An automobile radiator may be viewed as 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.
a) If the overall heat transfer coefficient is 200 W/ m2 .K, what is the required heat transfer surface area?
b) Repeat part (a), except assume that the air flow is mixed.
c) Suppose that the heat exchange area of the radiator is 1.0 m2 , and that U and the inlet temperatures are the same as in part (a) (both outlet temperatures are unknown). If the mass flow rates are also the same as in part (a), what is the water outlet temperature? Assume that the air flow is mixed.

The term "temperature" describes how hot or cold a body is. It is, specifically, a method of calculating the kinetic energy of the particles that make up an item. When particles travel more quickly, the temperature rises, and vice versa.

Nearly every aspect of daily life as well as every scientific field, from physics to geology, depend heavily on 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

and

ε=q/qmax

= 14,732W / 21,045W = 0.700

NTU≅1.5, hence

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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