Question

The oil (cp = 1759 W-s/kg K ) from an oil-cooled electric transformer is cooled from 79.4oC to 29.4oC at the rate of 1,360.5kg/hr. This is done in an oil-water heat exchanger that receives 2,948kg/hr of water at 15.6oC. For the exchanger, U= 295 W/m2K. Find the exit water temperature and heating area required a) for counter flow b) for parallel flow

Answer 1

The question requires an application of thermodynamics for calculating the heat transfer rates and exit temperatures of a fluid in a heat exchanger, as well as determining the necessary heating area for efficient operation.

Step-by-step explanation:

The student's question pertains to the field of thermodynamics, specifically the calculation of heat transfer rates and exit temperatures in a heat exchanger system, with considerations of both counter flow and parallel flow.

Calculating heat exchange efficiency and exit temperatures involves the principles of energy conservation and thermodynamic properties such as specific heat (cp).

The area required for a certain heat transfer rate is found using the heat transfer coefficient (U) and the temperature difference between the fluids.

While the question does not provide all necessary information to calculate the exact answers, an analysis would typically use formulas

such as Q = m ⋅ cp ⋅ ΔT (heat transfer) and A = Q/(U ⋅ ΔT_ln) (area for heat exchange),

where ΔT_ln is the logarithmic mean temperature difference.

Knowledge of heat exchanger design and operation is crucial for such calculations.