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3 years ago

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A cylindrical bar of metal having a diameter of 20.9 mm and a length of 205 mm is deformed elastically in tension with a force o

f 45200 N. Given that the elastic modulus and Poisson's ratio of the metal are 68.8 GPa and 0.32, respectively, determine the following: (a) The amount by which this specimen will elongate in the direction of the applied stress. (b) The change in diameter of the specimen. Indicate an increase in diameter with a positive number and a decrease with a negative number.

1 answer:

natita [175]3 years ago

8 0

Answer:

a. 1.91 b. -8.13 mm

Explanation:

Modulus =stress/strain; calculating stress =F/A, hence determine the strain

Poisson's ratio =(change in diameter/diameter)/strain

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A building wall has dimensions of 3 m tall and 10 m wide. It is constructed of 2 cm. wallboard (k = 0.5 W/m-C) on the inside, 3

Art [367]

Answer: heat loss through wall is 16.58034kW

Temperature of inside wall surface is 47°c

Temperature of outside wall surface is -2.7°c

Explanation:detailed calculation and explanation is shown in the image below.

4 0

3 years ago

A vertical piston-cylinder device initially contains 0.2 m3 of air at 20°C. The mass of the piston is such that it maintains a c

Ann [662]

Answer:

Amount of air left in the cylinder=m $_{2}$ =0.357 Kg

The amount of heat transfer=Q=0

Explanation:

Given

Initial pressure=P1=300 KPa

Initial volume=V1=0.2 $m^{3}$

Initial temperature=T $_{1}$ =20 C

Final Volume= $V_{2}$ =0.1 $m^{3}$

Using gas equation

$m_{1}=((P_{1}*V_{1})/(R*T_{1}))$

m1==(300*0.2)/(.287*293)

m1=0.714 Kg

Similarly

m2=(P2*V2)/R*T2

m2=(300*0.1)/(0.287*293)

m2=0.357 Kg

Now calculate mass of air left,where me is the mass of air left.

me=m2-m1

me=0.715-0.357

mass of air left=me=0.357 Kg

To find heat transfer we need to apply energy balance equation.

$Q=(m_{e}*h_{e})+(m_{2}*h_{2})-(m_{1}*h_{1})$

Where me=m1-m2

And as the temperature remains constant,hence the enthalpy also remains constant.

h1=h2=he=h

Q=(me-(m1-m2))*h

me=m1-me

Thus heat transfer=Q=0

6 0

3 years ago

A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the

raketka [301]

Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Intertia constant = H = 8 MJ/MVA

Mechanical power = Pmech = 80 MW

Electrical power = Pelec = 50 MW

Number of poles = P = 4

No. of cycles = 10

Required Information:

(a) stored energy = ?

(b) rotor acceleration = ?

(c) change in torque angle = ?

(c) rotor speed = ?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

(c) change in torque angle (in elec deg) = 6.75 elec deg

(c) change in torque angle (in rmp/s) = 28.12 rpm/s

(c) rotor speed = 1505.62 rpm

Explanation:

(a) Find the stored energy in the rotor at synchronous speed.

The stored energy is given by

$E = G \times H$

Where G represents complex rated power and H is the inertia constant of turbo-generator.

$E = 100 \times 8 \\\\E = 800 \: MJ$

(b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses.

The rotor acceleration is given by

$P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2}$

Where M is given by

$M = \frac{E}{180 \times f}$

$M = \frac{800}{180 \times 50}$

$M = 0.0889 \: MJ \cdot s/ elec \: \: deg$

So, the rotor acceleration is

$P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2}$

$30 = 0.0889 \frac{d^2 \delta}{dt^2}$

$\frac{d^2 \delta}{dt^2} = \frac{30}{0.0889}$

$\frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2$

(c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.

