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scarlettp13
20.10.2020 •
Engineering
Create a Verilog module called ssdec that has three ports: • in: a four-bit input wire • enable: a single input wire • out: a seven-bit output wire The module should implement a hexadecimal decoder for a seven-segment display. It should be similar to the work you did for Lab Experiment 6. This time, however, the decoder will only light LED segments if the enable wire is asserted. The decoder outputs will all be low if the enable wire is low. Instantiate the ssdec module into the standard top module with a statement like this: ssdec ss(.in (pb[3:0]), .enable (pb[19]), .out(ss0[6:0])); The end result should be Verilog code that will display the four-bit value placed on buttons 3..0 on the ss0 display when the 'Z' button is pressed. Submit the Verilog code for both the top and ssdec modules below. Reuse the ssdec module from (1), and build a new top module that instantiates ssdec eight times---one for each of the eight seven-segment displays. Connect each one to its own display. Connect all of their enable wires to pb[19] (the 'Z') button. Connect all of their input wires to pb[3:0]. The end result should be that all eight seven-segment displays show a hexadecimal representation of the binary pattern entered with buttons 3..0. (It is possible to do this with only a single instantiation of ssdec with its outputs connected to all the displays. However, we really want you to make eight separate instances of ssdec this time. Update the design you created in question (2). Modify the enable port for each ssdec instance such that pb[18:16] (which is the set of buttons Y, X and W) indicates which display should turn on. pb[18:16] is a 3-bit binary value, which can be used to represent (-7, so for example if you press pb[18] and pb[16] (or Y and W, which represents the number 5, or 3'b101) the digit should be displayed on ss5. You should be able to change the digit using pb[3:0]. To hold down buttons in the simulator, make sure your editor cursor is not active by clicking anywhere inside the page other than the editor, and pressing the keys 0-9, A-F, or W/X/Y/Z while holding down Shift. This allows you to test multiple digits on multiple displays. Submit the Verilog code for the top and ssdec modules below.
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Ответ:
W= 8120 KJ
Explanation:
Given that
Process is isothermal ,it means that temperature of the gas will remain constant.
T₁=T₂ = 400 K
The change in the entropy given ΔS = 20.3 KJ/K
Lets take heat transfer is Q ,then entropy change can be written as
Now by putting the values
Q= 20.3 x 400 KJ
Q= 8120 KJ
The heat transfer ,Q= 8120 KJ
From first law of thermodynamics
Q = ΔU + W
ΔU =Change in the internal energy ,W=Work
Q=Heat transfer
For ideal gas ΔU = m Cv ΔT]
At constant temperature process ,ΔT= 0
That is why ΔU = 0
Q = ΔU + W
Q = 0+ W
Q=W= 8120 KJ
Work ,W= 8120 KJ