Dear Sir,
Cp of water= 4.187 kJ/kg-degC.
Flow of chilled water can be taken around 10 m3/h.
Area of both is to be calculated.
Pressure drop allowable is 70 kPa.
This is the only information that I have.
CALCULATION:
Using heat balance of m(cp)dT, Cp of the hydrocarbon viscous fluid comes out to be 2.524 kJ/kg-degC. LMTD value is 53.1 degC.
NTU based on hot fluid is 1.789, based on cold fluid is 0.433. Using curves of pressure drop vs. h as a function of viscosity, we get:
h(hot) = 350 W/m2-K & h(cold) = 13500 W/m2-K.
Stainless steel plates SS316 has conductivity= 14.9 W/m-K and assumed thickness of the plates is 0.5 mm. Using these, U= 337.294 W/m2-K. Dirt factor of 0.0002 (FPS units) gives Ud= 333.334 W/m2-K.
Therefore, Area= Q/U(LMTD)F, F= 0.96
Area = 7.85 m2.
What should I do next? Don't feel it's the full solution.