![]() ![]() ![]() Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 8-1. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983). In this case of the throttling process (1.15MPa to 1MPa) the vapor quality increases from 87% to 87.4% and the temperature decreases from 186☌ to 179.9☌. H T, wet = h T,vapor x + (1 – x ) h T,liquid From steam tables we have to find the vapor quality using the same equation and solving the equation for vapor quality, x: Since it is an isenthalpic process, we know the enthalpy for point T. H D, wet = h D,vapor x + (1 – x ) h D,liquid = 2782. The enthalpy for the state D must be calculated using vapor quality: Assume the process is adiabatic and no work is done by the system. Determine the vapor quality of the steam when throttled from 1.15 MPa to 1.0 MPa. ![]() Steam leaves this stage of turbine at a pressure of 1.15 MPa, 186☌ and x = 0.87 (point D). However in special cases, the temperature may remain the same or it may increase.Ī high-pressure stage of steam turbine operates at steady state with inlet conditions of 6 MPa, t = 275.6☌, x = 1 (point C). Normally the temperature of the fluid will drop. If there is a change in the internal energy, u, then there must be a temperature change. Because mass flow is constant, the change in specific volume is observed as an increase in gas velocity, and this is also verified by observations. Therefore, if pressure decreases then specific volume must increase if enthalpy is to remain constant (assuming u is constant). The specific enthalpy is equal to the specific internal energy of the system plus the product of pressure and specific volume. We can also observe that specific enthalpies remains the same, i.e. From experience we can observe that: p in > p out, v in < v out, where p is the pressure and v is the specific volume. The minor losses are roughly proportional to the square of the flow rate and therefore they can be easy integrated into the Darcy-Weisbach equation through resistance coefficient K.įor example, consider a throttling of an ideal gas flowing through a valve that is partially open. Such pressure losses are generally termed minor losses, although they often account for a major portion of the head loss. This restriction is commonly done by means of a partially open valve or a porous plug. A throttling can be achieved simply by introducing a restriction into a line through which a gas or liquid flows. Characteristics of throttling process:Ī throttling of the flow causes significant reduction in pressure, becauses a throttling device causes a local pressure loss. On the other the throttling process cannot be isentropic, it is a fundamentally irreversible process. ![]() During the throttling process no work is done by or on the system (dW = 0), and usually there is no heat tranfer ( adiabatic) from or into the system (dQ = 0). In fact, the throttling process is one of isenthalpic processes. A throttling process is a thermodynamic process, in which the enthalpy of the gas or medium remains constant (h = const). ![]()
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