Question

A double-pipe parallel-flow heat exchanger is used to heat cold tap water with hot water. Hot water $\left(c_{p}=4.25 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\( enters the tube at \)85^{\circ} \mathrm{C}$ at a rate of \(1.4 \mathrm{~kg} / \mathrm{s}\) and leaves at \(50^{\circ} \mathrm{C}\). The heat exchanger is not well insulated, and it is estimated that 3 percent of the heat given up by the hot fluid is lost from the heat exchanger. If the overall heat transfer coefficient and the surface area of the heat exchanger are \(1150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and $4 \mathrm{~m}^{2}$, respectively, determine the rate of heat transfer to the cold water and the log mean temperature difference for this heat exchanger.

Short answer

The rate of heat transfer to the cold water in this heat exchanger is 201,922.5 W. 2. What is the log mean temperature difference (LMTD) for this heat exchanger? The log mean temperature difference (LMTD) for this heat exchanger is 43.76°C.

Step-by-step solution

Calculate heat transfer rate from the hot fluid

Calculate the heat transfer rate from the hot fluid using the mass flow rate, specific heat, and temperature difference. Formulation: $$q_\text{hot} = m_\text{hot} \cdot c_{p,\text{hot}} \cdot \Delta T_\text{hot} $$ Where: \(q_\text{hot}\): Heat transfer rate from the hot fluid (in Watts) \(m_\text{hot}\): Mass flow rate of the hot fluid, which is \(1.4\) kg/s \(c_{p,\text{hot}}\): Specific heat of the hot fluid, which is \(4.25\) kJ/kg·K (convert it to \(4250\) J/kg·K) \(\Delta T_\text{hot}\): Hot fluid temperature difference, which is \(85 - 50 = 35\) °C Calculating, we have: $$q_\text{hot} = 1.4 \cdot 4250 \cdot 35$$ $$q_\text{hot} = 208250~W$$

Calculate heat transfer rate to the cold water

Calculate the heat transfer rate to the cold water by accounting for the 3% heat loss from the hot fluid: $$q_\text{cold} = 0.97 \times q_\text{hot}$$ Where, \(q_\text{cold}\): Heat transfer rate to the cold water (in Watts) Calculating, we have: $$q_\text{cold} = 0.97 \times 208250$$ $$q_\text{cold} = 201922.5~W$$

Calculate the log mean temperature difference (LMTD)

The LMTD for a parallel-flow heat exchanger is given by the formula: $$\text{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln(\frac{\Delta T_1}{\Delta T_2})}$$ Where: \(\Delta T_1\): Temperature difference at one end of the heat exchanger \(\Delta T_2\): Temperature difference at the other end of the heat exchanger We know, \(q_\text{cold} = U \cdot A \cdot \text{LMTD}\), where \(U\) and \(A\) are the overall heat transfer coefficient and the surface area of the heat exchanger (\(1150~W/m^2·K\) and \(4~m^2\), respectively). Now we can solve for LMTD: $$\text{LMTD} = \frac{q_\text{cold}}{U \cdot A}$$ Substituting, $$\text{LMTD} = \frac{201922.5}{1150 \cdot 4}$$ $$\text{LMTD} = 43.76^{\circ}\mathrm{C}$$ So the rate of heat transfer to the cold water is \(201,922.5\) W, and the log mean temperature difference (LMTD) for this heat exchanger is \(43.76^{\circ}\mathrm{C}\).