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A coaxial tube counter flow heat exchanger is used to cool 0.03 kg/s of benzene from 360 K to 310 K with a counter flow of 0.02 kg/s of water initially at . If the inner tube outside diameter is 2 cm and the overall heat transfer coefficient based on outside area is 650 , determine the required length of the exchanger. Take the specific heats of benzene and water as 1880 and 4175 J/kg‐K, respectively.
(B) 2.60
Correct answer is between ','. Can you explain this answer?
Verified Answer
A coaxial tube counter flow heat exchanger is used to cool 0.03 kg/s ...
By energy balance,
Q = mhcp(Th1 - Th2) = mccc(Tc2 - Tc1)
Q = 0.03 x 1880(360 - 310)
= 0.02 x 1475(Tc2 - 290)
Q = 2820 w,Tc2 = 323.8 K
ΔTi = 310 -290 ΔTe = 360 - 323.8
= 20 = 36.2
LMTD(ፀm) = 36.2 - 20/ In 36.2/20 = 27.3 K
Q = UoAo(θm) = UoπdoL(θm)
L = 2820/650 x π x 0.02 x 27.3 = 2.53 m
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A coaxial tube counter flow heat exchanger is used to cool 0.03 kg/s ...
Solution:
Given data:
Mass flow rate of benzene, m1 = 0.03 kg/s
Inlet temperature of benzene, T1 = 360 K
Outlet temperature of benzene, T2 = 310 K
Mass flow rate of water, m2 = 0.02 kg/s
Inlet temperature of water, T3 = 290 K
Outer diameter of the inner tube, d0 = 2 cm = 0.02 m
Overall heat transfer coefficient based on outside area, Uo = 650 W/m2-K
Specific heat of benzene, Cp1 = 1880 J/kg-K
Specific heat of water, Cp2 = 4175 J/kg-K
We can use the following equation to calculate the length of the heat exchanger:
Q = Uo x Ao x LMTD
Where,
Q = Heat transfer rate = m1 x Cp1 x (T1 - T2) = m2 x Cp2 x (T3 - T4)
Ao = Outside surface area of the inner tube = π x d0 x L
LMTD = Logarithmic mean temperature difference = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)
ΔT1 = T1 - T3
ΔT2 = T2 - T4
Assuming that the outlet temperature of water, T4 is equal to the inlet temperature, T3, we can solve for L as follows:
m1 x Cp1 x (T1 - T2) = m2 x Cp2 x (T3 - T4)
L = Q / (Uo x Ao x LMTD)
ΔT1 = 360 K - 290 K = 70 K
ΔT2 = 310 K - 290 K = 20 K
LMTD = (70 - 20) / ln(70 / 20) = 39.2 K
Ao = π x 0.02 m x L = 0.04πL
Substituting the values in the above equations, we get:
L = (0.03 kg/s x 1880 J/kg-K x 50 K) / (650 W/m2-K x 0.04π x 39.2 K)
L = 2.60 m
Therefore, the required length of the heat exchanger is 2.60 m.
Given data:
Mass flow rate of benzene, m1 = 0.03 kg/s
Inlet temperature of benzene, T1 = 360 K
Outlet temperature of benzene, T2 = 310 K
Mass flow rate of water, m2 = 0.02 kg/s
Inlet temperature of water, T3 = 290 K
Outer diameter of the inner tube, d0 = 2 cm = 0.02 m
Overall heat transfer coefficient based on outside area, Uo = 650 W/m2-K
Specific heat of benzene, Cp1 = 1880 J/kg-K
Specific heat of water, Cp2 = 4175 J/kg-K
We can use the following equation to calculate the length of the heat exchanger:
Q = Uo x Ao x LMTD
Where,
Q = Heat transfer rate = m1 x Cp1 x (T1 - T2) = m2 x Cp2 x (T3 - T4)
Ao = Outside surface area of the inner tube = π x d0 x L
LMTD = Logarithmic mean temperature difference = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)
ΔT1 = T1 - T3
ΔT2 = T2 - T4
Assuming that the outlet temperature of water, T4 is equal to the inlet temperature, T3, we can solve for L as follows:
m1 x Cp1 x (T1 - T2) = m2 x Cp2 x (T3 - T4)
L = Q / (Uo x Ao x LMTD)
ΔT1 = 360 K - 290 K = 70 K
ΔT2 = 310 K - 290 K = 20 K
LMTD = (70 - 20) / ln(70 / 20) = 39.2 K
Ao = π x 0.02 m x L = 0.04πL
Substituting the values in the above equations, we get:
L = (0.03 kg/s x 1880 J/kg-K x 50 K) / (650 W/m2-K x 0.04π x 39.2 K)
L = 2.60 m
Therefore, the required length of the heat exchanger is 2.60 m.
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A coaxial tube counter flow heat exchanger is used to cool 0.03 kg/s of benzene from 360 K to 310 K with a counter flow of 0.02 kg/s of water initially at . If the inner tube outside diameter is 2 cm and the overall heat transfer coefficient based on outside area is 650 , determine the required length of the exchanger. Take the specific heats of benzene and water as 1880 and 4175 J/kg‐K, respectively.(B) 2.60Correct answer is between ','. Can you explain this answer?
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