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Olegator [25]
3 years ago
5

An environmental engineer is considering three methods for disposing of a nonhazardous chemical sludge: land application, fluidi

zed-bed incineration, and private disposal contract. The estimates for each method are below. (a) Determine which has the least cost on the basis of an annual worth comparison at 10% per year. (b) Determine the equivalent present worth value of each alternative using its AW value
Engineering
1 answer:
padilas [110]3 years ago
5 0

Answer:

Explanation:

The annual worth of land application = $ 110,000 ( A/P , 10% , 5 years ) + $ 95,000 - $ 15,000 (A/F , 10% , 5 years )

The annual worth of land application = $ 110,000 X 0.263797 + $ 95,000 - $ 15,000 X 0.163797

The annual worth of land application = $ 29,017.54 + $ 95,000 - $ 2,456.95

The annual worth of land application = $ 121,560.59

The annual worth of land Incineration = $ 800,000 ( A/P , 10% , 6 years ) + $ 60,000 - $ 250,000 X (A/F , 10% , 6 years )

The annual worth of land Incineration = $ 800,000 X 0.229607 + $ 60,000 - $ 250,000 X 0.129607

The annual worth of land Incineration = $ 211,283.85

The annual worth of contract = $ 190,000

The annual worth of contract = $ 190,000

The land application has the least cost , hence it is preferred .

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solong [7]

The guy named Kirby didn't touch the power button on the projector. If he touched means the projector will work fine.

<u>Explanation</u>:

  • Kirby touched the gear on the projector. He should check the projector carefully before operating the projector. But he didn't check properly.
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7 0
3 years ago
Suppose there are 93 packets entering a queue at the same time. Each packet is of size 4 MiB. The link transmission rate is 1.4
Ghella [55]

Answer:

0.19s

Explanation:

Queueing delay is the time a job waits in a queue before it can be executed. it is the difference in time betwen when the packet data reaches it destination and the time when it was executed.

Queueing delay =(N-1) L /2R

where N = no of packet =93

L = size of packet = 4MB

R = bandwidth = 1.4Gbps = 1×10⁹ bps

4 MB = 4194304 Bytes

(93 - 1)4194304 / 2× 10⁹

queueing delay =192937984 ×10⁻⁹

=0.19s

5 0
3 years ago
Suppose you must remove an average of 3.9×108J of thermal energy per day to keep your house cool during the summer. Part A If yo
noname [10]

Answer:

The required mechanical work is required to reduce each day by 1.05×10^8 Joules.

Explanation:

Coefficient of Performance (COP) = Q/W

Q is thermal energy absorbed by the air conditioner

W is mechanical work done

Q = 3.9×10^8 J

COP of old air conditioner = 2.3

W = Q/COP = 3.9×10^8/2.3 = 1.70×10^8 J

COP of new air conditioner = 6

W = Q/COP = 3.9×10^8/6 = 6.5×10^7 J

Reduction in mechanical work = (1.7×10^8) - (6.5×10^7) = 1.05×10^8 J

3 0
3 years ago
A TV USE 75 WATTS WHILE IN USED ASSMING THAT ITIS USED 4 HOURS EVERY DAY HOW MUCH ENERGY IN 4 IN KWH WOULD THE TV CONSUME ANNUAL
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Answer:

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Explanation:

7 0
2 years ago
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A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls
Inessa05 [86]

Answer:

the percent increase in the velocity of air is 25.65%

Explanation:

Hello!

The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.

m1=m2

Now remember that mass flow is given by the product of density, cross-sectional area and velocity

(α1)(V1)(A1)=(α2)(V2)(A2)

where

α=density

V=velocity

A=area

Now we can assume that the input and output areas are equal

(α1)(V1)=(α2)(V2)

\frac{V2}{V1} =\frac{\alpha1 }{\alpha 2}

Now we can use the equation that defines the percentage of increase, in this case for speed

i=(\frac{V2}{V1} -1) 100

Now we use the equation obtained in the previous step, and replace values

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the percent increase in the velocity of air is 25.65%

6 0
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