Question

Oil is being cooled from \(180^{\circ} \mathrm{F}\) to \(120^{\circ} \mathrm{F}\) in a oneshell and two-tube heat exchanger with an overall heat transfer coefficient of $40 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\(. Water \)\left(c_{p c}=1.0 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\( enters at \)80^{\circ} \mathrm{F}$ and exits at \(100^{\circ} \mathrm{F}\) with a mass flow rate of $20,000 \mathrm{lbm} / \mathrm{h}\(. Determine \)(a)\( the NTU value and \)(b)$ the surface area of the heat exchanger.

Short answer

Answer: We are unable to determine the NTU value and surface area of the heat exchanger because the given conditions result in an effectiveness of 1, which is impossible for a heat exchanger that doesn't have an infinite NTU. Additionally, we lack information about the mass flow rate or specific heat capacity of the oil, which is necessary to proceed with the calculations.

Step-by-step solution

Determine the heat transfer rate

Using the given conditions and the water cp value, we can determine the heat transfer rate by using the water mass flow rate and the temperature difference between the water inlet and outlet temperatures: $$ \dot{q}=m_{c} \cdot c_{p c} \cdot\left(T_{c \mathrm{out}}-T_{c \mathrm{in}}\right) $$ Using the given values: $$ \dot{q}=20,000 \frac{\mathrm{lbm}}{\mathrm{h}}\left(1.0 \frac{\mathrm{Btu}}{\mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}}\right)\left(100^{\circ} \mathrm{F} -80^{\circ} \mathrm{F}\right) $$ Calculate the heat transfer rate: $$ \dot{q}= 400,000 \, \mathrm{Btu} / \mathrm{h} $$

Determine the heat transfer capacity rates

For the hot fluid (the oil): $$ C_{h} = m_{h} \cdot{ } c_{p h} $$ For the cold fluid (the water): $$ C_{c} = m_{c} \cdot{ } c_{p c} = 20,000 \, \mathrm{lbm} / \mathrm{h} \cdot{ } 1 \, \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F} $$ We are not given the mass flow rate or the specific heat capacity of the oil, so we have to find it using the heat transfer rate. Since the heat transfer from the hot fluid equals the heat transfer to the cold fluid: $$ q_{h} = q_{c} = \dot{q} $$ The heat transfer rate from the hot fluid can be expressed as: $$ q_{h} = m_{h} \cdot{ } c_{p h} \cdot{ }(T_{h \mathrm{in}} - T_{h \mathrm{out}}) $$ $$ 400,000 \, \mathrm{Btu} / \mathrm{h} = m_{h} \cdot{ } c_{p h} \cdot{ }(180^{\circ} \mathrm{F} - 120^{\circ} \mathrm{F}) $$ We can't proceed here until m_h and/or c_ph are given, or we can assume it's the same rate as the cold flow: $$ C_{h} \approx C_{c} $$

Calculate the heat exchanger effectiveness

Using the heat transfer rate, we can find the effectiveness of the heat exchanger. The effectiveness can be found by dividing the heat transfer rate by the minimum capacity rate: $$ \varepsilon = \frac{\dot{q}}{C_{min}} $$ Since we have assumed that the hot fluid has the same rate as the cold flow, the minimum capacity rate is either of them: $$ \varepsilon = \frac{400,000 \, \mathrm{Btu} / \mathrm{h}}{20,000 \, \mathrm{lbm} / \mathrm{h} \cdot{ } 1 \, \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}} = \frac{20^{\circ} \mathrm{F}}{100^{\circ} \mathrm{F} - 80^{\circ} \mathrm{F}} = 1 $$

Calculate the NTU value

Assuming a 1-1 shell-and-tube heat exchanger, the NTU-ε relation for both fluids mixed is: $$ \varepsilon = 1 - e^{-NTU \left(1 + \frac{C_{min}}{C_{max}}\right)} $$ Since the fluids have the same capacity rate (C_min = C_max), we can simplify the formula to: $$ \varepsilon = 1 - e^{-2NTU} $$ Solving for NTU: $$ 1 = 1 - e^{-2NTU} $$ $$ 0 = e^{-2NTU} $$ NTU can't be found in this case, as the effectiveness is 1, which is impossible for a heat exchanger that doesn't have an infinite NTU.

Calculate surface area of the heat exchanger

Assuming we found NTU, we could calculate the surface area of the heat exchanger using the overall heat transfer coefficient and the heat transfer rate: $$ A = \frac{\dot{q}}{U \cdot{ } \Delta T_{lm}} $$ Since we don't have the logarithmic mean temperature difference (ΔT_lm), we can't proceed to find the surface area of the heat exchanger. In conclusion, we couldn't find the NTU value or surface area of the heat exchanger with the given information. Additional information, such as mass flow rate or specific heat capacity of the oil, is required to proceed with this problem.