Answer:
2.0%
Explanation:
Percentage of aggregate = 94%
Specific gravity = 2.65
Specific gravity of asphalt = 1.9
Density of mix = 147pcf = 147lb/ft³
Total weight of mix: (volume = 1ft³)
= (147lb/ft³)(1ft³)
= 147lb
Percentage weight of asphalt in<u> mix:</u>
100% - 94%
= 6%
Weight of asphalt binders
= 6% x 147lb
= 8.82lb
Weight of aggregate in mix:
= 94% x 147
= 138.18lb
Specific weight of asphalt binder:
(Gab)(Yw)
Yw = specific Weight of water
= 62.4lb
Gab = specific gravity of asphalt binder
= 1.0
(62.4lb)(1.0)
= 62.4 lb/ft³
Volume of asphalt in binder:
8.82/62.4
= 0.14ft³
Specific weight of binder in mix:
2.65 x 62.4lb/ft³
= 165.36 lb/ft³
Volume of aggregate:
= 138.18/165.36
= 0.84ft³
Volume of void in the mix:
1ft³ - 0.84ft³ - 0.14ft³
= 0.02ft³
<u>The percentage of void in total mix:</u>
VTM = (0.02ft³/1ft³)100
= 2.0%
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 answer is "
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Explanation:
Please find the complete question in the attached file.
Its change in temperature in pipes depends on rate heads and loss in pipes owing to pipe flow, contractual loss, etc.
The temperature change thus relies on V1 v2 p d D L.
Answer:
hello your question is incomplete attached below is the missing part and also attached is the solution
answer: a) 0.4801
b) 5.398 kw
c) 2.14
d) 12.72
Explanation:
The quality of the refrigerant at the evaporator inlet
h4 = hf4 + x4(hfx4)
Refrigeration load
Ql = m(h1-h4)
COP of the refrigerator
Ql / m(h2-h1) - Qm
Theoretical maximum refrigeration load
( Ql )max = COPr.rev * [m(h2-h1) - Qin]
