Thermodynamics

1. Chemical and Mechanical Engineering 2300 / Thermodynamics I
Solution to Homework Assignment 1 (Lectures 1 - 3)
Prof. Geof Silcox
Chemical Engineering
University of Utah
Due Monday, 2014 September 8, by 17:00
Problem 1
The energy content of 1-liter of whole milk is 680 Cal. Using that data, determine how
high can you climb (in meters) on a liter of whole milk. Assume that your mass is 65 kg
and that your body can convert 20 percent of the energy in the milk to mechanical work.
Approximate answer: 900 m.
Solution
Let h = energy content of 1-liter of milk. Let  = fraction of energy in milk that is
converted to mechanical work. Then the change in the potential energy of your body as
you climb a distance z is
PE  mgz  h
Solving for the change in elevation gives
  2 2
0.20 680 Cal 4.1868 kJ 1000 J 1kg m /s

z h m
1Cal 1kJ 1 J 893 (65)kg(9.81) m
   
2
s
mg
The energy content of the milk that is not used to increase the potential energy your body
will be converted to thermal energy. The energy in the chemical bonds in the milk is
conserved. Note how important the units are in solving this problem. Always check your
units to make sure they are consistent.
Problem 2
Suppose that you put some dry ice (solid CO2) into an empty, 2-liter plastic pop bottle,
and screw on the lid. When the dry ice sublimes (vaporizes) and the gases warm to room
temperature, the bottle will explode if you have added enough dry ice. Calculate the
minimum number of grams of dry ice required to cause an explosion given that the bottle
will rupture when its internal pressure reaches 11 bar (gage).

2. Solution

3. Problem 3
The barometer of a mountain hiker reads 930 mbars at the beginning of a hiking trip and
780 mbars at the end. Neglecting the effect of altitude on the local gravitational
acceleration, determine the vertical distance climbed. Assume an average air density of
1.20 kg/m3.
Solution
The definition of a bar is
1 bar  105 Pa
The relationship between mbars and Pa is
103 mbar  105 Pa
The change in pressure with elevation is approximately
P  gz
Then
1270 m
z P 3
10 Pa
10 mbar
930 mbar 780 mbar
9.80665 m
s
1.20 kg
m
g
5


3 2


 
 







 
Problem 4
When buildings have large exhaust fans, exterior doors can be difficult to open due to a
pressure difference between the inside and outside. Do you think you could open a 3- by
7-ft door if the inside pressure were 1 in. of water (vacuum)?
Solution
The pressure difference will make it difficult to open the door. The force holding the
door closed will be
F  PA   gzA
where

4. 3




1000 kg / m
2
g 9.81m/s
z 0.0254 m
A 1.951m
2


The force is
F 1000 kg 9.81 m 0.0254 m 1.951 m 486 N
     
 2 
3 2
m s
 
This corresponds to the force exerted by a mass of 486/9.81 = 49.6 kg or 109 lb.
Problem 5
A twin-bladed wind turbine has rotors with diameter 91 m. The reported average power
output of the turbine is 2.5 MW. How does this compare with the best possible at your
location? The theoretical maximum utilization of the kinetic energy in the wind that
approaches such turbines is 16/27 (called the Betz limit). Assume that for your location
the average air temperature is 10 C and the average wind velocity, V, is 24 km/hr.
The rate of flow of kinetic energy (W) is
1 2
2

 
KE mV
The mass flow rate of air is
m   AV
where the area is A = D2/4 (D is the diameter of the rotors) and  is the density of air.
Solution
We can assume Patm = 100 kPa and calculate the density of air from the ideal gas law:
100 1.232
P kPa kg
RT kPa m K m
   
3 3
0.287 283.15
kg K
The rate of flow of kinetic energy is

5.            
1 1 1.232 kg  91m 
24 km 1h 1000 m 1.19 MW
2 2 m 4 h 3600 s 1km
3
3 2
3
KE AV


   
The maximum power output is
16 1.19 
0.703 MW
27
The reported output, 2.5 MW, is far too high.
Problem 6
(a) Calculate the specific internal energy (kJ/kg) of helium (He) at 300 K using the
formula given in Lesson 3 of the notes,
3
2
u  RT
and compare your values to those obtained from
v u  c T
where cv is the specific heat (kJ/kgK) from Table A-2a, and T is the temperature in
degrees kelvin. Explain any differences that you see.
Solution
(a) From the kinetic theory of gases,
8.314 kJ
3 3 kmol K 300 K 934.68 kJ 2 2 4.003 kg kg
u  RT  
kmol
From Table A-2a,
3.1156 kJ
v kg K c 
and
3.1156 kJ 300 K 934.68 kJ
v kg K kg u  c T  

6. The theoretical and measured values of u agree perfectly.
(b) Because He is a monatomic element, cv will not change with temperature. We
conclude that the value of cv for monatomic ideal gases is a constant given by
3 or 3
v 2 v 2 u c  R c  R