Question

Water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the \(2.5\)-cm-internaldiameter tube of a double-pipe counterflow heat exchanger at \(20^{\circ} \mathrm{C}\) at a rate of $2.2 \mathrm{~kg} / \mathrm{s}\(. Water is heated by steam condensing at \)120^{\circ} \mathrm{C}\left(h_{f g}=2203 \mathrm{~kJ} / \mathrm{kg}\right)$ in the shell. If the overall heat transfer coefficient of the heat exchanger is $700 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, determine the length of the tube required in order to heat the water to \(80^{\circ} \mathrm{C}\) using \((a)\) the LMTD method and \((b)\) the \(\varepsilon-\mathrm{NTU}\) method.

Short answer

Based on the given data for the double-pipe counterflow heat exchanger, a tube length of approximately 19.97 meters using the LMTD method and 20.02 meters using the effectiveness-NTU method is required to heat the water from 20°C to 80°C.

Step-by-step solution

Calculate the mass flow rate and specific heat of water

Firstly, we have to find the mass flow rate and the specific heat of water. For water, the specific heat \(c_{p} = 4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and the flow rate is given as \(2.2 \mathrm{~kg} / \mathrm{s}\). The mass flow rate is then \(m_{w} = (2.2 \mathrm{~kg}) / (1 \mathrm{s}) = 2.2 \mathrm{~kg/s}\).

LMTD method (a) - Calculate the heat required to heat water

Using the specific heat and mass flow rate, we can find the heat required to heat the water \((Q)\) to the desired temperature. The formula to calculate it is: \(Q = m_{w} \cdot c_{p} \cdot (T_{out} - T_{in})\) Where \(T_{out} = 80^{\circ} \mathrm{C}\) and \(T_{in} = 20^{\circ} \mathrm{C}\). Substituting the values, we get: \(Q = (2.2 \mathrm{~kg/s}) \cdot (4180 \mathrm{~J/kg \cdot K}) \cdot (80 - 20) \mathrm{K} = 55144 \mathrm{~W}\).

LMTD method (a) - Calculate the LMTD for the heat exchanger

Next, we need to find the LMTD of the heat exchanger. The formula for the LMTD in a counterflow exchanger is: \(LMTD = \dfrac{(T_{h,in} - T_{c,out}) - (T_{h,out} - T_{c,in})}{\ln \left[\frac{T_{h,in} - T_{c,out}}{T_{h,out} - T_{c,in}}\right]}\) For our problem, the values are \(T_{h,in} = 120^{\circ} \mathrm{C}\), \(T_{c,out} = 80^{\circ} \mathrm{C}\), \(T_{h,out} = 120^{\circ} \mathrm{C}\) (as it remains constant in condensation) and \(T_{c,in} = 20^{\circ} \mathrm{C}\). Plugging in these values, we get: \(LMTD = \dfrac{(120 - 80) - (120 - 20)}{\ln \left[\frac{120 - 80}{120 - 20}\right]} = 50.41^{\circ} \mathrm{C}\).

LMTD method (a) - Calculate the heat transfer area required

Now that we have the heat required and the LMTD, we can calculate the heat transfer area \((A)\) required using the overall heat transfer coefficient \((U)\), which is given as \(700 \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The formula to calculate the area is: \(A = \dfrac{Q}{U \cdot LMTD}\) Substituting the values, we get: \(A = \dfrac{55144 \mathrm{W}}{(700 \mathrm{W/m^{2} \cdot K}) \cdot (50.41 \mathrm{K})} = 1.5685 \mathrm{m^2}\).

LMTD method (a) - Calculate the tube length required

Finally, to find the tube length \((L)\) required, we will use the internal diameter \((D_{i})\) of the pipe, which is given as \(2.5 \mathrm{cm} = 0.025 \mathrm{m}\). The formula to calculate the length is: \(L = \dfrac{A}{\pi D_{i}}\) Plugging in the values, we get: \(L = \dfrac{1.5685 \mathrm{m^2}}{(3.1416)(0.025 \mathrm{m})} = 19.97 \mathrm{m}\). Hence, for the LMTD method, a tube length of approximately \(19.97 \mathrm{m}\) is required. We will continue with the effectiveness-NTU (b) method.

Effectiveness-NTU method (b) - Calculate the capacity rate

For this approach, we start by determining the capacity rate of the water, which is represented by \(C_{min}\) since it is the fluid being heated. The formula for capacity rate is: \(C_{min} = m_{w} \cdot c_{p}\) Substituting the values, we get: \(C_{min} = (2.2 \mathrm{~kg/s}) \cdot (4180 \mathrm{~J/kg \cdot K}) = 9196 \mathrm{~W/K}\).

Effectiveness-NTU method (b) - Calculate the effectiveness

Next, we determine the effectiveness \((\varepsilon)\) of the heat exchanger using the inlet and outlet temperatures of the cold and hot fluids: \(\varepsilon = \dfrac{T_{c,out} - T_{c,in}}{T_{h,in} - T_{c,in}}\) Substituting the values, we get: \(\varepsilon = \dfrac{80 - 20}{120 - 20} = 0.6\).

Effectiveness-NTU method (b) - Calculate the NTU

Now, we calculate the Number of Transfer Units (NTU) for a counterflow double-pipe heat exchanger using the formula: \(NTU = \dfrac{\sqrt{1 + 4(\varepsilon^2)/(1-\varepsilon)^2} - 1}{2}\) Substituting the effectiveness, we get: \(NTU = \dfrac{\sqrt{1 + 4(0.6^2)/(1-0.6)^2} - 1}{2} \approx 1.203\).

Effectiveness-NTU method (b) - Calculate the heat transfer area required

Using the NTU and capacity rate of the water, we can now find the heat transfer area required: \(A = \dfrac{NTU \cdot C_{min}}{U}\) Substituting the values, we get: \(A = \dfrac{(1.203)(9196 \mathrm{W/K})}{(700 \mathrm{W/m^{2} \cdot K})} \approx 1.5731 \mathrm{m^2}\).

Effectiveness-NTU method (b) - Calculate the tube length required

Lastly, as before, we use the internal diameter of the pipe to calculate the tube length required: \(L = \dfrac{A}{\pi D_{i}}\) Plugging in the values, we get: \(L = \dfrac{1.5731 \mathrm{m^2}}{(3.1416)(0.025 \mathrm{m})} = 20.02 \mathrm{m}\). Hence, for the effectiveness-NTU method, a tube length of approximately \(20.02 \mathrm{m}\) is required. In conclusion, a tube length of approximately \(19.97 \mathrm{m}\) as per the LMTD method and \(20.02 \mathrm{m}\) based on the effectiveness-NTU method is required to heat the water to the desired temperature.