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Olin [163]
4 years ago
9

For a bronze alloy, the stress at which plastic deformation begins is 274 MPa and the modulus of elasticity is 118 GPa. (a) What

is the maximum load that may be applied to a specimen having a cross-sectional area of 316 mm2 without plastic deformation? (b) If the original specimen length is 132 mm, what is the maximum length to which it may be stretched without causing plastic deformation?
Engineering
1 answer:
likoan [24]4 years ago
7 0

Answer:

86584N

132.306 mm

Explanation:

Q = 274

Modulus of elasticity = 118 gpa

1.

Area = 316mm² without plastic deformation

F = QA

= 274x10⁶x316x10^-6

= 274000000 x 0.000316

= 86584 N

This is the maximum load.

2.

Max length =

L = 132(1 + 274x10⁶/118x10⁹)

L = 132(1+274000000/118000000000)

L = 132(1+0.002322)

L = 132(1.002322)

L = 132.306

This is the maximum length to which it may be stretched without causing plastic deformation.

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An isentropic steam turbine processes 2 kg/s of steam at 3 MPa, which is exhausted at50 kPa and 100C. Five percent of this flow
borishaifa [10]

Answer:

2285kw

Explanation:

since it is an isentropic process, we can conclude that it is a reversible adiabatic process. Hence the energy must be conserve i.e the total inflow of energy must be equal to the total outflow of energy.

Mathematically,

\\ E_{inflow} = E_{outflow}

Note: from the question we have only one source of inflow and two source of outflow (the exhaust at a pressure of 50kpa and the feedwater at a pressure of 5ookpa). Also the power produce is another source of outgoing energy    \\ E_{inflow} = m_{1} h_{1} .

\\

E_{outflow} = m_{2} h_{2} + m_{3} h_{3} + W_{out}

\\

Where m_{1} h_{1} are the mass flow rate and the enthalpies at the inlet  at a pressure of 3Mpa \\,

m_{2} h_{2} are the mass flow rate and the enthalpies  at the outlet 2 where we have a pressure of 500kpa respectively.\\,

and  m_{3} h_{3}   are the mass flow rate and the enthalpies  at the outlet 3 where we have a pressure of 50kpa respectively.\\,

We can now express write out the required equation by substituting the new expression for the energies \\

m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}   \\

from the above equation, the unknown are the enthalpy values and  the mass flow rate. \\

first let us determine the enthalpy values at the inlet and the out let using the Superheated water table.  \\

It is more convenient to start from outlet 3 were we have a temperature 100^{0}C and pressure value of (50kpa or 0.05Mpa ). using double interpolation method  on the superheated water table to determine the enthalpy value with careful calculation we have  \\

h_{3}  = 2682.4 KJ/KG , at this point also from the table the entropy value ,s_{3} value is 7.6953 KJ/Kg.K. \\

Next we determine the enthalphy value at outlet 2. But in this case, we don't have a temperature value, hence we use the entrophy value since the entropy  is constant at all inlet and outlet. \\

So, from the superheated water table again, at a pressure of 500kpa (0.5Mpa) and entropy value of  7.6953 KJ/Kg.K with careful  interpolation we arrive at a enthalpy value of 3206.5KJ/Kg.\\

Finally for inlet one at a pressure of 3Mpa, interpolting with an entropy value of 7.6953KJ/Kg.K  we arrive at enthalpy value of 3851.2KJ/Kg. \\

Now we determine the mass flow rate at each inlet and outlet. since  mass must also be balance, i.e  m_{1} = m_{2} + m_{3} \\

From the question the, the mass flow rate at the inlet m_{1}}  is 2Kg/s \\

Since 5% flow is delivered into the feedwater heating,  \\

m_{2} = 0.05m_{1} = 0.05 *2kg/s = 0.1kg/s \\

Also for the outlet 3 the remaining 95% will flow out. Hence

m_{3} = 0.95m_{1} = 0.95 *2kg/s = 1.9kg/s \\

Now, from m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}   \\ we substitute values

W_{out} = m_{1} h_{1}-m_{2} h_{2}-m_{3} h_{3}

W_{out} = (2kg/s)(3851.2KJ/Kg) - (0.1kg/s)(3206.5kJ/kg)- (1.9)(2682.4kJ/kg)

\\

W_{out} = 2285.19 kW.

Hence the power produced is 2285kW

7 0
3 years ago
A 750-turn solenoid, 24 cm long, has a diameter of 2.3 cm . A 19-turn coil is wound tightly around the center of the solenoid. P
oksian1 [2.3K]

Answer: 2.26x10^-4 v

Explanation:

Lenght of the selonoid = 24x10^-2m

Diameter of the selonoid = 2.3cm

The radius will then be = 1.15cm = 1.15x10^-2m

The area of the selonoid = ¶r^2 = 3.142 x (1.15x10^-2)^2 = 0.000415m^2.

Number of turns on selonoid N1 is 750

For the small center coil, number of turns N2 is 19.

There is a change in current dI/dt from 0 to 5.1 in 0.7s, dI/dt = (5.1-0)/0.7

dI/dt = 7.29A/s.

Induced EMF on selonoid due to magnetic Flux due to changing current in small coil is given as;

E = -M(dI/dt), where M is the mutual inductance of the coils.

but M = (u°AN1N2)/L, where u°= 4¶x10^-7,

A = area of selonoid,

L = Lenght of selonoid.

M = (4¶X10^-7X0.000415X750X19)/(24X10^-2)

M = 3.096X10^-5H

Induced EMF E = 3.096X10^-5 x 7.29

E = 2.26x10^-4V

6 0
4 years ago
2. Other igneous rock forms from lava that cools quickly on Earth’s surface. Classify the rock as either intrusive or extrusive,
Darya [45]

Answer:

extrusive,

Explanation:

lava exlodes outwards, making it extrusive and not intrusive.

5 0
3 years ago
Sarah is developing a Risk Assessment for her organization. She is asking each department head how long can they be without thei
Natali [406]

Answer:

Sarah is asking each department head how long they can be without their primary system. Sarah is trying to determine the Recovery Time Objective (RTO) as this is the duration of time within which the primary system must be restored after the disruption.

Recovery Point Objective is basically to determine the age of restoration or recovery point.

Business recovery and technical recovery requirements are to assess the requirements to recover by Business or technically.

Hence, Recovery Time Objective (RTO) is the correct answer.

8 0
3 years ago
Drivers killed in speed related accidents usually have a history of_______
bazaltina [42]
I would go with C but i am not 100 percent on that
3 0
3 years ago
Read 2 more answers
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