ASSIGNMENT 2
CHE 3473
Solution
#Problem 1: 3.3
a) One kilogram of steam contained in a horizontal frictionless piston and cylinder is heated at
constant pressure of 1.013 bar from 125 C to such a temperature that its volume doubles.
Calculate the amount of heat that must be added to accomplish this change, the final
temperature of the steam, the internal energy and enthalpy changes of the steam for this
process.
b) Repeat for heating at constant volume to a pressure that is twice the initial pressure
c) Repeat part a assuming ideal gas with heat capacity of 34.4 J/mol K
d) Repeat part b assuming ideal gas
Choose system to be the contents of the piston and cylinder
Stage 1: system at the initial state
2: system at the final state
a) Mass balance equation for closed system
M2 – M1 = 0  M 1 = M 2 = M
Energy balance equation for closed system, no shaft work with negligible kinetic and potential
energy
𝑈2 − 𝑈1 = 𝑄 − ∫ 𝑃𝑑𝑉 = 𝑄 − 𝑃(𝑉2 − 𝑉1 )
𝑄 = 𝐻2 − 𝐻1
At initial state
P1 = 1.013 bar ≈ 0.1 MPa
T1 = 125oC
Interpolation from data table in appendix AIII, we get
̂1 = 2726.3
𝐻
𝑘𝐽
𝑘𝑔
𝑉̂1 = 1.8161
𝑚3
𝑘𝑔
̂1 = 2544.8
𝑈
𝑘𝐽
𝑘𝑔
At final state
P2 = 1.013 bar ≈ 0.1 MPa
𝑚
V2 = 2V1  𝑉̂2 = 2𝑉̂1 = 3.6322
3
𝑘𝑔
With known values of P2 and 𝑉̂2 , we could interpolate other values as below
T2 = 514.5oC
̂2 = 3519.5
𝐻
𝑘𝐽
𝑘𝑔
̂2 = 3156.3
𝑈
𝑘𝐽
𝑘𝑔
̂2 − 𝐻
̂1 ) = 793.2 kJ
 Q = H2 – H1 = M(𝐻
W = – P(V2 – V1) = −105
𝑚3
𝑁
∗ 1.8161 ∗ 1𝑘𝑔 = −181.61 𝑘𝐽
𝑚2
𝑘𝑔
̂
̂
∆U = U2 – U1 = M(𝑈2 − 𝑈1 ) = 611.5 kJ
∆H = Q = 793.2 kJ
b) Mass balance equation for closed system
M2 – M1 = 0  M 1 = M 2 = M
Energy balance equation for closed system, no shaft work with negligible kinetic and potential
energy, no change of volume of system
U2 – U1 = Q
At final state
3
𝑚
P2 = 0.2 MPa ; 𝑉̂2 = 1.8161 (because V1 = V2; M1 = M2)
𝑘𝑔
Interpolation gives us
T2 = 514.98oC
̂2 = 3156.4
𝑈
𝑘𝐽
𝑘𝑔
̂2 = 3519.6
𝐻
𝑘𝐽
𝑘𝑔
Then we have
̂2 − 𝑈
̂1 ) = 611.6 kJ
Q = U2 – U1 = M(𝑈
W=0
∆U = Q = 611.6 kJ
̂2 − 𝐻
̂1 ) = 793.3 kJ
∆H = M(𝐻
c) Mass balance equation for closed system
M2 – M1 = 0  M1 = M2 = M  N1 = N2 = N
𝑃1 𝑉1 𝑃2 𝑉2
=
𝑅𝑇1
𝑅𝑇2
Since
P1 = P2; V2 = 2V1
 T2 = 2T1 = 2(125 + 273.15) = 796.3 K
Q = ∆H = NCp∆T =
1000
18
(𝑚𝑜𝑙) ∗ 34.4
𝑊 = −𝑃∆𝑉 = −𝑃 (
𝐽
𝑚𝑜𝑙𝐾
∗ (796.3 − 398.15)𝐾 = 760.9𝑘𝐽
𝑁2 𝑅𝑇2 𝑁1 𝑅𝑇1
−
) = −𝑁𝑅(𝑇2 − 𝑇1 ) = −183.9𝑘𝐽
𝑃2
𝑃1
∆U = Q + W = 577 kJ
d) Mass balance equation for closed system
M2 – M1 = 0  M1 = M2 = M  N1 = N2 = N
𝑃1 𝑉1 𝑃2 𝑉2
=
𝑅𝑇1
𝑅𝑇2
Since
P2 = 2P1; V2 = V1
 T2 = 2T1 = 2(125 + 273.15) = 796.3 K
Q = ∆U = NCv∆T =
1000
18
(𝑚𝑜𝑙) ∗ (34.4 − 8.314)
𝐽
𝑚𝑜𝑙𝐾
= 577 kJ
W = 0 due to ∆V = 0
∆H = NCp∆T = 760.9 kJ
∗ (796.3 − 398.15)𝐾
#Problem 2 3.4
In Joule’s experiments, the slow lowering of a weight turned a stirrer in an insulated container of
water. As a result of viscosity, the kinetic energy transferred from the stirrer to the water
eventually dissipated. In this process the potential energy of the water was first converted to
kinetic energy of the stirrer and the water, and then as a result of viscous forces, the kinetic
energy of the water was converted to thermal energy apparent as a rise in temperature.
Assuming no friction in the pulleys and no heat losses, how large a temperature rise would be
found in 1 kg of water as a result of a 1-kg weight being lowered 1 m?
𝑚𝑔∆ℎ = 𝑚𝐶𝑝 ∆𝑇
∆𝑇 =
𝑔∆ℎ
𝐶𝑝
2
9.816 𝑚 ⁄𝑠 2
∆𝑇 =
2
4184 𝑚 ⁄𝑠 2 𝐾
∆𝑇 = 0.002346 𝐾
#Problem 3 3.5
Steam at 500 bar and 600 C is to undergo a Joule-Thompson expansion to atmospheric pressure.
What will be the temperature of the steam after the expansion? What would be the downstream
temperature if the steam were replaced by an ideal gas?
Steam undergoes a Joule-Thomson expansion from 500bar, 600oC to atmospheric pressure (≈
1bar) would have
𝐻1 (500𝑏𝑎𝑟, 600𝑜 𝐶) = 𝐻2 (1𝑏𝑎𝑟, 𝑇2 )
From appendix A.III we have
𝐻1 (500𝑏𝑎𝑟, 600𝑜 𝐶) = 3247.6 (
𝑘𝐽
) = 𝐻2 (1𝑏𝑎𝑟, 𝑇2 )
𝑘𝑔
With known values of P2, 𝐻2 we could obtain T2 = 385oC from the steam table
If the steam were replaced by an ideal gas, since enthalpy of ideal gas is a function of temperature
only, we easily obtain T2 = T1 = 600oC
#Problem 4 3.6
Water in an open metal drum is to be heated from room Temperature (25 C) to 80 C by adding
steam slowly enough that all the steam condenses. The drum initially contains 100 kg of water,
and steam is supplied at 3.0 bar and 300 C. How many kilograms of steam should be added so
that the final temp of the water is exactly 80 C. Neglect all heat losses.
