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

A)
`\epsilon = 0.67122`

B)
`T_(h, out) = 44.795\ degree\  C`

Step-by-step explanation:

Heat capacity of oil
`C_h =m_h C_(ph) = 0.02 * 2 = 0.04` Kw/K

Heat capacity of air
`C_C = m_C C_(pc) = 0.2 * 1 = 0.2` Kw/K

therefore
`C_{min = C_h`

and
`C_(max) = C_C`

we know that capacity ratio is

`c = (C_(min))/(C_(max)) = 0.2`

`NTU = (U A_s)/(DC_(min)) = (0.05 * 1)/(0.04) 1.25`

effectiveness is given as

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

`\epsilon = 1 -e^{(1.25^0.22)/(0.2) [ exp (-0.2* 1.25^(0.78)) -1]}`

`\epsilon = 0.67122`

we knwo that actual heat Q is given

`Q = \epsilon * Q_(max)`

`Q_H  = \epsilon * 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`