A hypothetical accumulator processor uses one of the following 16-bit instruction formats, depending on the instruction.(a) (10
points.) Draw a block diagram of this CPU, showing the sequencer, ALU, memory, and control and data registers, and use arrows to indicate how the various parts interact with each other. Assume that instructions and data are accessed using an MAR and MDR register. (This will be the Von Neumann architecture diagram of this processor, much like the one we created in class.)(b) (20 points.) Give a control unit design for this processor using D-flip-flops similar to the one we designed in the lectures for Vesp 1.0. The design should include fetch, decode and execute steps in as much detail as possible.(c) (20 points) Give a datapath design, also using D-flip-flops for the MDR register of this processor, clearly describing all the transfers into the MDR with the help of a typical flip-flop as we did in class by making sure that you completely specify both the data and clock inputs for that flip-flop.
1 answer:
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Answer:
1)
2)
Explanation:
For isothermal process n =1
calculate pressure ratio to determine correction factor
correction factor for calculate dpressure ration for isothermal process is
c1 = 1.03
b) for adiabatic process
n =1.4
volume of hydraulic accumulator is given as
calculate pressure ratio to determine correction factor
correction factor for calculate dpressure ration for isothermal process is
c1 = 1.15
Answer:
Explanation:
The rate of flow in the pipe system in Figure P4.5.2 is 0.05 m3/s. The pressure at point 1 is measured to be 260 kPa. All the pipes are galvanized iron with roughness value of 0.15 mm. Determine the pressure at point 2. Take the loss coefficient for the sudden contraction as 0.05 and v = 1.141 × 10−6 m2/s.
The answer to the above question is
The pressure at point 2 = 75.959 kPa
Explanation:
Bernoulli's equation with losses gives
hL = z₁ - z₃ +(P₁-P₃)/(ρ×g) + (v₁²-v₃²)/(2×g)
Between points 1 and 2, z₁ = z₃ + 0.6 m therefore
hL = 0.6 m +(P₁-P₂)/(ρ×g) + (v₁²-v₃²)/(2×g)
hL = (f₁×L₁×v₁²)/(D₁×2×g) + (f₂×L₂×v₂²)/(D₂×2×g) + (f₃×L₃×v₃²)/(D₃×2×g) + k×V₃₂/(2×g) = 0.6 +(P₁-P₂)/(ρ×g) + (v₁²-v₃²)/(2×g)
But v = Q/A
or since A = π×D²/4 we have
A₁ = 1.77×10-2 m² , A₂ = 5.73×10-2 m², A₃ = 3.8×10-2 m²
Therefore from v = Q/A we have v₁ = 2.83 m/s v₂ = 0.87 m/s and v₃ = 1.315 m/s from there we find the friction coefficient from Moody Diagram as follows
ε = Which gives
the friction coefficients as f₁ = 0.02, f₂ = 0.017 and f₃ =0.0175
Substituting he above values into the equation we get
= 19.761 m
Combined head loss = 19.761 m
Hence 19.743 m = 0.6 m +(260 kPa-P₃)/(ρ×9.81) + (6.276)/(2×9.81)
or 260 kPa-18.82 m × 9.81 m/s²×ρ= P₃
Where ρ = density of water, we have
260000 Pa - 18.82 m×9.81 m/s²×997 kg/m³ = 75958.598 kg/m·s² = 75.959 kPa