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3 years ago

What's a disadvantage of highest MERV-rated filters?

Engineering

2 answers:

Arte-miy333 [17]3 years ago

3 0

Answer:

3) the pressure drop across high MERV filters is significant.

Explanation:

MERV (Minimum-Efficiency Reporting Value) is used to measure the efficiency of filter to remove particles. A filter of high MERV can filter smaller particles but this causes an increase in reduced air flow that is an increase in pressure drop. High MERV filters capture more particles causing them to get congested faster and thereby increasing pressure drop.

Excessive pressure drop can cause overheating and lead to damage of the filter. The pressure drop can be reduced by increasing the surface area of the filter.

QveST [7]3 years ago

3 0

Answer:

3) The pressure drop across high MERV filters is significant.

Explanation:

The higher the MERV filter rate is, the most efficient it will be when it comes to trapping small particles. This comes with a cost. Since the space between fibers is smaller, this translates into a higher pressure drop. This is a disadvantge since in air conditioning or ventilation systems, the higher the pressure drop, the biggest the equipment and the most expensive it is.

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Which system provides an easier way for people to communicate with a computer than a graphical user interface (GUI)?

Fed [463]

Answer:

Explanation:

Natural language processing or NLP can be defined as an artificial intelligence that aids computers to understand, interpret, and manipulate human language. NLP is subfiled of many disciplines, for instance, artifical intelligence, linguistics, computer science, etc. NLP helps computers to interact with humans in their own language.

So, the easier way through which people can communicate with computers apart from GUI is NLP. Thus option C is correct.

7 0

3 years ago

Types of technology include:

Anastaziya [24]

The answer is D-all choices

3 0

3 years ago

As the impurity concentration in solid solution of a metal is increased, the tensile strength:________.a) decreasesb) increasesc

valkas [14]

Answer:

Increases

Explanation:

By inhibiting the motion of dislocations by impurities in a solid solutions, is a strengthening mechanism. In solid solutions it is atomic level strengthening resulting from resistance to dislocation motion. Hence, the strength of the alloys can differ with respect to the precipitate's property. Example, the precipitate is stronger (ability to an obstacle to the dislocation motion) than the matrix and it shows an improvement of strength.

5 0

4 years ago

2. A well of 0.1 m radius is installed in the aquifer of the preceding exercise and is pumped at a rate averaging 80 liter/min.

hodyreva [135]

Question:

The question is not complete. See the complete question and the answer below.

A well that pumps at a constant rate of 0.5m3/s fully penetrates a confined aquifer of 34 m thickness. After a long period of pumping, near steady state conditions, the measured drawdowns at two observation wells 50m and 100m from the pumping well are 0.9 and 0.4 m respectively. (a) Calculate the hydraulic conductivity and transmissivity of the aquifer (b) estimate the radius of influence of the pumping well, and (c) calculate the expected drawdown in the pumping well if the radius of the well is 0.4m.

Answer:

T = 0.11029m²/sec

Radius of influence = 93.304m

expected drawdown = 3.9336m

Explanation:

See the attached file for the explanation.

8 0

3 years ago

A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is

-BARSIC- [3]

Answer:

$\mathbf{C_{10} = 137.611 \ kN}$

Explanation:

From the information given:

Life requirement = 40 kh = 40 $40 \times 10^{3} \ h$

Speed (N) = 520 rev/min

Reliability goal $(R_D)$ = 0.9

Radial load $(F_D)$ = 2600 lbf

To find C10 value by using the formula:

$C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}$

where;

$x_D = \text{bearing life in million revolution} \\ \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}$

$\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}$

The Weibull parameters include:

$x_o = 0.02$

$(\theta - x_o) = 4.439$

$b= 1.483$

∴

Using the above formula:

$C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}$