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3 years ago

What is Differential Analysis in fluid mechanics?

Engineering

1 answer:

OleMash [197]3 years ago

6 0

Answer:

Differential analysis is used when it is needed to determine the detail information of the flow i.e. pressure or stress variation along any point.

Explanation:

In fluid mechanics, sometime situation arise in which we need to determine in detail about flow characteristics like stress and pressure variation.

To find these flow characteristics some relationship need to imply either at a point or at very small volume and analysis of flow at very small point is known as differential analysis.

Example: pressure and shear stress variation in a line of the wing of a plane.

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A cold storage room is used to keep the temperature inside the room maintain at low temperature.

lianna [129]

Answer:

The power of the brick wall it may be how the soiled ness of the wall too keep in the cold

Explanation:

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3 years ago

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Assume a program requires the execution of 50 x 106 FP instructions, 110 x 106 INT instructions, 80 x 106 L/S instructions, and

Pavlova-9 [17]

Answer:

Part A:

$1.3568*10^{-5}=\frac{5300* New\ CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12$

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

$1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8$

CPI reduced by $1-\frac{0.8}{4} = 0.80$ =80%

Part C:

New Execution Time= $\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s$

Increase in speed= $1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%$

Explanation:

FP Instructions=50*106=5300

INT Instructions=110*106=11660

L/S Instructions=80*106=8480

Branch Instructions=16*106=1696

Calculating Execution Time:

Execution Time= $\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}$

Execution Time= $\frac{5300*1+11660*1+8480*4+1696*2}{2*10^9\ Hz}$

Execution Time= $2.7136*10^{-5}\ s$

Part A:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time= $\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s$

$1.3568*10^{-5}=\frac{5300* New\ CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12$

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time= $\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s$

$1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8$

CPI reduced by $1-\frac{0.8}{4} = 0.80$ =80%

Part C:

$New\ CPI_1=0.6*Old\ CPI_1=0.6*1=0.6\\New\ CPI_2=0.6*Old\ CPI_2=0.6*1=0.6\\New\ CPI_3=0.7*Old\ CPI_3=0.7*4=2.8\\New\ CPI_4=0.7*Old\ CPI_4=0.7*2=1.4$

New Execution Time= $\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}$

New Execution Time= $\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s$

Increase in speed= $1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%$

8 0

3 years ago

Select the correct answer.

Elodia [21]

I think balance

Can I get Brainlyist

3 0

3 years ago

Time complexity of quick short

Irina-Kira [14]

Answer:

The time complexity will be "O(n log n)".

Explanation:

Many realistic Quick sort implementations choose a randomized special edition. The time complexity variable O(n Logn) was predicted in the randomized edition.
Throughout the randomized version, probably the most disgusting case is also conceivable, but by far the worst scenario for something like a given pattern does not exist as well as randomized Quick sort performs well throughout the practice.

3 0

3 years ago

In alternating current, how often does the current alternate between negatively and postively?

Serggg [28]

In an alternating current (AC) circuit the two poles alternate between negative and positive and the direction of the current (electron flow) reverses periodically.

4 0

2 years ago

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