The change in torque angle is given by

$\Delta \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2$

Where t is given by

$1 \: cycle = 1/f = 1/50 \\\\10 \: cycles = 10/50 = 0.2 \\\\t = 0.2 \: sec$

So,

$\Delta \delta = \frac{1}{2} \cdot 337.46 \cdot (0.2)^2$

$$ \Delta \delta = 6.75 \: elec \: deg$

The change in torque in rpm/s is given by

$\Delta \delta = \frac{337.46 \cdot 60}{2 \cdot 360\circ }$

$\Delta \delta =28.12 \: \: rpm/s$

The rotor speed in revolutions per minute at the end of this period (10 cycles) is given by

$Rotor \: speed = \frac{120 \cdot f}{P} + (\Delta \delta)\cdot t$

Where P is the number of poles of the turbo-generator.

$Rotor \: speed = \frac{120 \cdot 50}{4} + (28.12)\cdot 0.2$

$Rotor \: speed = 1500 + 5.62$

$$ Rotor \: speed = 1505.62 \:\: rpm$

4 0

3 years ago

Determine the resistance of 3km of copper having a diameter of 0,65mm if the resistivity of copper is 1,7x10^8

LUCKY_DIMON [66]

Answer:

Resistance of copper = 1.54 * 10^18 Ohms

Explanation:

<u>Given the following data;</u>

Length of copper, L = 3 kilometers to meters = 3 * 1000 = 3000 m

Resistivity, P = 1.7 * 10^8 Ωm

Diameter = 0.65 millimeters to meters = 0.65/1000 = 0.00065 m

$Radius, r = \frac {diameter}{2}$

$Radius = \frac {0.00065}{2}$

Radius = 0.000325 m

To find the resistance;

Mathematically, resistance is given by the formula;

$Resistance = P \frac {L}{A}$

Where;

P is the resistivity of the material.
L is the length of the material.
A is the cross-sectional area of the material.

First of all, we would find the cross-sectional area of copper.

Area of circle = πr²

Substituting into the equation, we have;

Area = 3.142 * (0.000325)²

Area = 3.142 * 1.05625 × 10^-7

Area = 3.32 × 10^-7 m²

Now, to find the resistance of copper;

$Resistance = 1.7 * 10^{8} \frac {3000}{3.32 * 10^{-7}}$

$Resistance = 1.7 * 10^{8} * 903614.46$

<em>Resistance = 1.54 * 10^18 Ohms </em>

3 0

3 years ago

4. A 25 km2 watershed has a time of concentration of 1.6 hr. Calculate the NRCS triangular UH for a 10-minute rainfall event and

Alika [10]

Answer:

The NRCS triangular UH for a 10-minute rainfall is Qp = 49.84 m³/s

Peak flow of the aggregated runoff hydrograph is 420.58 m³/s

The total volume of runoff is 2125000 m³/s

Explanation:

We have

A = 25 km²

tr = 10 min = 1/6 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/12 + 0.96 = 1.043 hr

Qp = 2.08×25/1.043 = 49.84 m³/s

Tb = 8/3×Tp = 8/3×1.043 = 2.782 hr

Since the area is

Time (min) Runoff (cm) Volume of runoff m³

0 0 0

10 4 1000000 m³

20 2.5 625000 m³

30 2 500000 m³

Total volume of runoff = 1000000 + 625000 + 500000 = 2125000 m³/s

For the 1st 10 minutes, we have

A = 25 km²

tr = 30 min = 1/2 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr

Qp = 2.08×25×4/1.043 = 197.92 m³/s

Tb = 8/3×Tp = 8/3×1.21 = 3.227 hr

For the 2nd 10 minutes, we have

A = 25 km²

tr = 30 min = 1/2 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr

Qp = 2.08×25×2.5/1.043 = 123.7 m³/s

Tb = 8/3×Tp = 8/3×1.21 = 3.227 hr

For the 3rd 10 minutes, we have

A = 25 km²

tr = 30 min = 1/2 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr

Qp = 2.08×25×2.5/1.043 = 98.96 m³/s

Tb = 8/3×Tp = 8/3×1.21 = 3.227 hr

Peak flow of aggregate runoff is given by

Qp (total) = 98.96 + 123.7 +197.92 = 420.58 m³/s

Total volume of runoff is given by

Total volume of runoff = 1000000 + 625000 + 500000 = 2125000 m³/s

6 0

3 years ago