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E234 - Thermodynamics
Assignment – Week 3
1. The electrical power of some electronic chips may soon reach 75 W/cm3. Assume
the chip has a surface of 4 cm2. Could it be cooled by forced convection of
gases? (use data given in the class notes). If not, suggest a possible scheme for
Forced convection of gases: h = 25~250 [W/m2] and the heat flux = hA(Ts-Tsurr)
Forced convection won’t work.
Other possible way : increase surface area of the heat sink. What IBM is considering,
and this was discussed in class, is to use a liquid coolant that can carry away much more
energy than a gas.
2. The wall of a house can be modeled as a composite consisting of a layer of bricks
10 cm thick next to a layer of urethane foam 8 cm thick. The house has doublepane windows with the specifications (and the solution) given in Problem 6,
Assignment 3 Spring 2007. Assume the roof is well insulated and has negligible
heat losses. A typical Hoboken house is 6 m wide, 12 m long and 10 m tall. It
has 18 windows of size 1mx2m.
a) Determine the net heat losses of the house on a day when the temperature is
- 5 oC outside and 25 oC inside.
b) If the house is heated by electricity, what is the cost of heating it for a month
(assume the conditions above prevail during the entire 30 days)
(1) Double pane window
R " RA + glass + air + glass + 0.402[m 2 K / W ]
h K glass K air K glass h
a) Use Q as given in problem 6. Assignment 3, S2007. There is an extra term to account for
convection from the inside of the house to the inside wall. In problem 6, the temperature
of the inner pane is given whereas here we are given the temperature of the room.
1 0.1 0.08 1
R " RA + Brick + Urethane + +
0.458[m 2 K / W ]
h K Brick KUrethane h 0.72 0.26 175
T T A
R" Fall 2008 30[ K ] ﾴ 2[m 2 ] ﾴ 18 30[ K ] ﾴ 324[ m 2 ]
Q Qwindow +Qwall +
24,900W 24.9 KW
0.402[m 2 K / W ] 0.458[m 2 K / W ]
b) 0.10$ 24.9kW ﾴ 24h 30day
$1, 793 / month
1month 3. In a cyclic process, a gas undergoes the four following processes:
- From State 1 (V = 1 m3, P = 500 kPa, nRT = 500) to State 2, constant volume
process to P = 1000 kPa
- From State 2 to State 3, isothermal process to P = 500 kPa
- From State 3 to State 4, constant volume process such that the temperature in
State 4 is equal to that in State 1 (T4 = T1).
- From State 4 to State 1, isothermal process back to the initial state
a) Sketch the four processes (the cycle) on the same paper. (See previous
b) Calculate the work in each process and the total work
) for isothermal work
State 1►2, 3►4 : W=0 (constant volume)
Wtotal W23 +W41 1000 ln( ) +500 ln( ) 346.6kJ
2 b) W p dv nRT ln( 4. A balloon behaves so the pressure is P = C2V1/3, C2 = 100 kPa/m. The balloon is
blown up with air from a starting volume of 1 m3 to a volume of 3 m3.
a) Why is the unit of C2 in kPa/m
b) Find the final mass of air assuming it is at 25oC
c) Determine the work done by the air.
a) P = C2V1/3 , V=m3
(kPa is the unit of P)
Fall 2008 kPa= C2m
∴ C2= kPa/m
b) P = C2V1/3
=100 [kPa/m] x 31/3 [m] = 144.2 [kPa]
PV = mRT
144.2[kPa] 3[m 3 ]
RT 8.314[kPa m 3 / kmol K ] 298.15[ K ]
Molecular mass of air = 28.97 kg/kmol
174.5[mol ] 5.06kg
C 1/ 3
[ 4/3 3
c) W p dv 2V dv C2 V ]1 249.5 J
5. For a buffer storage of natural gas (CH4) a large bell in a container can move up
and down keeping a pressure of 105 kPa inside. The sun then heats the container
and the gas from 280 K to 300 K during 4 hours. What happens to the volume and
what is the sign of the work term?
nR (T300 K T280 K )
The volume increases as the temperature rises.
V V f Vi The work is positive since the pressure is constant and volume is increasing.
W P V Fall 2008 ...
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This note was uploaded on 05/25/2010 for the course E-234 E-234 taught by Professor Gallois during the Spring '09 term at Stevens.
- Spring '09