Question

Hot engine oil with a heat capacity rate of \(4440 \mathrm{~W} / \mathrm{K}\) (product of mass flow rate and specific heat) and an inlet temperature of \(150^{\circ} \mathrm{C}\) flows through a double-pipe heat exchanger. The double-pipe heat exchanger is constructed of a \(1.5\)-m-long copper pipe \((k=250 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with an inner tube of inside diameter \(2 \mathrm{~cm}\) and outside tube diameter of $2.25 \mathrm{~cm}$. The inner diameter of the outer tube of the double-pipe heat exchanger is \(6 \mathrm{~cm}\). Oil flowing at a rate of $2 \mathrm{~kg} / \mathrm{s}$ through the inner tube exits the heat exchanger at a temperature of \(50^{\circ} \mathrm{C}\). The cold fluid, i.e., water, enters the heat exchanger at \(20^{\circ} \mathrm{C}\) and exits at \(70^{\circ} \mathrm{C}\). Assuming the fouling factor on the oil side and water side to be $0.00015 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\( and \)0.0001 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}$, respectively, determine the overall heat transfer coefficient on the inner and outer surfaces of the copper tube.

Short answer

#tag_title#Step 5: Calculate the overall heat transfer coefficient (U)#tag_content#We will now calculate the resistances: \(Resistance_{oilside} = 1.273e-6\mathrm{~m^2} / \mathrm{K} \mathrm{W}\) \(Resistance_{copper} = 6.675e-6\mathrm{~m^2} / \mathrm{K} \mathrm{W}\) \(Resistance_{waterside} = 8.529e-7\mathrm{~m^2} / \mathrm{K} \mathrm{W}\) Adding up the resistances: Total resistance = \(1.273e-6 + 6.675e-6 + 8.529e-7\) Total resistance = \(8.600e-6\mathrm{~m^2} / \mathrm{K} \mathrm{W}\) Now, we will take the reciprocal of the total resistance to find the overall heat transfer coefficient, U: \(U = \dfrac{1}{8.600e-6}\) \(U = 116280 \mathrm{~W} / \mathrm{m^2} \mathrm{K}\) The overall heat transfer coefficient on the inner and outer surfaces of the copper tube is approximately 116280 W/m²K.

Step-by-step solution

Calculate the heat transfer rate (q) of the fluids

We will use the following equation for heat transfer rate: q = mcΔT where m is the mass flow rate, c is the specific heat capacity, and ΔT is the temperature difference. For the oil: \(q_{oil} = (4440 \mathrm{~W} / \mathrm{K})(150^{\circ} \mathrm{C} - 50^{\circ} \mathrm{C})\) \(q_{oil} = 4440 \times 100 = 444000 \mathrm{W}\) For the water: mc = q/ΔT ΔT = 70 - 20 = 50°C We know the mass flow rate is 2 kg/s, so we can write: \(q_{water} = (mc)(50^{\circ} \mathrm{C})\) Since the heat transferred from the oil to the water is the same, we can set these equal to each other: \((mc)(50^{\circ} \mathrm{C}) = 444000 \mathrm{W}\) Now we can solve for the heat capacity rate of the water.

Calculate the heat capacity rate (mc) for the water

We can use the heat transfer rate equation to determine the heat capacity rate: (mc)(50^{\circ} \mathrm{C}) = 444000 \mathrm{W} mc = 444000 / 50 mc = 8880 \mathrm{~W} / \mathrm{K}$

Determine the resistance for the heat transfer process

We will consider three resistances for the heat transfer process: the oil side fouling factor, the copper pipe, and the water side fouling factor. Resistance_oil side = \(r_f oil / (2 \pi L k_f oil)\) Resistance_copper = \(ln(D_o / D_i) / (2 \pi L k_copper)\) Resistance_water side = \(r_f water / (2 \pi L k_f water)\) where \(r_f\) is the fouling factor, L is the length of the pipe, k is the thermal conductivity, and \(D_i\) and \(D_o\) represent the inner and outer diameters respectively.

Calculate the overall heat transfer coefficient

Now, we will use the total resistance for the heat transfer process to find the overall heat transfer coefficient (U): \(1 / U = Resistance_{oilside} + Resistance_{copper} + Resistance_{waterside}\) We can substitute the resistance expressions and solve for U. Using the given values, we can calculate: \(Resistance_{oilside} = (0.00015) / (2 \pi (1.5)(250))\) \(Resistance_{copper} = ln(0.045 / 0.02) / (2 \pi (1.5)(250))\) \(Resistance_{waterside} = (0.0001) / (2 \pi (1.5)(250))\) Adding these resistances and taking their reciprocal will give us the overall heat transfer coefficient, U.