Question

A thin-walled double-pipe counterflow heat exchanger is to be used to cool oil \(\left(c_{p}=2200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) from \(150^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) at a rate of $2.5 \mathrm{~kg} / \mathrm{s}\( with water \)\left(c_{p}=4180\right.\( \)\mathrm{J} / \mathrm{kg} \cdot \mathrm{K})\( that enters at \)22^{\circ} \mathrm{C}$ at a rate of \(1.5 \mathrm{~kg} / \mathrm{s}\). The diameter of the tube is $2.5 \mathrm{~cm}\(, and its length is \)6 \mathrm{~m}$. Determine the overall heat transfer coefficient of this heat exchanger.

Short answer

Answer: The overall heat transfer coefficient (U) of the given double-pipe counterflow heat exchanger is 1420 W/(m²·K).

Step-by-step solution

Find the mass flow rate of oil and water

The mass flow rate of oil (m_oil) is given as 2.5 kg/s, and the mass flow rate of water (m_water) is given as 1.5 kg/s.

Calculate the specific heat capacity of oil and water

The specific heat capacity of oil (cp_oil) is given as 2200 J/(kg·K), and the specific heat capacity of water (cp_water) is given as 4180 J/(kg·K).

Determine the temperature change for oil and water

The initial temperature of oil (T_initial_oil) is 150°C and the final temperature (T_final_oil) is 50°C. The temperature change for oil can be calculated as follows: ΔT_oil = T_initial_oil - T_final_oil = 150 - 50 = 100°C The initial temperature of water (T_initial_water) is 22°C. Since it's a counterflow heat exchanger, the final temperature of water (T_final_water) can be determined by knowing the heat load and using the formula: Q = m_water * cp_water * (T_final_water - T_initial_water) where Q is the heat load.

Determine the heat load (Q)

We know that the heat load gained by water is equal to the heat load lost by oil. Therefore, we can write the equation: m_oil * cp_oil * ΔT_oil = m_water * cp_water * (T_final_water - T_initial_water) By substituting the given values, we get the following equation: 2.5 * 2200 * 100 = 1.5 * 4180 * (T_final_water - 22) Solve the equation for T_final_water: T_final_water = 110°C Now we can calculate the heat load (Q) using the oil data: Q = m_oil * cp_oil * ΔT_oil = 2.5 * 2200 * 100 = 550000 J/s

Find the heat transfer area (A)

The length of the tube (L) is given as 6 m, and the diameter (D) is given as 2.5 cm (0.025 m). The heat transfer area (A) can be found using the formula: A = π * D * L By substituting the given values, we get: A = π * 0.025 * 6 = 0.471 m²

Calculate the overall heat transfer coefficient (U)

The overall heat transfer coefficient (U) can be found using the formula: U = Q / (A * ΔT_LM) ΔT_LM is the logarithmic mean temperature difference, which can be calculated as follows: ΔT_LM = ((ΔT_in - ΔT_out) / ln(ΔT_in / ΔT_out)) where ΔT_in is the temperature difference at the inlet, and ΔT_out is the temperature difference at the outlet. ΔT_in = T_initial_oil - T_initial_water = 150 - 22 = 128°C ΔT_out = T_final_oil - T_final_water = 50 - 110 = -60°C Now, we can find ΔT_LM: ΔT_LM = ((128 - (-60)) / ln(128 / (-60))) = 81°C Finally, we can find the overall heat transfer coefficient (U): U = Q / (A * ΔT_LM) = 550000 / (0.471 * 81) = 1420 W/(m²·K) The overall heat transfer coefficient of this heat exchanger is 1420 W/(m²·K).