AREN 2110
Spring 2010
Homework #2: Due Friday, Jan 29, 6 PM Solutions
1. Consider a 70-kg woman who has a total footprint area of 400 cm2. She wishes to
walk on the snow, but the snow cannot withstand pressures greater than 0.5 kPa.
Determine the minimum size of the snowshoes (imprint area per snowshoe) to
enable her to walk on the snow without sinking.
Assumption: walking in snowshoes requires that one snowshoe at a time carries the entire
load.
Solution: FBD of snowshoe. Resistance of snow before loading by woman = atmospheric
pressure force, where P = net pressure of snow loaded by women on snowshoes = 0.5
kPa*A, where A = area of one snowshoe
70 kg * 9.81 m/s2
0.5 kPa (1000 Pa/kPa) *A
A = 70*9.81 N/500 N/m2 = 1.37 m2
(these would be enormous snowshoes – typically snowshoes have much smaller area and
snowshoer sinks somewhat into the snow, which in turn packs the snow and increases
resistance)
2. Determine the pressure exerted on a diver at 30 m below the free surface of the
ocean. Assume the barometric pressure is 101 kPa and the specific gravity of
seawater is 1.03.
Solution:
Patm = 101 kPa
Pw(z = 30m)
Pw = gz = 1.03*(1000 kg/m3)*9.81 m/s2*30m * 10-3 kPa/Pa = 303 kPa
Pdiver = Patm + Pw = 101 + 303 kPa = 404 kPa
3. Is it possible to have water vapor at -10 oC (263K)? Explain.
Yes. Saturated ice and saturated water vapor are in equilibrium when T is lower
than the triple point temperature for the triple point pressure of the system. From
appendix table A-8, at T = -10 oC, corresponding pressure for saturated ice and
vapor is 0.2599 kPa.
4. Does the amount of heat absorbed as 1 kg of saturated liquid water boils to
saturated vapor have to the same as the heat released when 1 kg saturated water
vapor condenses at 100 oC? Explain.
Yes. Absorbed energy during vaporization has to equal the energy released during
condensation at the same T and P. If you could release/absorb more energy in either
direction, then you would be creating or destroying energy – a violation of the first
law of thermodynamics.
5. Complete the following table for H2O
1
2
3
4
T (oC)
50
120.21
250
110
P (kPa)
12.352
200
400
600
v (m3/kg)
4.16
0.88578
0.59520
0.001052
phase
Saturated mixture
Saturated vapor
Superheated vapor
Compressed liquid
1: vf < v < vg, @ 50 oC means saturated mixture and P = Psat @ 50 oC (Table A-4)
2: T = Tsat @ 200 kPa and v = vg,@ 200 kPa (Table A-5)
3: T > Tsat @ 400 kPa (143.6 oC). Superheated vapor; get v from Table A-6.
4. T < Tsat @ 400 kPa (158.8oC), v ~ vf @ 110 oC (Table A-4)
6. Complete the following table for H2O:
1
P (kPa)
200
T (oC)
120.21
h (kJ/kg)
2045.8
x
0.7
2
361.53
140
1800
0.565
3
4
950
500
177.66
80
752.74
335.02
0.0
NA
5
800
350
3162.2
NA
phase
Saturated
mixture
Saturated
mixture
Saturated liquid
Compressed
liquid
Superheated
vapor
1: 0 < x < 1 means saturated mixture. h = x*hfg + hf @ 200 kPa (Table A-5)
2: hf < h < hg @ 140 oC, means saturated mixture and P = Psat @ 140 oC and
calculate x = (h – hf)/hfg @ 140 oC (Table A-4)
3: x = 0 is saturated liquid, T = Tsat and h = hf @ 950 kPa (Table A-5)
4: T < Tsat @ 500 kPa (151.8 oC) is compressed liquid. h ~ hf @80 oC (Table A-4)
5: h > hg @ 800 kPa is superheated vapor. Get h from Table A-6, P = 0.8 MPa,
T=350 oC.
7. Complete the following table for the various substances
substance P (kPa)
T (oC)
v (m3/kg)
x*
1
H2O
150
0.4
476.16
0.1576
2
H2O
150
0.4708
400
NA
3
R-134a
0
0.0500
293.01
0.719
4
R-134a
400
0
0.0007723 NA
* use "na" for "not applicable" where quality does not apply
phase
Saturated mixture
Superheated vapor
Saturated mixture
Compressed liquid
1: P = Psat, v = x(vfg) + vf @ 150 oC (Table A-4)
2: v > vg @ 150 oC (0.39248 m3/kg) is superheated vapor. Find P corresponding to v
= 0.4708 m3/kg @ 150 oC (Table A-6)
3. vf < v < vg, @ 0 oC C, is saturated mixture. P = Psat and x = v – vf/vfg @ 0 oC (Table
A-11)
4. T < Tsat, @ 400 kPa (143.6 oC) is compressed liquid (Table A-5) and v ~ vf @ 0 oC
(Table A-4)
8. A piston-cylinder device contains 0.85 kg of refrigerant 134a at -10 oC (263 K).
The piston has a mass of 12 kg and a diameter of 25 cm. The local atmospheric
pressure is 88 kPa. Now, heat is transferred to the refrigerant until the temperature
is 15 oC. Determine:
a. The final pressure
b. The change in volume of the cylinder space
c. The change in enthalpy of the refrigerant
Solution: piston cylinder keeps constant pressure during expansion process. Force
balance (FBD) on piston at state 1 to determine pressure of refrigerant at states 1 and 2.
