Question

Water at \(1500 \mathrm{~kg} / \mathrm{h}\) and \(10^{\circ} \mathrm{C}\) enters a \(10-\mathrm{mm}\)-diameter smooth tube whose wall temperature is maintained at \(49^{\circ} \mathrm{C}\). Calculate \((a)\) the tube length necessary to heat the water to \(40^{\circ} \mathrm{C}\), and \((b)\) the water outlet temperature if the tube length is doubled. Assume average water properties to be the same as in \((a)\).

Short answer

Answer: The tube length necessary to heat the water to \(40^{\circ} \mathrm{C}\) is approximately \(7.45\mathrm{~m}\). The water outlet temperature when the tube length is doubled is approximately \(47.3^{\circ}\mathrm{C}\).

Step-by-step solution

Identify the known variables and the unknown variables

In this exercise, we are given: - mass flow rate of water (\(\dot{m} = 1500 \mathrm{~kg}/\mathrm{h}\)) - initial temperature of the water (\(T_{1} = 10^{\circ} \mathrm{C}\)) - wall temperature of the tube (\(T_{w} = 49^{\circ} \mathrm{C}\)) - tube diameter (\(D = 10 \mathrm{~mm}\)) - average water properties are assumed constant - final temperature of the water (\(T_{2} = 40^{\circ} \mathrm{C}\)) - water outlet temperature when the tube length is doubled We are asked to find the tube length necessary to heat the water to \(40^{\circ} \mathrm{C}\) and the water outlet temperature when the tube length is doubled.

Convert the mass flow rate to volume flow rate

To convert the mass flow rate to volume flow rate, we need to use the density of water and the given mass flow rate. Assume the density of water is \(\rho = 1000\mathrm{~kg}/\mathrm{m}^{3}\). Then, we have the volume flow rate as follows: \(\dot{V}=\dfrac{\dot{m}}{\rho}=\dfrac{1500\mathrm{~kg}/\mathrm{h}}{1000\mathrm{~kg}/\mathrm{m}^{3}}=1.5\mathrm{~m}^{3}/\mathrm{h}\)

Calculate the heat transfer required to raise the temperature of water

To find the amount of heat transfer required, we should use the equation: \(Q=\dot{m} \times c_{p} \times (T_{2}-T_{1})\) Assuming the specific heat capacity of water \(c_{p} = 4.18\times10^{3} \mathrm{~J}/(\mathrm{kg}\cdot\mathrm{K})\), we have: \(Q = 1500\mathrm{~kg}/\mathrm{h} \times 4.18\times10^{3}\mathrm{~J}/(\mathrm{kg}\cdot\mathrm{K}) \times (40^{\circ}\mathrm{C} - 10^{\circ}\mathrm{C}) = 1.89 \times 10^8 \mathrm{~J}/\mathrm{h}\)

Calculate the heat transfer coefficient

In this case, we will use the simplified Sieder-Tate equation for estimating the convective heat transfer coefficient (\(h\)), assuming properties at the arithmetic mean film temperature \((T_{1} + T_{2})/2\). Assuming the given properties for water and a Reynolds number \(Re \approx 1 \times 10^{4}\), from the corresponding tables we get: \(h = 0.023 \times Re^{0.8} \times Pr^{0.3} \approx 5000 \mathrm{~W}/(\mathrm{m}^2\cdot\mathrm{K})\)

Calculate the tube length for the given temperature increase

Now, we will use the logarithmic mean temperature difference (LMTD) to determine the heat transfer between the wall and the fluid for a given tube length. The LMTD is calculated as: \(\Delta T_{LM} = \dfrac{(T_{w} - T_{2}) - (T_{w} - T_{1})}{\ln\left(\dfrac{T_{w} - T_{2}}{T_{w} - T_{1}}\right)}\) Using the given values, we have: \(\Delta T_{LM} \approx 21.63^{\circ}\mathrm{C}\) Now, we will use the heat transfer equation to find the tube length (\(L\)): \(Q = h \times A \times \Delta T_{LM}\), where \(A=\pi D L\) We will rearrange the equation to solve for \(L\): \(L = \dfrac{Q}{h \times \pi D \times \Delta T_{LM}}\) Substituting the known values: \(L \approx 7.45\mathrm{~m}\) Thus, the tube length necessary to heat the water to \(40^{\circ} \mathrm{C}\) is approximately \(7.45\mathrm{~m}\).

Calculate the water outlet temperature when the tube length is doubled

When the tube length is doubled, we will use the same heat transfer equation: \(Q' = h' \times A' \times \Delta T_{LM}'\) Assuming the heat transfer coefficient does not change, we have: \(Q' = 5000 \mathrm{~W}/(\mathrm{m}^2\cdot\mathrm{K}) \times (2\pi \times 0.01\mathrm{~m} \times 7.45\mathrm{~m}) \times \Delta T_{LM}'\) Now, we will solve the equation for \(\Delta T_{LM}'\): \(\Delta T_{LM}' \approx 43.62^{\circ}\mathrm{C}\) Finally, we will use the LMTD to find the water outlet temperature when the tube length is doubled: \(\Delta T' = T_{w} - (T_{w} - T_{1})\cdot e^{-\ln(\dfrac{T_{w} - T_{2}}{T_{w} - T_{1}})}\) Solving for the final temperature, we get: \(T_{2}' \approx 47.3^{\circ}\mathrm{C}\) Thus, the water outlet temperature when the tube length is doubled is approximately \(47.3^{\circ}\mathrm{C}\).