Answer:
the blank = life
Explanation:
:) happy holidays!
A brine solution is 26% salt with 70.0 kg of water evaporated per hour. To produce 195 kg of pure salt (0% moisture) per day, how long should the process operate each day and how much brine must be fed to the evaporator per hour?
The process should operate each day for ___ hours.
The amount of brine that must be fed to the evaporator is ___ kg/h.
Answer:
- the process should operate each day for 7.9286 hours
- the amount of brine that must be fed to the evaporator is 94.594 kg/hr
Explanation:
Given that;
concentration of brine = 26%
so water concentration will be 100% - 26% = 74%
for evaporation of 70kg water per hour, residual is pure salt( 0% moisture)
so mass flow rate of brine = 70/(74%) = 70/0.74 = 94.5945 kg/hr
Amount of pure dry salt produced(0%) = 94.5945 - 70 = 24.5945 kg/hr
Now for production of 195kg of pure dry salt, number of hours required will be
T = 195 / 24.5945 = 7.9286 hrs
Therefore the process should operate each day for 7.9286 hours.
Total brine solution required ( 26% salt conc.) = 195/0.26 = 750 kg
Feed rate of brine solution ( 26% salt conc.) = 750 / 7.9286 = 94.594 kg/hr
Therefore the amount of brine that must be fed to the evaporator is 94.594 kg/hr
A work element in a manual assembly task consists of the following MTM-1 elements: (1) R16C, (2) G4A, (3) M10B5, (4) RL1, (5) R14B, (6) G1B, (7) M8C3, (8) P1NSE, and (9) RL1.
(a) Determine the normal times in TMUs for these motion elements.
(b) What is the total time for this work element in sec
Answer:
a)
1) R16C ; Tn = 17 TMU
2) G4A ; Tn = 7.3 TMU
3) M10B5 ; Tn = 15.1 TMU
4) RL1 ; Tn = 2 TMU
5) R14B ; Tn = 14.4 TMU
6) G1B ; Tn = 3.5 TMU
7) M8C3 ; Tn = 14.7 TMU
8) P1NSE ; Tn = 10.4 TMU
9) RL1 ; Tn = 2 TMU
b) 3.1 secs
Explanation:
a) Determine the normal times in TMUs for these motion elements
1) R16C ; Tn = 17 TMU
2) G4A ; Tn = 7.3 TMU
3) M10B5 ; Tn = 15.1 TMU
4) RL1 ; Tn = 2 TMU
5) R14B ; Tn = 14.4 TMU
6) G1B ; Tn = 3.5 TMU
7) M8C3 ; Tn = 14.7 TMU
8) P1NSE ; Tn = 10.4 TMU
9) RL1 ; Tn = 2 TMU
b ) Determine the total time for this work element in seconds
first we have to determine the total TMU = ∑ TMU = 86.4 TMU
note ; 1 TMU = 0.036 seconds
hence the total time for the work in seconds = 86.4 * 0.036 = 3.1 seconds
Assume there exists some hypothetical metal that exhibits ferromagnetic behavior and that has (1) a simple cubic crystal structure, (2) an atomic radius of 0.154 nm, and (3) a saturation flux density of 0.83 tesla. Determine the number of Bohr magnetons per atom for this material. A: What is the volume, in m3, for this unit cell?B: How many atoms are there per m3 in this material?C: What is the number of Bohr magentons per atom for this material?
Answer:
a. 2.9218x10^(-29) m^3
b. 7.1230x10^(28) atoms/m^3
c. 2.0812 BM/atom
Explanation:
Atomic radius r = 0.154 nm
saturation flux density Bs = 0.83 tesla
- the formula for the volume of the simple cubic crystal (Vc) = a^3 = (2r)^3
= (2 x 0.154x10^-9)^3
= 2.9218x10^-29 m^3
- The formula for the Bohr magneton per atom with respect to VC, Bs, permeability of the vacuum Uo and magnetic moment per Bohr magneton Ub is;
Bohr magneton per atom nb = (Bs x VC) / (Ub x Uo)
= (0.83 x 2.9218x10^-29) / (9.27x10^-24) x(1.257x10^-6)
= 2.4251x10^-29 / 1.1652x10^-29
= 2.0812 BM/atom
- Number of atoms per Vc, N = nb / Vc
= 2.0812 / 2.9218x10^-29 = 7.1230x10^28 atoms
On a Test please help
When towing a trailer, you should consider all of the following except
the minimum speed limit on roadways.
your personal protective equipment.
the gross weight of the load.
the increased stopping distance
Answer:
I'm not 100%sure but I think it's the first one