Answer:
hello some data related to your question is missing attached below is the missing data
answer : 63700 Ib/hr
Explanation:
Given data :
Limestone mix : consists of 94% CaCO3 and 6% inert material
Actual feed rate = 36,000 Ib/hr
SO2 in flue gas = 20,314 Ib/hr
FGD efficiency = 97%
resulting sludge contains 58% solids
<u>Calculate the Total sludge production rate </u>
First : determine SO2 removed in sludge
= 0.97 * 20314
= 19704.58 Ib/hr
next : moles of SO2 removed
= 19704.58 / 64 Ib/ Ib mol
= 307.88 Ib mol / hr
also moles of CaSO3 produced = 307.88 Ib mol / hr
mass of CaSO3 = 307.88 * 120 = 36946.09 Ib/hr
Therefore Total sludge production rate
= 36946.09 / 0.58
= 63700.15 Ib/hr
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Answer:
if you're talking about the car b-post, the answer is "posts"
Explanation:
looked it up
Explanation:
I'm not exactly a master at coding, but I'm pretty sure that:
The farmer will remove dirt as long as there is a pile, then stop when the pile is done.
Answer: downward velocity = 6.9×10^-4 cm/s
Explanation: Given that the
Diameter of the smoke = 0.05 mm = 0.05/1000 m = 5 × 10^-5 m
Where radius r = 2.5 × 10^-5 m
Density = 1200 kg/m^3
Area of a sphere = 4πr^2
A = 4 × π× (2.5 × 10^-5)^2
A = 7.8 × 10^-9 m^2
Volume V = 4/3πr^3
V = 4/3 × π × (2.5 × 10^-5)^3
V = 6.5 × 10^-14 m^3
Since density = mass/ volume
Make mass the subject of formula
Mass = density × volume
Mass = 1200 × 6.5 × 10^-14
Mass M = 7.9 × 10^-11 kg
Using the formula
V = sqrt( 2Mg/ pCA)
Where
g = 9.81 m/s^2
M = mass = 7.9 × 10^-11 kg
p = density = 1200 kg/m3
C = drag coefficient = 24
A = area = 7.8 × 10^-9m^2
V = terminal velocity
Substitute all the parameters into the formula
V = sqrt[( 2 × 7.9×10^-11 × 9.8)/(1200 × 24 × 7.8×10^-9)]
V = sqrt[ 1.54 × 10^-9/2.25×10-4]
V = 6.9×10^-6 m/s
V = 6.9 × 10^-4 cm/s