# Example Problem with Complete Solution

2D-2 : Dew Point Calculations for Ammonia 4 pts
Ammonia is contained in sealed test tube at 25oC.  The test tube is slowly cooled until liquid ammonia droplets condense on the inside of the test tube. At this point, the temperature of the ammonia is -20oC.
Determine the initial pressure in the test tube, before the cooling process began.

Read : We know the initial temperature of the ammonia.  If we knew the intial temperature, we could look up the specific volume in the Superheated Vapor Tables for ammonia.  There is only one pressure that yields this value of the specific volume when the system is at 25oC.  Therefore, we could also look at the problem in the following way.  We know the intitial temperature of the ammonia and IF we also knew the specific volume of the ammonia, we could use the Superheated Vapor Tables to work backwards and determine the initial pressure!  That is what we are going to need to do in this problem.
When the 1st droplet of liquid appears on the wall of the glass vessel, the vapor inside the vessel is a satuated vapor.  We can look up the properties of this saturated vapor in the Ammonia Tables.  Since the vapor is saturated at -20oC, it must be superheated at 25oC.  But at both the initial and final state the specific volume must be the same because neither the mass nor the volume of the system changed! This is the key to the problem.  Because we know the values of 2 intensive variables at the initial state, specific volume and temperature, and the initial state is a pure substance in a single phase, we can determine the values of ALL other properties!  In this case we need to determine the pressure.
Given: T1 25 oC Find: P1 ??? kPA
T2 -20 oC
x2 1 kg vap/kg
Assumptions: 1- The system contains saturated vapor in the final state.
Equations / Data / Solve:
The specific volume of the system is equal to the specific volume of saturated ammonia vapor at T2.
We can look up this value in the Saturated Temperature table of the Ammonia Tables at -20oC :
Vsat vap 0.62373 m3/kg
Next, we scan the Superheated Ammonia Tables to determine the 2 pressures between which this value of specific volume falls, at the given temperature of 25oC.
Here are the data at 25oC: P (kPa) V (m3/kg)
100 1.1381
200 0.59465
400 0.30941
500 0.25032
600 0.21035
700 0.18145
We need to interpolate between 100 kPa and 200 kPa to determine the system pressure that corresponds to our value of specific volume at a temperature of 25oC. Eqn 1 Eqn 2
slope -184.0134 kPa/(m3/kg) P 194.6 kPa
Verify: The assumption cannot be verified. 