<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:
<em>Electric current is the movement of electrons through a wire. Electric current is measured in amperes (amps) and refers to the number of charges that move through the wire per second. If we want current to flow directly from one point to another, we should use a wire that has as little resistance as possible.</em><em>Current is directly proportional to voltage, inversely proportional to resistance. One of the most common electrical measurements you'll use is the watt, a unit of electrical power: W (Watts) = E (Volts) x I (Amperes). The quantity of electric charge is measured in coulombs.</em><em>They can even pass through bones and teeth. This makes gamma rays very dangerous. They can destroy living cells, produce gene mutations, and cause cancer.</em>
Explanation:
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Answer:
Load carried by shaft=9.92 ft-lb
Explanation:
Given: Power P=4.4 HP
P=3281.08 W
<u><em>Power: </em></u>Rate of change of work with respect to time is called power.
We know that P=
rad/sec
So that P=
So 3281.08=
T=13.45 N-m (1 N-m=0.737 ft-lb)
So T=9.92 ft-lb.
Load carried by shaft=9.92 ft-lb
Answer:
The volume flow rate necessary to keep the temperature of the ethanol in the pipe below its flashpoint should be greater than 1.574m^3/s
Explanation:
Q = MCp(T2 - T1)
Q (quantity of heat) = Power (P) × time (t)
Density (D) = Mass (M)/Volume (V)
M = DV
Therefore, Pt = DVCp(T2 - T1)
V/t (volume flow rate) = P/DCp(T2 - T1)
P = 20kW = 20×1000W = 20,000W, D(rho) = 789kg/m^3, Cp = 2.44J/kgK, T2 = 16.6°C = 16.6+273K = 289.6K, T1 = 10°C = 10+273K = 283K
Volume flow rate = 20,000/789×2.44(289.6-283) = 20,000/789×2.44×6.6 = 1.574m^3/s (this is the volume flow rate at the flashpoint temperature)
The volume flow rate necessary to keep the ethanol below its flashpoint temperature should be greater than 1.574m^3/s