# LMTD – PARALLEL & COUNTER FLOW HEAT EXCHANGER CLASS NOTES

**LMTD – PARALLEL & COUNTER**

** FLOW HEAT EXCHANGER**

** CLASS NOTES **

### In a heat exchanger, temperature

### difference is variable all the length of the

### heat exchanger. There is a difficulty in

### handling variable temperature

### difference. Efforts are made to find the

### mean temperature difference. In case of

### heat exchanger the mean temperature

### difference comes in the logarithmic form. Hence

### it is called logarithmic mean temperature

### difference. Thus, LMTD is an equivalent

### constant temperature difference in place

### of variable temperature difference in

### actual. It will give constant average rate

### of heat transfer.

### DERIVATION LMTD COUNTER FLOW HEAT EXCHANGER

##### Fig. Counter Flow Heat Exchanger

### Assumptions used

#### 1. U is constant all along the HEX.

#### 2. Steady flow conditions.

#### 3. Specific heats of hot and cold fluids are constant

#### 4. Flow rates of hot and cold fluids are constant.

#### 5.There is no heat loss to the surroundings. There is no heat gain from the surroundings.

#### 6.There is no phase change of the either fluid.

#### 7.Conduction in the tube and shell are negligible.

#### 8. Kinetic as well potential energies of fluids are constant.

#### Fig. Temp Distribution for a Counter Flow HEX

Consider a small area dA with dQ^{.} heat transfer and temperature differences on hot is dT_{h} and cold side is dT_{c} as shown in Fig.

#### dQ^{.} =U dA ΔT =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^{.}_{h }c_{ph} = heat capacity of hot fluid

#### C_{c} = m^{.}_{c }c_{pc} = heat capacity of cold fluid

#### dt_{h} — dt_{c} =–d Q^{.}(1/C_{h} — 1/C_{c})

#### dθ = –d Q^{.} (1/C_{h} — 1/C_{c})

#### dθ = — U dA (T_{h} – T_{c}) (1/C_{h} — 1/C_{c})

#### dθ = — U dA θ (1/C_{h} — 1/C_{c})

#### = –(1/C_{h} + 1/C_{c})

#### ln( θ_{2}/θ_{1}) = — U A (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( θ_{2}/θ_{1}) = — U A ( (T_{h1} –T_{h2}/ Q^{.}) — ( T_{c2} –T_{c1}/ Q^{.}) )

#### ln( θ_{2}/θ_{1}) = (U A/ Q^{.} ) ( ( T_{h2} –T_{c2}) — ( T_{h1} –T_{c1}) )

#### ln( θ_{2}/θ_{1}) = –(U A/ Q^{.} ) (θ_{1}—θ_{2})

#### ln( θ_{2}/θ_{1}) = (U A/ Q^{.} ) (θ_{2}— θ_{1})

#### Q^{.} = U A (θ_{2}— θ_{1})/ ln( θ_{2}/θ_{1})

#### Q^{.} = U A θ_{m}

**θ**_{m} = LMTD = (θ_{2}— θ_{1})/ ln( θ_{2}/θ_{1})

_{m}= LMTD = (θ

_{2}— θ

_{1})/ ln( θ

_{2}/θ

_{1})

** LMTD = (θ**_{1}—θ_{2})/ ln( θ_{1}/θ_{2})

_{1}—θ

_{2})/ ln( θ

_{1}/θ

_{2})

** LMTD = (θ**_{max}—θ_{min})/ ln( θ_{max}/θ_{min})

_{max}—θ

_{min})/ ln( θ

_{max}/θ

_{min})

**If θ**_{1}=θ_{2}=θ, then LMTD = θ

_{1}=θ

_{2}=θ, then LMTD = θ

**Then Q=UAθ**

**DERIVATION LMTD PARALLEL FLOW HEAT EXCHANGER**

##### Fig. Parallel Flow Heat Exchanger

**ASSUMPTIONS USED IN DERIVATION **

#### 1. U is constant all along the HEX.

#### 2. Steady flow conditions.

#### 3. Specific heats of hot and cold fluids are constant

#### 4. Flow rates of hot and cold fluids are constant.

#### 5.There is no heat loss to the surroundings. There is no heat gain from the surroundings.

#### 6.There is no phase change of the either fluid.

#### 7.Conduction in the tube and shell is negligible.

#### 8. Kinetic as well potential energies of fluids are constant.

#### PROOF OF LMTD FOR PARALLEL FLOW HEAT EX-CHANGER

#### Fig. Temp variation in a parallel flow heat exchanger

#### Consider a small area dA with dQ^{.} heat transfer rate. Temperature differences on hot fluid is dT_{h} and cold fluid is dT_{c} in length dx

#### dQ^{.} =U dA ΔT =-m^{.}_{h }c_{ph }dt_{h} = m^{.}_{c }c_{pc }dt_{c} =U dA (T_{h} – T_{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^{.}_{h }c_{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})

#### dθ = — dQ^{.} (1/C_{h} + 1/C_{c})

#### dθ = — U dA (T_{h} – T_{c}) (1/C_{h} + 1/C_{c})

#### dθ = — U dA θ (1/C_{h} + 1/C_{c})

#### dθ/θ = –UdA(1/C_{h} + 1/C_{c})

#### ln( θ_{2}/θ_{1}) = — U A (1/C_{h} + 1/C_{c})

#### But 1/C_{h} =T_{h1} –T_{h2}/ Q^{.} and 1/C_{c} =T_{c1} –T_{c2}/ Q^{.}

#### ln( θ_{2}/θ_{1}) = — U A ( T_{h1} –T_{h2}/ Q^{.} + T_{c1} –T_{c2}/ Q^{.} )

#### ln( θ_{2}/θ_{1}) = (U A/ Q^{.} ) (( T_{h2} –T_{c2}) — ( T_{h1} –T_{c1}) )

#### ln( θ_{2}/θ_{1}) = (U A/ Q^{.} ) (θ_{2}— θ_{1})

#### Q^{.} = U A (θ_{2}— θ_{1})/ ln( θ_{2}/θ_{1})

#### Q^{.} = U A θ_{m}

**θ**_{m} = LMTD = (θ_{2}— θ_{1})/ ln( θ_{2}/θ_{1})

_{m}= LMTD = (θ

_{2}— θ

_{1})/ ln( θ

_{2}/θ

_{1})

** ****LMTD = (θ**_{max}—θ_{min})/ ln( θ_{max}/θ_{min})

**LMTD = (θ**

_{max}—θ_{min})/ ln( θ_{max}/θ_{min})** **

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