Answer:
Explanation:
In electrical terms, is the ratio of time in which a load or circuit is ON compared to the time in which the load or circuit is OFF.
The duty cycle or power cycle, is expressed as a percentage of the activation time. For example, a 70% duty cycle is a signal that 70% of the time is activated and the other 30% disabled. Its equation can be expressed as:
Where:
Here is a picture that will help you understand these concepts.
Answer:
four seconds
Explanation:
lookin at a vehicle respectively at a second can cause accident
Answer:
835,175.68W
Explanation:
Calculation to determine the required power input to the pump
First step is to calculate the power needed
Using this formula
P=V*p*g*h
Where,
P represent power
V represent Volume flow rate =0.3 m³/s
p represent brine density=1050 kg/m³
g represent gravity=9.81m/s²
h represent height=200m
Let plug in the formula
P=0.3 m³/s *1050 kg/m³*9.81m/s² *200m
P=618,030 W
Now let calculate the required power input to the pump
Using this formula
Required power input=P/μ
Where,
P represent power=618,030 W
μ represent pump efficiency=74%
Let plug in the formula
Required power input=618,030W/0.74
Required power input=835,175.68W
Therefore the required power input to the pump will be 835,175.68W
Answer:
(a) 0.243 m3/day
(b) 96 mg/l
(c) 0.426 m3/min
Explanation:
The sludge has an average solids concentration of 4 percent and considering TSS concentration in the influent of 240 mg/L then solids in sludge will be 0.04*240= 9.6 mg/L
Considering the average flow of 0.3 m3/s then mass of sludge per day will be given by 0.3*1000*9.6*60*60*24/1000000=248.832 kg/day
To get volume, considering specific gravity given as 1.025 and taking density of water as 1000 kg/m3 then density of sludge is 1025 kg/m3
Volume is mass/density hence 248.832/1025=0.2427629268292 m3/day
Approximately, the volume of sludge is 0.243 m3/day.
(b)
Efficiency of 60 percent is equivalent to 0.6
Efficiency=(influent concentration- flow rate)/influent concentration
0.6=(240-flow rate)/240
Flow rate= 96 mg/l
(c)
Cycle time= 0.243/0.57=0.4263157894736 m3/min
Rounded off, cycle time is 0.426 m3/day
Answer: The complete part of the question is to find the exit velocity
Explanation:
Given the following parameters
Inlet pressure = 700kpa
outlet pressure = 40kpa
Temperature = 80°C = 353k
mass flow rate = 1 kg/s
The application of the continuity and the bernoulli's equation is employed to solve the problem.
The detailed steps and the appropriate formula is as shown in the attached file.
