<u>Solution and Explanation:</u>
Volume of gas stream = 1000 cfm (Cubic Feet per Minute)
Particulate loading = 400 gr/ft3 (Grain/cubic feet)
1 gr/ft3 = 0.00220462 lb/ft3
Total weight of particulate matter = 
Cyclone is to 80 % efficient
So particulate remaining = 
emissions from this stack be limited to = 10.0 lb/hr
Particles to be remaining after wet scrubber = 10.0 lb/hr
So particles to be removed = 685.7136- 10 = 675.7136
Efficiency = output multiply with 100/input = 98.542 %
Answer:
HIGH from the supply voltage
LOW from ground
Explanation:
The answer depends on the kind of system and the purpose of the signal. But for practical reasons, in a DIGITAL system where 5V is HIGH and 0 V is LOW, 5 volts can be taken from the supply voltage (usually the same as high, BUT must be verified), and the LOW signal from ground.
If the user has a multimeter, it must be set to continuous voltage on 0 to 20 V range. Then place the probe in the ground of the circuit (must be a big copper area). Finally leave one probe in the circuit ground and place the other probe in some test points to identify 5 v.
Answer:
A piece of code hidden in spread sheet
Answer:
see vous se to pe a he ko off a nack u
Answer:
12.5%
Explanation:
Compaction ratio= Volume before reduction/volume after reduction
Compaction ratio= 8/1
% reduction in volume= Volume after reduction/Volume before reduction× 100= 1/compaction ratio × 100
% reduction in volume= 1/(8/1) × 100
=12.5
