Fluid Mechanics and Thermodynamics

Laminar pipe flow

Laminar flow in a circular pipe is a fully developed, viscosity-dominated flow in which fluid moves in concentric layers without mixing. The velocity profile is parabolic, u(r)=u_{max}\left(1-\frac{r^2}{R^2}\right), so the centreline velocity is the maximum velocity. The mean velocity is exactly half of it, V_{avg}=u_{max}/2, which is one of the most tested relations in this topic.

• Hagen–Poiseuille flow — laminar, fully developed pipe flow driven by a pressure gradient. Pressure drop is \Delta p=\frac{32\mu L V_{avg}}{D^2}. 예) For a given pressure drop, doubling the radius raises the flow rate by a factor of 16.
• Transition — flow is generally laminar for Re<2300 in a pipe, so a computed Reynolds number below this value supports the laminar assumption.

For heat transfer, fully developed laminar flow in a circular tube with uniform wall heat flux has a constant Nusselt number Nu_D=4.36, so the convective coefficient is h=Nu_D k/D.

Reynolds number

The Reynolds number Re=\frac{\rho V L}{\mu}=\frac{VL}{\nu} is the ratio of inertial force to viscous force. It is dimensionless and separates laminar from turbulent flow. Other named numbers measure different force ratios: the Mach number compares inertia with compressibility (speed of sound), the Froude number compares inertia with gravity, and the Weber number compares inertia with surface tension. In pipe flow, L is the diameter and V is the mean velocity. If only the maximum velocity of a laminar profile is given, convert it to the mean velocity (V_{avg}=V_{max}/2) before computing Re. Example: air at 2 m/s in a 0.05 m pipe with \nu=1.5\times10^{-5} m^2/s gives Re\approx 6667.

Otto cycle thermal efficiency

The ideal Otto cycle consists of two isentropic processes and two constant-volume processes. Heat is added at constant volume, q_{in}=c_v(T_3-T_2), and rejected at constant volume, q_{out}=c_v(T_4-T_1), so the heat quantities can be read as internal energy changes. The efficiency is \eta=1-\frac{q_{out}}{q_{in}}, which for an air-standard cycle gives \eta=1-\frac{1}{r^{\gamma-1}}, where r is the compression ratio. The efficiency depends only on r and \gamma, not on the amount of heat added. Working backwards, a given efficiency fixes r=(1-\eta)^{-1/(\gamma-1)}; for \eta=0.5 and \gamma=4/3, r=8. Then T_2=T_1 r^{\gamma-1} and p_2=p_1 r^{\gamma}, and the maximum pressure follows from the constant-volume heat addition.

• Constant-volume heat addition — heat input at fixed volume, so \Delta T = q_{in}/c_v. 예) A heat input of 500 kJ/kg with c_v=0.7 kJ/kg-K raises the temperature by about 714 K.

Fourier law heat flux

Fourier's law states that the heat flux is proportional to the temperature gradient, \mathbf{q}=-k\nabla T. The negative sign means heat flows from high to low temperature. The net heat leaving a closed surface is Q=\oint \mathbf{q}\cdot\mathbf{n}\,dA, and by the divergence theorem this equals -k\int_V \nabla^2 T\,dV. If \nabla^2 T is uniform, the net outflow is simply -k\,\nabla^2 T times the volume. For steady one-dimensional conduction through a plane wall with no generation, the heat flux is uniform, q=k(T_1-T_2)/L. When the wall surface is exposed to a fluid, the convective flux equals the conductive flux, h(T_\infty-T_s)=\frac{k}{L}(T_s-T_1), which allows the unknown surface temperature to be found.

• Net outflow sign — a negative Laplacian of temperature means net heat leaves the volume. 예) For T=x^2+y^2+z^2, \nabla^2T=6, so heat enters the volume.

Refrigerator coefficient of performance

The coefficient of performance of a refrigerator is \text{COP}=\frac{Q_L}{W}, the heat removed from the cold space divided by the work input. Heat removal rates given per minute must be converted to kW (kJ/s) before dividing by a power in kW. For a heat pump, the corresponding ratio is \text{COP}_{HP}=\frac{Q_H}{W}, and the two are related by \text{COP}_{HP}=\text{COP}_R+1. The cold-space temperature does not enter the direct definition; it appears only in the Carnot upper bound \text{COP}_{max}=\frac{T_L}{T_H-T_L}. 예) Removing 60 kJ/min with 1 kW input gives \text{COP}=1.

Tank draining time

Torricelli's theorem gives the exit velocity of a small orifice at the bottom of a tank, neglecting losses, as v=\sqrt{2gh}, where h is the free-surface height. A mass balance on the tank gives A\,\frac{dh}{dt}=-a\sqrt{2gh}, with A the tank area and a the hole area. Separating variables and integrating from h_0 to zero gives the draining time t=\frac{A}{a}\sqrt{\frac{2h_0}{g}}. The draining time depends on the area ratio A/a and on the initial head, not on the absolute size of the tank alone. 예) With A/a=100 and h_0=2 m, t=100\sqrt{4/9.81}\approx 64 s. Unit consistency should be checked by writing g in m/s^2.

Entropy change of ideal gas

Entropy is a property, so the entropy change between two states depends only on the end states, even if the actual process is irreversible. For an ideal gas the change is \Delta s=c_p\ln\frac{T_2}{T_1}-R\ln\frac{p_2}{p_1}, or equivalently \Delta s=c_v\ln\frac{T_2}{T_1}+R\ln\frac{v_2}{v_1}. For a calorically perfect gas, c_p=\frac{\gamma R}{\gamma-1}. Temperatures must be in kelvin in the logarithm, and the pressure term has a negative sign. 예) For air heated at constant pressure from 300 K to 600 K, \Delta s=c_p\ln 2\approx 0.69\,c_p, which is positive because the temperature rises.

Radiation shield heat transfer

Two infinite black parallel surfaces exchange radiation q=\sigma(T_1^4-T_2^4) per unit area, and the total radiation resistance of the pair is 1 for black surfaces. Placing a radiation shield midway between them, with the same geometry, adds a second radiation resistance, so the total resistance doubles and the heat transfer is halved. For N black shields in parallel, the heat transfer ratio is \frac{1}{N+1} compared with the unshielded case. Shields made of low-emissivity material are more effective, because each surface contributes a resistance of \frac{1}{\varepsilon}-1 in the network. 예) A single black shield at the midpoint halves the radiation flow.