Answer :
Answer:
47.4889 m^2
Explanation:
Calculate the total heat transfer as
[tex]q= \.{m}c_p(T_2-T_1)[/tex]
= 3×4184(80-30)= 627600 J/s
Now calculating logarithmic mean temp. difference
[tex]LMTD= \frac{\theta_1-\theta_2}{\ln\frac{\theta_2}{\theta_1} }[/tex]
[tex]LMTD= \frac{(225-80)-(100-30)}{\ln\frac{145}{70} }[/tex]
soloving we get
LMTD= 103°C
To calculate the effectiveness we use
ε= [tex]\frac{225-100}{225-30}[/tex]
=0.641
Now we know that
Q= εAsU LMTD
627600 = 0.641×A_s×200×103
[tex]A_s= \frac{Q}{\epsilon U LMTD}[/tex]
putting values and calculating we get
As= 47.4889 m^2
The required surface area is
= 47.4889 m^2
Answer:
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