Choose system to be the contents inside the drum
Mass balance equation for an open system
M2 – M1 = Msteam
In which M1: mass in the drum at initial time
M2: mass in the drum at final time
Msteam: mass of steam into the drum
Energy balance equation with neglect of kinetic energy and potential energy change
̂𝑠𝑡𝑒𝑎𝑚 + Q + Ws – ∫ 𝑃𝑑𝑉
U2 – U1 = Msteam𝐻
with
Q = 0; Ws = 0,
Liquids are incompressible, so
∫ 𝑃𝑑𝑉 = 𝑃∆𝑉 ≈ 0
̂𝑠𝑡𝑒𝑎𝑚 = 3069.3 (𝑘𝐽/𝑘𝑔) (from appendix AIII)
𝐻
We assume that internal energy of liquid water is independent of pressure, so that we have
̂1 (25 𝑜 𝐶, 1.013𝑏𝑎𝑟) = 𝑈
̂1 (25 𝑜 𝐶, 3.169𝑘𝑃𝑎) = 104.88 (𝑘𝐽/𝑘𝑔)
𝑈
̂2 (80 𝑜 𝐶, 1.013𝑏𝑎𝑟) = 𝑈
̂2 (80 𝑜 𝐶, 47.39𝑘𝑃𝑎) = 334.86 (𝑘𝐽/𝑘𝑔)
𝑈
Substitute into energy balance equation, we have
̂2 − 𝑀1 𝑈
̂1 = 𝑀𝑠𝑡𝑒𝑎𝑚 𝐻
̂𝑠𝑡𝑒𝑎𝑚
𝑀2 𝑈
(100 + 𝑀𝑠𝑡𝑒𝑎𝑚 ) ∗ 334.86 − 100 ∗ 104.88 = 𝑀𝑠𝑡𝑒𝑎𝑚 ∗ 3069.3
 Msteam = 8.41 kg
#Problem 5 3.7
Consider the following statement: “The adiabatic work necessary to cause a given change of state
in a closed system is independent of the path by which that change occurs.
a) Consider a process that changes system from state 1 to state 2 (state 1 and state 2 are fixed)
Energy balance equation for closed system
𝑣2
𝑣2
{𝑈 + 𝑀( + 𝜑)} − {𝑈 + 𝑀 ( + 𝜑)} = 𝑄 + 𝑊
2
2
2
1
State 1 and state 2 are fixed
 Left hand side of the above equation = constant
 Q + W = constant
But Q = 0 for any adiabatic path
 W = constant for any adiabatic path
In illustration 3.4-6, we found that (Q + W) is the same for all path. This statement must also be
hold in any adiabatic path. However, in adiabatic path, Q + W = W, thus W is the same for all
adiabatic path. Because of this, this conclusion is not in contradict with illustration 3.4-16
b) Consider a system receives work Wa to change from state 1 to state 2 in an adiabatic path. Then
it releases work Wb to get back to state 1 in another adiabatic path
For these two different paths, energy balance equations for closed system, no heat exchange, with
neglect of kinetic and potential energy change give us (assume that U2 > U1)
U2 – U1 = Wa (Wa > 0)
U1 – U2 = Wb (Wb < 0)
 Wa + Wb = 0 or |𝑊𝑏 | = |𝑊𝑎 |
If |𝑊𝑏 | > |𝑊𝑎 | then Wa + Wb < 0: we have generated more energy than we put in.
If |𝑊𝑏 | < |𝑊𝑎 | then Wa + Wb > 0: we just add energy to the system that goes nowhere.
In the first case, if we add the work Wa to the system to change from state 1 to state 2, and then
get back to state 1 on the path generating work Wb, then we would obtain a machine that generates
energy.
#Problem 6 3.8
A nonconducting tank of negligible heat capacity and 1 m3 volume is connected to a pipeline
containing steam at 5 bar and 370 C, filled with steam to a pressure of 5 bar, and disconnected
from the pipeline.
Mass balance equation for open system
M2 – M1 = Msteam
Energy balance equation for open system with neglect of change in kinetic and potential energy,
no shaft work, constant volume
̂𝑖𝑛 + 𝑄
𝑈2 − 𝑈1 = 𝑀𝑠𝑡𝑒𝑎𝑚 𝐻
Since tank is nonconducting and has negligible heat capacity  Q = 0
̂𝑖𝑛
𝑈2 − 𝑈1 = 𝑀𝑠𝑡𝑒𝑎𝑚 𝐻
a)
If the tank is initially evacuated: M1 = 0
 M2 = Msteam
Energy equation balance becomes
̂2 = 𝑀𝑠𝑡𝑒𝑎𝑚 𝐻
̂𝑖𝑛
𝑀2 𝑈
But
M2 = Msteam, thus
̂2 = 𝐻
̂𝑖𝑛 = 3209.38 (
𝑈
𝑘𝐽
) 𝑎𝑡 370𝑜 𝐶 𝑎𝑛𝑑 5𝑏𝑎𝑟
𝑘𝑔
̂2 we can interpolate to get other data and obtain
With known values of P2 and 𝑈
T2 = 547.3oC ; 𝑉̂2 = 0.755
𝑀2 =
𝑚3
𝑘𝑔
𝑉2
1
=
= 1.325(𝑘𝑔)
𝑉̂2 0.755
 Msteam = M2 = 1.325 kg
b)
Now we have
T1 = 150oC; P1 = 1bar
̂1 = 2582.8 kJ/kg
 𝑉̂1 = 1.9364 m3/kg ; 𝑈
thus
𝑀1 =
𝑉1
= 0.5164 (𝑘𝑔)
𝑉̂1
Mass balance equation becomes
M2 – M1 = Msteam  M2 = Msteam + M1
Energy balance equation is
̂2 − 𝑀1 𝑈
̂1 = 𝑀𝑠𝑡𝑒𝑎𝑚 𝐻
̂𝑖𝑛
(𝑀𝑠𝑡𝑒𝑎𝑚 + 𝑀1 )𝑈
̂2 − 0.5164 ∗ 2582.8 = 𝑀𝑠𝑡𝑒𝑎𝑚 ∗ 3209.38
(𝑀𝑠𝑡𝑒𝑎𝑚 + 0.5164)𝑈
̂2 − 3209.38) ∗ 2582.8 = 1333.76 − 0.5164𝑈
̂2
𝑀𝑠𝑡𝑒𝑎𝑚 (𝑈
̂2 , 𝑉̂2 . Plug
Due to lack of equations, we have to guess an initial value of T2, then infer values of 𝑈
̂
back value of 𝑈2 into the above equation, we can calculate Msteam, and then M2
With values of V2 and M2 we calculate 𝑉̂2′ =
𝑉2
𝑀2
. This value should be close enough to 𝑉̂2
(difference should be less than 3%)
Finally we obtain T2 = 425oC
Check: with T2 = 425oC; P2 = 5bar, we get
̂2 = 3004.5
𝑈
𝑘𝐽
𝑘𝑔
𝑉̂2 = 0.6407
𝑚3
𝑘𝑔
 Msteam = 1.06288 kg
 M2 = 1.57928 kg
𝑉
𝑚3
 𝑉̂2′ = 2 = 0.633
𝑀2
𝑘𝑔
𝑉̂2 − 𝑉̂2 ′
(∆𝑉 =
= 1.2%)
𝑉̂2
#Problem 7 3.11
A frictionless piston-and-cylinder system shown here is subjected to 1.013 bar external pressure.
