AREN 2110
Spring 2010
Homework #1 SOLUTIONS: Due Friday, Jan. 22
1. A closed rigid-wall tank with volume = 0.1 m3 contains 1 kg wet steam including
liquid water and vapor phases in equilibrium, as shown below. The specific
volumes of the liquid and vapor phases are 1.127 x 10-3 m3/kg and 0.1943 m3/kg,
respectively. (Sketch is not to scale)
a. Calculate the volume occupied by the vapor phase
b. Calculate the mass fraction of liquid water in the water-steam mixture in
the system.

MT = Mg + Mf
Where MT = total mass (kg), Mg = mass sat. vapor (kg),
Water
and Mf = mass sat. liquid (kg)
vapor
(steam)

VT = Vg + Vf
Where VT = total volume (m ), Vg = volume sat. vapor
(m3), and Vf = volume sat. liquid (m3)
3
Liquid
water
 V = M*v
where v = specific volume (m3/kg)
combine  and : MTvT = Mgvg + Mf vf
from  substitute for Mf = MT - Mg
MTvT = Mgvg + (MT - Mg)vf
Solve for Mg =
solve for M g
M T vT
vg
MT v f
vf
0.1 0.001127
0.512 kg
0.1943 0.001127
a) Vg = Mgvg = = 0.512kg(0.1943m3/kg) = 0.0994 m3
Mf
MT
1 0.512
1
0.488
2. The temperature, T in oC on a thermometric scale is defined in terms of a
property, x, by the relationship.
T = alog10x + b
At the ice point, x = 2.3 and at the steam point, x = 6.8. Evaluate the temperature
when x = 3.
Solution: find a and b from two points given:
0 = a(log10(2.3)) + b
and
100 = a(log10(6.8)) + b
a = 212.5 and b = -76.94
T(x=3) = 212.5(log10(3)) – 76.94
T = 24.5 oC
3. A CU engineering student is tired of the fact that the ice point of water in existing
absolute temperature scales is an awkward number. She proposes a new absolute
temperature scale, called the Buff scale, with the Buff degree symbolized by B.
The ice point of water is 500 B.
a. Is this a valid temperature scale? Why/Why not?
Yes as long as the lowest value on the Buff scale = 0 and the units (degrees) are of
equal size.
b. What is the boiling (steam) point of water in B?
Solution: find degrees B in terms of degrees K:
500B = 273K
1K = (500/273)B
Boiling temperature of water at 1 atm. = 373K = 373K(500/273)B/K = 683 B
c. “Room temperature” is generally considered to be 20 degrees Celsius (oC).
Calculate room temperature in B.
Solution: use relation from b to solve for room temperature in degrees B
Room temperature = 20 oC = 293K = 293K(500/273)B/K = 537B
4. Calculate the work done by the systems (denoted in italics) in the following
processes:
a. A body of mass 10 kg falls freely through a vertical distance of 30 m in a
gravitational field with g = 9.81 m/s2. The drag force of the atmosphere on
the body is 4 N. (system consists of the body and the atmosphere)
Solution:
Work = W = potential energy exerted - work against drag force
W = mg( h) - Fd( h) = 10kg(9.81 m/s2)30m – 4N(30m)
W = 2,823 J
b. A mechanical stirrer driven by an electric motor mixes a fluid for 10 min.
The motor exerts a torque of 0.006 N-m and the stirrer rotates at 900 rpm.
(The system is the stirrer and the fluid stirred.)
Solution:
W=
t = 0.006 N-m (900 min-1) 10 min
W = 54 J
c. A 2-kW electric kettle (includes heating element and water) is turned on
for 5 min.
Solution:
W = Pt = 2 kJ/s 5 min (60 s/min)
W = 600 KJ
5. Specify:
a. A closed system with a fixed boundary and define any two intensive
thermodynamic properties of the system
Rigid tank filled with air at 400K and 200 KPa
b. A closed system with a movable boundary and define any two extensive
thermodynamic properties of the system
Piston-cylinder containing 0.1 kg steam in a volume = 0.01 m3
c. An open system and define any two independent intensive properties of
the system
Pump with the pressure of liquid water at the outlet = 8,000 kPa, and temperature =
200 oC. (or P-v, or T-v, etc.)
6. Gas is trapped in two horizontal cylinders by frictionless pistons, shown below.
The cylinders are connected by a valve that is closed. The system is the gas in the
cylinders. What is the magnitude of the force, F, in kN, required to maintain static
equilibrium in the system with the valve closed? (area of piston face = 0.03 m2)?
F
Patm =
100 kPa
Gas, P =
200 kPa
Gas, P =
100 kPa
Patm =
100 kPa
Solution: Equilibrium of forces at piston in left cylinder determines required force
when valve is closed (right cylinder is at equilibrium with only force from
atmospheric pressure):
F + 100 kPa(Ap) = 200kPa(Ap)
F = 100 kPa(0.03 m2) = 3 kN
7. The device below is used to raise the pressure of air in the rigid tank above the
initial value of 100 kPa. When a force, F, is applied to the frictionless piston and
the pressure in the cylinder exceeds 200 kPa, the one-way valve opens and air is
pushed into the rigid tank. Assume that during this process, the air pressure in the
cylinder remains at just barely above 200 kPa while the piston moves 0.2 m to the
right. Cross sectional area of the piston (face) is 0.03 m2. Consider the system as
the cylinder and the rigid tank.
a. Is this system open or closed? Justify your answer.
b. Calculate the work of the system, applying the sign convention.
c. Calculate the work of the surroundings, applying the sign convention.
Rigid Tank
Air, P =
200 kPa
F
Patm =
100 kPa
Air, P
= 100
kPa
Solution
a) total system is the piston cylinder and the tank. It is a closed system since no
mass crosses the boundary during the process with the valve open.
b) Work of the system = boundary work at constant pressure
W = P(V2 – V1) = P*Ap(x2 – x1) where x = linear displacement of the
system and Ap = piston area.
Wsys = 200 kPa (0.03) m2 (-0.02 m) = - 1.2 kJ (work done on system)
c) work of surroundings = - Wsys = + 1.2 kJ (work done by surroundings)
8. Find the atmospheric pressure in KPa when the height of Hg in a barometer
column is 750 mm.
Solution: apply manometer equation:
Patm = gh = 13,600 kg/m3 (9.81 m/s2) 0.75 m (1 kPa/1000 Pa) = 100.1 kPa
9. The barometer of a mountain hiker reads 930 mbars at the beginning of the hike
and 780 mbars at the end. Neglecting the effect of altitude on local gravitational
acceleration, determine the vertical distance climbed, assuming an average air
density of 1.2 kg/m3.
P2, z 2
Solution: P1 > P2, and , g constant, 1 bar = 105 Pa
P1 – P2 = g (z2 – z1) = 93,000 - 78,000 Pa
15,000 Pa = 1.2 kg/m3 (9.81 m/s2) (z2 – z1) m
(z2 – z1) = 1,274 m
P1, z 1
10. Pressure cooker with inside gage pressure of 100 kPa, atmospheric pressure = 101
kPa. Petcock opening = 4 mm2. Find mass of petcock.
Solution: petcock in static equilibrium @ P = 201 kPa
PatmAp
a. Free body diagram of petcock:
b. calculate m from force balance:
(P-Patm)Ap = Pg =100 kPa (Ap) = m (9.81 m/s2)
m = 100 kPa(1000 Pa/kPa) (4 mm2) (10-6 m2/mm2)/9.81 m/s2
m = 0.0408 kg
mg
PAp