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otez555 [7]
3 years ago
15

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.​
Engineering
1 answer:
Citrus2011 [14]3 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

brainly.com/question/11623795

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Read 2 more answers
Water is the working fluid in an ideal Rankine cycle. Steam enters the turbine at 1400 lbf/in2 and 1000°F. The net power output
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Answer:

(a) The mass flow rate of the steam is approximately 1.803×10⁶ lb/h

(b) The rate of heat transfer is approximately 2.52×10⁹ BTU/h

(c) The thermal efficiency  is approximately 39.68%

(d) The mass flow rate of cooling water is approximately 9.478 × 10⁷ lb/h

Explanation:

(a) The parameters are;

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s₁ = 1.61 BTU/(lb·R) =  6.741 kJ/(kg·K)

Therefore, due to isentropic expansion from state 1 to state 2, we have;

s₁ = s₂ = 1.61 BTU/(lb·R)

P₂ = 2 psi

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h_{f2} = 94.02 BTU/(lb)

h_g = 1116 BTU/lb

s_g = 1.919 BTU/(lb·R)

We have;

x₂ = (1.61 - 0.17498)/(1.919 - 0.17498) ≈ 0.823

h₂ = h_f + x₂×(h_g -

P₃ = P₂ = 2 psi

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v₃ = 0.01605 ft³/lb

h₄ = h₃ + v₃ × (P₄ - P₃)

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\dot m ≈ 1.803×10⁶ lb/h

The mass flow rate of the steam ≈ 1.803×10⁶ lb/h

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\dot Q_{in}  ≈ 2.52×10⁹ BTU/h

(c) The thermal efficiency =  \dot W_{cyc}/\dot Q_{in} = 1×10⁹/(2.52×10⁹) = 0.3968 ≈ 39.68%

The thermal efficiency ≈ 39.68%

(d) The mass flow rate of cooling water \dot m_w = \dot m(h₂- h₃)/(c_w \Delta T)

c_w = 1 BTU/(lb·°F)

\dot m_w =  1.803×10⁶ (935.11- 94.02)/(1 * (76 - 60)) ≈ 9.478 × 10⁷ lb/h

\dot m_w ≈ 9.478 × 10⁷ lb/h

The mass flow rate of cooling water ≈ 9.478 × 10⁷ lb/h.

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