The piston’s mass is 200 kg, it has an area of 0.15 m2 and the initial volume of the entrapped ideal
gas is 0.12 m3. The piston and cylinder do not conduct heat, but heat can be added to the gas by
a heating coil. The gas has a constant-volume heat capacity of 30.1 J/(mol K) and an initial
temperature of 298 K, and 10.5 kJ of energy are to be supplied to the gas through the heating
coil.
a) Initial pressure of the system (contents in piston-cylinder)
𝑚
200𝑘𝑔 ∗ 9.81 2
𝑠
5
𝑃1 = 1.013 ∗ 10 𝑃𝑎 +
= 114380 𝑃𝑎
0.15𝑚2
For ideal gas we have
𝑃1 𝑉1 114380 ∗ 0.12
𝑁1 =
=
= 5.54 𝑚𝑜𝑙
𝑅𝑇1
8.314 ∗ 298
Mass balance equation is
M1 – M2 = 0  M1 = M2  N1 = N2
Energy balance equation is
𝑈2 − 𝑈1 = 𝑄 + 𝑊𝑠 − ∫ 𝑃𝑑𝑉
Ws = 0
∆V = 0
 ∆U = Q = 10.5 kJ
Thus
∆𝑇 =
∆𝑈
10500𝐽
=
𝑁𝐶𝑣 5.54𝑚𝑜𝑙 ∗ 30.1
𝐽
𝑚𝑜𝑙𝐾
= 62.97𝐾
 T2 = T1 + ∆T = 360.97 K
𝐽
𝑁2 𝑅𝑇2 5.54𝑚𝑜𝑙 ∗ 8.314 𝑚𝑜𝑙𝐾 ∗ 360.97𝐾
𝑃2 =
=
= 138550 𝑃𝑎
𝑉2
0.12𝑚3
b) Piston is allowed to move freely
 P2 = P1 = 114380 Pa
From mass balance equation
N1 = N2 = 5.54 mol
Energy balance equation for closed system with neglect of change in kinetic and potential
energy is
U2 – U1 = Q – P(V2 – V1)
 Q = H2 – H1 = NCp(T2 – T1)
With
Cp = Cv + R = 30.1 + 8.314 = 38.414 J/molK
Thus
𝑄
10500𝐽
𝑇2 = 𝑇1 +
= 298 +
= 347.34𝐾
𝐽
𝑁𝐶𝑝
5.54𝑚𝑜𝑙 ∗ 38.414
𝑚𝑜𝑙𝐾
𝐽
𝑁2 𝑅𝑇2 5.54𝑚𝑜𝑙 ∗ 8.314 𝑚𝑜𝑙𝐾 ∗ 347.34𝐾
𝑉2 =
=
= 0.1399𝑚3
𝑃2
114380𝑃𝑎
#Problem 8 3.15
An isolated chamber with rigid walls is divided into two equal compartments, one containing gas
and the other evacuated. The partition between the compartments ruptures. After the passage
of a sufficiently long period of time, the temperature and pressure are found to be uniform
throughout the chamber.
a) If the filled compartment initially contains an ideal gas of constant heat capacity at 1MPa and
500 K. what are the final temperature and pressure in the chamber?
Choose a system to be contents of both two compartments, named a and b.
Mass balance equation for this system is
M2 – M1 = 0  M1 = M2  N1 = N2
With
N1 = N1a + N1b = N1a  N1a = N2
Energy balance equation is
U2 – U1 = 0  U2 = U1
With
U1 = U1a + U1b = U1a  U1a = U2  T1a = T2 = 500K
Since ideal gas is only a function of temperature.
We have
P1aV1a = N1aRT1a
P2V2 = N2RT2
With
N1a = N2; T1a = T2; V2 = 2V1a
 P2 = 0.5P1a = 0.5 MPa
b) If the filled compartment initially contains steam at 1 MPa and 500 K, what are the final
temperature and pressure in the compartment?
For steam, we still have the following results
N2 = N1a; M2 = M1a
U2 = U1a
V2 = 2V1a
Thus
̂2 = 𝑈
̂1𝑎
𝑈
𝑉̂2 = 2𝑉̂1𝑎
Initially, at 1MPa and 500K
̂1𝑎 = 2669.1 𝑘𝐽/𝑘𝑔
𝑈
𝑚3
𝑘𝑔
𝑉̂1𝑎 = 0.2203
Thus
̂2 = 2669.1 𝑘𝐽/𝑘𝑔
𝑈
𝑉̂2 = 0.4406
𝑚3
𝑘𝑔
From these two intensive properties, state 2 is totally defined
From steam table  T2 ≈ 216.1oC; P2 ≈ 0.5 MPa
c) Repeat part (a) if the second compartment initially contains an ideal gas, but at half the pressure
and 100 K higher temperature.
Mass balance equation for this system is
M2 – M1 = 0  M2 = M1  N2 = N1  N2 = N1a + N1b
so
𝑃2 𝑉2 𝑃1𝑎 𝑉1𝑎 𝑃1𝑏 𝑉1𝑏
=
+
𝑇2
𝑇1𝑎
𝑇1𝑏
With
V2 = 2V1a = 2V1b
2𝑃2 𝑃1𝑎 𝑃1𝑏
1
0.5
=
+
=
+
= 2.833 ∗ 10−3
𝑇2
𝑇1𝑎 𝑇1𝑏 500 600
Energy balance equation is
U2 – U1 = 0
 U2 – U1a – U1b = 0
For ideal gas, this equation becomes
𝑃1𝑎 𝑉1𝑎
𝑃1𝑏 𝑉1𝑏
𝐶𝑣 (𝑇2 − 𝑇1𝑎 ) +
𝐶 (𝑇 − 𝑇1𝑏 ) = 0
𝑅𝑇1𝑎
𝑅𝑇1𝑏 𝑣 2
1
0.5
(𝑇2 − 500) +
(𝑇 − 600) = 0
500
600 2
 T2 = 529.4K
 P2 = 529.4*2.833*10-3/2 = 0.75 MPa
d) Repeat part (b) if the second compartment initially contains steam, but at half the pressure
and 100 K higher temperature.
From mass balance equation, we have
𝑀2 = 𝑀1𝑎 + 𝑀1𝑏
𝑉2 𝑉1𝑎 𝑉1𝑏
=
+
𝑉̂2 𝑉̂1𝑎 𝑉̂1𝑏
With
V2 = 2V1a = 2V1b
𝑚3
𝑎𝑡 1𝑀𝑃𝑎, 500𝐾
𝑘𝑔
𝑚3
= 0.5481
𝑎𝑡 0.5𝑀𝑃𝑎, 600𝐾
𝑘𝑔
𝑉̂1𝑎 = 0.2203
𝑉̂1𝑏
Plug back into the above equation and solve for 𝑉̂2 , we obtain
𝑉̂2 = 0.3143
𝑚3
𝑘𝑔
From energy balance equation, we have
U2 – U1 = 0  U2 = U1
2𝑉2
𝑉
𝑉
̂2 = 1𝑎 𝑈
̂1𝑎 + 1𝑏 𝑈
̂1𝑏
𝑈
𝑉̂2
𝑉̂1𝑎
𝑉̂1𝑏
̂2
2𝑈
2669.1 2845.7
=
+
0.3143 0.2203 0.5481
̂2 = 2719.9
𝑈
𝑘𝐽
𝑘𝑔
̂2 , from steam data table, we obtain
From two values of 𝑉̂2 and 𝑈
T2 = 284oC; P2 = 0.8 MPa
Another estimate results in T2=253C and P2=0.76MPa. Use your judgment.
