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:
A worn inner CV joint often makes a clunking noise during starts and stops.
Answer:
a. 0.28
Explanation:
Given that
porosity =30%
hydraulic gradient = 0.0014
hydraulic conductivity = 6.9 x 10⁻4 m/s
We know that average linear velocity given as
The velocity in m/d ( 1 m/s =86400 m/d)
v= 0.27 m/d
So the nearest answer is 'a'.
a. 0.28
