Answer:
0.96kg/s
Explanation:
Hello! To solve this exercise we must use the first law of thermodynamics, which states that the sum of the energies that enter a system is the same amount that must go out. We must consider the following!
state 1 : is the first flow in the input of the chamber
h1=entalpy=335.02KJ/kg
m1=mass flow=0.56kg/s
state 2 : is the second flow in the input of the chamber
h2=entalpy=83.915KJ/kg
state 3:is the flow that comes out
h3=entalpy=175.90 kJ/kg
now use the continuity equation that states that the mass flow that enters is the same as the one that comes out
m1+m2=m3
now we use the first law of thermodynamics
m1h1+m2h2=m3h3
335.02m1+83.915m2=175.9m3
as the objective is to find the cold water mass flow(m2) we divide this equation by 175.9
1.9m1+0.477m2=m3
now we subtract the equations found in the equation of continuity and first law of thermodynamics
m1 + m2 = m3
-
1.9m1 + 0.477m2=m3
----------------------------------
-0.9m1+0.523m2=0
solving for m2
the mass flow rate of the cold-water is 0.96kg/s
Answer:
17.71 veh/mi/ln
Explanation:
Given,
Free Flow Speed, FFS = 45 mph
Number of lanes in each direction = 2
Peak Flow, V = 1300 veh/hr
Peak-hour factor = 0.85
HV = 8 %
Level Terrain
fHV = 1/ (1 + 0.08 (1.5-1)) = 1/1.04 = 0.96
fP = 1.0
Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = 1300/(0.96*2*0.85*1.0) = 796.57 ~ 797 veh/hr/ln
Vp < 1400
S = FFS = 45 mph
Density = Vp/S = (797) / (45) = 17.71 veh/mi/ln
Density of LOS B should lie between 11 – 18 veh/mi/ln
False you have to do the work and perricapate
Answer:
I hot u
Explanation:
it makes things function better and more fluently than a manual version of the certain object