Link AB is to be made of a steel for which the ultimate normal stress is 65 ksi. Determine the cross-sectional area of AB for which the factor of safety will be 3.20. Assume that the link will be adequately reinforced around the pins at A and B.

Answers

Answer 1

Answer:

Explanation:

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At full load, a commercially available 100hp, three phase induction motor operates at an efficiency of 97% and a power factor of 0.88 lag. The motor is supplied from a three-phase outlet with a line voltage rating of 208V.a. What is the magnitude of the line current drawn from the 208 V outlet? (1 hp = 746 W.) b. Calculate the reactive power supplied to the motor.

Answers

Answer:

I = Line Current = 242.58 A

Q = Reactive Power = 41.5 kVAr

Explanation:

Firstly, converting 100 hp to kW.

Since, 1 hp = 0.746 kW,

100 hp = 0.746 kW x 100

100 hp = 74.6 kW

The power of a three phase induction motor can be given as:

$P_(in) = √(3) VI Cos\alpha\n$

where,

P in = Input Power required by the motor

V = Line Voltage

I = Line Current

Cosα = Power Factor

Now, calculating Pin:

$efficiency = \frac{{P_(out)} }{P_(in) }\n0.97 = (74.6)/(P_(in) ) \nP_(in) = (74.6)/(0.97)\n P_(in) = 76.9 kW$

a) Calculating the line current:

$P_(in) = √(3)VICos\alpha \n76.9 * 1000= √(3)*208*I*0.88\nI = (76.9*1000)/(√(3)*208*0.88 )\nI = 242.58 A$

b) Calculating Reactive Power:

The reactive power can be calculated as:

Q = P tanα

where,

Q = Reactive power

P = Active Power

α = power factor angle

Since,

$Cos\alpha =0.88\n\alpha =Cos^(-1)(0.88)\n\alpha=28.36$

Therefore,

$Q = 76.9 * tan (28.36)\nQ = 76.9 * (0.5397)\nQ = 41. 5 kVAr$

The barrel of a bicycle tire pump becomes quite warm during use. Explain the mechanisms responsible for the temperature increase.

Answers

Answer:

The air heats up when being compressed and transefers heat to the barrel.

Explanation:

When a gas is compressed it raises in temperature. Assuming that the compression happens fast and is done before a significant amount of heat can be transferred to the barrel, we could say it is an adiabatic compression. This isn't exactly true, it is an approximation.

In an adiabatic transformation:

$P^(1-k) * T^k = constant$

For air k = 1.4

$P0^(-0.4) * T0^(1.4) = P1^(-0.4) * T1^(1.4)$

$T1^(1.4) = (P1^(0.4) * T0^(1.4))/(P0^(0.4))$

$T1^(1.4) = (P1)/(P0)^(0.4) * T0^(1.4)$

$T1 = T0 * (P1)/(P0)^(0.4/1.4)$

$T1 = T0 * (P1)/(P0)^(0.28)$

SInce it is compressing, the fraction P1/P0 will always be greater than one, and raised to a positive fraction it will always yield a number greater than one, so the final temperature will be greater than the initial temperature.

After it was compressed the hot air will exchange heat with the barrel heating it up.

A cylindrical specimen of cold-worked steel has a Brinell hardness of 250.(a) Estimate its ductility in percent elongation.(b) If the specimen remained cylindrical during deformation and its original radius was 5 mm (0.20 in.), determine its radius after deformation.

Answers

Answer:

A) Ductility = 11% EL

B) Radius after deformation = 4.27 mm

Explanation:

A) From equations in steel test,

Tensile Strength (Ts) = 3.45 x HB

Where HB is brinell hardness;

Thus, Ts = 3.45 x 250 = 862MPa

From image 1 attached below, for steel at Tensile strength of 862 MPa, %CW = 27%.

Also, from image 2,at CW of 27%,

Ductility is approximately, 11% EL

B) Now we know that formula for %CW is;

%CW = (Ao - Ad)/(Ao)

Where Ao is area with initial radius and Ad is area deformation.

Thus;

%CW = [[π(ro)² - π(rd)²] /π(ro)²] x 100

%CW = [1 - (rd)²/(ro)²]

1 - (%CW/100) = (rd)²/(ro)²

So;

(rd)²[1 - (%CW/100)] = (ro)²

So putting the values as gotten initially ;

(ro)² = 5²([1 - (27/100)]

(ro)² = 25 - 6.75

(ro) ² = 18.25

ro = √18.25

So ro = 4.27 mm

B. Suppose R1 is a fuse which burns out due to a sudden surge of current, thus, it essentially becomes an open switch. How do the currents change after this?

Answers

Answer:

The currents becomes 0

Explanation:

when the fuse burns out due to a sudden surge of current and becomes an open switch (with a resistance of Infinity ∞) this automatically reduces the currents through it to zero

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