Question

Derive the LMTD equation for counter flow heat exchanger?

Answer 1

We will perform a heat balance for the hot fluid and the cold fluid.
Consider a small area dA with dQ.

dQ =U dA ?T =U dA (T_h – T_c) = -m_h c_ph dt_h = -m_c c_pc dt_c

dt_h= – dQ/m_h c_ph = –dQ/C_h

dt_c= – dQ/m_c c_pc = –dQ/C_c

where C_h = m_hc_ph = heat capacity of hot fluid

C_c = m_c c_pc = heat capacity of cold fluid

dt_h - dt_c = –dQ(1/C_h - 1/C_c)

d? = - dQ (1/C_h -1/C_c)

d? = - U dA (T_h - T_c) (1/C_h - 1/C_c)

d? = - U dA ? (1/C_h - 1/C_c)

On integrating this equation and applying limits,

ln(?₂/?₁) = -UA (1/C_h –1/C_c)

Q = C_h(T_h1 - T_h2)= C_c(T_c1 - T_c2 )

But 1/C_h =T_h1 –T_h2/ Q and 1/C_c =T_c2 –T_c1/ Q

ln(?₂/?₁) = - UA ((T_h1 –T_h2/ Q) - ( T_c2 –T_c1/ Q))

ln(?₂/?₁) = (UA/ Q) (( T_h2 -T_c2) - ( T_h1 -T_c1))

ln(?₂/?₁) = –(UA/Q) (?1-?2)

ln(?₂/?₁) = (UA/Q) (?₂ - ?₁)

Q = UA(?₂ - ?₁)/ ln(?₂/?₁)

Q = U A ?_m

?_m = LMTD = (?₂ - ?₁)/ ln(?₂/?₁)

= (?₁ - ?₂)/ ln(?₁/?₂)

= (?_max - ?_min)/ ln(?_max/?_min)