#Problem 9 3.22
Nitrogen gas is being withdrawn at the rate of 4.5 g/s from a 0.15-m3 cylinder, initially containing
the gas at a pressure of 10 bar and 320 K. The cylinder does not conduct heat, nor does its
temperature change during the emptying process. What will be the temperature and pressure of
the gas in the cylinder after 5 minutes? What will be the rate of change of the gas temperature
at this time? Nitrogen can be considered to be an ideal gas with C*p = 30 J/(mol K)
Choose a system as contents inside the cylinder
From Mass balance equation, we have
𝑑𝑁
= 𝑁̇
𝑑𝑡
𝑃
𝑉 𝑑(𝑇 )
= 𝑁̇
𝑅 𝑑𝑡
4.5 𝑚𝑜𝑙
𝐽
𝑃
−
∗ 8.314
𝑑( )
𝑚𝑜𝑙𝐾 = −8.908 ∗ 10−5 𝑏𝑎𝑟
𝑇 = 𝑁̇ 𝑅 = 28 𝑠
𝑑𝑡
𝑉
0.15𝑚3
𝐾. 𝑠
Integrate this equation, we obtain
𝑃
𝑃
10
( )𝑓𝑖𝑛𝑎𝑙 = ( )𝑖𝑛𝑖𝑡𝑖𝑎𝑙 − 8.908 ∗ 10−5 𝑡 =
− 8.908 ∗ 10−5 𝑡
𝑇
𝑇
320
From illustration 3.4-5 in textbook, we have
𝑇𝑓𝑖𝑛𝑎𝑙 𝐶𝑝
𝑃𝑓𝑖𝑛𝑎𝑙
)𝑅 =
𝑇𝑖𝑛𝑖𝑡𝑖𝑎𝑙
𝑃𝑖𝑛𝑖𝑡𝑖𝑎𝑙
(
30
8.314
𝑇𝑓𝑖𝑛𝑎𝑙
𝑃𝑓𝑖𝑛𝑎𝑙
30
30
8.314
𝑇𝑖𝑛𝑖𝑡𝑖𝑎𝑙
3208.314
=
=
𝑃𝑖𝑛𝑖𝑡𝑖𝑎𝑙
10
(2)
From (1)&(2)  Tfinal = 152.57K; Pfinal = 0.6907bar
We have
𝑃
𝐶𝑣 𝑑𝑙𝑛𝑇 𝑑𝑙𝑛(𝑇 ) 𝑑𝑙𝑛𝑃 𝑑𝑙𝑛𝑇
𝑑𝑃 𝑑𝑙𝑛𝑇
=
=
−
=
−
𝑅 𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑃𝑑𝑡
𝑑𝑡
𝑃
𝐶𝑣 𝑑𝑙𝑛𝑇 𝑇 1 𝑑𝑃 𝑃𝑑𝑙𝑛𝑇
𝑇 𝑑(𝑇 ) 𝑁̇𝑅𝑇
= [ ( −
)] =
=
𝑅 𝑑𝑡
𝑃 𝑇 𝑑𝑡
𝑑𝑡
𝑃 𝑑𝑡
𝑃𝑉
𝐶𝑣 𝑑𝑇 𝑁̇𝑅𝑇
=
𝑅𝑇 𝑑𝑡
𝑃𝑉
(1)
𝑑𝑇 𝑁(̇𝑅𝑇)2
=
𝑑𝑡
𝐶𝑣 𝑃𝑉
At t = 5 mins
P = 0.6907 bar
T = 152.57 K
Cv = Cp – R = 30 – 8.314 = 21.686 J/molK
Plub back into the above equation, we get
𝑑𝑇
𝐾
= −1.151
𝑑𝑡
𝑠
Alternatively, here is a little different way of looking at the problem:
Open system, adiabatic, no shaft work, and no expansion work. So, the gas leaving the cylinder
must have the same properties as the gas entering the valve. From the energy balance, we get
the following:
𝒅(𝑵𝑼)
̇
= −𝑵𝑯
𝒅𝒕
d ( NU )
 NH
dt
d (U )
d ( N ) dN
N
U

H
dt
dt
dt
NdU  UdN  HdN
NdU  UdN  HdN
NdU  ( H  U )dN
Now, we can use definitions of enthalpy and internal energy for ideal gas:
H  Cp(T  TR ) H  (Cv  R)(T  TR ) U  Cv(T  TR )  RTR
 H  U  (Cv  R)(T  TR )  Cv(T  TR )  RTR   RTR
Now, plugging into equation above, and cancelling out terms, we get:
NCvdT dN(RT)
and finally, using separation of variables, we get:
Cv dT dN

R T
N
integrating, we get:
𝐶𝑣
𝑇2
𝑁2
ln ( ) = ln ( )
𝑅
𝑇1
𝑁1
Now, we rearrange and the final number of moles can be computed by multiplying the rate at
which nitrogen is being withdrawn, times 5 minutes.
𝑅
𝑁2 (𝐶𝑣)
𝑇2 = 𝑇1 ( )
𝑁1
𝑁2 = 𝑁1 −
𝑑𝑁
∗𝑡
𝑑𝑡
𝑅
𝑁2 (𝐶𝑣)
𝑇2 = 𝑇1 ( )
= 𝟏𝟓𝟐. 𝟓𝟗𝑲
𝑁1
using the ideal gas law, we get that: P=0.691bar
For the change, since T is a function of time, the above equation also can be written
𝑁(𝑡)
𝑇(𝑡) = 𝑇1 (
)
𝑁1
𝑅
(𝐶𝑣)
So to find dT/dt, we take the derivative of both sides with respect to t. Everything but N(t) is a
constant, so we use our excellent Calculus I skills to find that
𝑅
𝑅
𝑑𝑇
𝑅
1 (𝐶𝑣)
𝑑𝑁
= 𝑇1 ( ) ( )
𝑁(𝑡)(𝐶𝑣−1)
𝑑𝑡
𝐶𝑣 𝑁1
𝑑𝑡
We know all values except dT/dt, so we can find that
𝒅𝑻
𝑲
= −𝟏. 𝟏𝟓
𝒅𝒕
𝒔
#Problem 10 3.32
Nitrogen can be liquefied using a Joule-Thomson expansion process. This is done by rapidly and
adiabatically expanding cold nitrogen gas from high pressure to a low pressure. If Nitrogen at 135
K and 20 MPa undergoes a Joule-Thomson expansion to 0.4 MPa:
a) Extimate the fraction of vapor and liquid present after the expansion and the temperature
of this mixture using the pressure-enthalpy diagram for nitrogen
As nitrogen goes through a Joule-Thomson process
̂1 = 𝐻
̂2
𝐻
With
T1 = 135K
P1 = 20MPa
̂1 = 153 kJ/kg
 𝐻
̂
̂2 = 153 𝑘𝐽/𝑘𝑔
 𝐻1 = 𝐻
We also have
P2 = 0.4MPa
 T2 = 90K
From Pressure-Enthalpy diagram for nitrogen, we find that 55% N2 is in vapor phase, 45% is in
liquid phase
b) Repeat the calculation assuming nitrogen to be an ideal gas with Cp*=29.3 J/(mol K)
As nitrogen is an ideal gas, we have
∆H = NCp∆T
With ∆H = 0  ∆T = 0  T2 = T1 = 135K: 100% N2 is in vapor phase
#Problem 11
A piston-cylinder assembly contains 6kg of steam at a pressure of 100 bar and a temperature of
400oC. It undergoes a process whereby it expands against a constant pressure of 20 bar, until the
forces balance. During the process, the piston generates 1497480 J of work. Steam is not an ideal
gas under these conditions. Determine the final temperature and the heat transferred during the
process.
