Question

A shell-and-tube heat exchanger with two shell passes and four tube passes is used for cooling oil $\left(c_{p}=2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\( from \)125^{\circ} \mathrm{C}\( to \)55^{\circ} \mathrm{C}$. The coolant is water, which enters the shell side at \(25^{\circ} \mathrm{C}\) and leaves at \(46^{\circ} \mathrm{C}\). The overall heat transfer coefficient is \(900 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). For an oil flow rate of \(10 \mathrm{~kg} / \mathrm{s}\), calculate the cooling water flow rate and the heat transfer area.

Short answer

$$q = 140,000 \mathrm{~W}$$ #tag_title# Step 2: Calculate the cooling water flow rate#tag_content# To calculate the cooling water flow rate, we can use the energy balance equation for the water side (cold side), assuming the specific heat capacity of water, \(c_{p\_water} = 4.18 \mathrm{~kJ/kg \cdot K}\): $$q = m_{cold}c_{p\_water}\left(T_{cold\_outlet} - T_{cold\_inlet}\right)$$ Solve for \(m_{cold}\) (cooling water flow rate): $$m_{cold} = \frac{q}{c_{p\_water}(T_{cold\_outlet} - T_{cold\_inlet})}$$ Plug in the given values and calculate the cooling water flow rate: $$m_{cold} = \frac{140,000 \mathrm{~W}}{4.18 \times 10^3 \mathrm{~J/kg \cdot K}(46 - 25) \mathrm{~K}}$$ $$m_{cold} = 6.35 \mathrm{~kg/s}$$ #tag_title# Step 3: Calculate the heat transfer area#tag_content# The heat transfer area can be calculated using the heat transfer equation: $$q = UA\Delta T_{lm}$$ where \(\Delta T_{lm}\) is the logarithmic mean temperature difference, and A is the heat transfer area. First, we need to calculate the logarithmic mean temperature difference: $$\Delta T_{lm} = \frac{\left(T_{hot\_inlet} - T_{cold\_outlet}\right) - \left(T_{hot\_outlet} - T_{cold\_inlet}\right)}{\ln\left(\frac{T_{hot\_inlet} - T_{cold\_outlet}}{T_{hot\_outlet} - T_{cold\_inlet}}\right)}$$ Plug in the given values and calculate \(\Delta T_{lm}\): $$\Delta T_{lm} = \frac{(125 - 46) - (55 - 25)}{\ln\left(\frac{125 - 46}{55 - 25}\right)}$$ $$\Delta T_{lm} ≈ 47.24 ^{\circ} \mathrm{C}$$ Next, calculate the heat transfer area, A, by rearranging the heat transfer equation: $$A = \frac{q}{U\Delta T_{lm}}$$ Plug in the given values and calculate the heat transfer area: $$A = \frac{140,000 \mathrm{ ~W}}{900 \mathrm{~W/m^2 \cdot K} * 47.24 \mathrm{~K}}$$ $$A ≈ 3.29 \mathrm{~m^2}$$ #Answer# The cooling water flow rate is 6.35 kg/s and the heat transfer area of the shell-and-tube heat exchanger is 3.29 m².

Step-by-step solution

Calculate the heat transfer rate

To calculate the heat transfer rate, we can use the energy balance equation for the oil side (hot side): $$q = m_{hot}c_{p}\left(T_{hot\_inlet} - T_{hot\_outlet}\right)$$ where \(q\) is the heat transfer rate. Plug in the given values and calculate the heat transfer rate: $$q = 10 \mathrm{~kg/s} * 2.0 \mathrm{~kJ/kg \cdot K} * (125 - 55) \mathrm{~K} = 10^3* 2.0 * 70 \mathrm{~W}$$