A heat exchanger is to be designed to heat water flowing at a rate of 1720 kg/hr

Question

A heat exchanger is to be designed to heat water flowing at a rate of 1720 kg/hr from 20 c to 45 c using steam condensing at 110 c on the outside surface of 4 meter long brass tubes (thermal conductivity = 111.65 w/m.k, outside diameter = 25 mm, inside diameter = 22.5 mm)

A heat exchanger is to be designed to heat water flowing at a rate of 1720 kg/hr from 20°C to 45 °C using steam condensing at 110 °C on the outside surface of 4-meter long brass tubes (thermal conductivity 11 1.65 W/m·K, outside diameter 25 mm, inside diameter= 22.5 mm). The velocity of water is 61.2 m/min and the weight of steam condensed is 1.25 kg/s. Find the number of brass tubes required to heat water from 20 °C to 45 °C 3) Density of water = 9957 kg/m3 Thermal conductivity of water = 0.617 W/mK Kinematic viscosity of water = 6.59 x 10.7 m 2/s Specific heat capacity of water = 4.174 kJ/kg. K Latent heat of vaporization of water 2230 kJ/kg Steam side film heat transfer coefficient 4650 W/m2K DATA:

Explanation / Answer

Rate of enrgy given by steam condensing= massflow rate x Latent heat = 1.25 kg/s x 2230 kJ/kg = 2787.5 kW

For water flowing inside,

Reynolds no. Re = velocity xDia/(kinematic viscosity) = 1.02 m/s x 22.5 x 10-3m/(6.59 x 10-7m2/s) = 34825.5

Prandtl no. Pr= Cp/k = 4.174 x 103 x 6.562 x 10-4/0.617 = 4.44

Since the flow is turbulent, we can use Dittus Boeler correlation for innertube.

Nusselt no. Nu=0.023 Re0.8Pr0.4 = 179.54

But, Nu= hd/k or hin = Nu.k/d = 179.54 x 0.617/(22.5 x 10-3) = 4923.35 W/m2K

Now, we have 1/UA =1/hinAin + ln(r2/r2)/(2kcul) + 1/houtAout

For unit length

Ain = (22.5 x 10-3 ) = 0.07068 m2

Similarly, for D=25 mm, A out = 0.07854 m2

Subsituting all values, we get UA= 173.55 W/K

LMTD = (110-45) - (110-20)/[ln((110-45)/(110-20))] = 76.823 C

Now, heat transfer from 1 metre length of tube = UA x LMTD = 173.55X 76.823 = 13332.67 W

Required rate = mass flow rate x heat capacity of water x temp inc = 0.4777 kg/s x 4184 J/kg x 25C = 49967.42 W

Thus length required = 49967.42/13332.67 = 3.74 m --- one rod is sufficient

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