Thermal Fluid Sciences Spring 2013 First Test

Name ______(100/104)

Note: be fairly tough on grading – not like the quizzes

Make a rubric whenever you give partial credit (x points for the equation, y points for correctly getting parameter z, k points for getting the units correct, etc)

Make sure they include units

1.  (4) Mark each property below as intensive (I), extensive (E):

a)  Temperature ___I____

b)  Pressure ____I____

c)  Mass ___E______

d)  Molar specific volume _I______

2.  (4) ( True / False ) The continuum idealization allows us to treat properties as point functions and to assume the properties vary continually in space with no jump discontinuities.

3. (12) Jane, the engineering wannabe who didn’t study her thermodynamics, pushes down with force F1 of 50 N at location (1), on the piston with area (A1) of 0.01 m2. Her effort is just sufficient to support the weight of your new Subaru station wagon with a mass of 1000 kg. The bottom of each cylinder is 1 meter above the floor. Find P1, P2, A2 and F2. Ignore the mass of the cylinders and platform the car is on.

Write out all relevant equations:

P1=P2, F1=P1A1 F2=P2A2

Solve for P1

P1=F1/A1 = 50 N/(0.01 m^2) = 5,000 Pa or N/m2

Solve for F2

F2 = P2*A2 = P1*A2 = mg = 1000 kg *9.81 m/s2 = 9810 N

Solve for P2 and A2

P2=P1=5,000 Pa A2=F2/P2 = 9810 N/5000 Pa = 1.96 m2

4. (8) At an elevation of 9,800 ft in the mountains the temperature is 20 C and the pressure is 70.2 kPa. What is the boiling point of water at this elevation? Explain
Look in the saturated water and find 70.2 kPa, it corresponds to 90C
at boiling the vapor pressure of the water equals the total atmospheric precssure
5. (8) What point the highest pressure in the tank? __H___ or copout
At Point B a) absolute pressure is > gage pressure, b) absolute pressure < gage pressure, c) absolute pressure = gage pressure, d) cop out


6. (8) (also refers to figure above) If h = 10 m, T = 50 Celsius, atmospheric pressure is 101.3 kPa. Find:

a)  Specific volume of the water
0001012 m^3/kg from the saturated T table

b)  Density of the water
= 1/specific volume = 1/0.001012 kg/m3 = 988.14 kg/m3

c)  Gage pressure at A
=rho g h = 988 kg/m3 * 9.81 m/s2 10 m = 96923 N/m2 = 96923 Pa

d)  Absolute pressure at A

= Patm + Pgage = 101.3 kPa + 96.923 kPa = 198 kPa

7. A closed vial of water and water vapor has a pressure of 150 kPa and a quality of 0.10. Find the temperature and specific internal energy of the water.

Go to the saturated water pressure chart, at 150 kPa, T is 111.35 C

And along the same line: uf = 466.97, ufg=2052.3 u = uf + x ufg = 466.97 + 0.1*2052.3=672.2 kJ/kg

8. (12) After his fifth wife left him and he was fired, J-- moved into a shed set up along the border. Regular government checks allow him to purchase cheap wine which comes in clear glass bottles. Occasionally he even gets to flirt with the ladies in the border patrol – life could be worse. Perhaps, if the shed wasn’t so stinking hot, he could get his favorite to come inside and stay awhile. One morning he has a delirium tremor derived crazy idea based upon refilling the wine bottles with water and sticking them through the roof. Explain how his air conditioning system works. Specifically:

a.  How does his system depend upon day/night temperature related changes in density of the water? How does he prevent the heat on top of the roof from coming down the water in the bottle? How does the heat escape at night?
during the night the bottle tops radiate thermal energy to the clear desert sky and thus (potentially) cool below atmospheric temperature. The cooled water descends to the bottom of the bottles, cooling the shed, since cooing increases the density of the water

during the day the sun and outside air heat the water in the bottles, since hot water has a lower density, it doesn’t descend to the bottom of the bottle; the remaining cold water at the bottom of the bottle left over from the nigth absorbs thermal energy from inside the hot shed during the day

b.  If each bottle holds 1 kg of water, how much thermal energy will each bottle remove from the shed as the water in the bottle changes from 25 degrees C to 30 degrees C?
Table A-3 gives the specific heat of water at this T as 4.18 J/Kg/K
m Cp (T2-T1) = 1 kg 4.18 kJ/kg/K (30-25) C = 20.9 kJ

c.  If the bottles contain only water and water vapor, what is the pressure of the water vapor in the bottles in the morning when the water is 25 degrees C?
from the table, Psat = 3.17 kPa

9. (12) Is this cooling or heating the house?

Based on the direction of the arrows, it is heating the house, that’s why the arrow showing heat flux is going into the house not out of it

The COP is:

COPHP = QH/(QH-QL) = 7/(7-5) = 7/2 = 3.5

The Wnet, in is:

COP = QH/W or W = QH/COP = 7 kJ/3.5 = 2 kJ

10. (12) Reasonable claims obey the laws of thermodynamics, does this one?

Draw and label an energy flow diagram (e.g., like in the previous problem) for the system in this problem. Label W, QH, QL and put on the arrows. Don’t put in numbers

The best you can do is the Carnot refrigerator and for that COPcarnot = 1/(TH/TL-1) = 1/(298/261)-1) = 7.05

Thus this system is theoretically possible and very good if true.

Note: TH = 25+273 = 298 K TL = -12+273 = 261

Picture can look like below, arrows can go up or down but they must go from refrigerator to the room

11. (4) Label the diffuser and nozzle (upper is nozzle, lower is diffuser)

12. (12) Compare the specific energy in the water velocity and in the 90 m drop:

Specific energy in velocity (J/kg = m2/s2)
V^2/2 = 3^2/2 = 9/2 = 4.5 m2/s2
Specific energy in elevation drop:

g z = 9.81 m/s2 90 m = 882.9 m2/s2

Prove that J/kg = m2/s2 :

m2/s2 * kg/kg = kg m/s2 * m / kg = N*m/kg = J/kg or you can work back from J/kg and cancel the kg