Question

A single-pass, cross-flow heat exchanger uses hot exhaust gases (mixed) to heat water (unmixed) from 30 to 80C at a rate of 3 kg/s. The exhaust gases, having thermophysical properties similar to air, enter and exit the exchanger at 225 and 100C, respectively. If the overall heat transfer coefficient is 200 W/m2 K, estimate the required surface area.

Answer 1

Step-by-step explanation:

The specific heat of the given gas similar to air is

Q= m´Cp(t₂ - t₁)

m´ is the mass rate which is 3kg/s

and

Cp is the specific heat capacity of the air which 1 kJ/kg.

and

t₂= final temperature =80C⁰=273+80= 353 K

t₁= initial temperature= 50C⁰= 323K

(t₂ - t₁)= 30K

thus

Q= 3x1x30= 90J= 25kJ/h= 25 W

now the heat exchange calculations for the exchangers are given as

∆Tᵢ = ( T₁ - t₂) - (T₂ - t₁) /ln( ( T₁ - t₂) / (T₂ - t₁))

where T₁ is the inlet temperature= 100C⁰= 373K

T₂= is the outlet temperature= 225C⁰= 598K

now

∆Tᵢ = (373-353)-(598-323)/ln((373-353)/(598-323))

= - 255/ ln(-255)= -255/ln(20/275)

=97.28

now the surface area of the exchanger

A= Q/ U x ∆Tᵢ

where U = 200Wm²/K

A= 25/ 200x 97.28

A= 0.00128m²= 12. 8cm²

Answer 2

47.4889 m^2

Step-by-step explanation:

Calculate the total heat transfer as

`q= \.{m}c_p(T_2-T_1)`

= 3×4184(80-30)= 627600 J/s

Now calculating logarithmic mean temp. difference

`LMTD= (\theta_1-\theta_2)/(\ln(\theta_2)/(\theta_1) )`

`LMTD= ((225-80)-(100-30))/(\ln(145)/(70) )`

soloving we get

LMTD= 103°C

To calculate the effectiveness we use

ε=
`(225-100)/(225-30)`

=0.641

Now we know that

Q= εAsU LMTD

627600 = 0.641×A_s×200×103

`A_s= (Q)/(\epsilon U LMTD)`

putting values and calculating we get

As= 47.4889 m^2

The required surface area is

= 47.4889 m^2