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:
The solution to this question is 5.153×10⁻⁴(kmol)/(m²·s)
That is the rate of diffusion of ammonia through the layer is
5.153×10⁻⁴(kmol)/(m²·s)
Explanation:
The diffusion through a stagnant layer is given by
Where
= Diffusion coefficient or diffusivity
z = Thickness in layer of transfer
R = universal gas constant
= Pressure at first boundary
= Pressure at the destination boundary
T = System temperature
= System pressure
Where = 101.3 kPa
,
,
0.5×101.3 = 50.65 kPa
Δz = z₂ - z₁ = 1 mm = 1 × 10⁻³ m
R = T = 298 K and
= 1.18
= 1.8×10⁻⁵
= 5.153×10⁻⁴
Hence the rate of diffusion of ammonia through the layer is
5.153×10⁻⁴(kmol)/(m²·s)