Question

Consider a crossflow engine oil heater that uses ethylene glycol flowing at a temperature of \(110^{\circ} \mathrm{C}\) to heat the oil initially at \(10^{\circ} \mathrm{C}\). The ethylene glycol enters a tube bank consisting of copper tubes \((k=250 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with staggered arrangement in a \(0.5-\mathrm{m} \times 0.5-\mathrm{m}\) plenum. The outside diameter of the \(0.5\)-m-long copper tubes is \(25 \mathrm{~mm}\), and the wall thickness is \(2 \mathrm{~mm}\). The longitudinal and transverse pitch of the rod bundles is \(0.035 \mathrm{~m}\) each. The engine oil to be heated flows inside the tubes with a mass flow rate of $4.05 \mathrm{~kg} / \mathrm{s}\(. Take the heat transfer coefficient of the oil to be \)2500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. If the minimum desired exit temperature of oil is \(70^{\circ} \mathrm{C}\) and the measured exit temperature of ethylene glycol is \(90^{\circ} \mathrm{C}\), determine (a) the mass flow rate of ethylene glycol and \((b)\) the number of tube rows. In your calculation, use the following properties for the ethylene glycol.

Short answer

(b) How many tube rows are there in the heater?

Step-by-step solution

Apply energy balance to oil and ethylene glycol

Since the mass flow rate of oil and the temperature difference of oil are given, we can calculate the energy balance of the oil: \(\dot{Q}=\dot{m}_{o} C_{p, o} \left(T_{o, exit} - T_{o, init}\right)\) Where, \(\dot{Q}\) = heat transfer rate (\(W\)) \(\dot{m}_{o}\) = mass flow rate of oil (\(4.05 \mathrm{~kg/s}\)) \(C_{p, o}\) = specific heat of oil \((2300 \mathrm{~J/kg \cdot K})\) \(T_{o, exit} = 70^{\circ} \mathrm{C}\) \(T_{o, init} = 10^{\circ} \mathrm{C}\) Calculate the heat transfer rate: \(\dot{Q} = 4.05 \times 2300 \times (70 - 10)\)

Apply energy balance to ethylene glycol

The energy balance of ethylene glycol can be calculated using the formula: \(\dot{Q} = \dot{m}_{e} C_{p, e} \left(T_{e, init} - T_{e, exit}\right)\) Where, \(\dot{m}_{e}\) = mass flow rate of ethylene glycol (\(kg/s\)) \(C_{p, e}\) = specific heat of ethylene glycol (\(2520 \mathrm{~J/kg \cdot K}\)) \(T_{e, init} = 110^{\circ} \mathrm{C}\) \(T_{e, exit} = 90^{\circ} \mathrm{C}\) Rearrange the formula to find the mass flow rate of ethylene glycol: \(\dot{m}_{e} = \frac{\dot{Q}}{C_{p, e}(T_{e, init} - T_{e, exit})}\) Calculate the mass flow rate of ethylene glycol using the heat transfer rate we found in step 1: \(\dot{m}_{e} = \frac{\dot{Q}}{2520 \times (110 - 90)}\)

Calculate heat transfer rate on the outside of the tubes

Now that the mass flow rate of ethylene glycol is known, we can calculate the heat transfer rate on the outside of the tubes using the heat transfer coefficient: \(q'' = h_o (T_{e, init} - T_{o, exit})\) Where, \(q''\) = heat transfer rate per unit area (\(\mathrm{W/m^2}\)) \(h_o\) = heat transfer coefficient of oil (\(2500 \mathrm{~W/m^2 \cdot K}\)) Calculate the heat transfer rate on the outside of the tubes: \(q'' = 2500 \times (110 - 70)\)

Calculate the overall heat transfer coefficient

We can calculate the overall heat transfer coefficient, \(U\), using the heat transfer rate per unit area and the heat transfer rates between the fluids: \(U = \frac{q''}{\dot{Q}}\) Calculate the overall heat transfer coefficient: \(U = \frac{q''}{\dot{Q}}\)

Calculate the number of tube rows

Finally, we can calculate the number of tube rows in the system by using the overall heat transfer coefficient and the dimensions of the tube bank: \(n = \frac{U A_{total}}{\dot{Q}}\) Where, \(n\) = number of tube rows \(A_{total}\) = total surface area of the tube bank (\(0.5 \times 0.5\) \(m^2\)) Calculate the number of tube rows: \(n = \frac{U \times (0.5 \times 0.5)}{\dot{Q}}\) In conclusion, we have determined the mass flow rate of ethylene glycol and the number of tube rows required for the given crossflow engine oil heater.