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zysi [14]
2 years ago
14

A student lives in an apartment with a floor area of 60 m2 and ceiling height of 1.8 m. The apartment has a fresh (outdoor) air

exchange rate of 0.5/hr. The stove in the apartment heats by natural gas. The student cooks a meal using two gas burners that each emit carbon monoxide (CO) at a rate of 100 mg/hr. The outdoor CO concentration can be assumed to be negligible (0 ppm). The initial (time = 0) indoor CO concentration can be assumed to be 0 ppm (except for problem 4). Carbon monoxide can be considered as an inert gas, i.e., it does not stick to or react with any surfaces or other gases in air.
1. Assume that the student cooks for a long enough period of time to achieve a steady-state CO concentration in the apartment. What is that concentration in ppb?
2. Assume that the student cooks for only 45 minutes and turns off both burners at that time. What is the CO concentration in ppb at the end of 45 minutes?
3. Repeat problem 2 for air exchange rates that vary from 0.1 to 1/hr and plot the concentration at 45 minutes (in ppb) versus air exchange rate.
4. Assume that for the conditions of problem 2, the student waits 25 minutes after turning the burners off and then starts cooking again with two burners on. How long will it take to reach a concentration that is 95% of steady-state under this condition?
Note that you can actually address this question with an eloquent mathematical derivation (preferred) or simply by crunching the concentration profile in a spreadsheet.
What is the concentration at 95% of steady-state?
Compare your result with the time that would be required to reach 95% of steady-state had the initial indoor CO concentration been 0 ppm.
Engineering
1 answer:
USPshnik [31]2 years ago
5 0

Answer:

4

Explanation:

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Vilka [71]

Answer:

<em>The direction of ball will be Number 4 (as can be seen in attached picture) ---- the path of ball will be making some angle when it leaves the tube. </em>

Explanation:

The question is incomplete. So the picture, which is missing in question, is attached for your review.

As it can be seen in the picture, the ball coming out of the tube will have two components of velocity. One is along the length of tube (because ball is moving in that direction and is coming out from the hole), other is velocity component will be perpendicular to the tube (because the ball is made to move in that direction as the tube is rolling on the surface).

<em>So, taking the resultant of two vectors of velocity, the resultant direction of ball will be Number 4 (as can be seen in attached picture) ---- the path of ball will be making some angle when it leaves the tube. </em>

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2 years ago
An error was reported in a software program by a user. Before that error can be diagnosed and fixed
Bas_tet [7]

Answer:

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Explanation:

6 0
3 years ago
Read 2 more answers
Compare and contrast ""centralized"" and ""decentralized"" routing algorithms. (What are the advantages and disadvantages of eac
pantera1 [17]

Answer:

Comparison between centralized and decentralized routing algorithms:

The major similarity between both centralized and decentralized routing algorithms is that they are both communication serving systems. They both utilize node system(e.g, computer) and are both a communication liking system( e.g, Cable).

Contrasting between centralized and decentralized routing algorithms:

There are few differences between these two type of communication link or path way system but to name a couple of them,

For centralized, there is a single client server distribution node which simply means that one or more client server system are connected to a central processing server.

This also means that if the central processing server or pathway fails, it leads to the failure of the entire system. That is, there is no sending, responding or general processing of any form of requests. Example of a system that uses the centralized routing algorithm is the google search engine.

WHILE:

for the decentralized routing algorithms, there are multiple client server distribution pathway and each server makes its own decision. Here,there is no single entity that receives and responds to the request therefore,failure of any form of central path way processing node does not lead to the failure of the whole system unlike for the centralized system.

Advantages of centralized routing algorithm:

It can be easily protected or secured due to the nature of the system. If the central node is been secured, it generally translate to the different client node being secured.

It is easy to disconnect a connected client node from the central pathway node or central server as the case may be.

Disadvantages of centralized routing algorithm:

The client nodes are totally dependent on the central node or server so if there is a failure in the central server, the client node is then totally shut down.

Advantages of decentralized routing algorithm:

There is a random distribution of data on all the processing node or server which automatically creates a form of balance within the system. This leads to minimal or no down time processing client request.

