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The rate of gain for the high reservoir would be 780 kj/s.
A. η = 35%
W =
W = 420 kj/s
Q2 = Q1-W
= 1200-420
= 780 kJ/S
<h3>What is the workdone by this engine?</h3>B. W = 420 kj/s
= 420x1000 w
= 4.2x10⁵W
The work done is 4.2x10⁵W
c. 780/308 - 1200/1000
= 2.532 - 1.2
= 1.332kj
The total enthropy gain is 1.332kj
D. Q1 = 1200
T1 = 1000
= 1200 - 369.6/1200
= 69.2 percent
change in s is zero for the reversible heat engine.
Read more on enthropy here: brainly.com/question/6364271
Answer:
•Estimated density = 39.685Pc/mi/en
•Level of service, LOS frequency = LOSC
Explanation:
We are given:
•Freeway current lane width,B = 12ft
• freeway current shoulder width,b = 6ft
• percentage of heavy vehicle, Ptb = 10℅
• peak hour factor, PHF = 0.9
Let's consider,
•Number of lanes N = 4
• flow of traffic V = 7500vph
• percentage of Rv = 0, therefore the freeflow speed in freeway FFS = 70mph
• cars equivalent for recreational purpose Er= 2
•cars to be used for trucks and busses Etb= 2.5
Let's first calculate for the heavy adjustment factor.
We have:
Substituting figures in the equation we have:
= 0.75
Let's now calculate equivalent flow rate of the car using:
= 2777.7 pc/h/en
Calculating for traffic density, we have:
D = 39.685 Pc/mi/en
Using the table for LOS criteria of basic frequency segment, the level of service LOS of frequency is LOSC