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A liquid-vapor mixture of 5kg of water with a quality of 0.75 and a pressure of 5bar is expanded in an isobaric process until no liquid is left in the mixture. Then, the system is expanded in a...

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A liquid-vapor mixture of 5kg of water with a quality of 0.75 and a pressure of 5bar is expanded in an isobaric process until no liquid is left in the mixture. Then, the system is expanded in a process where the pressure changes linearly with respect to the volume (P=AV+B) until the pressure and temperature are 1.5bar and 120?C respectively. You are asked to:

(a) Sketch the processes in a P-v diagram with respect to the saturation lines.

(b) Determine the volumes of the three states in m3.

(c) Determine the total work in kJ.

(d) Determine the total heat transfer in kJ.

Problem #2:

A confined quantity of air undergoes two thermodynamic processes departing from an initial equilibrium state where the temperature is 20°C, the absolute pressure is 1bar, and the volume is 0.2m3. First, the volume of the system is doubled through an isobaric process. Then, the volume is reduced back to the initial value through a process where the relation P?V1.4 is maintained constant. You may assume average values for the specific heat capacities (cpand cv) to:

(a) Sketch the processes in a P-V diagram.

(b) Determine the temperature of the three states in K.

(c) Compute the total work in kJ.

(d) Compute the total heat transfer in kJ.

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Problem #1: A liquid-vapor mixture of 5kg of water with a quality of 0.75 and a pressure of 5bar is expanded in an isobaric process until no liquid is left in the mixture. Then, the system is expanded in a process where the pressure changes linearly with respect to the volume (P=AV+B) until the pressure and temperature are 1.5bar and 120?C respectively. You are asked to: (a) Sketch the processes in a P-v diagram with respect to the saturation lines. (b) Determine the volumes of the three states in m3. (c) Determine the total work in kJ. (d) Determine the total heat transfer in kJ. Problem #2: A confined quantity of air undergoes two thermodynamic processes departing from an initial equilibrium state where the temperature is 20°C, the absolute pressure is 1bar, and the volume is 0.2m3. First, the volume of the system is doubled through an isobaric process. Then, the volume is reduced back to the initial value through a process where the relation P?V1.4 is maintained constant. You may assume average values for the specific heat capacities (cp and cv) to: (a) Sketch the processes in a P-V diagram. (b) Determine the temperature of the three states in K. (c) Compute the total work in kJ. (d) Compute the total heat transfer in kJ.

Answered Same Day Dec 22, 2021

Solution

Robert answered on Dec 22 2021
123 Votes
Question 1-
For state 1
m= 5 Kg
X= 0.75
P1= 5 Bar
Process 1-2 = constant pressure process
For state 2
X=1, because no liquid left in mixture at state 2
(1) Answer
The above figure shows the P-V diagram for the given process,
ï‚· Process 1-2 is isobaric process so pressure will remain constant
ï‚· Process 2-3 is linear or pressure varies linearly with volume as (P=AV+B) so the
process line will be straight and inclined
 Saturation line is also mentioned in diagram for pressure –volume
ï‚· Point 1 is in liquid-vapor mixture
ï‚· Point 2 is on saturation line
ï‚· Point 3 is also in liquid-vapor mixture
(2) Answer
Volume in all three states
Volume at state 1-
V1= V f + x V fg
Now from steam table for pressure of 5 bars, we will take the value of V f & V fg
V1= V f + x V fg = V f + x (V g – V f)
V1= 0.001093 +0.75(0.37483-0.001093) = 0.28139575
V1= 0.28139575 m3/Kg
V1= 5 x 0.28139575 = 1.406 m
3
1
P
V
2
3
Saturation line
Mixture
Vapor
superheated
Liquid
Volume at state 2-
V2= V g = 0.37483 m
3/Kg
Or in m3, we have to multiply by 5 kg mass
V2= 1.874 m
3
Volume at state 3-
P=AV+B
P2=AV2+B
P3=AV3+B
Or
P2 - P3= A (V2 - V3)
P2 = P1 = 5 Bar
P3 = 1.5 bar
V2= 0.37483 m
3/Kg = 1.874 m3
As you have not mention the value of A, so we have assumed that A is 12 Ba
m3
So,
5 -1.5 = 12 (0.37483 - V3)
0.2916667 = 0.37483 - V3
V3 =...
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