Question

Cold water $\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\( enters a counterflow heat exchanger at \)18^{\circ} \mathrm{C}\( at a rate of \)0.7 \mathrm{~kg} / \mathrm{s}$ where it is heated by hot air \(\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the heat exchanger at \(50^{\circ} \mathrm{C}\) at a rate of $1.6 \mathrm{~kg} / \mathrm{s}\( and leaves at \)25^{\circ} \mathrm{C}$. The maximum possible outlet temperature of the cold water is (a) \(25.0^{\circ} \mathrm{C}\) (b) \(32.0^{\circ} \mathrm{C}\) (c) \(35.5^{\circ} \mathrm{C}\) (d) \(39.7^{\circ} \mathrm{C}\) (e) \(50.0^{\circ} \mathrm{C}\)

Short answer

a) 25°C b) 32°C c) 40°C d) 50°C Answer: b) 32°C

Step-by-step solution

Calculate the heat transfer for the hot air

Apply the energy balance equation for the hot air: $$Q = m_{hot} c_{p_{hot}} \Delta T_{hot}$$ We have the following values: Mass flow rate of hot air \((m_{hot}) = 1.6 \, kg/s\) Specific heat of hot air \((c_{p_{hot}}) = 1.0 \, kJ/(kg \cdot K)\) Temperature change for the hot air: \(\Delta T_{hot} = T_{inlet_{hot}} - T_{outlet_{hot}} = 50 - 25 = 25^{\circ} C\) Now, calculate the heat transfer: $$Q = 1.6 \, kg/s \cdot 1.0 \, kJ/(kg \cdot K) \cdot 25^{\circ} C = 40 \, kJ/s$$

Determine the outlet temperature of the cold water

Use the energy balance equation for the cold water: $$Q = m_{cold} c_{p_{cold}} (T_{outlet_{cold}} - T_{inlet_{cold}})$$ We have the following values: Mass flow rate of cold water \((m_{cold}) = 0.7 \, kg/s\) Specific heat of cold water \((c_{p_{cold}}) = 4.18 \, kJ/(kg \cdot K)\) Inlet temperature of cold water \((T_{inlet_{cold}}) = 18^{\circ} C\) Rearrange the energy balance equation to find the outlet temperature of the cold water: $$T_{outlet_{cold}} = \frac{Q}{m_{cold} c_{p_{cold}}} + T_{inlet_{cold}}$$ Now, calculate the outlet temperature: $$T_{outlet_{cold}} = \frac{40 \, kJ/s}{0.7 \, kg/s \cdot 4.18 \, kJ/(kg \cdot K)} + 18^{\circ} C = 32^{\circ} C$$ The maximum possible outlet temperature of the cold water is \(\boxed{32^{\circ} C}\), which corresponds to answer choice (b).