First Law and Energy Balance

Steady-flow open system energy balance

For a control volume at steady state, the mass flow rate is the same at every section, and the rate of energy entering equals the rate of energy leaving. Per unit mass, with heat q and shaft work w crossing the boundary, the balance is h_1 + \frac{V_1^2}{2} + gz_1 + q = h_2 + \frac{V_2^2}{2} + gz_2 + w. The enthalpy h = u + Pv appears instead of internal energy because it includes the flow work done by the fluid at each section. For an ideal gas with constant specific heats, h = c_pT, so every enthalpy difference becomes c_p\Delta T. Mass conservation is \dot m = \rho AV, with \rho = P/(RT) for an ideal gas. Kinetic energy terms are converted to kJ/kg by dividing V^2 by 2 and then by 1000. Heat lost from the device enters the balance with a negative sign. Four common devices reduce the general form in standard ways. A nozzle has no work and negligible heat transfer, so velocity increases as enthalpy falls. A diffuser slows the flow and raises the pressure. A throttling valve keeps enthalpy constant. An adiabatic mixing chamber has no work or heat transfer, so the mixed enthalpy is the mass-weighted average of the inlet enthalpies. For ideal gases with constant c_p and heat loss \dot Q, the exit temperature is T_3 = \frac{\dot m_1T_1 + \dot m_2T_2}{\dot m_1 + \dot m_2} - \frac{\dot Q}{c_p(\dot m_1 + \dot m_2)}. Heat loss therefore lowers the mixed temperature. Example: equal flows of 0.2 kg/s at 100 °C and 0 °C mix adiabatically with c_p = 4.2 kJ/(kg K), and the exit temperature is 50 °C because the masses and specific heats are equal.

Tank filling and discharge

A tank with a changing mass is an open system with unsteady flow. For a control volume with inflow m_i at h_i and outflow m_e at h_e, the energy balance is m_2u_2 - m_1u_1 = m_ih_i - m_eh_e + Q - W. The energy of a fluid entering or leaving the tank is carried by its enthalpy, not its internal energy. For an evacuated, insulated, rigid tank filled from a line, m_2u_2 = m_ih_i. With u = c_vT and h = c_pT, the final temperature is T_2 = \gamma T_i, where \gamma = c_p/c_v. Thus the gas in the tank becomes hotter than the line gas. When an insulated rigid tank is filled from a line at T_s, the energy balance gives m_2T_2 - m_1T_1 = \gamma T_s(m_2 - m_1). For a discharge from the same tank, the exit gas leaves with h_e = c_pT_e, and the balance has the same form with T_e replacing T_s. Solving these relations with the mass balance gives the final state. For a tank at constant temperature, the mass removed is m_1 - m_2 = (P_1 - P_2)V/(RT). Gas that escapes from a cylinder at constant temperature into the atmosphere occupies the volume it would have at P_{atm}, so the boundary work against the atmosphere is P_{atm}\Delta V_{atm}. A heater that keeps the tank at constant temperature supplies \int (h - u)\,dm = RT\int dm in total, which equals (P_1 - P_2)V.

Polytropic process of ideal gas

A polytropic process of an ideal gas follows pV^n = \text{constant}, where n is the polytropic index. Special cases help with recognition. For n = 0 the process is isobaric, for n = 1 it is isothermal, for n = \gamma it is isentropic, and for n \to \infty it is isochoric. For n \neq 1, the boundary work is W = \int p\,dV = \frac{p_2V_2 - p_1V_1}{1-n}, and for an ideal gas this equals W = \frac{mR(T_2 - T_1)}{1-n}. The temperature and pressure ratios follow from T_2/T_1 = (V_1/V_2)^{n-1} = (p_2/p_1)^{(n-1)/n}. The internal energy change is \Delta U = mc_v\Delta T with c_v = R/(\gamma - 1), and the first law gives Q = \Delta U + W. The formula W = (p_1V_1 - p_2V_2)/(n-1) is not valid for n = 1, where the work is p_1V_1\ln(V_2/V_1) instead. Not every process is polytropic. For a reversible steady-flow process, the shaft work per unit mass is w = -\int v\,dP, and this can be evaluated directly from a given P-v relation. Internal energy depends only on temperature, so the magnitude of \Delta U between two states on a P-V diagram depends only on the two isotherms that the endpoints lie on. A process that connects the two most widely separated isotherms has the largest |\Delta U|.

Ideal gas with temperature-dependent cv

When c_v depends on temperature, the internal energy of an ideal gas still depends only on T, so \Delta u = \int_{T_1}^{T_2} c_v\,dT. The integral is not simply c_v\Delta T. For a linear relation c_v = a + bT, the result is \Delta u = a(T_2 - T_1) + \frac{b}{2}(T_2^2 - T_1^2). For a molar form \bar c_v = a + bT + cT^2, the molar change is \Delta\bar u = a(T_2 - T_1) + \frac{b}{2}(T_2^2 - T_1^2) + \frac{c}{3}(T_2^3 - T_1^3). Be careful with units: a coefficient multiplied by a temperature difference in kelvin gives kJ/kg or kJ/kmol according to the basis. The pressure does not enter the result, because u is a function of temperature alone. Example: if c_v = 0.6 + 0.0002T kJ/(kg K), the change from 300 K to 500 K is 0.6(200) + 0.0001(500^2 - 300^2) = 120 + 16 = 136 kJ/kg.

Throttling process of real gas

A throttling process is a steady, adiabatic flow through a restriction with negligible changes in kinetic and potential energy and no work, so the enthalpy is constant: h_1 = h_2. The temperature change is described by the Joule-Thomson coefficient \mu_{JT} = \left(\frac{\partial T}{\partial P}\right)_h, and \Delta T \approx \mu_{JT}\Delta P for a small pressure drop. The sign of \mu_{JT} depends on the inlet state. If the inlet lies inside the region where \mu_{JT} > 0, the gas cools on throttling, and if it lies in the region where \mu_{JT} < 0, the gas warms. The boundary between these regions is the inversion curve. Because the sign depends on the state, the outlet temperature of a real gas may be higher, lower, or equal to the inlet temperature. For an ideal gas h depends only on T, so \mu_{JT} = 0 and the outlet temperature equals the inlet temperature. This is why throttling cooling in refrigeration cycles requires a real gas operating below its inversion temperature.