The total number of trips that the vehicle has to make based on the given sequence of operation is 120 trips.
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The sequence of operation is A - E - D - C - B - A - F
The given parameters;
- <em>number of pieces that will flow from the first machine A to machine F, = 2,000 pieces</em>
- <em>initial unit load specified in the first machine, L₁ = 50</em>
- <em>final unit load, L₂ = 100 </em>
- <em>the capacity of the vehicle = 1 unit load</em>
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The given sequence of operation of the vehicle;
A - E - D - C - B - A - F
<em>the vehicle makes </em><em>6 trips</em><em> for </em><em>100</em><em> unit </em><em>loads</em>
The total number of trips that the vehicle has to make, in order to transport the 2000 pieces of the load given, is calculated as follows.
100 unit loads ----------------- 6 trips
2000 unit loads --------------- ?
Thus, the total number of trips that the vehicle has to make based on the given sequence of operation is 120 trips.
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Answer:
A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.
Explanation:
Answer:
Explanation:
Cop of reversible refrigerator = TL / ( TH - TL)
TL = low temperature of freezer = 20 °F
TH = temperature of air around = 75 °F
Heat removal rate QL = 75 Btu/min
W actual, power input = 0.7 hp
conversion on F to kelvin = (T (°F) + 460 ) × 5 / 9
COP ( coefficient of performance) reversible = (20 + 460) × 5/9 / (5/9 ( ( 75 +460) - (20 + 460) ))
COP reversible = 480 / 55 = 8.73
irreversibility expression, I = W actual - W rev
COP r = QL / Wrev
W rev = QL / COP r where 75 Btu/min = 1.76856651 hp where W actual = 0.70 hp
a) W rev = 1.76856651 hp / 8.73 = 0.20258 hp is reversible power
I = W actual - W rev
b) I = 0.7 hp - 0.20258 hp = 0.4974 hp
c) the second-law efficiency of this freezer = W rev / W actual = 0.20258 hp / 0.7 hp = 0.2894 × 100 = 28.94 %