Answer:
8.5 days
Explanation:
Given data :
Flow ( Q ) = 2.6 MGD = 11819.834 m^3/day
BOD = 131 mg/L
BOD loading rate = 35 Ibs/1000 ft^3 per day = 0.5606 kg/m^3/day
<u>Calculate the sludge age of the facility </u>
Given the BOD applied to the aeration tank = 11819.834 m^3/day * 131mg/l
= 1548.398 kg/day
first calculate the volume of the aeration tank
V = BOD applied / BOD loading rate
V = 1548.398 / 0.5606 = 2762.03 m^3
Hydraulic Detention time = V / Q
= 2762.03 / 11819.834 = 0.2336 day = 5.6 hour
next : determine the mass rate of the waste
= 44% * 0.5606 kg/m^3/day
= 0.2466 kg/m^2/day
finally determine the sludge age
= V * Xt / ∅w * R
= ( 2762.03 m^3 * 2100 * 10^-3 ) / ( 0.2466 * 2762.03 kg/day )
= 8.5 days
Answer:
The work done or the heat transfer for air and hydrogen are 91.71 kj/kg and 1317.89 kj/kg respectively.
Explanation:
Given:
Initial pressure is 7.6 bar.
Initial temperature is 77 °C.
Final pressure is 3.05 bar.
Calculation:
(a)
Take air as an ideal gas with gas constant 0.287 kj/kgK.
Step1
For isothermal process work done and heat transfer is the same because of no change in internal energy.
Work done is calculated as follows:


W = 91.71 kj/kg.
Thus, the work done or the heat transfer for isothermal process is 91.71 kj/kg.
Step2
For isothermal process work done and heat transfer is the same because of no change in internal energy. Take Hydrogen as an ideal gas with gas constant 4.1242 kj/kgK.
Work done is calculated as follows:


W = 1317.89 kj/kg.
Thus, the work done or the heat transfer for isothermal process is 1317.89 kj/kg.
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Answer:
Power needed to pump=4.79 KW.
Explanation:
Given that:
We know that coefficient of performance of heat pump
COP=
So COP=
COP=10.43
COP=
10.43 =
=4.79 KW
So power needed to pump=4.79 KW.