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:
L= 50000 lb
D = 5000 lb
Explanation:
To maintain a level flight the lift must equal the weight in magnitude.
We know the weight is of 50000 lb, so the lift must be the same.
L = W = 50000 lb
The L/D ratio is 10 so
10 = L/D
D = L/10
D = 50000/10 = 5000 lb
To maintain steady speed the thrust must equal the drag, so
T = D = 5000 lb
Answer:
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