True
the answer to this would be true
Answer:
14.506°C
Explanation:
Given data :
flow rate of water been cooled = 0.011 m^3/s
inlet temp = 30°C + 273 = 303 k
cooling medium temperature = 6°C + 273 = 279 k
flow rate of cooling medium = 0.02 m^3/s
Determine the outlet temperature
we can determine the outlet temperature by applying the relation below
Heat gained by cooling medium = Heat lost by water
= ( Mcp ( To - 6 ) = Mcp ( 30 - To )
since the properties of water and the cooling medium ( water ) is the same
= 0.02 ( To - 6 ) = 0.011 ( 30 - To )
= 1.82 ( To - 6 ) = 30 - To
hence To ( outlet temperature ) = 14.506°C
Answer:
(N-1) × (L/2R) = (N-1)/2
Explanation:
let L is length of packet
R is rate
N is number of packets
then
first packet arrived with 0 delay
Second packet arrived at = L/R
Third packet arrived at = 2L/R
Nth packet arrived at = (n-1)L/R
Total queuing delay = L/R + 2L/R + ... + (n - 1)L/R = L(n - 1)/2R
Now
L / R = (1000) / (10^6 ) s = 1 ms
L/2R = 0.5 ms
average queuing delay for N packets = (N-1) * (L/2R) = (N-1)/2
the average queuing delay of a packet = 0 ( put N=1)
Answer:
A)
D = 158.42 kmol/h
B = 191.578 kmol/h
B) Rmin = 1.3095
Explanation:
<u>a) Determine the distillate and bottoms flow rates ( D and B ) </u>
F = D + B ----- ( 1 )
<em>Given data :</em>
F = 350 kmol/j
Xf = 0.45 mole
yD ( distillate comp ) = 0.97
yB ( bottom comp ) = 0.02
back to equation 1
350(0.45) = 0.97 D + 0.02 B ----- ( 2 )
where; B = F - D
Equation 2 becomes
350( 0.45 ) = 0.97 D + 0.02 ( 350 - D ) ------ 3
solving equation 3
D = 158.42 kmol/h
resolving equation 2
B = 191.578 kmol/h
<u>B) Determine the minimum reflux ratio Rmin</u>
The minimum reflux ratio occurs when the enriching line meets the q line in the VLE curve
first we calculate the value of the enriching line
Y =( Rm / R + 1 m ) x + ( 0.97 / Rm + 1 )
q - line ; y = ( 9 / 9-1 ) x - xf/9-1
therefore ; x = 0.45
Finally Rmin
= (( 0.97 / (Rm + 1 )) = 0.42
0.42 ( Rm + 1 ) = 0.97
∴ Rmin = 1.3095
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