Final ExamSpring 2017Thermal Fluid Sciences
Full Name ______(100/110)
Closed book, one page of notes one
You must show the work for all problems to get credit.
Casio: All fx-115 models. Any Casio calculator must contain fx-115 in its model name.
Hewlett Packard: The HP 33s and HP 35s models, but no others.
Texas Instruments: All TI-30X and TI-36X models. Any Texas Instruments calculator must contain either TI-30X or TI-36X in its model name.
By my signature below, I hereby swear that I have neither received aid from any other student, nor have I given aid to any other student on this exam.Moreover, I swear that I have not committed any form of academic dishonesty and have maintained ethical integrity in accordance with the UTEP Handbook of Operating Procedures, the UTEP Civil Engineering Honor Code, and the ASCE Code of Ethics.
Signature: ______
By my signature below, I swear that I have not witnessed any other student commit any form of academic dishonesty with regard to this exam.
Signature: ______
Potentiallyuseful information:
286.9 J/kg/K is the ideal gas law constant for air, the molecular weight of air is 29
The specific heat of water is 4.19 kJ/kg/K.
1. (9, 3 each) Find the boiling point of water at 5,000 meters elevation to the nearest 3 degrees (i.e., don’t bother to interpolate). Explain your logic. Specifically,
a) what is the pressure at this point?
54.05 kPa
b) describe in words how you will go about finding the boiling point (i.e., what’s your logic):
at boiling, fluid pressure = atmospheric pressure, take 54.06 and go to the tables; from pressure table, 81.32 < T < 91.76; from T table 80<T<85
c) find the approximate boiling point (a range between two table numbers is OK, don’t bother to interpolate)
at boiling, fluid pressure = atmospheric pressure, take 54.06 and go to the tables; from pressure table, 81.32 < T < 91.76; from T table 80<T<85
2. (15)
Since the same problem was on the practice test, we will just change the question. Assuming the velocity in the pipe is 16.2 ft/s. Find:
a) (3) the velocity head at a point in the middle of the pipe between the two tanks (i.e., after 225 ft. of pipe). The elevation at this location is 250 feet.
b) (6, half credit for equation) the pressure head at the middle point
z1-z2= p2/rho g + v2^2/2g + hloss or p2/rho g = z1-z2- v^2/2g (f L/D + K+1)
p/rho g = 300-250 + 4.075(0.02 *225/0.0833 + 0.5 + 1) = 21.88 ft
c) (3) the location of the HGL and EGL relative to the center of the pipe at the middle point. (draw on figure
correct as long as consistent with prior answers; EGL is 26 ft above pipe, HGL is 21.9 ft above pipe
d) (3) the minor head loss at the entrance of the pipe (from tank A into the pipe) in feet
3. (6, 3 each) The piston moves in the direction shown.
a) If the system is isothermal, show the path from V1 to V2 on a P-V (pressure on y-axis, volume on x axis) diagram.
Path must be downward, otherwise don’t be picky
b) Graphically show the work done during the expansion.
Area under curve shown above
answer: PV =n RT, if its isothermal then P = nRT/V, so pressure must drop inversely with changes in V; full credit if the path shows pressure declining
4. (9)
a) (3) Draw a figure showing the plate and all the relevant heat transfer mechanisms (with basic governing equations) to from and through the plate; define all the terms in the equations with units and show where each applies
show radiation and convection on walls, conduction through
b) (3) Solve the appropriate equation for the thermal conductivity with no numbers (only symbols)
see below, conduction equation solved for k
c) (3) Plug appropriate numbers into the equation to get the thermal conductivity
5. (12) The pipe in the figure was erased when Joe Marketer spilled his coffee on the paper, fortunately engineers can get all this information from the EGL and HGL – no problem. The pipe went from the opening at the bottom of the tank to just above point B and then to point C. The nozzle that causes a jet of water is still shown near point C. Note: EGL is labeled as E.L. in this figure
a) Which way does water flow? Right to left, left to right, water is not flowing
left to right
b) What is the water velocity in ft/s at point C, the jet caused by the nozzle
V^2/2g = 13.7 ft V = sqrt(2 g 13.7) = 29.7 ft/s
c) Which of the two pipe sections has a greater velocity a) one before the pump or b) one after the pump
after because more space between EGL and HGL
6. (15) Explain how the thermal syphon water heater below uses changes in water density to create flow without a pump. The solar collector captures radiation from sunlight. The solar collector warms the water from 50 to 75C. Note that the tank is actually full of water, the white area on top is supposed to show the tank wall.
a) show flow direction with arrows; explain why it flows this direction based upon water density changes
flow counterclockwise because hot fluid is less dense and rises
b) Where in the figure does the water have the greatest density?
blue section at bottom of tank and bottom of panel
c) what is that greatest density in units of kg/m3?
1/0.001012 from temperature table = 988.142 also available from Table A-3
d) How much enthalpy is added by the solar collector in units of kJ/kg of water?
e) How could you modify the piping system shown to increase the flow rate of water through the system (i.e., lower the head loss)? Give at least two types of changes to the pipe system that would increase flow.
Larger pipes and more gradual bends and shorter pipes
7. (12) In the post final exam celebration Uriel Utepper throws his 0.5 kg unopened favorite beverage can across the room at a velocity of 10 m/s. The can hits the wall and the kinetic energy turns to thermal energy.
a) (4) find the momentum of the can before it hits the wall
mass * velocity = 0.5 kg *10 m/s = 5 kg m/s
b) (4) find the kinetic energy of the can before it hits the wall
½ m v^2 = 0.5 kg/2 * (10 m/s)^2 = .25 * 100 = 25 kg m^2/s^2 = 25 N m = 25 J
c) (4) The can smashes into the opposite wall and all the kinetic energy turns to heat. How much does the temperature of the can change assuming we can use the properties of water to approximate the thermal properties of the can (i.e., its mostly water).
Energy = m Cp (T2 - T1)
(T2 - T1) = energy /m Cp = 25 J/ 0.5 kg / 4.19 kJ/kg/K. = 0.12 C or K
8. (12) Below is the coefficient of performance of commercial heat pumps based on the cold outdoor temperature (Tevap) and the warm indoor temperature (Tcond).
a) (3) It is better to have a HIGH or LOW coefficient of performance? Why?
High, more energy efficient COP = QH/Work
b) (3) Provide an equation and define all the terms in it with units that proves your answer in part (a)
COP = QH/Work Q is in Watts or joules W is watts or joules depending upon if it is rates or overall amounts
c) (6) For Tcond=70C and Tevap=30C, the actual COP = 6. What would be the COP of a reversible heat pump at these temperatures?
COP hp = 1/(1-TL/TH)
9. (12, 4 each) Air is heated from 50 to 100 C at a constant pressure. If the initial specific volume is 1 cubic meter/kg.
a) what equation is used to describe the pressure/volume relationships of ideal gasses? Write it out and define terms with units.
Pv = R T
P = pressure, Pa, v = specific volume, m3/kg, R = gas constant for air , 286.9 J/kg/K, T, absolute temperature, K
b) Find the pressure:
P = R T/v = 286.9 * (50+273.15)/1.
c) Find the final specific volume:
count correct if they used pressure above, even if wrong
v = R T/P =
10. (8) For the system below:
a) Mark the location of greatest pressure with a P
just before turbine
b) Which direction does water flow, mark with arrow
right to left
c) At point 1, in the center of the pipe, mark and label, elevation head, pressure head, velocity head
d) Circle any pipe area(s) where the pressure is negative.
Left of turbine