Question

Recall that the maximum heat transfer possible is the product of the minimum heat capacity (mass flow rate times specific heat) times the difference between the hot and cold side inlet temperatures. Address the following questions related to heat exchanger effectiveness. a. Sketch the temperature profiles for a 50% and 80% effectiveness heat exchanger. (As with Problem 1, make sure you have the correct shape, which, including degree of steepness at each end). b. What is the effectiveness for a heat exchanger with the following characteristics: the hot side is water entering at 160 F and 2lb/s, and the cold side is air entering at 20 F and 4 lb/s. The HX has been designed to the correct shape, which, including degree of steepness at each end). b. achieve an air outlet of 110 F.

Answer 1

The effectiveness for a heat exchanger is 0.6428.

Step-by-step explanation:

a) Sketch the temperature profiles for a 50% and 80% effectiveness heat exchanger.

The heat transfer type was not mentioned.

Assume the heat exchanger which is counter flow and the heat exchanger performance is high.

In the cold fluid stream, the 80% of effective heat exchanger compares the 50% of heat exchanger.

The diagram was attached below.

b) The effectiveness for a heat exchanger is calculated.

Given data

`m_(h)` = 2 lb/s

`T_(hi)` = 160 F

`m_(c)` = 4 lb/s

`T_(ae)` = 110 F

`T_(ai)` = 20 F

ε =
`(Actual heat transfer)/(Maximum possible heat transfer)`

From tables,

`(C_(p) )water` = 1.001 Btu / lb ° F

`C_(water) = C_(p) water`×
`m_(h)`

= 1.001 × 2

= 2.002

`(C_(P)) air` = 0.24 Btu / lb ° F

`C_(air)` =
`(C_(P)) air` ×
`m_(c)`

= 0.24 × 4

= 0.96

ε =
`( C_(air) )/(C_(air) )`
`((T_(ae) - T_(ai) ) )/((T_(hi) - T_(ai) ))`

=
`((110-20))/((160-20))`

ε = 0.6428