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Feb 20, 2024

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Name: Date: 1/31/24 PTEC C116 Instrumentation 1 – Module 01 Written Exercise – 5 points Process variables affect each other on a continuous basis and processes are designed to accounts for these dependences. The exercises below will help you to apply what you’ve learned about the relationships between process variables, such as pressure, temperature, level, and flow. Exercise #1: The relationship between temperature and pressure in a closed vessel The temperature inside a closed vessel is 70 degrees F and the pressure of the contained gas is atmospheric pressure (14.7 PSIA). 1. If the temperature increases to 100 degrees, what is the pressure (measured in PSIA) inside the vessel? Show the equation and your calculations below. P 1 = P 2  P 2 = P 1 T 2  P 2 =(14.7 PSIA)(100 F)  P 2 = (14.7 PSIA)(560 R) T 1 T 2 T 1 70 F (530 R) P 2 = 15.5 PSIA 2. What is the pressure measured in PSIG at 70 degrees? PSIG=14.7 PSIA- 14.7 PSIG= 0 3. What is the pressure measured in PSIG at 100 degrees? PSIG= 15.5 PSIA -14.7 PSIG= 0.8 4. A fluid is pumped into a closed vessel at 65 degrees F. After the vessel is filled, the pressure is increased from 14.7 PSIA to 29.4 PSIA. What is the temperature of the pressurized fluid? Show the equation and your calculations below. P 1 = P 2  T 2 = P 2 T 1  (29.4 PSIA)(65 F)  (29.4 PSIA)(525 R)  T 1 T 2 P 1 14.7 PSIA 14.7 PSIA T 2 = 1050 R – 460 R= 590 F

Exercise #2: The relationship between pressure and head Concepts to remember:  Static head, or "head" is a common term for liquid level or height and is usually expressed in feet or in PSI.  Static pressure depends on gravity, the density of the fluid (sometimes expressed as specific gravity), and the height of the fluid.  As the fluid height increases, static pressure increases.  As the fluid density/specific gravity increases, static pressure increases. To find static pressure at any point in a column of liquid:  Calculate the specific gravity of the liquid. Specific gravity is the ratio of the density of liquid to the density of water.  Multiply the specific gravity by 0.433.  Multiply by the liquid head in feet above the point being measured.  The following formula can be used to calculate static pressure in PSI: Formula Description P=(SG) (0.433)(h) Where: P = the pressure at the point being measured in the tank (in PSI) h = height of the fluid in feet above the measured point SG = the specific gravity of the fluid (0.433) is a constant and represents the fact that one foot of water exerts .433 lbs of pressure at one foot Exercise #3 – Practical Problem A vented tank is filled with 15 feet of oil that has specific gravity of .75. What is the pressure of the liquid at each point in the tank in PSI? Show your calculations below.

5. Pressure at Point A P= 0.75(0.433)(10) P= 3.2475 6. Pressure at Point B P= (0.75)(0.433)(5) P= 1.62375 7. Pressure at Point C P= (0.75)(0.433)(0) P= 0 Exercise #4: Practical Questions 8. The pressure gauge on a closed vessel reads 17.2 PSIG. Within a few minutes, the pressure has increased to 20.5 PSIG. What could be the cause for the pressure increase? Pressure increase can be due to an increase in temperature or there being a chemical reaction. A chemical reaction would produce more gas molecules in the closed vessel. A B C 10 Feet 5 Feet Tank Bottom specific gravity of oil =.75

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9. Why do long sections of piping require the use of pressure safety relief valves? The use of pressure safety relief valves on long sections of piping protects it from overpressure. 10.The boiling point of a “light product” such as Propane is below ambient temperature (-44 degrees F) at standard atmospheric pressure (14.7 psia). What will happen to liquid propane entering a closed vessel at 14.7 psia that is at standard atmospheric pressure? Liquid propane would not change in temperature or pressure when entering a closed vessel at 14.7 PSIA because Propane at -44 degrees F is used to that environment.