An engine oil flows through a 25-mm diameter tube at a rate of 0.5 kg/s. The oil enters the tube at a temperature of 25 °C, while the tube surface temperature is maintained at 100 °C.
(a) Determine the oil outlet temperature for a 5-m and for 100-m long tube. For each case, compare the log mean temperature difference to arithmetic mean temperature difference.
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Answers
Explanation:
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Given:
Diameter of tube = 25 mm.
Maas flow rate = 0.5 kg/sec.
Temperature of oil, T(i) = 25°C.
Surface temperature, T(w) = 100°C.
To Find:
Determine the oil outlet temperature for a 5-m and 100-m long tube. For each case, compare the log mean temperature difference to the arithmetic means temperature difference.
Solution:
(i), When L = 5m.
Reynold's number, Re = v D / ν.
Re = 4m° / πDμ.
Here, engine oil properties are taken as follows,
ρ = 865.8 kg/m³.
Cp = 2035 J/kg.K.
K = 0.141 W/m.K.
μ = 0.0836 N.s/m².
Pr = 1205.
Now,
Re = (4 × 0.5) / (π×0.025×0.0836).
Re = 304.6.
Here, Re < 2300.
So, the flow is laminar.
Therefore, for constant wall temperature,
T = (K/D)[ 3.66 + (0.0688[D/L]Re D Pr) / (1 + 0.04 [(D/L)Re D Pr]∧2/3]
When L = 5m,
T = 5.64 (3.66 + 17.51).
T = 119 W/m².K.
So by equation,
(To - Tw) / (Ti - Tw) = e∧(-hAs/mCp).
To = 28.4°C.
28.4°C is the outlet temperature of the oil.
Now,
Log mean temperature difference, LMTD = (∅2 - ∅1) / ln(∅2/∅1).
LMTD = 73.29°C.
Then,
Arithmetic mean temperature difference, AMTD = ΔT(am) = (∅2 + ∅1) / 2.
ΔT(am) = 73.3°C.
Hence, the oil outlet temperature for a 5-m and 100-m long tube is 28.4°C and Log mean temperature difference, LMTD and Arithmetic mean temperature difference, AMTD are 73.29°C, and 73.3°C.