Choose system as the steam inside cylinder
Mass balance equation for this closed system is
M2 = M 1 = M
Energy balance for this closed system with neglect of change of kinetic and potential energy,
with no shaft work
𝑈2 − 𝑈1 = 𝑄 + 𝑊 = 𝑄 − ∫ 𝑃𝑑𝑉 = 𝑄 − 𝑃(𝑉2 − 𝑉1 )
From appendix AIII
𝑉̂1 = 0.02641
𝑚3
(𝑎𝑡 100 𝑏𝑎𝑟, 400𝑜 𝐶)
𝑘𝑔
𝑉1 = 𝑀1 𝑉̂1 = 6 ∗ 0.02641 = 0.15846 (𝑚3 )
Thus
𝑉2 = 𝑉1 −
𝑊
−1497480
= 0.15846 −
= 0.9072 𝑚3
𝑃
20 ∗ 105
𝑉̂2 =
𝑉2
= 0.1512 (𝑚3 /𝑘𝑔)
𝑀2
With known values of 𝑉̂2 and P2, we can get T2 = 400oC (20 bar, 0.1512 m3/kg)
Also from appendix AIII
̂2 = 2945.2
𝑈
𝑘𝐽
𝑘𝑔
̂1 = 2832.4
𝑈
𝑘𝐽
𝑘𝑔
̂2 − 𝑈
̂1 ) = 6 ∗ (2945.2 − 2832.4) = 676.8 (𝑘𝐽)
∆𝑈 = 𝑀(𝑈
Q = ∆U – W = 676.8 + 1497.48 = 2174.28 kJ
#Problem 12 3.20
A clever chemical engineer has devised the thermally operated elevator shown in the
accompanying diagram. The elevator compartment is made to rise by electrically heating the air
contained in the piston-and-cylinder drive mechanism, and the elevator is lowered by opening a
valve at the side of the cylinder, allowing the air in the cylinder to slowly escape. Once the
elevator compartment is back to the lower level, a small pump forces out the air remaining in the
cylinder and replaces it with air at 20 C and a pressure just sufficient to support the elevator
compartment. This cycle can then be repeated. There is no heat transfer between the piston,
cylinder, and the gas; the weight of the piston, elevator, and the elevator contents is 4000 kg;
the piston has a surface area of 2.5 m2; and the volume contained in the cylinder when the
elevator is at its lowest level is 25 m3. There is no friction between the piston and the cylinder,
and the air in the cylinder is assumed to be an ideal gas with Cp*=30 J/(mol K)
a) What is the pressure in the cylinder throughout the process?
𝐹
𝑃=
𝐴
𝑃𝑡𝑜𝑡 = 𝑃𝑎𝑡𝑚 + 𝑃𝑝𝑖𝑠𝑡𝑜𝑛
𝑃𝑡𝑜𝑡 = 101,325 +
4000 𝑘𝑔 ∗ 9.816 𝑚⁄𝑠 2
= 117.03 𝑘𝑃𝑎
2.5𝑚2
b) How much heat must be added to the air during the process of raising the elevator 3 m, and
what is the final temperature of the gas?
We can use the ideal gas law to calculate the number of moles.
𝑃𝑉
𝑅𝑇
117031 𝑃𝑎 ∗ 25 𝑚3
𝑁=
= 1200.44 𝑚𝑜𝑙
3
8.314 𝑚 𝑃𝑎⁄𝐾 𝑚𝑜𝑙 ∗ 293.15 𝐾
𝑁=
We can also use the ideal gas law to calculate the final temperature of the gas.
∆𝑉 = 2.5 ∗ 3 = 7.5𝑚3
𝑉𝑓 = 25𝑚3 + 2.5𝑚2 ∗ 3𝑚 = 32.5𝑚3
𝑃𝑉
𝑇𝑓 =
𝑁𝑅
117031 𝑃𝑎 ∗ 32.5 𝑚3
𝑇𝑓 =
= 381.1 𝐾
3
1200.44 𝑚𝑜𝑙 ∗ 8.314 𝑚 𝑃𝑎⁄𝐾 𝑚𝑜𝑙
Now the gas does work both against the atmosphere and the piston
𝑊𝑎𝑡𝑚 = −𝑃𝑎𝑡𝑚 ∗ ∆𝑉 = −877.7325 𝑘𝐽
𝑊𝑝𝑖𝑠𝑡𝑜𝑛 = −𝑚 ∗ 𝑔 ∗ ∆ℎ = −117.792 𝑘𝐽
𝑊𝑇 = 𝑊𝑎𝑡𝑚 + 𝑊𝑝𝑖𝑠𝑡𝑜𝑛 = −995.9245 𝑘𝐽
Using a simplified version of the first law,
𝑄 + 𝑊 = 𝑁𝐶𝑝 (𝑇𝑓 − 𝑇𝑖 )
𝑄 = 𝑁𝐶𝑝 ∆𝑇 − 𝑊
𝐽
𝑄 = 1200.44 𝑚𝑜𝑙 ∗ 30 ⁄𝑚𝑜𝑙 𝐾 ∗ (381.1 𝐾 − 293.15 𝐾) + 995924.5 𝐽 = 4160.8 𝑘𝐽
c) What fraction of the heat added is used in doing work, and what fraction is used in raising
the temperature of the gas?
995.9245
%𝑄𝑊 =
= 24%
4160.8
1200.44 ∗ 30 ∗ 87.95
= 76%
4160800
d) How many moles of air must be allowed to escape in order for the elevator to return to the
lowest level?
%𝑄∆𝑇 =
Using the ideal gas law
𝑁=
𝑁𝑓 =
𝑃𝑉
𝑅𝑇
117031 𝑃𝑎 ∗ 25 𝑚3
= 923.4 𝑚𝑜𝑙
3
8.314 𝑚 𝑃𝑎⁄𝐾 𝑚𝑜𝑙 ∗ 381.1 𝐾
∆𝑁 = 𝑁𝑓 − 𝑁𝑖 = 1200.44 𝑚𝑜𝑙 − 923.4 𝑚𝑜𝑙 = 277.04 𝑚𝑜𝑙
##Problem 13
An air compressor is designed to compress atmospheric air (assumed to be at 100 kPa, 20oC) to a
pressure of 1 MPa. The heat transfer rate to the environment is anticipated to be about equal to
10% of the power input to the compressor. The air enters at 50 m/s where the inlet area is 9x10-3
m2 and leaves at 120m/s through an area 5x10-4m2. Determine the exit-air temperature and the
power input to the compressor. The compressor is working at steady state, and air could be
assumed to be an ideal gas.
