Answer:
look below at illustrated diagrams and solution for Mz=1.13kN
Answer: hydraulic retention time,τ=28.67 hours
Explanation:
The hydraulic retention time τ (tau), is given as The volume of the settling tank(V) divided by the influent flowrate(Q)
τ =V/Q
But Volume is not known and is given as
Volume = surface area x depth of the tank
= 8000 ft² X 12 ft
= 96,000 ft³
Also, the influent flow rate is in mgd ( million gallons per day), we change it to ft³/sec so as to be in same unit with the volume in ft³
1 million gallons/day = 1.5472286365101 cubic feet/second
0.6mgd = 1.5472286365101 cubic feet/second x 0.6
=0.93cubic feet/second
τ =V/Q
96,000 ft³/0.93 ft³/sec
τ=103,225.8 secs
changing to hours
103,225.8 /3600 =28.67 hours
The hydraulic retention time =28.67 hours
Answer:
2.95 approximately 3
Explanation:
For a heat pump,
COP = Q/W
Where Q = power needed for heating process
W = power input into heat pump.
Power for heating Q = 6.85 kW
Proposed power input to heat pump W = 2.32 kW
Minimum COP = 6.85/2.32 = 2.95
Approximately 3
Answer:
Part A:
In W-h:
Energy Stored=1440 W-h
In Joules:
Part B:
In W-h:
Energy left=240 W-h
In Joules:
Energy left= 8.64*10^5 J
Explanation:
Part A:
We are given rating 120A-h and voltage 12 V
Energy Stored= Rating*Voltage (Gives us units W-h)
Energy Stored=120A-h*12V
Energy Stored=1440 W-h
Converting it into joules (watt=joules/sec)
Energy Stored=
Energy Stored=5184000 Joules
Part B:
Energy used by lights for 8h=150*8
Energy used by lights for 8h=1200W-h
Energy left= Energy Stored(Calculated above)- Energy used by lights for 8h
Energy left=1440-1200
Energy left=240 W-h
Energy left=
Energy left=864000 Joules
Energy left= 8.64*10^5 J