CEE 230 quiz 1 practice problems solutions

CEE 230: Quiz 1 Study Problems NOTE: Question 1 is much harder and longer than the other problems that are provided. Question 1: This will incorporate a lot of different skills required to create T-V and P-V diagrams, analyze the state of a substance, and use the property Tables. This question can be LONG and very CHALLENGING, but it should be good practice! Given: A system consisting of 1 kg of H 2 O undergoes a power cycle of the following processes. Process 1-2: Constant Pressure Heating at 10 bar from saturated vapor Process 2-3: Constant-Volume Cooling to P 3 = 5 ¯ ¿ , T 3 = 160 ℃ Process 3-4: Isothermal compression with Q 3 − 4 =− 815.8 kJ Process 4-1: Constant volume heating Find: Create a T-V diagram and P-V diagram for the cycle of processes and find the Pressure, specific volume (v), and specific internal energy (u) at each state in the table below: State 1 State 2 State 3 State 4 Unit Pressure [bar] Specific Volume, v [m 3 /kg] Specific Internal Energy, u [kJ/kg]

Start with T-V diagram basics: Process 1-2: Constant Pressure Heating at 10 bar from saturated vapor Because you have saturated vapor, state 1 will be located somewhere on the right side of the curve. To get to state 2, we undergo constant pressure heating, so we follow an isobar (a line of constant pressure) to the right to increase the pressure. Think about how a pressure cooker works. Adding heat, increases pressure.

We know that our given T 3 = 160 ℃ is greater than T sat =151.83 o C. So our point 3 will be above this since it’s a higher temperature, we will just follow the y-axis! Since it’s still above the saturated liquid-vapor dome, we remain only vapor! If T3 was less than 151.83, we would be a mixture of liquid and vapor like the diagram on the left, above. Process 3-4: Isothermal compression with Q 3 − 4 =− 815.8 kJ To get to state 4, we know that we undergo isothermal (constant temperature) compression (decrease in volume). Because it’s constant temperature, we follow an isotherm, and because we decrease volume, we go to the left. BUT we have another problem. How far to the left (decrease in volume) do we go???

Because we are a CYCLE, we must begin and end at the same state. So we can look at process 4-1 to guide us. Process 4-1 is Constant volume heating, so increasing the temperature (going up on y-axis) without changing volume (not moving on the x-axis). This is just a vertical line! SO, to answer how far left we go from Process 3-4 (aka how much we decrease volume), we decrease the volume to be immediately below state 1. And that’s our T-V diagram! We can remove the rest of the isobar to clean it up:

Isotherms switch angles! Process 1-2: Constant Pressure Heating at 10 bar from saturated vapor Because you have saturated vapor, state 1 will be located somewhere on the right side of the curve.

To get to state 2, we undergo constant pressure heating, so we follow an isobar (a line of constant pressure) to the right to increase the pressure (in this case, this is just following the x-axis to the right. If you are confused as to why we go to the right, we are adding heat and increasing the temperature of our vapor so we have to “jump isotherms” from T2 to T1).

BUT we have another problem. How far to the left (decrease in volume) do we go??? Because we are a CYCLE, we must begin and end at the same state. So we can look at process 4-1 to guide us. Process 4-1 is Constant volume heating, so increasing the temperature (going up on y-axis) without changing volume (not moving on the x-axis). This is just a vertical line! And that’s our P-V diagram! We can remove the rest of the isotherm to clean it up:

Now for the table: Fill in given information! State 1 State 2 State 3 State 4 Unit Pressure [kPa] Specific Volume, v [m3/kg] Specific Internal Energy, u [kJ/kg] Process 1-2: Constant Pressure Heating at 10 bar is given! Remember to convert units. This gives us P1 and P2 because P stays constant State 1 State 2 State 3 State 4 Unit Pressure 1000 1000 [kPa] Specific Volume, v [m 3 /kg] Specific Internal Energy, u [kJ/kg] Process 2-3: Constant-Volume Cooling to P 3 = 5 ¯ ¿ is given! Remember to convert units. This gives us P3 State 1 State 2 State 3 State 4 Unit Pressure 1000 1000 500 [kPa] Specific Volume, v [m 3 /kg] Specific Internal Energy, u [kJ/kg]

State 1 State 2 State 3 State 4 Unit Pressure 1000 1000 500 618 [kPa] Specific Volume, v 0.194 [m 3 /kg] Specific Internal Energy, u 2583 [kJ/kg] Now for state 4 data since it’s the next easiest to find. Looking at our T-V diagram, we know that v1=v4. State 1 State 2 State 3 State 4 Unit Pressure 1000 1000 500 618 [kPa] Specific Volume, v 0.194 0.194 [m 3 /kg] Specific Internal Energy, u 2583 [kJ/kg] We know that we are a mix of liquid and vapor, so we need to find the quality. From the given data we know that T 3 = 160 C = T4. Use Table A-4 and look up v for T=160 x = ( v 4 − v liquid , sat ) v vap,sat − v liquid , sat = .194 − 0.001102 0.30680 − 0.001102 = 0.63 or 63% Vapor! Using this, we find u4. x = ( u 4 − u liquid ,sat ) u vap, sat − u liquid ,sat rearrange to solve for u4 u 4 = u liq ,sat + x ( u vap, sat + u liq ,sat ) = 674.49 + 0.63 ( 2567.8 − 674.49 ) = 1871 State 1 State 2 State 3 State 4 Unit Pressure 1000 1000 500 618 [kPa] Specific Volume, v 0.194 0.194 [m 3 /kg] Specific Internal Energy, u 2583 1871 [kJ/kg]

Now let’s look again at state 3, since there’s more given information that we haven’t used. Using Table A-6 let’s look up v at T=160C We must interpolate from the Tsat of 151.83C and 200C to find the volume at 160C v 160 = 0.383 Now for v2, States 2 and 3 have the same v. State 1 State 2 State 3 State 4 Unit Pressure 1000 1000 500 618 [kPa] Specific Volume, v 0.194 0.383 0.383 0.194 [m 3 /kg] Specific Internal Energy, u 2583 1871 [kJ/kg] Use A-6 let’s look up u at T=160 o C and 5 bar = 0.5 MPa We must interpolate from the Tsat of 151.83C and 200C to find the internal energy at 160C u 160 = 2574.7 State 1 State 2 State 3 State 4 Unit Pressure 1000 1000 500 618 [kPa] Specific Volume, v 0.194 0.383 0.383 0.194 [m 3 /kg] Specific Internal Energy, u 2583 2574.7 1871 [kJ/kg] Now last is u2: We are given P1 = P2 = 10 bar = 1000kPa = 1MPa Use A-6. We don’t know the temperature, but we have the specific volume at P2! So let’s use the specific volumes at P2 and interpolate to find the internal energy u. U2= 3231! State 1 State 2 State 3 State 4 Unit Pressure 1000 1000 500 618 [kPa]

Question 3: A cylinder-piston system contains 2 kg of H 2 O at 150 kPa and has a volume of 0.35 m 3 . The piston is moved so that the final volume of the inside the cylinder is 2.319 m 3 . During the piston motion from the initial to the final state, there is heat transfer to the cylinder in such a way as to hold the temperature constant. (a) What is the final pressure in the cylinder? (b) How much work was done by the steam? (c) Evaluate the heat transfer during the process. *)Typo: both W and V are absolute work and volume (“upper case”) in the equation above.