A car radiator is a cross-flow heat exchanger with both fluids unmixed. Water, which has a flow rate of 0.05 kg/s, enters the radiator at 400 K and is to leave at 330 K. The water is cooled by air that enters at 0.75 kg/s and 300 K. If the overall heat transfer coefficient is 200 W/m2-K, what is the required heat transfer surface area?
Known: flow rate and inlet temperature for automobile radiator.
Overall heat transfer coefficient.
Find: Area required to achieve a prescribed outlet temperature.
Assumptions: (1) Negligible heat loss to surroundings and kinetic and
potential energy changes, (2) Constant properties.
Analysis: The required heat transfer rate is
q = (m c)h (T h,i - T h,o) = 0.05 kg/s (4209J / kg.K) 70K = 14,732 W
Using the ε-NTU method,
Cmin = Ch = 210.45 W / K
Cmax = Cc = 755.25W / K
Hence, Cmin/Cmx(Th,i - Th,o) = 210.45W / K(100K) = 21,045W
ε=q/qmax = 14,732W / 21,045W = 0.700
A=NTU(cmin / U) = 1.5 x 210.45W / K(200W) / m² .K) = 1.58m²
1. the air outlet is..
Tc,o = Tc,i + q / Cc = 300K + (14,732W / 755.25W / K) = 319.5K
2. using the LMTD approach ΔTlm = 51.2 K,, R=0.279 and P=0.7
hence F≅0.95 and
A = q/FUΔTlm = (14,732W) / [0.95(200W / m².K) 51.2K] = 1.51m²
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