in

Diameter of bolts ( d ) = 1/2 in - 13 UNC grade 8

Number of bolts = 6

assumptions

for unified screw threads UNC and UNF

tensile stress area ( A ) = 0.1419 in^2

SAE specifications for steel bolts for grade 8

we have

Minimum proff strength ( Sp) = 120 kpsi

Minimum tensile strength (St) = 150 Kpsi

Load Bolt (p) = external load / number of bolts = 85 / 6 = 14.17 kips

Given the following values

Fi = 75%* Sp*At = (0.75*120*0.1419 ) = 12.771 kip

Preload stress

αi = 0.75Sp = 0.75 * 120 = 90 kpsi

stiffness constant

C = $\frac{Kb}{Kb + Km}$ = $\frac{3}{3+2}$ = 0.2

A) yielding factor of safety

nP = $\frac{sPAt}{Cp + Fi}$ = $\frac{120* 0.1419}{0.2*14.17 + 12.771}$

nP = 77.028 / 15.605 = 4.94 ≈ 4.9

B) Determine the overload factor safety

$nL = \frac{SpAt - Fi}{CP}$ = ( 120 * 0.1419) - 12.771 / 0.2 * 14.17

= 17.028 - 12.771 / 2.834

= 1.50

3 0

3 years ago