Question

A single-pass crossflow heat exchanger with both fluids unmixed has water entering at \(16^{\circ} \mathrm{C}\) and exiting at \(33^{\circ} \mathrm{C}\), while oil $\left(c_{p}=1.93 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right.\( and \)\left.\rho=870 \mathrm{~kg} / \mathrm{m}^{3}\right)$ flowing at \(0.19 \mathrm{~m}^{3} / \mathrm{min}\) enters at $38^{\circ} \mathrm{C}\( and exits at \)29^{\circ} \mathrm{C}$. If the surface area of the heat exchanger is \(20 \mathrm{~m}^{2}\), determine \((a)\) the NTU value and \((b)\) the value of the overall heat transfer coefficient.

Short answer

Question: Calculate the NTU value and the overall heat transfer coefficient for a single-pass crossflow heat exchanger with both fluids unmixed, given the inlet and outlet temperatures of water and oil, the flow rate of oil, and the surface area of the heat exchanger. Inlet temperatures: Water: \(T_{wi} = 20 ^{\circ}C\) Oil: \(T_{oi} = 70 ^{\circ}C\) Outlet temperatures: Water: \(T_{wo} = 45 ^{\circ}C\) Oil: \(T_{oo} = 42 ^{\circ}C\) Flow rate of oil: \(\text{0.19 m}^3\text{/min}\) Density of oil: \(870\,\text{kg/m}^3\) Specific heat capacity of oil: \(c_{p_{o}} = 2.1\,\text{kJ/kg}\cdot^{\circ}C\) Heat exchanger surface area: \(A = 20\,\text{m}^2\) Calculate (a) NTU value, and (b) the value of the overall heat transfer coefficient.

Step-by-step solution

Find the capacity rates of water and oil

To find the capacity rates (C), we first need to find the mass flow rate of oil and the specific heat capacities of water and oil. We are given the specific heat capacity of oil, but we need to find the specific heat capacity of water. The specific heat capacity of water is approximately \(c_{p_{w}} = 4.18 \mathrm{kJ/kg \cdot K}\). Next, we need to find the mass flow rate of oil (m_o) using the given volumetric flow rate and density: $$m_o = \rho \cdot \text{flow rate} = 870 \mathrm{kg/m^3} \cdot 0.19 \mathrm{m^3/min} \cdot \frac{1 \mathrm{min}}{60 \mathrm{s}} = 2.744 \mathrm{kg/s}$$ Now, we can calculate the capacity rates of water and oil (C_w and C_o, respectively): $$C_w = m_{w}c_{p_{w}} \hspace{1cm} \text{and} \hspace{1cm} C_o = m_{o}c_{p_{o}}$$

Calculate the heat transfer effectiveness

Next, we need to find the heat transfer effectiveness (ε) using the given temperature values: $$ε = \frac{q_{actual}}{q_{max}}$$ Here, the actual heat transfer (\(q_{actual}\)) is the product of the capacity rates and the change in temperature for either the water or the oil: $$q_{actual} = m_{w}c_{p_{w}}(T_{wo} - T_{wi}) = m_{o}c_{p_{o}}(T_{oi} - T_{oo})$$ The maximum possible heat transfer (\(q_{max}\)) is given as the product of the minimum capacity rate and the change in inlet temperatures: $$q_{max} = C_{min}(T_{oi}-T_{wi})$$

Calculate the NTU value

Now, we can use the effectiveness-NTU method to find the NTU value: $$NTU = \frac{-\log{(1-ε)}}{1-\frac{C_{min}}{C_{max}}}$$

Calculate the rate of heat transfer (Q)

We will now calculate the rate of heat transfer: $$Q = C_{min} (T_{oi} - T_{wi}) \cdot ε$$

Calculate the log mean temperature difference (LMTD)

Next, we need to find the LMTD by using the inlet and outlet temperatures: $$LMTD = \frac{(ΔT_1 - ΔT_2)}{\log{\frac{ΔT_1}{ΔT_2}}}$$ where $$ΔT_1 = T_{wi} - T_{oo} \hspace{1cm} \text{and} \hspace{1cm} ΔT_2 = T_{wo} - T_{oi}$$

Determine the overall heat transfer coefficient (U)

Finally, we can now calculate the overall heat transfer coefficient (U) using the surface area (A), rate of heat transfer (Q), and LMTD: $$U = \frac{Q}{A \cdot LMTD}$$