Question

Inside a condenser, there is a bank of seven copper tubes with cooling water flowing in them. Steam condenses at a rate of \(0.6 \mathrm{~kg} / \mathrm{s}\) on the outer surfaces of the tubes that are at a constant temperature of \(68^{\circ} \mathrm{C}\). Each copper tube is \(5 \mathrm{~m}\) long and has an inner diameter of \(25 \mathrm{~mm}\). Cooling water enters each tube at \(5^{\circ} \mathrm{C}\) and exits at \(60^{\circ} \mathrm{C}\). Determine the average heat transfer coefficient of the cooling water flowing inside each tube and the cooling water mean velocity needed to achieve the indicated heat transfer rate in the condenser.

Short answer

Answer: The average heat transfer coefficient of the cooling water flowing inside each copper tube is approximately 1.916 kJ/s.m².K, and the cooling water mean velocity needed to achieve the indicated heat transfer rate in the condenser is approximately 0.936 m/s.

Step-by-step solution

Determine the mass flow rate and the specific heat of cooling water

We are given that the steam condenses at a rate of 0.6 kg/s on the outer surfaces of the tubes. The specific heat of water (cooling medium) can be assumed as a constant value cp = 4.18 kJ/kg.K.

Calculate the heat transfer rate from the steam to the cooling water

We can use the formula for heat transfer rate, Q: Q = mc ΔT where m is the mass flow rate of the cooling water (kg/s), c is the specific heat of the cooling water (kJ/kg.K), and ΔT is the difference in temperature between the inlet and outlet of the cooling water (K). Using the given values for the steam condensation rate (0.6 kg/s), specific heat (4.18 kJ/kg.K), and temperatures (5°C and 60°C), we can calculate Q: Q = (0.6 kg/s) * (4.18 kJ/kg.K) * (60°C - 5°C) = 138.06 kJ/s

Calculate the heat transfer surface area

The heat transfer surface area can be calculated using the formula: A = NπDL where N is the number of tubes, D is the inner diameter of the tubes (m), and L is the length of the tubes (m). We have N = 7, D = 25 mm (0.025 m), and L = 5 m: A = 7 * π * 0.025 m * 5 m ≈ 2.745 m²

Calculate the average heat transfer coefficient

The average heat transfer coefficient (h) can be calculated using the heat transfer equation: Q = hAΔTlm where ΔTlm is the logarithmic mean temperature difference (LMTD) for the condenser, which can be calculated using the equation: ΔTlm = (ΔTi - ΔTo) / ln(ΔTi / ΔTo) where ΔTi is the temperature difference at the cooling water inlet, and ΔTo is the temperature difference at the cooling water outlet. In our case, ΔTi = 68°C - 5°C = 63°C, and ΔTo = 68°C - 60°C = 8°C, so we can calculate ΔTlm: ΔTlm = (63°C - 8°C) / ln(63°C / 8°C) ≈ 26.32°C Now we can solve for h: h = Q / (AΔTlm) = 138.06 kJ/s / (2.745 m² * 26.32°C) ≈ 1.916 kJ/s.m².K Therefore, the average heat transfer coefficient of the cooling water flowing inside each tube is approximately 1.916 kJ/s.m².K.

Calculate the cooling water mean velocity

To calculate the cooling water mean velocity (v), we can use the formula: v = 4Q / (ρπD²N ΔTlc) where ρ is the density of cooling water (kg/m³) which can be considered as 1000 kg/m³ for water, ΔTlc is the linear mean temperature difference which can be assumed as (ΔTi + ΔTo) / 2 and Q is the heat transfer rate in m³/s (we will convert the units). First, we need to convert Q to m³/s: Q = 138.06 kJ/s * (1 m³ / 4.18e6 kJ) ≈ 3.3e-5 m³/s Now substituting the values, v = 4 * 3.3e-5 m³/s / (1000 kg/m³ * π * (0.025 m)² * 7 * ((63°C + 8°C) / 2)) ≈ 0.936 m/s The cooling water mean velocity needed to achieve the indicated heat transfer rate in the condenser is approximately 0.936 m/s.