Lubricating oil is cooled in a tubular heat exchanger to maintain its viscosity and effectiveness in the journal bearings used in a large steam turbine of an electric power plant. Oil flows at the rate of 0.1 kg/s inside a 12.5-mm-diameter circular tube, which is maintained at a uniform surface temperature of 25°C, and the oil is cooled from 80°C to 40°C. Calculate the length of tube required for this heat exchanger, the total heat flux dissipated from its surface in the cooling process, and the pressure drop across the tube length.

Question

GIVEN

• Lubricating oil cooled in tubular heat exchanger.

• Oil flow rate, m = 0.1kg/s.

• Oil inlet temperature, Tb,in = 80°C.

• Oil outlet temperature Tb,out=400C

• Inner or inside diameter of pipe in which oil flows, D = 12.5 mm = 0.0125 m.

• Temperature of pipe surfaceTs = 25°C.

FIND

• Length of tube required for heat exchanger

• Total heat flux dissipated from its surface in cooling process.

• Pressure drop across the tube length.

ASSUMPTIONS

• The temperature of wall is constant and uniform across the length of pipe.

• The thermal resistance of the pipe is negligible, and hence the inside surface temperature of the pipe is Tw = Ts, this represents a uniform pipe surface temperature condition.

SKETCH



PROPERTIES AND CONSTANTS

for unused engine oil at Tb= 60°C

Density, ? = 864.0 kg/m3

Thermal conductivity, k = 0.140 W/(m K)

Absolute viscosity, ?b = 72.5 × 10–3 (Ns)/m2

Prandtl number, Pr = 1050 Specific heat, cp = 2047 J/(kg K)

At the pipe surface temperature of 25°C, the absolute viscosity ?s = 652 × 10–3 (Ns)/m2

Answer 1

The Reynolds number for oil flow inside the pipe is



Total heat dissipated from the tube is given by



Also we have



Calculating thermal entrance length we have

For, laminar flow we have entrance length given as



using we have





Substituting this value of L in equation (II) above we get





Now, solving the problem iteratively further we get to



On further iterating we reach the final solution of L=140.2 m which is the required length of the pipe.

Total heat flux dissipated is given by q a= 8188 W

The pressure drop in pipe flow is given by



COMMENTS

We can see that the length required for the pipe is extremely large. This is due to the high temperature

difference and high mass flow rate. In practical cases, the flowrate is distributed among the bundles of

tubes arranged in rows and columns in parallel. As the thermal entrance length is quite high, the high

heat transfer rate can be achieved by doing so due to high heat transfer coefficient.