Solution Manual for Engineering and Chemical Thermodynamics 2nd Edition by Koretsky PDF (7 Files Merged) | Complete Step-by-Step Solutions and Worked Examples | Covers the Laws of Thermodynamics, Energy Balances, Entropy, Phase Equilibria, Chemical Reacti

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SOLUTION MANUAL

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1.2
An approximate solution can be found if we combine Equations 1.4 and 1.5:


_!_ mJ7 2 = e;olecular
2
kT = e;olecular
2


.-. v l:
Assume the temperature is 22 °C. The mass of a single oxygen molecule is m = 5.14 x 10-
26
kg . Substitute and solve:


V = 487.6 [mis]
The molecules are traveling really, fast (around the length of five football fields every

second). Comment:
We can get a better solution by using the Maxwell-Boltzmann distribution of speeds that
is sketched in Figure 1.4. Looking up the quantitative expression for this expression, we
have:


f ( v)dv = 4;r(_!!!_) 312
2 2
exp{ -_!!! v }v dv
2;rkT 2kT

where.f(v) is the fraction of molecules within dv of the speed v. We can find the average
speed by integrating the expression above


Jf ( v)vdv =
00




-=
V 0
8kT = 449 [m/s ]
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1.3
Derive the following expressions by combining Equations 1.4 and 1.5:




Therefore,


Va 2
mb
V-2b ma


Since mb is larger than ma , the molecules of species A move faster on average.




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, 1.4
We have the following two points that relate the Reamur temperature scale to the
Celsius scale:

(o °C, 0 °Reamur) and (100 °C, 80 °Reamur)

Create an equation using the two points:

T (0Reamur) = 0.8 T(° Celsius)

At 22 °C,

T = 17.6 °Reamur




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1.5

(a)
After a short time, the temperature gradient in the copper block is changing (unsteady
state), so the system is not in equilibrium.

(b)
After a long time, the temperature gradient in the copper block will become constant
(steady state), but because the temperature is not uniform everywhere, the system is not
in equilibrium.

(c)
After a very long time, the temperature of the reservoirs will equilibrate; The system is
then homogenous in temperature. The system is in thermal equilibrium.




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, 1.6
We assume the temperature is constant at 0 °C. The molecular weight of air is

MW = 29 g/mol = 0.029 kg/mol

Find the pressure at the top of Mount Everest:



P= -(0.029 kg/molX9.8l [m!s]Xss48 m)
(1atm)exp '- --
- ----'-'----=-----=-'--------'-


8.314 [ J ]](273.15 K)
( mol·K
P = 0.330 atm = 33.4 kPa

Interpolate steam table data:

T sat = 71.4 oc for p sat = 33.4 kPa


Therefore, the liquid boils at 71.4 °C. Note: the barometric relationship given assumes
that the temperature remains constant. In reality the temperature decreases with height
as we go up the mountain. However, a solution in which T and P vary with height is
not as straight-forward.




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1.7
To solve these problems, the steam tables were used. The values given for each part
constrain the water to a certain state. In most cases we can look at the saturated table, to
determine the state.

(a) Subcooled liquid
Explanation: the saturation pressure at T = 170 [0C] is 0.79 [MPa] (see page
508);
Since the pressure of this state, 10 [bar], is greater than the
saturation pressure, water is a liquid.
(b) Saturated vapor-liquid mixture
Explanation: the specific volume of the saturated vapor at T = 70 [0C] is
5.04 [m3/kg] and the saturated liquid is 0.001 [m3/kg] (see page 508); Since
the volume of this state, 3 [m3/kg], is in between these values we have a
saturated vapor- liquid mixture.
(c) Superheated vapor
Explanation: the specific volume of the saturated vapor at P = 60 [bar] = 6
[MPa], is 0.03244 [m3/kg] and the saturated liquid is 0.001 [m3/kg] (see
page 511); Since the volume of this state, 0.05 [m3/kg], is greater than this
value, it is a vapor.
(d) Superheated vapor
Explanation: the specific entropy of the saturated vapor at P = 5 [bar] = 0.5
[MPa], is 6.8212 [kJ/(kg K)] (see page 510); Since the entropy of this state,
7.0592 [kJ/(kg K)], is greater than this value, it is a vapor. In fact, if we go
to the superheated water vapor tables for P = 500 [kPa], we see the state is
constrained to T = 200 [0C].




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, 1.8
From the steam tables in Appendix B.1:


vcritical 7;]
= 0.003155 [ (T = 374.15 °C, P = 22.089 MPa)


At 10 bar, we find in the steam tables

3
vsat = 0.001127 [m ]
l kg


vi"' = o. 19444 [ 7;J
Because the total mass and volume of the closed, rigid system remain constant as the
water condenses, we can develop the following expression:




where x is the quality of the water. Substituting values and solving for the quality, we
obtain

x = 0.0105 or 1.05 %

A very small percentage of mass in the final state is vapor.




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1.9
The calculation methods will be shown for part (a), but not parts (b) and (c)

(a)
Use the following equation to estimate the specific volume:

v(l.9 MPa, 250 °C) = v(l.8 MPa, 250 °C)+ 0.5[v(2.0 MPa, 250 °C)-v(l.8 MPa, 250
°C)]
Substituting data from the steam tables,



V(I.9 MPa, 250 °C) = 0.11821 [ ;]



From the NIST website:

VNisr (l.9 MPa, 250 °C) = 0.11791 [ :]


Therefore, assuming the result from NIST is more accurate

J
lv - vN1sr l x 100 % = 0.254 %
% Difference (
= VNJST


(b)
Linear interpolation:

V(I.9 MPa, 300 °C) = 0.13284 [ :]


NIST website:


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VN1.11" (1.9 MPa, 300 "C)= 0.13249 [ :]
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