Solution
Choose system to be the control volume inside the compressor
Mass balance equation for this opened system is
𝑑𝑀
= 𝑀̇𝑖𝑛 − 𝑀̇𝑜𝑢𝑡 = 0
𝑑𝑡
𝑀̇𝑖𝑛 = 𝑀̇𝑜𝑢𝑡 = 𝑀̇
𝐴𝑖𝑛 𝑣𝑖𝑛 𝐴𝑖𝑛 𝑣𝑖𝑛 𝑃𝑖𝑛 𝑀𝑎𝑖𝑟
=
𝑅𝑇𝑖𝑛
𝑉̂𝑖𝑛
9 ∗ 10−3 ∗ 50 ∗ 100 ∗ 28.97
𝑘𝑔
=
= 0.535
8.314 ∗ 293
𝑠
𝑀̇ = 𝑀̇𝑖𝑛 =
𝑀̇ = 𝑀̇𝑖𝑛
We have
𝑀̇𝑖𝑛 = 𝑀̇𝑜𝑢𝑡 = 𝑀̇
which is equivalent to
𝐴𝑖𝑛 𝑣𝑖𝑛 𝑃𝑖𝑛 𝑀𝑎𝑖𝑟 𝐴𝑜𝑢𝑡 𝑣𝑜𝑢𝑡 𝑃𝑜𝑢𝑡 𝑀𝑎𝑖𝑟
=
𝑅𝑇𝑖𝑛
𝑅𝑇𝑜𝑢𝑡
𝐴𝑜𝑢𝑡 𝑣𝑜𝑢𝑡 𝑃𝑜𝑢𝑡
𝑇𝑜𝑢𝑡 = 𝑇𝑖𝑛
𝐴𝑖𝑛 𝑣𝑖𝑛 𝑃𝑖𝑛
𝑇𝑜𝑢𝑡 = 293 ∗ (
5 ∗ 10−4 120 1000
)(
)(
) = 391 𝐾
9 ∗ 10−3
50
100
Energy balance equation for this system in steady state is
𝑣2
𝑣2
̂ + + 𝜑)𝑖𝑛 − 𝑀̇𝑜𝑢𝑡 (𝐻
̂ + + 𝜑)𝑜𝑢𝑡 + 𝑄̇ + 𝑊̇
0 = 𝑀̇𝑖𝑛 (𝐻
2
2
With 𝑄̇ = −0.1𝑊̇ (< 0)
Neglect potential energy change, we have
2
2
𝑣𝑖𝑛
− 𝑣𝑜𝑢𝑡
)=0
2
2
2
𝑀̇
𝑣𝑖𝑛
− 𝑣𝑜𝑢𝑡
̇
̂
̂
𝑊=−
(𝐻 − 𝐻𝑜𝑢𝑡 +
)
0.9 𝑖𝑛
2
̂𝑖𝑛 − 𝐻
̂𝑜𝑢𝑡 +
0.9𝑊̇ + 𝑀̇ (𝐻
With
𝑘𝐽
𝑘𝑔
𝑘𝐽
= 392.1
𝑘𝑔
̂𝑖𝑛 = 293.3
𝐻
̂𝑜𝑢𝑡
𝐻
Solving the above equation, we obtain
𝑊̇ = 62.27 𝑘𝑊
#Problem 14
Steam enters a turbine with a pressure and temperature of 15MPa and 600oC and leaves at 100kPa
as a saturated vapor. The flow area at the turbine inlet is 0.045m2 and at the exit it is 0.31 m2. The
steam flows steadily through the turbine at a mass flow rate of 30kg/s. Calculate the power that
can be produced by the turbine, assuming negligible heat transfer from the system. Specific volume
and specific enthalpy of inlet air are 0.0249 m3/kg and 3581.5 kJ/kg.
Choose system as a control volume inside the turbine
Mass balance equation for this system is
𝑑𝑀
= 𝑀̇𝑖𝑛 − 𝑀̇𝑜𝑢𝑡 = 0
𝑑𝑡
𝑀̇𝑖𝑛 = 𝑀̇𝑜𝑢𝑡 = 𝑀̇
𝑀̇ =
𝐴𝑖𝑛 𝑣𝑖𝑛 𝐴𝑜𝑢𝑡 𝑣𝑜𝑢𝑡
=
𝑉̂𝑖𝑛
𝑉̂𝑜𝑢𝑡
𝑉̂𝑖𝑛 = 0.0249
𝑉̂𝑜𝑢𝑡 = 1.694
𝑚3
(𝑎𝑡 15 𝑀𝑃𝑎, 600𝑜 𝐶)
𝑘𝑔
𝑚3
(100𝑘𝑃𝑎, 𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑒𝑑 𝑣𝑎𝑝𝑜𝑟)
𝑘𝑔
𝑀̇ = 30
𝑘𝑔
𝑠
Thus we have
30 ∗ 0.0249
𝑚
= 16.6
0.045
𝑠
30 ∗ 1.694
𝑚
=
= 163.9
0.31
𝑠
𝑣𝑖𝑛 =
𝑣𝑜𝑢𝑡
Energy balance equation for this system is
𝑣2
𝑣2
̂ + + 𝜑)𝑖𝑛 − 𝑀̇(𝐻
̂ + + 𝜑)𝑜𝑢𝑡 + 𝑄̇ + 𝑊̇
0 = 𝑀̇(𝐻
2
2
If heat transfer is negligible, and neglect change in potential energy, we have
2
𝑣 2 − 𝑣𝑖𝑛
̂𝑜𝑢𝑡 − 𝐻
̂𝑖𝑛 + 𝑜𝑢𝑡
𝑊̇ = 𝑀̇(𝐻
)
2
With
𝑘𝐽
𝑘𝑔
𝑘𝐽
= 2675.1
𝑘𝑔
̂𝑖𝑛 = 3581.5
𝐻
̂𝑜𝑢𝑡
𝐻
Thus
𝑊̇ = 26793 𝑘𝑊
#Problem 15
An insulated vessel has two compartments separated by a membrane. On one side is 2kg of steam
at 500oC and 200 bar. The other side is evacuated. The membrane ruptures, filling the entire
volume. The final pressure is 100 bar. Determine the final temperature of the steam and the volume
of the vessel.
