Question

Oil in an engine is being cooled by air in a cross-flow heat exchanger, where both fluids are unmixed. Oil (cp = 2000 J/kg. K) flowing with a flow rate of 0.020 kg/s enters the heat exchanger at 75°C, while air (cp = 1000 J/kg. K) enters at 30°C with a flow rate of 0.20 kg/s. The overall heat transfer coefficient of the heat exchanger is 50 W/m2 . K and the total surface area is 1 m2 . Determine: (a) the heat transfer effectiveness and (15 points) (b) the outlet temperature of the oil. (Use the effectiveness-NTU method) (15 points)

Answer 1

Answer:

A) $\epsilon = 0.67122$

B) $T_{h, out} = 44.795\ degree\ C$

Explanation:

Heat capacity of oil $C_h =m_h C_{ph} = 0.02 \times 2 = 0.04$ Kw/K

Heat capacity of air$C_C = m_C C_{pc} = 0.2 \times 1 = 0.2$ Kw/K

therefore $C_{min = C_h$

and $C_{max} = C_C$

we know that capacity ratio is

$c = \frac{C_{min}}{C_{max}} = 0.2$

$NTU = \frac{U A_s}{DC_{min}} = \frac{0.05 \times 1}{0.04} 1.25$

effectiveness is given as

$\epsilon = 1 -e^{\frac{NTU^0.22}{c} [ exp (-cNTU^{0.78}) -1]}$

$\epsilon = 1 -e^{\frac{1.25^0.22}{0.2} [ exp (-0.2\times 1.25^{0.78}) -1]}$

$\epsilon = 0.67122$

we knwo that actual heat Q is given

$Q = \epsilon \times Q_{max}$

$Q_H = \epsilon \times Q_{max}$

$Q_H = C_h (T_{h, in} -T_{h, out})$

$Q_H = \epsilon C_{min} (T_{h, in} - T_{c, in})$

$T_{h, out} = T_{h, in} - \epsilon (T_{h, in} - T_{c, in})$

$T_{h, out} = 75 - 0.67122(72 - 30)$

$T_{h, out} = 44.795\ degree\ C$