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A square isothermal chip is of width w = 5 mm on a side and is mounted in a substrate such that its side and back surfaces are w

adelina 88 [10]

Answer:

Maximum allowable chip power is 0.35 W

Explanation:

This question is incomplete. The complete question is

A square isothermal chip is of width w = 5 mm on a side and is mounted in a substrate such that its side and back surfaces are well insulated; the front surface is exposed to the flow of a coolant at t[infinity] = 15°c. from reliability considerations, the chip temperature must not exceed t = 85°c. f the coolant is air and the corresponding convection 200 w/m2 k, what is the maximum allowable chip power?

<u>ANSWER:</u>

The heat transfer through convection, we have the equation:

q = hA(T - T∞)

where,

q = power transfer through convection = ?

h = convection coefficient = 200 W/m²K

A = Area of convection surface = (0.005 m)² = 0.000025 m²

T = Chip surface temperature = 85° C

T∞ = Fluid temperature = 15° C

Therefore,

q = (200 W/m².K)(0.000025 m²)(85° C - 15° C)

Since, difference in temperature is same on both Celsius and kelvin scale. Therefore, Celsius is written as kelvin for difference and they shall be cancelled.

3 0

4 years ago

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

Interpret the Blame responsibility and causation in your own words in the light of Columbia Accident.

Licemer1 [7]

Answer:

Proposed Improvements and Generic Lessons

Within 2 h of losing the signal from the returning spacecraft, NASA’s Administrator established the Columbia Accident Investigation Board (CAIB) to uncover the conditions that had produced the disaster and to draw inferences that would help the US space program to emerge stronger than before (CAIB, 2003). Seven months later, the CAIB released a detailed report that included its recommendations (Starbuck and Farjoun, 2005).

The CAIB (2003) report attempted to seek answers to the following four crucial questions:

1.

Why did NASA continue to launch spacecraft despite many years of known foam debris problems?

2.

Why did NASA managers conclude, despite the concerns of their engineers, that the foam debris strike was not a threat to the safety of the mission?

3.

How could NASA have forgotten the lessons of Challenger?

4.

What should NASA do to minimize the likelihood of such accidents in the future?

Although the CAIB’s comprehensive report raised important questions and offered answers to some of them, it also left many major questions unanswered (Starbuck and Farjoun, 2005).

1.

Why did NASA consistently ignore the recommendations of several review committees that called for changes in safety organization and practices?

2.

Did managerial actions and reorganization efforts that took place after the Challenger disaster contribute, both directly and indirectly, to the Columbia disaster?

3.

Why did NASA’s leadership fail to secure more stable funding and to shield NASA’s operations from external pressures?

By examining, with respect to the Columbia disaster, the case of NASA as an organization, one can try to extract generalizations that could be useful for other organizations, especially those engaged in high-risk activities—such as nuclear power plants, oil and gas, hospitals, airlines, armies, and pharmaceutical companies—and such generic principles may also be salutary for any kind of organization.

The CAIB (2003) report recommended developing a plan to inspect the condition of all RCC systems, the investigation having found the existing inspection techniques to be inadequate. RCC panels are installed on parts of the shuttle, including the wing leading edges and nose cap, to protect against the excessive temperatures of reentry. They also recommended that taking images of each shuttle while in orbit should be standard procedure as well as upgrading the imaging system to provide three angles of view of the shuttle, from liftoff to at least SRB separation. “The existing camera sites suffer from a variety of readiness, obsolescence, and urban encroachment problems.” The board offered this suggestion because NASA had had no images of the Columbia shuttle clear enough to determine the extent of the damage to the wing. They also recommended conducting inspections of the TPS, including tiles and RCC panels, and developing action plans for repairing the system. The report included 29 recommendations, 15 of which the board specified must be completed before the shuttle returned to flight status, and also made 27 “observations” (CAIB, 2005).

7 0

3 years ago

Tech A says you can find the typical angle of a V-block engine by dividing the number of cylinders by 720

Lady_Fox [76]

Answer:

Tech A is correct

Explanation:

Tech A is right as its V- angle is identified by splitting the No by 720 °. Of the piston at the edge of the piston.

Tech B is incorrect, as the V-Angle will be 720/10 = 72 for the V-10 motor, and he says 60 °.

8 0

3 years ago

Which of the following statements is correct?

Reika [66]

Answer:

b. rivets are tempory fasteners that bind two plate of metals together

3 0

3 years ago