Answer:
the current consumed is 3.3 A
Explanation:
Given;
resistance, R = 30 ohms
inductance, L = 200 mH
Voltage supply, V = 230 V
frequency of the coil, f = 50 Hz
impedance, Z = 69.6 Ohms
The current consumed is calculated as;
Therefore, the current consumed is 3.3 A
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
Answer:
The shaft work generated per kilogram is
Explanation:
Given:
Temperature K
Initial Pressure MPa
Final pressure MPa
From the table superheated,
and
Work done by shaft is,
But here efficiency is 0.56,
So work generated per kg is,
Work =
Work =
Therefore, the shaft work generated per kilogram is