Throttling Process – Isenthalpic Process
A throttling process is a thermodynamic process in which the enthalpy of the gas or medium remains constant (h = const). The throttling process is one of the isenthalpic processes. During the throttling process, no work is done by or on the system (dW = 0), and usually, there is no heat transfer (adiabatic) from or into the system (dQ = 0). On the other, the throttling process cannot be isentropic. It is a fundamentally irreversible process. Characteristics of throttling process:
- No Work Transfer
- No Heat Transfer
- Irreversible Process
- Isenthalpic Process
Throttling of the flow causes a significant reduction in pressure because a throttling device causes a local pressure loss. Throttling can be achieved simply by introducing a restriction into a line through which a gas or liquid flows. This restriction is commonly done using a partially open valve or a porous plug. Such pressure losses are generally termed minor losses, although they often account for a major portion of the heat loss. The minor losses are roughly proportional to the square of the flow rate, and therefore they can be easily integrated into the Darcy-Weisbach equation through resistance coefficient K.
Throttling of Wet Steam
Wet steam is characterized by the vapor quality, ranging from zero to unity – open interval (0,1). Throttling of the wet steam is also associated with the conservation of enthalpy. Enthalpy is conserved because no work is done by or on the system (dW = 0), and usually, there is no heat transfer (adiabatic) from or into the system (dQ = 0). But in this case, a reduction in pressure causes an increase in vapor quality. As the pressure drops, some of the liquid in the wet steam vaporizes and increases the vapor quality (i.e., dryness fraction). This process takes place because the saturation temperature is lower at the lower pressure. The lower temperature, lower pressure, higher quality steam contains the same enthalpy as the original steam.
Example: Throttling of Wet Steam
A high-pressure steam turbine stage operates at a steady state with inlet conditions of 6 MPa, t = 275.6°C, x = 1 (point C). Steam leaves this turbine stage at a pressure of 1.15 MPa, 186°C, and x = 0.87 (point D). Determine the vapor quality of the steam when throttled from 1.15 MPa to 1.0 MPa. Assume the process is adiabatic and no work is done by the system.
See also: Steam Tables
The enthalpy for the state D must be calculated using vapor quality:
hD, wet = hD,vapor x + (1 – x ) hD,liquid = 2782 . 0.87 + (1 – 0.87) . 790 = 2420 + 103 = 2523 kJ/kg
Since it is an isenthalpic process, we know the enthalpy for point T. From steam tables we have to find the vapor quality using the same equation and solving the equation for vapor quality, x:
hT, wet = hT,vapor x + (1 – x ) hT,liquid
x = (hT, wet – hT, liquid) / (hT, vapor – hT, liquid) = (2523 – 762) / (2777 – 762) = 0.874 = 87.4%
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°C to 179.9°C.