A piston/cylinder has 2 lbm of R-134 a at state with 200 F, 90 lbf / in^2, and is then brought t

step 1 All right, guys, we are given the following data for R134A. We've got B1 is equal to 90 LBF per

A piston/cylinder has 2 lbm of R-134 a at state with 200 F,...

A piston/cylinder has 2 lbm of $R-134 a$ at state with $200 \mathrm{F}, 90 \mathrm{lbf} / \mathrm{in}^{2},$ and is then brought to saturated vapor, state $2,$ by cooling while the piston is locked with a pin. Now the piston is balanced with an additional constant force and the pin is removed. The cooling continues to state $3,$ where the $R-134 a$ is saturated liquid. Show the processes in a $P-V$ diagram and find the work in each of the two steps, 1 to 2 and 2 to 3

Key Concepts

Saturation Properties and Phase Change

The concept of saturation properties involves the thermodynamic conditions at which a substance exists in equilibrium between two phases, typically as a saturated liquid and a saturated vapor. Understanding these properties is critical for analyzing phase change processes, where energy transfer leads to changes in the state of the substance without a change in temperature until the transition is complete.

Constant Volume (Isochoric) Process

A constant volume or isochoric process is one in which the volume remains fixed throughout the process. Since there is no change in volume, the work done by or on the system is zero, even though other state properties like pressure or temperature may change during the process.

Constant Pressure (Isobaric) Process

A constant pressure or isobaric process is one during which the external pressure acting on the system remains constant. In this type of process, any work done by the system is directly proportional to the change in volume, as expressed by the relation W = P?V.

Pressure-Volume (P-V) Diagram

A P-V diagram is a graphical representation of the changes in pressure and volume of a system during a thermodynamic process. It helps visualize different types of processes (e.g., constant volume, constant pressure) and is particularly useful in understanding work interactions, phase changes, and overall energy transfers within a cycle.

Work in Thermodynamic Processes

In thermodynamics, work is defined as the area under the process curve on a P-V diagram. For different processes, the calculation of work varies; for example, in constant volume processes no work is done since there is no change in volume, whereas in constant pressure processes, work is computed by multiplying the external pressure by the change in volume.

Transcript

00:01 All right, guys, we are given the following data for r134a.

00:04 We've got b1 is equal to 90 lbf per square inch.

00:12 Then we've got m is equal to 2lbm.

00:17 We've got t1 is equal to 200 fahrenheit.

00:21 Then we've got v1 is equal to v2.

00:26 So from superheated r134a table f10.

00:31 Corresponding to pressure p is equal to 80 psi and temperature t1 that is 200 degrees fahrenheit.

00:44 We can obtain specific volume and internal energy as v is equal to 0 .8205 cubic feet per lbm then we've got internal energy u is equal to 194 .83 btu per lbm.

01:09 Then from superheated r134a table f10 .2 corresponding to p 100 and temperature 200 degrees fahrenheits.

01:23 We can obtain specific volume and internal energy as v is equal to v is equal to 0 .646 cubic fit per lbm.

01:41 And then we've got u is equal to 194 .38 btu per lbm.

01:50 So calculating stage 1 specific volume and internal energy corresponding to p1, we've got that is 90 psi.

02:01 So we have v1 is equal to 0 .825 plus 0 .646 divided by 2 is equal to 0 .73355 cubic fit per lbm.

02:21 Then we've got it's lbm.

02:25 So then we've got u1 that is equal to 194 .83 plus 194 .38.

02:33 Divided by 2 is equal to 194 .605 btu per lbm.

02:44 So guys, in process 1 to 2 piston is locked with bin.

02:48 So this is a constant volume process.

02:50 So the work done is equal to 0.

02:53 Hence, v2 is equal to v1 and that is equal to 0 .73355 cubic fit per lbm.

03:05 We know that the work of stage 1 to 2 is equal to 0.

03:11 So from superheated r134a table f 10 .2 corresponding to vg, that is 0 .6632 cubic fit per lbm.

03:29 We can obtain it's lbm that we write this with clarity it's lbm so we can obtain specific internal energy and pressure that is u is equal to 166 .28 b t u per lbm and then we've got p is equal to 72 .271 p s ia so from our a 134a table f 10 .2 corresponding to vg that is 0 .7921 cubic fit per lbm.

04:14 We can obtain specific internal energy and pressure that is u is equal to 164 .945 btu per lbm.

04:26 And then we've got p is equal to 60 .311 p .m.

04:31 So calculating stage 2 specific internal energy and pressure by using method of linear interpolation that is u2 is equal to 166 .28 plus 0 .73375 minus 0 .6632 divided by 0 .79632 divided by 0 .79632 multiplied by 0 .762 multiplied by 0 .7632 multiplied by 0 .662, multiplied by 2.

05:03 By 164 .9 .95 minus 166 .28...

Please give Ace some feedback