Choose system as a whole volume inside the vessel
Mass balance equation for this closed system is
M2 = M1 = M1a + M1b = M1a (b is the evacuated compartment at beginning)
Energy balance equation with neglect of change of kinetic and potential energy, with no heat
transfer and work done for this closed system is
U2 – U1a – U1b = U2 – U1a = 0
thus
̂2 = 𝑈
̂1𝑎
𝑈
From appendix AIII
̂1𝑎 = 2942.9
𝑈
𝑘𝐽
(500𝑜 𝐶, 200 𝑏𝑎𝑟)
𝑘𝑔
Thus
̂2 = 2942.9
𝑈
𝑘𝐽
𝑘𝑔
We also have
P2 = 100 bar
Thus we can determine
T2 = 449.7oC
𝑉̂2 = 0.02973
𝑚3
𝑘𝑔
𝑉𝑣𝑒𝑠𝑠𝑒𝑙 = 𝑀2 𝑉̂2 = 2 ∗ 0.02973 = 0.05946 (𝑚3 )
#Problem 16
Consider a piston-cylinder assembly containing 10kg of steam. Initially the gas has a pressure of
20 bar and occupies a volume of 1.0 m3. Under these conditions, steam does not behave as an ideal
gas.
A) The system now undergoes a compression process in which it is compressed to 100 bar, the
external pressure is slightly larger than and could be assumed to be equal to internal pressure. The
pressure-volume relationship is given by 𝑃𝑉 1.5 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡. What is the final temperature and
internal energy of the system? Calculate the work done during this process. How much heat was
exchanged?
B) Consider a different process by which the system get to the same final state in part (A). In this
case, a large block is placed on the piston, forcing it to compress. Calculate the work done during
this process. How much heat was exchanged?
a) Choose system as the steam inside the assembly.
Mass balance equation for this closed system is
M2 – M1 = 0  M1 = M 2
Energy balance equation for this closed system with neglect of change of kinetic and potential
energy, no shaft work is
2
𝑈2 − 𝑈1 = 𝑄 − ∫ 𝑃𝑑𝑉
1
We have
𝑃1 𝑉̂11.5 = 𝑃2 𝑉̂21.5
P1 = 20 bar, P2 = 100 bar
and
𝑉̂1 =
1
𝑚3
= 0.1
10
𝑘𝑔
we can obtain
𝑚3
𝑘𝑔
𝑘𝐽
̂1 = 2602.8
𝑈
𝑘𝑔
With know values of P2 and 𝑉̂2 , from appendix AIII we get
𝑉̂2 = 0.0342
T2 = 524.7 K
𝑘𝐽
̂2 = 3094.6
𝑈
𝑘𝑔
̂2 = 10 ∗ 3094.6 = 30946 𝑘𝐽
𝑈2 = 𝑀2 𝑈
𝑃1 𝑉̂11.5
𝑊 = − ∫ 𝑃𝑑𝑉 = −𝑀 ∫
𝑑𝑉̂ = 2839 𝑘𝐽
1.5
̂
𝑉
1
1
2
2
Q = ∆U – W = U2 – U1 – W = 10(3094.6 – 2602.8) – 2839 = 2079 (kJ)
b)
2
2
𝑊 = − ∫ 𝑃𝑑𝑉 = −𝑀𝑃2 ∫ 𝑑𝑉̂ = −10 ∗ 100 ∗ (0.0342 − 0.1)
1
1
W = 6580 kJ
Q = ∆U – W = – 1662 kJ
#Problem 17
You wish to measure the temperature of steam flowing in a pipe at pressure of 9MPa. To do this
task you connect a well-insulated tank of volume 0.4m3 to this pipe through a valve. This tank
initially is at vacuum. The valve is opened, and the tank fills with steam until the pressure is 9
MPa. At this point the pressure of the pipe and tank are equal, and no more steam flows through
the valve. The valve is then closed. The temperature right after the valve is closed is measured to
be 800oC. The process takes place adiabatically. Determine the temperature of the system flowing
in the pipe. You may assume the steam in the pipe stays at the same temperature and pressure
throughout this process.
Mass balance equation for this open system with only one inlet stream is
𝑀2
𝑑𝑀
= 𝑀̇𝑖𝑛 − 𝑀̇𝑜𝑢𝑡 = 𝑀̇𝑖𝑛
𝑑𝑡
𝑡
∫ 𝑑𝑀 = ∫ 𝑀̇𝑖𝑛 𝑑𝑡 = 𝑀2 − 𝑀1 = 𝑀2
𝑀1
0
Energy balance equation for this system with neglect of change of kinetic and potential energy is
𝑑𝑈
̂𝑖𝑛 − 𝑀̇𝑜𝑢𝑡 𝐻
̂𝑜𝑢𝑡 + 𝑄̇ + 𝑊̇
= 𝑀̇𝑖𝑛 𝐻
𝑑𝑡
Since we have no outlet stream, adiabatic process, constant volume with no shaft work
2
𝑡
𝑑𝑈
̂𝑖𝑛
= 𝑀̇𝑖𝑛 𝐻
𝑑𝑡
𝑡
̂𝑖𝑛 𝑑𝑡 = 𝐻
̂𝑖𝑛 ∫ 𝑀̇𝑖𝑛 𝑑𝑡
∫ 𝑑𝑈 = ∫ 𝑀̇𝑖𝑛 𝐻
1
0
0
We can infer
̂𝑖𝑛
𝑈2 = 𝑀2 𝐻
̂2
𝑏𝑢𝑡 𝑈2 = 𝑀2 𝑈
̂
̂
𝑡ℎ𝑢𝑠 𝑈2 = 𝐻𝑖𝑛
From appendix AIII we get
𝑘𝐽
̂𝑖𝑛
(𝑎𝑡 9𝑀𝑃𝑎, 800𝑜 𝐶) = 𝐻
𝑘𝑔
̂𝑖𝑛 and Pin we can obtain Tin = 600oC
Now we have values of 𝐻
̂2 = 3632.5
𝑈
#Problem 18
Refrigerant 12 flows steadily through a 40 mm diameter horizontal pipe. At a point where the
velocity is 40m/s the temperature and pressure of the refrigerant are 40oC and 300kPa,
respectively. As a result of heat transfer from the surroundings, the temperature at a point
downstream reaches 50oC.
A) Assuming a negligible pressure drop, determine the heat transfer rate to the refrigerant 12.
Specific volumes of inlet and outlet streams are 0.06821m3/kg and 0.07077 m3/kg. Specific
enthalpy of inlet and outlet streams are 214.31 kJ/kg and 220.77 kJ/kg.
B) Do part (a) with consideration of 50kPa of pressure drop. Specific volume and specific enthalpy
of outlet stream now are 0.0856 m3/kg and 221.33 kJ/kg. Mass flow rate in pipe is constant.
