Question

Consider a closed-loop heat exchanger that carries exit water $\left(c_{p}=1 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right.$ and \(\left.\rho=62.4 \mathrm{lbm} / \mathrm{ft}^{3}\right)\) of a condenser side initially at \(100^{\circ} \mathrm{F}\). The water flows through a 500 -ft-long stainless steel pipe of 1 in inner diameter immersed in a large lake. The temperature of lake water surrounding the heat exchanger is $45^{\circ} \mathrm{F}$. The overall heat transfer coefficient of the heat exchanger is estimated to be $250 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}$. What is the exit temperature of the water from the immersed heat exchanger if it flows through the pipe at an average velocity of \(9 \mathrm{ft} / \mathrm{s}\) ? Use the \(\varepsilon-N T U\) method for analysis.

Short answer

Answer: The exit temperature of the water in the heat exchanger is \(45^{\circ} \mathrm{F}\).

Step-by-step solution

Determine the mass flow rate of water

We can calculate the mass flow rate of the water by multiplying the density, velocity, and cross-sectional area of the pipe. The mass flow rate is given by: \(m_{\text{flow}}=\rho v_{\text{flow}} A\) where - \(m_{\text{flow}}\) is the mass flow rate (\(\mathrm{lbm} / \mathrm{s}\)) - \(\rho\) is the density of water (\(62.4 \mathrm{lbm} / \mathrm{ft}^{3}\)) - \(v_{\text{flow}}\) is the water velocity (\(9 \mathrm{ft} / \mathrm{s}\)) - \(A\) is the cross-sectional area of the pipe (\((1/2)^2\pi \mathrm{ft}^{2}\)), considering 1 inch equals to 1/12 ft.

Calculate the mass flow rate of water

Now, we will plug in the given values to find the mass flow rate of the water: \(m_{\text{flow}}=(62.4 \mathrm{lbm} / \mathrm{ft}^{3})(9 \mathrm{ft} / \mathrm{s})[(1/12)^2\pi \mathrm{ft}^{2}] \approx 8.97 \mathrm{lbm} / \mathrm{s}\)

Determine the NTU

The Number of Transfer Units (NTU) is given by: \(NTU=\frac{U A_{s}}{m_{\text{flow}} c_{p}}\) where - \(U\) is the overall heat transfer coefficient (\(250 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\)) - \(A_{s}\) is the surface area of the pipe, which can be calculated as \(A_{s}=\pi D L\), where \(D\) is the inner diameter of the pipe and \(L\) is the length of the pipe. - \(m_{\text{flow}}\) is the mass flow rate of water (\(8.97 \mathrm{lbm} / \mathrm{s}\)) - \(c_{p}\) is the specific heat of water (\(1 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\))

Calculate the NTU

Now, we will calculate the NTU: \(NTU=\frac{(250 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F})(\pi(1/12)(500))}{(8.97 \mathrm{lbm} / \mathrm{s})(1 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F})} \approx 1.24\)

Calculate the effectiveness of the heat exchanger

For a counter-flow heat exchanger (here the water flows through the immersed pipe and the lake is stationary), the effectiveness (\(\varepsilon\)) is given by: \(\varepsilon=1-\mathrm{exp}\left(-NTU\cdot(1-C_{min}/C_{max})\right)\) Where \(C_{min}\) and \(C_{max}\) are the minimum and maximum heat capacities of the fluids, which in this case can be considered equal: \(C_{min}=C_{max}=m_{\text{flow}} c_{p} = 8.97 \mathrm{Btu} / \mathrm{s} \cdot{ }^{\circ} \mathrm{F}\) Calculating the effectiveness, we get: \(\varepsilon=1-\mathrm{exp}\left(-1.24(1-1)\right) = 1\)

Calculate the exit temperature of the water

Now, calculate the exit temperature of the water using the following relation: \(T_{\text{exit}}=T_{\text{inlet}}-\varepsilon(T_{\text{inlet}}-T_{\text{lake}})\) Where - \(T_{\text{exit}}\) is the exit temperature of the water - \(T_{\text{inlet}}\) is the initial water temperature in the pipe (\(100^{\circ} \mathrm{F}\)) - \(T_{\text{lake}}\) is the temperature of the lake water (\(45^{\circ} \mathrm{F}\)) Plugging in the values, we get: \(T_{\text{exit}}=100-1(100-45) = 45^{\circ} \mathrm{F}\) The exit temperature of the water in the heat exchanger is \(45^{\circ} \mathrm{F}\).