How Can I Solve These Thermodynamics Problems in Building Systems Engineering?

In summary, the conversation is about someone needing help with thermodynamics in their building systems engineering technology program. They are having trouble with two problems in their book and are looking for someone to show them the steps on how to solve them. The first problem involves finding the gage pressure using the formula pv = mrt. The second problem involves expanding gas in a rigid container and requires converting units from pounds per square inch to kelvin/rankin. The key to solving these problems is to always include units in equations and to carefully consider the given information.
  • #1
dealz
4
0
Need help with thermodynamics!

im studying building systems engineering technology. yeh i know it's only a technology program at the cegep level but when I'm done with it, I'm going to do software engineering in university. anyways I'm having trouble to understand two problems in the book. if anyone can help me with this, it would be greatly appreciated.

1) Determine the gage pressure if the atmospheric pressure is 14.7 psia, the gas constant is 96 ft-lb/lb R, and the specific volume is 10 ft^3/lbm.
The answer is 31.9 psig

2) 70 lbs mass of gas are contained in a rigid container at 200 psia and 80 F. The gas is then expanded to fill a 2000 ft^3 volume at a pressure pf 20 psia and a temperature of 70 F. Determine the volume of the rigid container.
The answer is 203.7 ft^3.

Can someone show me the steps on how they got the answer because i already know the answer. i just need help on how to solve the problem.

for the first question, the formula is pv = mrt. we'll i know right away to find P = mRT. i would multiply 96ft-lb*700 * R (I don't know how to get that)/ Volume (I also don't know how to get that but i know it's something to do with the specific volume of 10 ft^3)
for the second question, i multiplied (70)(80)/200. and from there I'm lost and don't know what to do.

specific volume is the volume per unit mass. (ft^3/slug)
for number 1ok i think when it says the specific volume is 10 ft^3/lbm, i did 1/(10ft^3/lbm) to get the density.
then i mutiplied (0.1)(96)(700) = 6720. but the answer is 31.9 psig so i know I'm really off even though i didn't even finished my answer.
 
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  • #2
The biggest trick to solving thermodynamics problems is to always include units in your equations so that you can see if the end result makes sense. You are throwing around so many numbers without units that you don't realize you have an answer that doesn't make sense. The pressure is given in pounds per square inch and the volume is given in feet cubed. Your method for the first one is okay, but it looks like you have some converting to do before you plug numbers in.

For the second one you have similar problems mixing units. Also always make sure never to use celcius/farenheit, you want to convert them to kelvin/rankin. After you get passed units, you have 2 states described in the problem each which you can describe with an equation. After you describe each state with an equation I'm sure you'll be able to see where to go from there.
 
  • #3


Hello,

I can definitely help you with these thermodynamics problems! Let's start with the first question.

1) Determine the gage pressure if the atmospheric pressure is 14.7 psia, the gas constant is 96 ft-lb/lb R, and the specific volume is 10 ft^3/lbm.

To solve this problem, we will use the ideal gas law, which is PV = mRT. In this equation, P is pressure, V is volume, m is mass, R is the gas constant, and T is temperature. We also need to keep in mind that the atmospheric pressure is acting on the gas, so we have to subtract it from the total pressure to get the gage pressure.

First, we need to rearrange the ideal gas law to solve for pressure (P). This gives us P = mRT/V. Then, we can plug in the given values: m = 1 lbm (since the specific volume is given as 10 ft^3/lbm), R = 96 ft-lb/lb R, T = 700 R (since 80 F = 540 R and we need to add 460 to convert to absolute temperature), and V = 10 ft^3.

So, P = (1 lbm)(96 ft-lb/lb R)(700 R) / (10 ft^3) = 6720 ft-lb/ft^3.

However, this is in units of ft-lb/ft^3, not pressure. To convert to pressure, we need to divide by the gravitational acceleration (32.2 ft/s^2). This gives us a final answer of 208.7 psf (pounds per square foot).

But, we need to subtract the atmospheric pressure (14.7 psia) to get the gage pressure. So, the final answer is 208.7 psf - 14.7 psia = 194 psig.

For the second question, we will use the same ideal gas law, but we need to use it twice - once before the gas is expanded and once after.

2) 70 lbs mass of gas are contained in a rigid container at 200 psia and 80 F. The gas is then expanded to fill a 2000 ft^3 volume at a pressure pf 20 psia and a temperature of 70 F. Determine the volume of the rigid container.

 

Related to How Can I Solve These Thermodynamics Problems in Building Systems Engineering?

What is thermodynamics?

Thermodynamics is the branch of physics that deals with the study of energy and its transformation from one form to another.

What are the laws of thermodynamics?

The laws of thermodynamics are fundamental principles that govern the behavior of energy in a given system. The first law states that energy cannot be created or destroyed, only transformed. The second law states that in any energy transformation, some energy will be lost as heat. The third law states that as temperature approaches absolute zero, the entropy (degree of disorder) of a system also approaches zero.

How is thermodynamics used in everyday life?

Thermodynamics has many practical applications in our daily lives. It helps us understand and improve the efficiency of engines, refrigerators, air conditioners, and other machines. It also plays a crucial role in determining the feasibility and sustainability of different energy sources.

What is the difference between thermodynamics and heat transfer?

Thermodynamics focuses on the macroscopic behavior of energy in a system, while heat transfer deals with the transfer of thermal energy between objects. Thermodynamics is a more general and theoretical approach, while heat transfer is more specific and practical.

What are some common misconceptions about thermodynamics?

One common misconception is that thermodynamics only applies to heat and temperature, when in fact it also encompasses other forms of energy. Another misconception is that energy can be created or destroyed, which goes against the first law of thermodynamics. Additionally, some people may believe that thermodynamics only applies to large systems, but it can also be applied to microscopic systems such as individual molecules.

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