How to Solve a Basic Heat Transfer Problem in Thermodynamics

Last Updated: July 6, 2024

Thermodynamics is a difficult subject for anyone. This wikiHow hopes to help instruct thermodynamics students in the basics of ideal gas law and heat transfer. This will be going over solving an energy balance problem that can be used in heat transfer. Almost all ideas and laws applied in this problem can be used in other questions too and is a good example for the basics of thermodynamics.

Part 1

Part 1 of 9:

Reading the Question

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Read the question. For example, your question might be the following. Two tanks are connected by a valve. One tank contains 2 kg of carbon monoxide gas at 77 °C (171 °F) and 0.7 bar. The other tank holds 8 kg of the same gas at 27 °C (80.6 °F) and 1.2 bar. The valve is opened and the gases are allowed to mix while receiving energy by heat transfer from the surroundings. The final equilibrium temperature is 42 °C (108 °F). Using the ideal gas model, determine the final equilibrium pressure, in bar; the heat transfer for the process in kJ
- Note it is easier to solve the problem by working with just variables, and then at the last step plug in the values. This same method will be followed here.
Using the information from the problem we know that both tanks have the same gas, one tank has 2 kg of gas at 77 °C (171 °F) at 0.7 bar. The other tank has 8 kg of gas at 27 °C (80.6 °F) and 1.2 bar. We also know the final temperature of the system is 42 °C (108 °F).

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To solve the problem in a closed system, 0.25kg of air initially at 1.034bar with a specific volume of 0.849 meter (2.8 ft)-cube/kg is compressed reversibly according to the law PV RAISE TO POWER 1.3 EQUALS CONSTANT until its pressure is 2.068bar.the specific internal energy of the air is 1.58pv where p is in KN/METERSQUARE and v is in meter-cube per kilogram determine the heat transfer.
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Write the assumptions needed to solve. These assumptions are obtained through using the problem information and inferring ways that can be applied to this problem. For this problem the assumptions are as follows:
- The total amount of carbon monoxide gas is a closed system (the carbon monoxide gas cannot leave or enter the system)
- The gas is modeled as an ideal gas with constant c _v . (This was assumed because the problem stated that the ideal gas model can be used and cv can only be used in an ideal situation)
- The gas initially in each tank is in equilibrium. The final state is an equilibrium state too. (This is assumed because the problem wants us to analyze the final equilibrium state)
- No energy is transferred to, or from the gas by work. (This assumption is that energy is conserved because there are no external forces doing work on the system)
- There is no change in kinetic or potential energy. (Assumption based on conservation of energy due to the assumption above)
Use the ideal gas law. P _f is the final equilibrium pressure, V is the total volume of the system after the valve is release, m is the total mass of the system, R is the universal gas constant with a known value, and T _f is the final equilibrium temperature that is given.
Rework equation 1 to solve for P _f , by dividing by volume.
The mass is the total mass of the two tanks because now both tanks are mix in this final state. The total mass is used because we are evaluating the final pressure in the final state. This is the state in which the gas is mixed together so the mass of the whole system needs to be considered.
The volume V is the total amount of volume from both tanks for the same reason as the mass. Unfortunately, the volume of the tanks is not given so we need to solve for it.
Since the initial pressure, temperature, and mass are given the initial volume of each tank can be calculated using the ideal gas equation shown in equation 1. This is where V ₁ , P ₁ , and T ₁ denote the conditions in tank 1, and V ₂ , P ₂ , and T ₂ denote the initial conditions in tank 2. Reworking the ideal gas law to solve for V, by dividing by pressure:
Substituting Values into Equation 3 Solving for Pf. Equation
This can be done by using the universal gas constant.
These known values should be:m ₁ = 2 kg, m ₂ = 8 kg, T ₁ = 77 °C (171 °F) , T ₂ = 27°C, P ₁ = 0.7 bar, P ₂ = 1.2 bar, T _f = 42°C
Solving the equation gives a final pressure of 1.05 bar.

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Part 2

Part 2 of 9:

Solving for the Heat Transfer

Set up an energy balance equation for the system using the general energy balance equation shown below, where ∆U is the change in internal energy, Q is the energy produce by heat transfer, and W is the work.
This simplifies the equation above by setting work to zero.
Simplifying ∆U gives us: where U _f is the final internal energy, and Ui is the initial internal energy.

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Part 3

Part 3 of 9:

Finding the Initial Internal Energy of System

The initial internal energy is a summation of the internal energy in each tank at the beginning of the process. The general internal energy equation is shown below, where m is the total mass, and u(T) is the internal energy evaluated at temperature T.
Using the equations above we find the initial internal energy, where m1 is the mass in tank 1, m2 is the mass in tank 2, and T1 and T2 are the initial temperatures in tank one and tank two respectively.

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Part 4

Part 4 of 9:

Finding the Final Internal Energy

The final internal energy is found the same way as the initial internal energy. Use equation 16, yet use the total mass of the system. The final internal energy is as follows:

Part 5

Part 5 of 9:

Substituting Known Information into Energy Balance Equation

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Part 6

Part 6 of 9:

Applying the Law of Specific Heat Constants

The law of specific heats allows for a simplification in the difference of the internal energies at two temperatures. The use of a specific heat constant, c _v , allows for the simplification of the difference of the internal energies at two states to just the temperatures at these states. This law applies to only ideal gases, and can be used due to our assumption of ideal gas. The relationship is seen below in equation 23.
Applying this to equation 22 we get

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Part 7

Part 7 of 9:

Looking in the Table T-10 to find the c_v constants

Convert the temperature from Celsius to Kelvin by adding 273 to both initial temperatures. 273 is the conversion factor from Celsius to Kelvin. The temperatures will be at 300 K, and 350 K.
Look in the table for carbon monoxide gas at the values for the temperatures at 300 K and 350 K. Pay attention to only look at the table for the cv constant as the cp is for enthalpy. What you should look at is shown below:

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Part 8

Part 8 of 9:

Finding the specific heat constant for the final equilibrium temperature

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The final equilibrium temperature is 315 Kelvin. The c _v constant needs to be evaluated at that temperature to be accurate. This is done through interpolation. Interpolation is the technique of assuming that the data is related linearly, and with two points one can calculate the value at a third point. Yet for this case, the difference between the c _v values is extremely small. Thus this interpolation can be assumed to just be an average of the two numbers. The mean is calculated below in equation 25.
- This c _v value ends up being 0.745 kJ/kg*K

Part 9

Part 9 of 9:

Beginning to input the information into equation 24

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Input all temperatures in Kelvin. To keep units consistent the temperatures must be input in kelvin
- The values that should be input are: m ₁ = 2 kg, m ₂ = 8 kg, T ₁ = 350 K, T ₂ = 300 K, T _f = 315 K, c _v = 0.745 kJ/kg*K
- The final solution should be that the total heat transfer of the system is 37.25 kJ. The plus sign informs us that there is heat transfer into the system.
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A plus sign for heat transfer indicates transfer into the system, while negative is transfer out of the system.

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Remember that this problem assumes the gases are ideal, and this is not true for high pressures and low temperatures.

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Always solve the problem as variables only first, and then input values.

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