Overcurrent protective devices, or OCPDs
Answer:
The time necessary to purge 95% of the NaOH is 0.38 h
Explanation:
Given:
vfpure water(i) = 3 m³/h
vNaOH = 4 m³
xNaOH = 0.2
vfpure water(f) = 2 m³/h
pwater = 1000 kg/m³
pNaOH = 1220 kg/m³
The mass flow rate of the water is = 3 * 1000 = 3000 kg/h
The mass of NaOH in the solution is = 0.2 * 4 * 1220 = 976 kg
When the 95% of the NaOH is purged, thus the NaOH in outlet is = 0.95 * 976 = 927.2 kg
The volume of NaOH in outlet after time is = 927.2/1220 = 0.76 m³
The time required to purge the 95% of the NaOH is = 0.76/2 = 0.38 h
Answer:
8 to 10 times
Explanation:
For dry road
u= 15 mph ( 1 mph = 0.44 m/s)
u= 6.7 m/s
Let take coefficient of friction( μ) of dry road is 0.7
So the de acceleration a = μ g
a= 0.7 x 10 m/s ² ( g=10 m/s ²)
a= 7 m/s ²
We know that
v= u - a t
Final speed ,v=0
0 = 6.7 - 7 x t
t= 0.95 s
For snow road
μ = 0.4
de acceleration a = μ g
a = 0.4 x 10 = 4 m/s ²
u= 30 mph= 13.41 m/s
v= u - a t
Final speed ,v=0
0 = 30 - 4 x t'
t'=7.5 s
t'=7.8 t
We can say that it will take 8 to 10 times more time as compare to dry road for stopping the vehicle.
8 to 10 times
Answer:
composition of alpha phase is 27% B
Explanation:
given data
mass fractions = 0.5 for both
composition = 57 wt% B-43 wt% A
composition = 87 wt% B-13 wt% A
solution
as by total composition Co = 57 and by beta phase composition Cβ = 87
we use here lever rule that is
Wα = Wβ ...............1
Wα = Wβ = 0.5
now we take here left side of equation
we will get
= 0.5
= 0.5
solve it we get
Ca = 27
so composition of alpha phase is 27% B
Answer:
250.7mw
Explanation:
Volume of the reservoir = lwh
Length of reservoir = 10km
Width of reservoir = 1km
Height = 100m
Volume = 10x10³x10³x100
= 10⁹m³
Next we find the volume flow rate
= 0.1/100x10⁹x1/3600
= 277.78m³/s
To get the electrical power output developed by the turbine with 92 percent efficiency
= 0.92x1000x9.81x277.78x100
= 250.7MW