Choose system as a control volume inside the pipe.
a) Mass balance equation:
𝑑𝑀
= 𝑀̇𝑖𝑛 − 𝑀̇𝑜𝑢𝑡 = 0
𝑑𝑡
𝑀̇𝑖𝑛 = 𝑀̇𝑜𝑢𝑡 = 𝑀̇
Energy balance equation with neglect of change of potential energy because pipe is horizontal is
𝑣2
𝑣2
̇
̂
̇
̂
0 = 𝑀𝑖𝑛 (𝐻 + )𝑖𝑛 − 𝑀𝑜𝑢𝑡 (𝐻 + )𝑜𝑢𝑡 + 𝑄̇ + 𝑊̇
2
2
Since no work was done
̂+
𝑄̇ = 𝑀̇[(𝐻
Since 𝑀̇𝑖𝑛 = 𝑀̇𝑜𝑢𝑡
𝑣2
𝑣2
̂ + )𝑖𝑛 ]
)𝑜𝑢𝑡 − (𝐻
2
2
𝐴𝑜𝑢𝑡 𝑣𝑜𝑢𝑡 𝐴𝑖𝑛 𝑣𝑖𝑛
=
𝑉̂𝑜𝑢𝑡
𝑉̂𝑖𝑛
Pipe has constant cross section  Aout = Ain
We now have
𝑣𝑜𝑢𝑡 𝑣𝑖𝑛
=
𝑉̂𝑜𝑢𝑡 𝑉̂𝑖𝑛
With
𝑚3
𝑘𝑔
𝑚3
= 0.07077
𝑘𝑔
𝑉̂𝑖𝑛 = 0.06821
𝑉̂𝑜𝑢𝑡
Substitute into the above equation, we obtain
vout = 1.037vin
thus we may neglect change in kinetic energy in the energy balance equation
0.042
1
̂𝑜𝑢𝑡 − 𝐻
̂𝑖𝑛 ) = 40 ∗ 𝜋 ∗
𝑄̇ = 𝑀̇(𝐻
∗
∗ (220.77 − 214.31)
4
0.06821
𝑄̇ = 4.76 𝑘𝑊
b) The exit velocity now is
𝑣𝑜𝑢𝑡 =
𝑉̂𝑜𝑢𝑡 𝑣𝑖𝑛
𝑚
= 50
𝑠
𝑉̂𝑖𝑛
Since there is big difference between vin and vout, we cannot neglect the change in kinetic energy.
Energy balance equation now becomes
̂+
𝑄̇ = 𝑀̇[(𝐻
𝑄̇ = 40 ∗ 𝜋 ∗
𝑣2
𝑣2
̂
) − (𝐻 + )𝑖𝑛 ]
2 𝑜𝑢𝑡
2
0.042
1
502 − 402
∗
(221.33 − 214.31 +
)
4
0.06821
2 ∗ 1000
𝑄̇ = 5.505 𝑘𝑊
#Problem 19
Air at 1 atm and 20oC occupies an initial volume of 1000 cm3 in a cylinder. The air is confined by
a piston, which has a constant restraining force so the gas pressure always remains constant. Heat
is added to the air until its temperature reaches 260oC. Calculate the heat added, the work the gas
does on the piston, and the change in internal energy of the gas. Air could be assumed to be ideal
gas with Cp* = 1.005 x 103 J/kgoC
Choose system as the air inside the cylinder
Heat added to system is calculated as
Q = mcp∆T
Since air is an ideal gas, we have
𝑀𝑃𝑉 28.97 ∗ 105 ∗ 10−3
𝑚=
=
= 1.189 𝑔
𝑅𝑇
8.314 ∗ 293
Q = 1.189*10-3 * 1.005 * 103 * (260 – 20) = 287 J
Mass balance equation for this system is
m1 = m2
thus
N1 = N2
𝑃1 𝑉1 𝑃2 𝑉2
=
𝑇1
𝑇2
𝑃1 𝑇2
533
𝑉2 = 𝑉1 ( ) ( ) = 10−3 ∗ 1 ∗ (
) = 1.819 ∗ 10−3 (𝑚3 )
𝑃2 𝑇1
293
2
𝑊 = − ∫ 𝑃𝑑𝑉 = −𝑃(𝑉2 − 𝑉1 ) = −105 ∗ (1.819 − 1) ∗ 10−3 = −81.9𝐽
1
The change in internal energy is
∆U = Q + W = 287 – 81.9 = 205.1 (J)
#Problem 20
Water is to be heated from its pipeline temperature of 20oC to 90oC using superheated steam at
450oC and 2.5 MPa in a steady-state process to produce 10kg/s of heated water. In each of the
processes below, assume there is no heat loss.
A) The heating is to be done in a mixing tank by direct injection of the system, all of which
condenses. Determine the two inlet mass flows needed to meet the desired hot water flow rate.
B) Instead of direct mixing, a heat exchanger will be used in which the water to be heated will
flow inside copper tubes and the steam will partially condense on the outside of the tubes. In this
case heat will flow from the steam to the water, but the two streams are not mixed. Calculate
the steam flow rate if the steam leaves the heat exchanger at 50 percent quality at 100oC
A) define the system as the water and steam
𝑀 = 𝑀𝑤𝑎𝑡𝑒𝑟 + 𝑀𝑠𝑡𝑒𝑎𝑚
Because the system is a closed adiabatic system,
10 ∗ 𝑈𝑓 (90𝑜 𝐶, 2.5𝑀𝑃𝑎) = (10 − 𝑀𝑠𝑡𝑚 ) ∗ 𝑈(20𝑜 𝐶, 2.5𝑀𝑃𝑎) + 𝑀𝑠𝑡𝑚 ∗ 𝑈(450𝑜 𝐶, 2.5 𝑀𝑃𝑎)
Plugging in given values from NIST or appendix A.III
𝑘𝑔
𝑘𝑔
𝑘𝐽
𝑘𝐽
𝑘𝐽
10 ⁄𝑠 ∗ 376.33 ⁄𝑘𝑔 = (10 ⁄𝑠 − 𝑀𝑠𝑡𝑚 ) ∗ 86.76 ⁄𝑘𝑔 + 𝑀𝑠𝑡𝑚 ∗ 3026.2 ⁄𝑘𝑔
Using some simple algebra and solving for 𝑀𝑠𝑡𝑚
𝑀𝑠𝑡𝑚 = 0.985
𝑘𝑔⁄
𝑠
B) Once again, the system is defined as both the water and the steam
𝑀 = 𝑀𝑤𝑎𝑡𝑒𝑟 + 𝑀𝑠𝑡𝑒𝑎𝑚
Since the system is still a closed adiabatic system,
10 ∗ 𝑈𝑓 (90𝑜 𝐶, 2.5𝑀𝑃𝑎) + 0.5𝑀𝑠𝑡𝑚
∗ (𝑈𝑔𝑎𝑠 (100𝑜 𝐶, 0.10135 𝑀𝑃𝑎) + 𝑈𝑙𝑖𝑞 (100𝑜 𝐶, 0.10135 𝑀𝑃𝑎))
= 10 ∗ 𝑈𝑖 (20𝑜 𝐶, 2.5𝑀𝑃𝑎) + 𝑀𝑠𝑡𝑚 ∗ 𝑈𝑖 (450𝑜 𝐶, 2.5 𝑀𝑃𝑎)
Plugging in values from NIST or appendix A.III
𝑘𝑔⁄
𝑘𝐽
𝑘𝐽
𝑘𝐽
𝑠 ∗ 376.33 ⁄𝑘𝑔 + 0.5𝑀𝑠𝑡𝑚 ∗ (2505.5 ⁄𝑘𝑔 + 418.94 ⁄𝑘𝑔)
𝑘𝑔
𝑘𝐽
𝑘𝐽
= 10 ⁄𝑠 ∗ 86.76 ⁄𝑘𝑔 + 𝑀𝑠𝑡𝑚 ∗ 3026.2 ⁄𝑘𝑔
Using some simple algebra and solving for 𝑀𝑠𝑡𝑚
10
𝑀𝑠𝑡𝑚 = 1.85
𝑘𝑔⁄
𝑠