Answer:
a) The Net power developed in this air-standard Brayton cycle is 43.8MW
b) The rate of heat addition in the combustor is 84.2MW
c) The thermal efficiency of the cycle is 52%
Explanation:
To solve this cycle we need to determinate the enthalpy of each work point of it. If we consider the cycle starts in 1, the air is compressed until 2, is heated until 3 and go throw the turbine until 4.
Considering this:
Now we can calculate the enthalpy of each work point:
h₁=281.4KJ/Kg
h₂=695.41KJ/Kg
h₃=2105KJ/Kg
h₄=957.14KJ/Kg
The net power developed:
The rate of heat:
The thermal efficiency:
The answer is D I’m 90% sure
Answer:
water sample have more water content
Explanation:
given data
soil 1 is saturated with water
unit weight of water = 1 g/cm³
soil 2 is saturated with alcohol
unit weight of alcohol = 0.8 g/cm³
solution
we get here water content that is express as
water content = ....................1
here soil is full saturated so is 100% in both case
so put here value for water
water content = 100 % × 1
water content = 1 g
and
now we get for alcohol that is
water content = 100 % × 0.8
water content = 0.8 g
so here water sample have more water content
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