PDAs with two stacks are strictly more powerful than PDAs with one stack. Prove that 2-stack PDAs are not a valid model for CFLs
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Answer:
a) , b)
Explanation:
a) The uniform dresser can be modeled using specific equilibrium equations:
Following some algebraic manipulations, the formulated equation is derived:
b) Similarly, the man can be represented by a set of equilibrium equations:
After some algebraic changes, the expression for the coefficient of static friction comes out as:
1) The three possible assumptions are
a) All processes are internally reversible
b) Air, as the working fluid, circulates in a closed-loop
cycle
c) Combustion is represented as a heat-adding process
2) Diagrams are included
5) The net work per cycle is 845.88 kJ/kg
The horsepower produced is approximately 45374 hP
Explanation:
1) The three valid assumptions are
a) All processes are internally reversible
b) Air, the working fluid, moves continuously in a closed-loop
cycle
c) The combustion process is set as a heat addition step
2) Diagrams for illustration are provided
5) The cylinder bore diameter measures 3.7 in., equating to 0.09398 m
The stroke length is 3.4 in., approximately 0.08636 m.
The cylinder volume is calculated as v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³
The clearance volume amounts to 16% of the cylinder volume = 0.16×5.99×10⁻⁴ m³
The clearance volume, v₂ is 9.59 × 10⁻⁵ m³
p₁ is 14.5 lbf/in.² = 99973.981 Pa
T₁ equals 60 F = 288.706 K
For the Otto cycle's T-S diagram,
T₂ calculates to 288.706* = 592.984 K
The peak temperature is T₃ = 5200 R = 2888.89 K
T₄ resolves to 2888.89 / = 1406.5 K
Work performed, W = ×(T₃ - T₂) -
×(T₄ - T₁)
0.718×(2888.89 - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg
The power generated in the Otto cycle = W×Cycle per second
= 845.88 × 2400 / 60 = 33,835.377 kW = 45373.99 ≈ 45374 hP.