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2 years ago

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Write a Nios II assembly program that reads binary data from the Slider Switches, SW11-0, on the DE2-115 Simulator, and display

the decimal equivalent on the seven-segment displays HEX3-0.

1 answer:

Citrus2011 [14]2 years ago

6 0

Algorithm of the Nios II assembly program.

Attain data for simulation from the SW11-0, on the DE2-115 Simulator
The data will be read from the switches in loop.
The decimal output is displayed using the seven-segment displays and done using the loop.
The program is ended by the user operating the SW1 switch

and

The decimal equivalent on the seven-segment displays HEX3-0 is

DE2-115
DE2-115_SW11
DE2-115_HEX3
DE2-115_HEX4
DE2-115_HEX5
DE2-115_HEX6
DE2-115_HEX7

<h3>The Algorithm and decimal equivalent on the seven-segment displays HEX3-0</h3>

Generally, the program will be written using a cpulator simulator in order to attain best result.

We are to

Attain data for simulation from the SW11-0, on the DE2-115 Simulator
The data will be read from the switches in loop.
The decimal output is displayed using the seven-segment displays and done using the loop.
The program is ended by the user operating the SW1 switch

This will be the Algorithm of the Nios II assembly program .

Hence, the decimal equivalent on the seven-segment displays HEX3-0 is

DE2-115
DE2-115_SW11
DE2-115_HEX3
DE2-115_HEX4
DE2-115_HEX5
DE2-115_HEX6
DE2-115_HEX7

For more information on Algorithm

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ololo11 [35]

Answer:

A) W' = 15680 KW

B) W' = 17113.87 KW

Explanation:

We are given;

Temperature at state 1; T1 = 290 K

Temperature at state 3; T3 = 1100 K

Rate of heat transfer; Q_in = 35000 kJ/s = 35000 Kw

Pressure of air into compressor; P_c = 95 kPa

Pressure of air into turbine; P_t = 760 kPa

A) The power assuming constant specific heats at room temperature is gotten from;

W' = [1 - ((T4 - T1)/(T3 - T2))] × Q_in

Now, we don't have T4 and T2 but they can be gotten from;

T4 = [T3 × (r_p)^((1 - k)/k)]

T2 = [T1 × (r_p)^((k - 1)/k)]

r_p = P_t/P_c

r_p = 760/95

r_p = 8

Also,k which is specific heat capacity of air has a constant value of 1.4

Thus;

Plugging in the relevant values, we have;

T4 = [(1100 × (8^((1 - 1.4)/1.4)]

T4 = 607.25 K

T2 = [290 × (8^((1.4 - 1)/1.4)]

T2 = 525.32 K

Thus;

W' = [1 - ((607.25 - 290)/(1100 - 525.32))] × 35000

W' = 0.448 × 35000

W' = 15680 KW

B) The power accounting for the variation of specific heats with temperature is given by;

W' = [1 - ((h4 - h1)/(h3 - h2))] × Q_in

From the table attached, we have the following;

At temperature of 607.25 K and by interpolation; h4 = 614.64 KJ/K

At T3 = 1100 K, h3 = 1161.07 KJ/K

At T1 = 290 K, h1 = 290.16 KJ/K

At T2 = 525.32 K, and by interpolation, h2 = 526.12 KJ/K

Thus;

W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000

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Answer:

$Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})$

Explanation:

According to the first thermodynamic law, the energy must be conserved so:

$dQ = dU - dW$

Where Q is the heat transmitted to the system, U is the internal energy and W is the work done by the system.

This equation can be solved by integration between an initial and a final state:

(1) $\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU - \int\limits^1_2 {} \, dW$

As per work definition:

$dW = F*dr$

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$dW = PA*dx$

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$dW = - P*dV$

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$\int\limits^1_2 {- P} \, dV$

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$\int\limits^1_2 {- P} \, dV = \int\limits^1_2 {- \frac{n*R*T}{V}} \, dV$

In this particular case as the systems is closed and the temperature constant, n, R and T are constants:

$\int\limits^1_2 {- \frac{n*R*T}{V}} \, dV = -nRT \int\limits^1_2 {\frac{1}{V}} \, dV$

Replacion this and solving equation (1) between state 1 and 2:

$\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU + nRT \int\limits^1_2 {\frac{1}{V}} \, dV$

$Q_{2} - Q_{1} = U_{2} - U_{1} + nRT(ln V_{2} - ln V_{1})$

$Q_{2} - Q_{1} = U_{2} - U_{1} + nRT ln \frac{V_{2}}{V_{1}}$

The internal energy depends only on the temperature of the gas, so there is no internal energy change $U_{2} - U_{1} = 0$ , so the heat exchanged to the system equals the work done by the system:

$Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})$

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Answer:

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we know that $\frac{x_1^2}{Dt}=constant$ where x is distance and t is time .As the temperature is constant so D will be also constant

So $\frac{x_1^2}{t}=constant$

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