# Exercise 3 Solutions

# 10.3. Exercise 3 Solutions#

```
#5.3.2 Exercise 1
#A closed rigid container of 0.5 m3 is placed on a hot plate. Initially it contains a two phase mixture of saturated liquid and saturated vapor H2O at 𝑝1 = 1 bar with x = 0.5. After heating 𝑝2 = 1.5 bar.
#Determine T at states 1 and 2. Determine the heat transfer during the process.
V1 = 0.5 #m3
Q1 = 0.5
P1 = 100*1000 #Pa
P2 = 150*1000 #Pa
#Using Cantera:
import cantera as ct
#define state 1
species1=ct.Water()
species1.PQ=P1,Q1
print(f"Initial temperature: {round(species1.TQ[0],2)} K")
v1 = v2 = species1.TV[1] # Initial and final specific volume since it is a rigid container
#define state 2
species2=ct.Water()
species2.PV=P2,v2
print(f"Final temperature: {round(species2.TQ[0],2)} K")
m = V1/v1 # kg
u1 = species1.UV[0]
u2 = species2.UV[0]
Q = m*(u2-u1) # From Delta_U = Q-W, work is zero because volume is fixed
print(f"Heat Transfer: {round(Q/1000,2)} kJ")
```

```
Initial temperature: 372.81 K
Final temperature: 384.55 K
Heat Transfer: 298.1 kJ
```

```
#5.3.2 Exercise 2
#A piston cylinder assembly initially (State 1) contains 10 kg of H2O at 1200 kPa.
# a) If 8 kg is in the liquid form and the rest vapor, determine the temperature, specific volume and specific enthalpy.
# b) The cylinder continues to be heated at constant pressure until the temperature reaches 300 °C (State 2). Determine the final state (compressed liquid, saturated mixture, superheated vapor), specific volume and specific enthalpy.
# c) Determine the heat transfer and work for this process.
# part a
m=10 #kg
P1 = 1200*1000 #Pa
mg = 2 #kg
Q1 = mg/m
#Using Cantera:
import cantera as ct
#define state 1
species1=ct.Water()
species1.PQ=P1,Q1
T1 = species1.TV[0]
v1 = species1.TV[1]
h1 = species1.HP[0]
print(f"Initial temperature: {round(T1,2)} K")
print(f"Initial specific volume: {round(v1,6)} K")
print(f"Initial specific enthalpy: {round(h1,2)} K")
# part b
T2 = 300+273.15 # K
P2=P1
#define state 2
species2=ct.Water()
species2.TP=T2,P2
v2 = species2.TV[1]
phase = species2.phase_of_matter
h2 = species2.HP[0]
print(f"Final state of matter: {phase}")
print(f"Final specific volume: {round(v2,6)} K")
print(f"Final specific enthalpy: {round(h2,2)} K")
# part c
Q = m*(h2-h1) # J, constant pressure process
W = P2*(v2-v1) # J,constant volume process
print(f"Work: {round(W/1000,2)} kJ")
print(f"Heat Transfer: {round(Q/1000,2)} kJ")
```

```
Initial temperature: 461.18 K
Initial specific volume: 0.033582 K
Initial specific enthalpy: -14774746.42 K
Final state of matter: gas
Final specific volume: 0.213816 K
Final specific enthalpy: -12925000.32 K
Work: 216.28 kJ
Heat Transfer: 18497.46 kJ
```

```
#5.3.2 Exercise 3
#Water initially at 200 kPa and 300C is contained in a piston-cylinder device fitted with stops. The water is allowed to cool at constant pressure until it exists as a saturated vapor and the piston rests on its stops. Then the water continues to cool until the pressure is 100 kPa.
import cantera as ct
P1 = 200e3 #Pa
T1 = 300 + 273.15 #K
P2 = P1
P3 = 100e3 #Pa
Q2 = 1
state1 = ct.Water()
state1.TP = T1, P1
state2 = ct.Water()
state2.PQ = P2, Q2
v2 = state2.TV[1]
v3 = v2
state3 = ct.Water()
state3.PV = P3, v3
# Part A
# Find internal energy change between states 1 and 3
state1 = ct.Water()
state1.TP = T1, P1
u1 = state1.UP[0]
# P2 = P1, Q2 = 1
state2 = ct.Water()
state2.PQ = P2, Q2
u2 = state2.UP[0]
deltau21 = u2 - u1
print(f"A) : The change in internal energy between states 1 and 2 is {deltau21/1000} kJ/kg")
```

```
A) : The change in internal energy between states 1 and 2 is -279.015197695978 kJ/kg
```

```
# Part B
v1 = state1.TV[1]
v2 = state2.TV[1]
print(f"B) : The change in specific volume between states 1 and 2 is {(v2 -v1)/1000} kJ/kg")
```

```
B) : The change in specific volume between states 1 and 2 is -0.00043035589952211537 kJ/kg
```

```
# Part C
h1 = state1.HP[0]
h2 = state2.HP[0]
print(f"C): The change in specific enthalpy between states 1 and 2 is {round((h2 - h1)/1000,2)} kJ/kg")
```

```
C): The change in specific enthalpy between states 1 and 2 is -365.09 kJ/kg
```

```
# Part D
Q = (u2 - u1) + (P1*(v2 - v1))
print(f"D) : The heat transfer between states 1 and 2 is {round(Q/1000,2)} kJ/kg")
```

```
D) : The heat transfer between states 1 and 2 is -365.09 kJ/kg
```

```
# Part E
# specific heat transfer from cooling to stops
print(f"E) : The heat transfer between states 1 and 2 is {round((h2 - h1)/1000,2)} kJ/kg")
```

```
E) : The heat transfer between states 1 and 2 is -365.09 kJ/kg
```

```
# Part F
state3 = ct.Water()
state3.PV = P3, v3
u3 = state3.UP[0]
deltau32 = u3 - u2
print(f"F) : The change in internal energy between states 2 and 3 is {deltau32/1000} kJ/kg")
```

```
F) : The change in internal energy between states 2 and 3 is -1020.621570467569 kJ/kg
```

```
# Part G (total specific heat transfer)
q12 = h2 - h1
q23 = u3 - u2
qtotal = q12 + q23
print(f"G) : The total specific heat transfer is {round(qtotal/1000, 2)} kJ/kg")
```

```
G) : The total specific heat transfer is -1385.71 kJ/kg
```