Ask question

Ask question

All categories

3 years ago

12

To be able to write an ss-domain equation for a circuit, use partial fraction decomposition to separate the terms in this equati

on, and then inverse transform the equation back into the time-domain. Using the mesh-current method to analyze the following circuit yields a set of two integro-differential equations which would be difficult to solve analytically. However, applying a Laplace transform allows for these equations to be rewritten as polynomial equations which can be manipulated algebraically.

1 answer:

Ann [662]3 years ago

4 0

Answer:

Assuming it is for the circuit attached below. To construct a domain equivalent for t > 0

Explanation:

See hand written solution

You might be interested in

Where does Elizabeth want John to do and what does she want him to do there?

ollegr [7]

What is the question?

8 0

2 years ago

Masja [62]

please give a better explanation of what you want to be answered.

3 0

3 years ago

You are a designer of a new processor. You have to choose between two possible implementations (called M1 and M2) of the same ar

Kaylis [27]

Answer:

A ) CPI : M1 = 2.4 , M2 = 2.65

B ) MIPS : M1 = 1083, M2 = 1056

C ) The machine that has a better performance based on MIPS is M1 and this is by 27 million number of instructions per sec

Explanation:

A) The CPI for each machine

CPI = ( Total number of execution cycles ) / ( instruction counter executed )

For Machine 1 ( M1 )

we have to make some assumptions : number of instructions = 10

number of times A was executed = 4 , Number of times B was executed = 2.5 , number of times C was executed = 2.5, Number of times D was executed = 1. and this was based on the frequency given above

hence CPI for M1 =[ ( 1 * 4 ) + ( 3 * 2.5 ) + ( 3 * 2.5 ) + ( 5 * 1 ) ] / 10

CPI for M1 = 2.4

For Machine 2 ( M2 )

we have to make some assumptions : number of instructions = 10

number of times A was executed = 4 , Number of times B was executed = 2. number of times C was executed = 1.5, Number of times D was executed = 2.5 times. and this was based on the frequency given above

Hence CPI for M1 = [ ( 2 * 4 ) + ( 2 * 2 ) + ( 3 * 1.5 ) + ( 4 * 2.5 ) ] / 10

CPI for M2 = 2.65

B ) Calculate the native MIPS ratings for M1 and M2

MIPS = ( instruction counts ) / ( Execution time * 10^6 )

For M1

Assumptions : number of instructions executed = 10

each clock cycle = 0.3846 * 10^-9. frequency = 2.6 Ghz

first we calculate the total execution time which is equal to :

= [ ( 1 * 4 ) + ( 3 * 2.5 ) + ( 3 * 2.5 ) + ( 5 * 1 ) ] * 0.3846 * 10 ^-9

= 9.2304 * 10 ^-9 secs

therefore the MIPS for M1

= 10 / ( 9.2304 * 10^-9 ) * 10^6 = 1083

For M2

Assumptions : number of instructions executed = 10

each clock cycle = 0.3846 * 10^-9. frequency = 2.8 Ghz

first we calculate the total execution time which is equal to :

= [ (2*4) + (2*2) + (3 * 1.5 ) + ( 4 * 2.5 ) ] * 0.3846 * 10^-9 = 9.4631 * 10^-9 secs

therefor the MIPS for M2

= 10 / ( 9.4631*10^-9) * 10^6 = 1056

C ) The machine that has a better performance based on MIPS is M1 and this is by 27 million number of instructions per sec

8 0

2 years ago

Hot carbon dioxide exhaust gas at 1 atm is being cooled by flat plates. The gas at 220 °C flows in parallel over the upper and l

sergeinik [125]

The local convection heat transfer coefficient at 1 m from the leading edge is $0.44 \frac{W}{m^{2} \times K}$ , the average convection heat transfer coefficient over the entire plate is $0.293 \frac{W}{m^{2} \times K}$ and the total heat flux transfer to the plate is $61.6 KJ$ .

Explanation:

It is case of heat and mass transfer in which due to temperature difference between gas and surface. Further temperature boundary layer will developed on flat plate in longitudinal direction.

Hot carbon dioxide exhaust gas

physical properties

$r= 1.05 \frac{kg}{m^{3}}$

$c_p = 1.02 \frac{kJ}{Kg \times K}$

$m= 231 \times 10^{7} \frac{N \times s }{m^2}$

υ = $21.8 \times 10^{6} \frac{m^2}{s}$

$k = 32.5 \times 10^{3} \frac{W}{m \times K}$

$\alpha = 30.1 \times 10^{6} \frac{m^{2}}{s}$

$Pr = 0.725$

Apart from these other data arr given below,

$v= 3 \frac{m}{s} \\ p= 1 atm \\ L_c = 1.5m \\T_g= 220 C \\ T_s = 80 C$

To find the local convection heat transfer coefficient at 1 m from the leading edge, we use correlation used for laminar flow over flat plate,

$Nu = \frac{ h \times L }{k} = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })$

where h= Average heat transfer coefficient

L= Length of a plate

k= Thermal Conductivity of carbon dioxide

Re = Reynold's Number

Pr = Prandtle Number

(a) Convection heat transfer coefficient at 1 m from the leading edge

is referred as local convection heat transfer coefficient.

To find convection heat transfer coefficient at 1 m from leading edge,

$Nu = \frac{ h_local \times L }{k} = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })$

Here, first we have to find Re and Pr,

$Re = \frac{r \times v \times L}{m}$

$Re = \frac{1.0594 \times 3 \times 1}{231 \times 10^{7}}$

$Re = 20.63 \times 10^{-10}$

Pr number is take from physical property data and Pr is 0.725.

Putting value of Re and Pr in main equation,

we get

$Nu = \frac{ h_local \times 1 }{32.5 \times 10^{3}} = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

$h_local = 32.5 \times 10^{3} \times 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

$h_local = 0.44 \frac{W}{m^{2} \times K}$

(b) To find average convection heat transfer coefficient,

it can be find out as case (a), only difference is that instead of L=1 m, L=1.5 m would come,

Therefore,

$Nu = \frac{ h \times 1.5 }{32.5 \times 10^{3}} = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

Finally,

$h = \frac{0.44}{1.5}$

$h = 0.293 \frac{W}{m^{2} \times K}$

(C) Total heat flux transfer to the plate is found out by,

$Q = h \times (T_g - T_s)$

$Q = 0.293 \times (220-80) \\ Q= 0.293 \times 140 \\ Q= 61.6 KJ$

8 0

2 years ago

Which of the following is NOT true concerning the color of minerals? A. Some minerals have a consistent color, but many have a r

mixer [17]

Answer:

Option D

A mineral’s color reflects the wavelengths of light that are absorbed by the mineral.

Explanation:

Color is one of the physical properties of minerals. Many minerals have a wide range of colors but there are some minerals with one consistent color and such minerals are referred as monochromatic minerals for example azurite. Normally, the streak color tends to be less variable than the color of the whole mineral and impurities or minor chemical components in a mineral react and often control the display color of resultant mineral. Option D is incorrect since mineral's color don't reflect wavelengths of light absorbed by such minerals.

5 0

3 years ago