Answer:
A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.
Explanation:
Answer: the mass flow rate of concentrated brine out of the process is 46,666.669 kg/hr
Explanation:
F, W and B are the fresh feed, brine and total water obtained
w = 2 x 10^4 L/h
we know that
F = W + B
we substitute
F = 2 x 10^4 + B
F = 20000 + B .................EQUA 1
solute
0.035F = 0.05B
B = 0.035F/0.05
B = 0.7F
now we substitute value of B in equation 1
F = 20000 + 0.7F
0.3F = 20000
F = 20000/0.3
F = 66666.67 kg/hr
B = 0.7F
B = 0.7 * F
B = 0.7 * 66666.67
B = 46,666.669 kg/hr
the mass flow rate of concentrated brine out of the process is 46,666.669 kg/hr
Don’t go on that file will give a virus! Sorry just looking out and I don’t know how to comment!
Answer:
(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.
(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.
Explanation:
(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.
(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.
Answer:
W= 8120 KJ
Explanation:
Given that
Process is isothermal ,it means that temperature of the gas will remain constant.
T₁=T₂ = 400 K
The change in the entropy given ΔS = 20.3 KJ/K
Lets take heat transfer is Q ,then entropy change can be written as
Now by putting the values
Q= 20.3 x 400 KJ
Q= 8120 KJ
The heat transfer ,Q= 8120 KJ
From first law of thermodynamics
Q = ΔU + W
ΔU =Change in the internal energy ,W=Work
Q=Heat transfer
For ideal gas ΔU = m Cv ΔT]
At constant temperature process ,ΔT= 0
That is why ΔU = 0
Q = ΔU + W
Q = 0+ W
Q=W= 8120 KJ
Work ,W= 8120 KJ
