Assuming the outer surface of the heat exchanger to be well insulated so that any heat transfer occurs between the two fluids, and disregarding any changes in kinetic and potential energy, an energy balance on each fluid in a differential section of the heat exchanger can be expressed as

Fig: Variation of the fluid temperatures in a parallel-flow double-pipe heat exchanger.

δQ=-$m_h$ $C_ph$ $dT_h$………(1)

And

δQ=$m_c$ $C_pc$ $dT_c$………(2)

That is, the rate of heat loss from the hot fluid at any section of a heat exchanger is equal to the rate of heat gain by the cold fluid in that section. The temperature change of the hot fluid is a negative quantity, and so a negativesign is added to

Eq. 1 to make the heat transfer rate Q a positive quantity.

Solving the equations above for dThand dTcgives

$dT_h$=-$\frac{δQ}{(m_h C_ph )}$………(3)

And

$dT_c$=$\frac{δQ}{(m_c C_pc )}$………(4)

Taking their difference we get

$d(T_h-T_c )$=-$δQ\frac{1}{(m_h C_ph )}+\frac{1}{(m_c C_pc )}$………(5)

The rate of heat transfer in the differential section of the heat exchanger can also be expressed as

δQ=$U(T_h-T_c )dA_s$

Substituting this equation into eq 5 and rearranging gives

Integrating from the inlet of the heat exchanger to its outlet, we obtain

ln$\frac{T_h,out -T_c,out}{T_h,in - T_c,in}=UA_s\frac{1}{m_hC_ph}+\frac{1}{m_cC_pc}$………(6)

From first law of thermodynamicsrequires that the rate of heat transfer from the hot fluid be equal to the rate of heat transfer to the cold one. That is,

Q=$m_c C_pc (T_(c,out)-T_(c,in) )$

And

Q=$m_h C_ph (T_(h,in)-T_(h,out) )$

Taking values of mCp from above equation and substituting in the integrating solution and solving it

Q=$UA_s ∆T_lm$

Where $∆T_lm$=$\frac{(∆T_1-∆T_2)}{ln⁡\frac{(∆T_1)}{(∆T_2 )}}$

Is the log mean temperature difference, which is the suitable form of the average temperature difference for use in the analysis of heat exchangers.