Consider a water-to-water counter-flow heat exchanger with these specifications. Hot water enters at \(95^{\circ} \mathrm{C}\) while cold water enters at \(20^{\circ} \mathrm{C}\). The exit temperature of hot water is \(15^{\circ} \mathrm{C}\) greater than that of cold water, and the mass flow rate of hot water is 50 percent greater than that of cold water. The product of heat transfer surface area and the overall heat transfer coefficient is \(1400 \mathrm{~W} / \mathrm{K}\). Taking the specific heat of both cold and hot water to be \(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), determine (a) the outlet temperature of the cold water, \((b)\) the effectiveness of the heat exchanger, \((c)\) the mass flow rate of the cold water, and \((d)\) the heat transfer rate.

Step by step solution

01

Use energy balance equation for hot water

Since the mass flow rate of hot water is 50 percent greater than that of cold water, the mass flow rate of hot water can be represented as \(\dot{m}_{h} = 1.5\dot{m}_{c}\). From the energy balance equation for hot water, we have: \(Q = \dot{m}_{h}c_{p}(T_{h, in} - T_{h, out})\) Using \(\dot{m}_{h} = 1.5\dot{m}_{c}\), we get: \(Q = 1.5\dot{m}_{c}c_{p}(95 - (T_{c, out}+15))\) ... (1)

02

Use energy balance equation for cold water

From the energy balance equation for cold water, we have: \(Q = \dot{m}_{c}c_{p}(T_{c, out} - 20)\) ... (2)

03

Solve for outlet temperature of cold water

From (1) and (2), we can solve for the outlet temperature of the cold water: \(1.5\dot{m}_{c}c_{p}(95 - (T_{c, out}+15)) = \dot{m}_{c}c_{p}(T_{c, out} - 20)\) Solving for \(T_{c, out}\), we get: \(T_{c, out} = 42^{\circ} C\)

04

Calculate the effectiveness of the heat exchanger

The effectiveness of the heat exchanger is defined as: \(\epsilon = \frac{Q}{\dot{m}_{c}c_{p}(T_{h, in} - T_{c, in})}\) Using the results from Step 2 and Step 3, we get: \(\epsilon = \frac{\dot{m}_{c}c_{p}(42 - 20)}{\dot{m}_{c}c_{p}(95 - 20)} = \frac{22}{75} \approx 0.293\)

05

Determine the mass flow rate of cold water

We are given that the product of heat transfer surface area and the overall heat transfer coefficient is \(U = 1400 \frac{W}{K}\). We can write the following equation: \(Q = U\Delta{T}\) We can use the results from Step 1 to obtain the value of \(\Delta{T}\): \(\Delta{T} = \frac{1.5\dot{m}_{c}c_{p}(60)}{1400} = 22\) Finally, we can solve for the mass flow rate of the cold water: \(\dot{m}_{c} = \frac{U\Delta{T}}{c_{p}(T_{c, out} - T_{c, in})} = \frac{1400 \times 22}{4180\times(42-20)} \approx 0.108 \frac{kg}{s}\)

06

Calculate the heat transfer rate

Now that we have determined the mass flow rate of the cold water, we can calculate the heat transfer rate: \(Q = \dot{m}_{c}c_{p}(T_{c, out} - T_{c, in}) = 0.108\times4180\times(42-20) \approx 997.92 W\) #Summary# In conclusion, we have determined the following values for this counter-flow heat exchanger problem: (a) Outlet temperature of cold water: \(T_{c, out} = 42^{\circ} C\) (b) Effectiveness of the heat exchanger: \(\epsilon \approx 0.293\) (c) Mass flow rate of cold water: \(\dot{m}_{c} \approx 0.108 \frac{kg}{s}\) (d) Heat transfer rate: \(Q \approx 997.92 W\)

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