Consider a single crystal of some hypothetical metal that has the FCC crystal structure and is oriented such that a tensile stre
1 answer:
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Answer:
1) The three possible assumptions are
a) All processes are reversible internally
b) Air, which is the working fluid circulates continuously in a closed loop
cycle
c) The process of combustion is depicted as a heat addition process
2) The diagrams are attached
5) The net work per cycle is 845.88 kJ/kg
The power developed in horsepower ≈ 45374 hP
Explanation:
1) The three possible assumptions are
a) All processes are reversible internally
b) Air, which is the working fluid circulates continuously in a closed loop
cycle
c) The process of combustion is depicted as a heat addition process
2) The diagrams are attached
5) The dimension of the cylinder bore diameter = 3.7 in. = 0.09398 m
Stroke length = 3.4 in. = 0.08636 m.
The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³
The clearance volume = 16% of cylinder volume = 0.16×5.99×10⁻⁴ m³
The clearance volume, v₂ = 9.59 × 10⁻⁵ m³
p₁ = 14.5 lbf/in.² = 99973.981 Pa
T₁ = 60 F = 288.706 K
Otto cycle T-S diagram
T₂ = 288.706* = 592.984 K
The maximum temperature = T₃ = 5200 R = 2888.89 K
T₄ = 2888.89 / = 1406.5 K
Work done, W = ×(T₃ - T₂) -
×(T₄ - T₁)
0.718×(2888.89 - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg
The power developed in an Otto cycle = W×Cycle per second
= 845.88 × 2400 / 60 = 33,835.377 kW = 45373.99 ≈ 45374 hP.
Answer:
Vab ≈ 3.426 V
Explanation:
First of all, it is convenient to find the equivalent parallel resistance of R5 and R6. That will be ...
R56 = (R5)(R6)/(R5 +R6) = (1000)(1500)/(1000 +1500) = 600
Then we can call V1 the voltage at the top of R2. The voltage at Va is a divider from V1:
Va = V1·(R4/(R3+R4)) = V1(560/1030) ≈ 0.543689V1
The voltage at Vb is also a divider from V1:
Vb = V1·(R7+R8)/(R2 +R56 +R7 +R8) = V1(780/1710) ≈ 0.456140V1
The parallel branches containing Va and Vb have an effective resistance of ...
(1030)(1710)/(1030+1710) = 642.81
That forms a divider with R1 to give V1:
V1 = (100 V)642.81/(1000 +642.81) ≈ 39.1287 V
The difference Va-Vb is ...
Vab = (39.1287 V)(0.543689 -0.456140) ≈ 3.426 V
_____
We have done this using parallel resistance and voltage divider calculations. You can also do it using node voltage equations. Using the same definition for V1 as above, we have ...
(Vs -V1)/R1 +(Vb -V1)/(R56+R2) +(Va-V1)/R3 = 0
(V1 -Vb)/(R56 +R2) -Vb/(R7+R8) = 0
(V1 -Va)/R3 -Va/R4 = 0
The solution of interest is the value of Vab, shown in the attachment. It computes as 154200/45013 V ≈ 3.42568 V.
