An oil having a specific heat of cph=2613 Jkg-K and flowing at =24 kgs is to be cooled from Thi=150°C to Tho=75°C with chilled water in a parallel flow concentric tube heat exchanger The chilled wate

To determine the diameter of the tube through which the oil flows, we need to use the equation for the overall heat transfer coefficient:

1/U = 1/hiA + ln(do/di)/2πk + 1/hoA

Where: U = overall heat transfer coefficient (W/m^2-K) hi = convective heat transfer coefficient on the inside of the tube (W/m^2-K) ho = convective heat transfer coefficient on the outside of the tube (W/m^2-K) A = surface area of the thin-walled tube (m^2) do = outer diameter of the tube (m) di = inner diameter of the tube (m) k = thermal conductivity of the tube material (W/m-K)

We are given that U = 750 W/m^2-K and A = surface area of the thin-walled tube. We need to solve for do and di.

We can assume that the oil is treated as a single-phase fluid, so the convective heat transfer coefficient on the inside of the tube is given by:

hi = Nu * k / di

Where: Nu = Nusselt number k = thermal conductivity of the oil (W/m-K)

The Nusselt number can be determined using the correlation for parallel flow in a concentric tube heat exchanger:

Nu = 0.023 * Re^0.8 * Pr^0.4

Where: Re = Reynolds number Pr = Prandtl number

The Reynolds number is given by:

Re = ρ * V * di / μ

Where: ρ = density of the oil (kg/m^3) V = velocity of the oil (m/s) μ = viscosity of the oil (kg/m-s)

The Prandtl number is given by:

Pr = cp,h * μ / k

We are given that V = 2.4 kg/s / (π/4 * di^2) = 9.65 / di^2 m/s.

Now we can substitute the equations for hi, Re, and Pr into the equation for Nu:

Nu = 0.023 * (ρ * V * di / μ)^0.8 * (cp,h * μ / k)^0.4

Similarly, we can assume that the chilled water is treated as a single-phase fluid, so the convective heat transfer coefficient on the outside of the tube is given by:

ho = Nu' * k / do

Where: Nu' = Nusselt number on the outside of the tube

The Nusselt number on the outside of the tube can be determined using the correlation for forced convection over a flat plate:

Nu' = 0.664 * Re^0.5 * Pr^0.33

The Reynolds number on the outside of the tube is given by:

Re' = ρ * V' * do / μ'

Where: V' = velocity of the chilled water (m/s) μ' = viscosity of the chilled water (kg/m-s)

We are given that V' = 3.0 kg/s / (π/4 * do^2) = 38.20 / do^2 m/s.

Now we can substitute the equations for ho, Re', and Pr into the equation for Nu':

Nu' = 0.664 * (ρ * V' * do / μ')^0.5 * (cp,c * μ' / k)^0.33

We can now substitute the equations for hi, ho, Nu, and Nu' into the equation for the overall heat transfer coefficient:

1/U = 1/(0.023 * (ρ * V * di / μ)^0.8 * (cp,h * μ / k)^0.4 * π * di * L) + ln(do/di)/(2πk) + 1/(0.664 * (ρ * V' * do / μ')^0.5 * (cp,c * μ' / k)^0.33 * π * do * L)

Simplifying and solving for do, we get:

do = exp((1/(2πk) - 1/U - ln(di)/(2πk)) / (1/(0.664 * (ρ * V' * do / μ')^0.5 * (cp,c * μ' / k)^0.33 * π * L)))

Plugging in the given values, we can solve for do