A ____ is marked by two sets of double yellow lines, with each set having a broken line on the inside, and a solid line on the o
utside. white arrows appear in this lane as well.
2 answers:
Answer:
center left-turn lane
Explanation:
A <em>center left turn lane</em> will be marked as described. The arrows, if present, generally indicate that left turns are permitted from the lane with these markings.
__
If the double yellow lines are solid, they are considered to be a "barrier" and are not to be crossed.
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Answer:
P2 = 3.9 MPa
Explanation:
Given that
T₁ = 290 K
P₁ = 95 KPa
Power P = 5.5 KW
mass flow rate = 0.01 kg/s
solution
with the help of table A5
here air specific heat and adiabatic exponent is
Cp = 1.004 kJ/kg K
and k = 1.4
so
work rate will be
W = m × Cp × (T2 - T1) ..........................1
here T2 = W ÷ ( m × Cp) + T1
so T2 = 5.5 ÷ ( 0.001 × 1.004 ) + 290
T2 = 838 k
so final pressure will be here
P2 = P1 × ..............2
P2 = 95 ×
P2 = 3.9 MPa
Answer:
a)31%
b)34MW
Explanation:
A rankine cycle is a generation cycle using water as a working fluid, when heat enters the boiler the water undergoes a series of changes in state and energy until generating power through the turbine.
This cycle is composed of four main components, the boiler, the pump, the turbine and the condenser as shown in the attached image
To solve any problem regarding the rankine cycle, enthalpies in all states must be calculated using the thermodynamic tables and taking into account the following.
• The pressure of state 1 and 4 are equal
• The pressure of state 2 and 3 are equal
• State 1 is superheated steam
• State 2 is in saturation state
• State 3 is saturated liquid at the lowest pressure
• State 4 is equal to state 3 because the work of the pump is negligible.
Once all enthalpies are found, the following equations are used using the first law of thermodynamics
Wout = m (h1-h2)
Qin = m (h1-h4)
Win = m (h4-h3)
Qout = m (h2-h1)
The efficiency is calculated as the power obtained on the heat that enters
Efficiency = Wout / Qin
Efficiency = (h1-h2) / (h1-h4)
For this problem, we will first find the enthalpies in all states
h1=3231kJ/Kg
h2=2310kJ/Kg
h3=h4=272kJ/Kg
A) using the eficiency ecuation
Efficiency = (h1-h2) / (h1-h4)
Efficiency =(3231-2310)/(3231-272)=0.31=<u>31%</u>
b)using ecuation for Wout
Wout = m (h1-h2)
Wout=37(3231-2310)=34077KW=<u>34.077MW</u>
