The following laboratory test results for Atterberg limits and sieve-analysis were obtained for an inorganic soil. [6 points] Si
1 answer:
Answer: hello the complete question is attached below
answer:
A) Group symbol = SW
B) Group name = well graded sand , fine to coarse sand
C) It is not a clean sand given that ≤ 50% particles are retained on No 200
Explanation:
<u>A) Classifying the soil according to USCS system</u>
( using 2nd image attached below )
<em>description of sand</em> :
The soil is a coarse sand since ≤ 50% particles are retained on No 200 sieve, also
The soil is a sand given that more than 50% particles passed from No 4 sieve
The soil can be a clean sand given that fines ≤ 12%
The soil can be said to be a well graded sand because the percentage of particles passing through decreases gradually over time
Group symbol as per the 2nd image attached below = SW
B) Group name = well graded sand , fine to coarse sand
C) It is not a clean sand given that ≤ 50% particles are retained on No 200
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Answer:
a) the heat exchanger area required for the evaporator is 11178.236 m²
b) the required flow rate is 1993630.38 kg/s
Explanation:
Given the data in the question;
Water temperature near the surface = 300 K
temperature at reasonable depths ( cold ) = 280 K
power plant output W' = 2 MW
efficiency η = 3% = 0.03
we know that; efficiency η = W' / Q
we substitute
0.03 = 2 / Q
Q = 2 / 0.03
Q = 66.667 MW = 66.667 × 10⁶ Watt
T = 300 K T
= 292 K
T = 290 K T
= 290 K
Now, Heat transfer in evaporator;
Q = UA( LMTD )
so
LMTD = (ΔT₁ - ΔT₂) / ln( ΔT₁ / ΔT₂ )
first we get ΔT₁ and ΔT₂
ΔT₁ = T - T
= 300 - 290 = 10 K
ΔT₂ = T - T
= 292 - 290 = 2 K
so we substitute into our equation;
LMTD = (10 - 2) / ln( 10 / 2 )
LMTD = 8 / ln( 5 )
LMTD = 8 / 1.6094379
LMTD = 4.97
a) Heat transfer Area will be;
Q = UA( LMTD )
we substitute
66.667 × 10⁶ = 1200 × A × 4.97
66.667 × 10⁶ = 5964 × A
A = (66.667 × 10⁶) / 5964
A = 11178.236 m²
Therefore, the heat exchanger area required for the evaporator is 11178.236 m²
b) Flow rate
we know that;
Q = m'C
(
-
)
specific heat capacity of water Cp = 4.18 (kJ/kg∙°C)
we substitute
66.667 × 10⁶ = m' × 4.18 × ( 300 - 292 )
66.667 × 10⁶ = m' × 33.44
m' = ( 66.667 × 10⁶ ) / 33.44
m' = 1993630.38 kg/s
Therefore, the required flow rate is 1993630.38 kg/s