Answer:
Explanation:
given data
loading rate = 8.00 m/h
filtration rate = 7.70 m/h
dimensions = 10 m × 8 m
filter cycle duration = 52 h
time = 20 min
to find out
flow rate and volume of water is used for back washing plus rinsing the filter
solution
we consider here production efficiency is 96%
so here flow rate will be
flow rate = area × rate of filtration
flow rate = 10 × 8 × 7.7
flow rate = 616 m³/h
and
we know back washing generally 3 to 5 % of total volume of water per cycle so
volume of water is = 616 × 52
volume of water is 32032 m³
and
volume of water of back washing is = 4% of 32032
volume of water of back washing is 1281.2 m³
The most accurate answer to that process is definitely precision. The Rotary encoder is an electro-mechanical device that converts the angular position or motion of a shaft or axle to analog or digital output signals. The efficiency of these devices is subject to the position and angle of the axis in front of the encoder.
Most cars use reduction systems in their gearboxes that convert a certain signal input into an output. Mechanically for example, a 20: 1 reduction box already infers that if there is a revolution in the input at the output there are 20. That same transferred to the encoder pulses would imply greater precision.
For example a decoder with 50 holes would have to read 1000 pulses (50 * 20) which is basically a degree of accuracy of 0.36 degrees. In this way it is possible to conclude that if the assembly of the encoder is carried out next to the motor and not at the output, it can be provided with greater precision at the time of reading.
Answer:
you never mentioned the options
Answer:
2.83 kg
Explanation:
Given:
Volume, V = 0.8 m³
gage pressure, P = 200 kPa
Absolute pressure = gage pressure + Atmospheric pressure
= 200 + 101 = 301 kPa = 301 × 10³ N/m²
Temperature, T = 23° C = 23 + 273 = 296 K
Now,
From the ideal gas equation
PV = mRT
Where,
m is the mass
R is the ideal gas constant = 287 J/Kg K. (for air)
thus,
301 × 10³ × 0.8 = m × 287 × 296
or
m = 2.83 kg