Question

Cold water $\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\( enters a crossflow heat exchanger at \)14^{\circ} \mathrm{C}\( at a rate of \)0.35 \mathrm{~kg} / \mathrm{s}$ where it is heated by hot air $\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\( that enters the heat exchanger at \)65^{\circ} \mathrm{C}$ at a rate of \(0.8 \mathrm{~kg} / \mathrm{s}\) and leaves at $25^{\circ} \mathrm{C}$. Determine the maximum outlet temperature of the cold water and the effectiveness of this heat exchanger.

Short answer

Question: Calculate the effectiveness of a heat exchanger if the mass flow rate of the cold water is 0.35 kg/s and the hot air is 0.8 kg/s. The specific heat capacity of the cold water is 4.18 kJ/(kg·K) and for the hot air is 1.0 kJ/(kg·K). The inlet temperature of the cold water is 14°C and the inlet and outlet temperature of the hot air is 65°C and 25°C, respectively. Answer: The effectiveness of the heat exchanger is approximately \(81.4\%\).

Step-by-step solution

Calculate the maximum possible outlet temperature of the cold water

First, we need to calculate the temperature when the cold water and hot air reach thermal equilibrium. This can be determined by the energy balance equation which is the mass flow rate times the specific heat capacity times the temperature change: \(\dot{m}_{c}c_{p,c}(T_{c,out}-T_{c,in})=\dot{m}_{h}c_{p,h}(T_{h,in}-T_{h,out})\) Now, plug in the given values and solve for the temperature equilibrium: \(0.35 \mathrm{~kg/s} \cdot 4.18 \mathrm{~kJ/(kg \cdot K)} \cdot (T_{c,max}-14^{\circ}\mathrm{C}) = 0.8 \mathrm{~kg/s} \cdot 1.0 \mathrm{~kJ/(kg\cdot K)} \cdot (65^{\circ}\mathrm{C}-25^{\circ}\mathrm{C})\)

Solve the equation for the maximum outlet temperature

Next, we will solve the above equation for \(T_{c,max}\). \((0.35)(4.18)(T_{c,max}-14)=(0.8)(1.0)(40)\) \((1.463)(T_{c,max}-14)=32\) \(T_{c,max}-14=21.876\) \(T_{c,max}=35.876^{\circ}\mathrm{C}\) The maximum outlet temperature of the cold water is approximately \(35.88^{\circ}\mathrm{C}\).

Calculate the heat transfer in the heat exchanger

Now, we need to find the heat transfer from the hot air to the cold water using \(\dot{Q} = \dot{m}_{h}c_{p,h}(T_{h,in}-T_{h,out})\) \(\dot{Q} = 0.8 \mathrm{~kg/s} \cdot 1.0 \mathrm{~kJ/(kg\cdot K)} \cdot (65^{\circ}\mathrm{C}-25^{\circ}\mathrm{C}) = 32\ \mathrm{kJ/s}\)

Determine the actual outlet temperature of the cold water

Using the heat transfer value and the mass flow rate of the cold water, we can find the actual outlet temperature. \(\dot{Q}=\dot{m}_{c}c_{p,c}(T_{c,out}-T_{c,in}) \Rightarrow 32=(0.35 \mathrm{~kg/s})(4.18 \mathrm{~kJ/(kg \cdot K)})(T_{c,out}-14^{\circ}\mathrm{C})\) First, divide both sides by \(0.35\ \mathrm{kg/s}\) and \(4.18\ \mathrm{kJ/(kg\cdot K)}\): \(32/(0.35 \cdot 4.18)=T_{c,out}-14\) \(T_{c,out} = 32/(0.35 \cdot 4.18) + 14\) \(T_{c,out} \approx 33.12^{\circ}\mathrm{C}\) The actual outlet temperature of the cold water is approximately \(33.12^{\circ}\mathrm{C}\).

Calculate the effectiveness of the heat exchanger

Finally, we will calculate the effectiveness of the heat exchanger using the following formula: \(Effectiveness=\frac{Actual \ heat \ transfer}{Max \ heat \ transfer}=\frac{\dot{m}_{c}c_{p,c}(T_{c,out}-T_{c,in})}{\dot{m}_{c}c_{p,c}(T_{c,max}-T_{c,in})}\) Plug in the given values and temperatures calculated in the previous steps: \(Effectiveness=\frac{(0.35)(4.18)(33.12-14)}{(0.35)(4.18)(35.88-14)}\) \(Effectiveness \approx \frac{27.924}{34.304}\) \(Effectiveness \approx 0.814\) The effectiveness of the heat exchanger is approximately \(81.4\%\).