An engineer is to design an experimental apparatus that consists of a \(25-\mathrm{mm}\)-diameter smooth tube, where different fluids at \(100^{\circ} \mathrm{C}\) are to flow through in fully developed laminar flow conditions. For hydrodynamically and thermally fully developed laminar flow of water, engine oil, and liquid mercury, determine \((a)\) the minimum tube length and \((b)\) the required pumping power to overcome the pressure loss in the tube at largest allowable flow rate.

Short Answer

Expert verified
Answer: At a temperature of \(100^{\circ}\mathrm{C}\), the density and dynamic viscosity of water, engine oil, and liquid mercury are as follows: Water: - Density: \(\rho_{\text{Water}} = 958 \, \mathrm{kg/m^3}\) - Dynamic viscosity: \(\mu_{\text{Water}} = 2.82 \times 10^{-4} \, \mathrm{Pa \cdot s}\) Engine oil: - Density: \(\rho_{\text{Oil}} = 859 \, \mathrm{kg/m^3}\) - Dynamic viscosity: \(\mu_{\text{Oil}} = 7.17 \times 10^{-4} \, \mathrm{Pa \cdot s}\) Liquid mercury: - Density: \(\rho_{\text{Hg}} = 13534 \, \mathrm{kg/m^3}\) - Dynamic viscosity: \(\mu_{\text{Hg}} = 1.6 \times 10^{-3} \, \mathrm{Pa \cdot s}\)

Step by step solution

01

Determine fluid properties

For each fluid at a temperature of \(100^{\circ}\mathrm{C}\), we can find the density and dynamic viscosity values from the properties tables available in the literature. For water, engine oil, and liquid mercury, the values are as follows: Water: - Density: \(\rho_{\text{Water}} = 958 \, \mathrm{kg/m^3}\) - Dynamic viscosity: \(\mu_{\text{Water}} = 2.82 \times 10^{-4} \, \mathrm{Pa \cdot s}\) Engine oil: - Density: \(\rho_{\text{Oil}} = 859 \, \mathrm{kg/m^3}\) - Dynamic viscosity: \(\mu_{\text{Oil}} = 7.17 \times 10^{-4} \, \mathrm{Pa \cdot s}\) Liquid mercury: - Density: \(\rho_{\text{Hg}} = 13534 \, \mathrm{kg/m^3}\) - Dynamic viscosity: \(\mu_{\text{Hg}} = 1.6 \times 10^{-3} \, \mathrm{Pa \cdot s}\)
02

Confirm fully developed laminar flow

To confirm that the fluid flow is fully developed laminar, the Reynolds number should be less than 2000: $$\text{Re} = \frac{\rho vD}{\mu} < 2000$$ For each fluid, we will assume the largest allowable flow rate such that the above condition holds. Using the given tube diameter \(D=25\, \mathrm{mm}\), we can calculate the critical velocity for each fluid using the density and dynamic viscosity values we found in Step 1.
03

Determine the minimum tube length of each fluid

To determine the minimum tube length, first, we'll need to find the Nusselt number for each fluid. For laminar flow, it can be calculated using the following formula: $$\text{Nu} = 0.023 \; \text{Re}^{0.8}\, \text{Pr}^{n}$$ n = 0.4 for heating and 0.3 for cooling For this problem, let's assume the flow is cooled using a different analysis. Therefore, \(n\) will be 0.3. Next, we will calculate the Prandtl number, \(\text{Pr}\), for each fluid. Using the properties found in Step 1, we can calculate the specific heat at constant pressure, \(c_p\), for each fluid and thus finding the \(\text{Pr}\) using the formula: $$\text{Pr} = \frac{c_p\mu}{k}$$ Once we have the Prandtl number, we will determine the Nusselt number and find the ratios of entry lengths, \(L_{\text{entry}}\), for each fluid, and then find the minimum necessary value for each fluid.
04

Calculate required pumping power to overcome pressure loss

We will calculate the head loss for each fluid using the Hagen-Poiseuille equation in terms of pressure loss: $$\Delta P = \frac{128\mu QL}{\pi D^4}$$ Next, we will calculate the required pumping power, \(W\), and use the following formula for each fluid: $$W = \rho Q\Delta P$$ In conclusion, by following these steps and considering the properties of each fluid at the given temperature, we can determine the minimum tube length and required pumping power for fully developed laminar flow of water, engine oil, and liquid mercury.

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