Answer:
41.5° C
Explanation:
Given data :
1025 steel
Temperature = 4°C
allowed joint space = 5.4 mm
length of rails = 11.9 m
<u>Determine the highest possible temperature </u>
coefficient of thermal expansion ( ∝ ) = 12.1 * 10^-6 /°C
Applying thermal strain ( Δl / l ) = ∝ * ΔT
( 5.4 * 10^-3 / 11.9 ) = 12.1 * 10^-6 * ( T2 - 4 )
∴ ( T2 - 4 ) = ( 5.4 * 10^-3 / 11.9 ) / 12.1 * 10^-6
hence : T2 = 41.5°C
Answer:
a)
(Ω-m)^{-1}
b) Resistance = 121.4 Ω
Explanation:
given data:
diameter is 7.0 mm
length 57 mm
current I = 0.25 A
voltage v = 24 v
distance between the probes is 45 mm
electrical conductivity is given as

![\sigma = \frac{0.25 \times 45\times 10^{-3}}{24 \pi [\frac{7 \times 10^{-3}}{2}]^2}](https://tex.z-dn.net/?f=%5Csigma%20%20%3D%20%5Cfrac%7B0.25%20%5Ctimes%2045%5Ctimes%2010%5E%7B-3%7D%7D%7B24%20%5Cpi%20%5B%5Cfrac%7B7%20%5Ctimes%2010%5E%7B-3%7D%7D%7B2%7D%5D%5E2%7D)
(Ω-m)^{-1}[/tex]
b)


![= \frac{57 \times 10^{-3}}{12.2 \times \pi [\frac{7 \times 10^{-3}}{2}]^2}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B57%20%20%5Ctimes%2010%5E%7B-3%7D%7D%7B12.2%20%5Ctimes%20%5Cpi%20%5B%5Cfrac%7B7%20%5Ctimes%2010%5E%7B-3%7D%7D%7B2%7D%5D%5E2%7D)
Resistance = 121.4 Ω
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brainly.com/question/24078882
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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