Answer:
The grinding machine is used for roughing and finishing flat, cylindrical, and conical surfaces; finishing internal cylinders or bores; forming and sharpening cutting tools; snagging or removing rough projections from castings and stampings; and cleaning, polishing, and buffing surfaces.
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
Answer:
t = 25.10 sec
Explanation:
we know that Avrami equation
![Y = 1 - e^{-kt^n}](https://tex.z-dn.net/?f=Y%20%3D%201%20-%20e%5E%7B-kt%5En%7D)
here Y is percentage of completion of reaction = 50%
t is duration of reaction = 146 sec
so,
![0.50 = 1 - e^{-k^146^2.1}](https://tex.z-dn.net/?f=0.50%20%3D%201%20-%20e%5E%7B-k%5E146%5E2.1%7D)
![0.50 = e^{-k306.6}](https://tex.z-dn.net/?f=0.50%20%3D%20e%5E%7B-k306.6%7D)
taking natural log on both side
ln(0.5) = -k(306.6)
![k = 2.26\times 10^{-3}](https://tex.z-dn.net/?f=k%20%3D%202.26%5Ctimes%2010%5E%7B-3%7D)
for 86 % completion
![0.86 = 1 - e^{-2.26\times 10^{-3} \times t^{2.1}}](https://tex.z-dn.net/?f=0.86%20%3D%201%20-%20e%5E%7B-2.26%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%20t%5E%7B2.1%7D%7D)
![e^{-2.26\times 10^{-3} \times t^{2.1}} = 0.14](https://tex.z-dn.net/?f=e%5E%7B-2.26%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%20t%5E%7B2.1%7D%7D%20%3D%200.14)
![-2.26\times 10^{-3} \times t^{2.1} = ln(0.14)](https://tex.z-dn.net/?f=-2.26%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%20t%5E%7B2.1%7D%20%3D%20ln%280.14%29)
![t^{2.1} = 869.96](https://tex.z-dn.net/?f=t%5E%7B2.1%7D%20%3D%20869.96)
t = 25.10 sec
Answer:
304.13 mph
Explanation:
Data provided in the question :
The Speed of the flying aircraft = 300 mph
Tailwind of the true airspeed = 50 mph
Now,
The ground speed will be calculated as:
ground speed = ![\sqrt{300^2+50^2}](https://tex.z-dn.net/?f=%5Csqrt%7B300%5E2%2B50%5E2%7D)
or
The ground speed = ![\sqrt{92500}](https://tex.z-dn.net/?f=%5Csqrt%7B92500%7D)
or
The ground speed = 304.13 mph
Hence, the ground speed is 304.13 mph