Disadvantages of decentralized routing algorithm:

Due to the nature or fact that there are multiple processing system for different client node, it is difficult to detect which client node or request processing server is faulty. This can lead to delay in the fixing of fault in the system should it arise.

Why do we prefer a ""decentralized"" algorithm for routing messages through the internet?

The major reason why decentralized algorithm routing is preferred is because of the level of security attached to it. Each processing servers are secured independently and there is privilege of utilizing independent networking system.

5 0
3 years ago
Assuming 800 W / m 2 solar irradiance and a 35 % efficient solar panel, how much roof area should be covered to supply 10 A at 1
Korvikt [17]

Answer:

a) 4.286 m² b) $ 55.44/mo

Explanation:

If we assume that the sun is behaving as an isotropic radiator, the power density that is arriving to the house, is constant and equal to the quoted solar irradiance.

If the energy conversion capability of the solar panels were 100%, the roof area needed to supply the power required, would be simply the quotient between the power required and the solar irradiance, as follows:

A = P / SI = 10 A* 120 V / 800 W/m² = 1200 W / 800 W/m²= 1,5 m²

As the solar panels are only 35% efficient in converting the solar energy to useful electrical energy, we will need more roof area, according to this expression:

Ae = At / 0.35 = 1,5 / 0.35 = 4.286 m²

b) If we can get 1200 W during 7 hs/day, the energy supplied by the solar panels will be the product of the power times the time, as follows:

E= 1200 W* 7 hs = 8.4 Kwh

If the cost per Kwh, is $0.22, assuming 7 hs. of use in average during a month (assumed to be of 30 days), we can have savings as follows:

Cost = 0.22($/Kwh)* 8.4 (Kwh/day)*30 (days/mo) = $ 55.44

8 0
3 years ago
I. The time till failure of an electronic component has an Exponential distribution and it is known that 10% of components have
drek231 [11]

Answer:

(a) The mean time to fail is 9491.22 hours

The standard deviation time to fail is 9491.22 hours

(b) 0.5905

(c) 3.915 × 10⁻¹²

(d) 2.63 × 10⁻⁵

Explanation:

(a) We put time to fail = t

∴ For an exponential distribution, we have f(t) = \lambda e^{-\lambda t}

Where we have a failure rate = 10% for 1000 hours, we have(based on online resource);

P(t \leq 1000) = \int\limits^{1000}_0 {\lambda e^{-\lambda t}} \, dt = \dfrac{e^{1000\lambda}-1}{e^{1000\lambda}} = 0.1

e^(1000·λ) - 0.1·e^(1000·λ) = 1

0.9·e^(1000·λ) = 1

1000·λ = ㏑(1/0.9)

λ = 1.054 × 10⁻⁴

Hence the mean time to fail, E = 1/λ = 1/(1.054 × 10⁻⁴) = 9491.22 hours

The standard deviation = √(1/λ)² = √(1/(1.054 × 10⁻⁴)²)) = 9491.22 hours

b) Here we have to integrate from 5000 to ∞ as follows;

p(t>5000) = \int\limits^{\infty}_{5000} {\lambda e^{-\lambda t}} \, dt =\left [  -e^{\lambda t}\right ]_{5000}^{\infty} = e^{5000 \lambda} = 0.5905

(c) The Poisson distribution is presented as follows;

P(x = 3) = \dfrac{\lambda ^x e^{-x}}{x!}  = \frac{(1.0532 \times 10^{-4})^3 e^{-3} }{3!}  = 3.915\times 10^{-12}

p(x = 3) = 3.915 × 10⁻¹²

d) Where at least 2 components fail in one half hour, then 1 component is expected to fail in 15 minutes or 1/4 hours

The Cumulative Distribution Function is given as follows;

p( t ≤ 1/4) CDF = 1 - e^{-\lambda \times t} = 1 - e^{-1.054 \times 10 ^{-4} \times 1/4} = 2.63 \times 10 ^{-5}.

4 0
3 years ago
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