88kPa*Ap (kN)
12 kg*9.81m/s2*10-3 kN/N
P1*Ap (kN)
Ap = *(0.25 m)2/4 = 0.05 m2
a) P1 = (88*0.05 kN + 12*9.81* 10-3 kN)/0.05 m2 = 90.4 kPa = P2 (final pressure)
b) 90.4 kPa < Psat @ -10 oC, so superheated refrigerant. Use Table A-13. Find v1 (@
-10 oC) by interpolation between v @ 60 kPa and 100 kPa.
v1 = 0.242 m3/kg
to find v2, must first interpolate to find v @ 15 oC at P = 60 kPa and P = 100 kPa.
v15, 60 = (0.39302+0.37893)/2 = 0.386 m3/kg
v15, 100 = (0.23373+0.22506)/2 = 0.2294 m3/kg
v2 = 0.267 m3/kg
V = m*(v2 – v1) = 0.85 kg*(0.267 – 0.242) m3/kg = 0.0212 m3
c) Similar procedure as b) for H
h1 = 247.75 kJ/kg
to find h2, must first interpolate to find h @ 15 oC at P = 60 kPa and P = 100 kPa.
h15, 60 = (272.94+264.66)/2 = 268.8 kJ/kg
h15, 100 = (272.17 + 263.81)/2 = 268.0 kJ/kg
h2 = 269.4 kJ/kg
H = m*(h2 – h1) = 0.85 kg*(269.4 – 247.75) kJ/kg = 18.4 kJ/kg
9. A rigid tank with a volume of 2.5 m3 initially contains 15 kg of saturated liquidvapor mixture of water at 75 oC. The water is slowly heated until all the water is
saturated vapor.
a. What is the quality of the mixture at the initial state (before heating)?
b. Determine the temperature at with the liquid in the tank is completely
vaporized to saturated vapor.
c. What is the pressure in the tank?
d. Show the process on the T-v diagram on the next page.
Solution: this is a constant volume (isochoric) phase change process. Both temperature
and pressure will increase as heat is added to closed system rigid tank.
a) v = V/m = 2.5 m3/15 kg = 0.1667 m3/kg. Since vf < v < vg @ 75 oC, system is
saturated mixture.
x = 0.04
b) at state 2, water is saturated vapor, and v2 = v1 = 0.1667 m3/kg = vg @ T2 where
185 < T2 < 190 oC (Table A-4). Interpolate
T2 = 187 oC
c) interpolate to find P2:
P2 = 1176 kPa
T (oC)
Water Liquid-Vapor PhaseT-v Diagram
400
350
300
250
200
150
100
50
0
0.0001
0.001
0.01
0.1
1
10
100
1000
v (m3/kg)
10. One kilogram (1 kg) of water vapor at 200 kPa fills the left chamber of a
partitioned system shown below. The volume of this chamber is 1.1989 m3. The
right chamber has twice the volume of the left chamber and is evacuated at the
initial state.
1 kg water
200 kPa
1.1989 m3
Now the partition is removed and heat is transferred so that the temperature of the
water is 3 oC.
a. What is the initial temperature of the water (before the partition is removed)?
b. What is the pressure of the water after the partition is removed and heat
transferred?
c. What is the quality of water at the final equilibrium state?
Solution: State 1 Calculate v1 from given: V and m: v1 = 1.1989 m3/1 kg = 1.1989 m3/kg.
At P1 = 200 kPa, v1 > vg (0.88578 m3/kg) so system is superheated vapor.
a)
From Table A-6, v1 corresponds to T = 250 oC
b) At T = 3 oC, v2 = V2/m where V2 = 3V1 = 3*1.1989 m3 and
v2 = 3*1.1989 m3/1kg = 3.597 m3/kg. Interpolate to find vg @ 3 oC:
vg @ 3 oC = 182.4 m3/kg
OR notice that 3.597 m3/kg is less than vg at either 0.01 or 5 oC, so system is saturated
mixture at state 2.
Interpolate to find Psat at 3 oC:
P2 = 0.768 kPa
c) Find x at state 2 using vf and vg at 3 oC, 0.001 and 182.4 m3/kg, respectively.
x = 0.02