Mechanical Engineering Department GOVERNMENT ENGINEERING COLLEGE, DAHOD. A PPT On Derivation of LMTD for Parallel flow Heat Exchanger Prepared By: PATEL JIGARKUMAR NATVARBHAI PATEL NISHANT YOGESHBHAI TANDEL KANAIYA KISHORBHAI TANDEL TEJASH AMRUTBHAI BE Sem-5 Heat Transfer ( )

Derivation of LMTD for Parallel Flow Figure below shows the flow arrangement and temperature distribution in a simple pass parallel flow heat exchanger.

ASSUMPTIONS: The overall heat transfer coefficient is uniform throughout the exchanger. The potential and kinetic energy changes are negligible. The specific heat of the fluids are constant. The heat exchange takes place between the two fluids. The temperature of both the fluids are constant over a given cross section and may be represented by their bulk temperature.

Let us consider an elementary dA of the heat exchanger. The rate of flow of heat through this elementary area is given by, In this case due to heat transfer dQ through the area dA, the hot fluid is cooled down by dT h whereas the cold fluid is heated by dT c. The energy balance over a differential area dA may be written as, dQ = - m h. C ph. dT h = - m c. C pc. dT c dQ = U. dA. ΔT = U. dA (T h - T c )

In a parallel flow system, the temperatures of one of the fluid decrease and one of the fluid increase in the direction of heat exchanger length, hence the –ve sign for decrease in temperature and +ve sign for increase in temperature. therefore ……..(1) and ……..(2) On subtracting, ………(3) dT h = - = - dT c = = d (ΔT) = dT h - dT c

OR, ………..(4) Inserting the value of dQ from equation (1), we get d(ΔT) = - U dA (ΔT) d(ΔT) = - dQ = - U dA

Integrating along the heat exchanger length between section 1 and 2 ………(5) Now, the total heat transfer rate between the two fluids is given by, ……….(6) therefore ………..(7) and, ………..(8) = - U ln = - U A Q = C h (T h1 –T h2 ) = C c (T c2 –T c1 )

Substituting the values of and into equation (4),we get

Since therefore, Since contains log term, it is called Logarithmic Mean Temperature Difference(L.M.T.D).

Example on the basis of Parallel Flow Heat Exchanger A parallel flow heat exchanger has its tubes of 5cm internal and 6cm external diameter. The air flows inside the tubes and receives heat from hot gases circulated in the annular space of the tube at the rate of 100kW.Inside and outside heat transfer coefficients are 250W/ metre sq K and 400 W/ metre sq K respectively. Inlet temperature of hot gases is 500 degree, outlet temperature of hot gases is 300 degree, inlet temperature of air is 50 degree. Exit temperature of air is 140 degree. Calculate: 1) Overall heat transfer coefficient based on outer surface area 2)Length of the tube required to affect the heat transfer rates. Neglect resistance of the tube. 3)If the length is 3m length find the number of tubes required.

Solution: Q=100kW,di=5 cm=0.05m, do=6 cm=0.06m,hi=250W/metre sq K, ho=400W/metre sq K 1)Overall heat transfer coefficient based on outer surface area,Uo:

2) Length of the tube, L required : 3)Number of tubes, N if each tube is l=3m long:

Reference :

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