A dual-fluid heat exchanger has 10 lbm/s water entering at 100 F, 20 psia and leaving at 50 F, 20 psia. The other fluid is glyco
1 answer:
Answer:
Rate of internal heat transfer = 23.2 Btu/Ibm
mass flow rate = 21.55 Ibm/s
Explanation:
using given data to obtain values from table F7.1
Enthalpy of water at temperature of 100 F = 68.04Btu/Ibm
Enthalpy of water at temperature of 50 F = 18.05 Btu/Ibm
from table F.3
specific constant of glycerin
<u>The rate of internal heat transfer ( change in enthalpy ) </u>
h4 - h3 = Cp ( T4 - T3 ) --------------- ( 1 )
where ; T4 = 50 F
T3 = 10 F
Cp = 0.58 Btu/Ibm-R
substitute given values into equation 1
change in enthalpy ( h4 - h3 ) = 23.2 Btu/Ibm
<u>Determine mass flow rate of glycol</u>
attached below is the detailed solution
mass flow rate of glycol = 21.55 Ibm/s
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Answer:
P = 0.490 kip
Explanation:
given data
allowable bearing stress = 2 ksi
allowable tensile stress = 18 ksi
diameter = 0.31 in
outer diameter = 0.75 in
inner diameter (hole) = 0.50 in
solution
we find here cross section area of shank that is express as
Area = ..................1
area =
area = 0.0754 in²
and
now we get here allowable load in bolt will be
...................2
P =
P = 18 × 10³ × 0.0754
P = 1357.2 = 1.357 kip
and
now find here area of washer is
Area = .......................3
put here value
Area =
area = 0.2454 in²
so now we get here allowable load of washer will be
.....................4
P = 2 × 10³ × 0.245
P = 490 = 0.490 kip