Sonntag, Borgnakke and van Wylen
4.28
A car drives for half an hour at constant speed and uses 30 MJ over a distance of
40 km. What was the traction force to the road and its speed?
Solution:
We need to relate the work to the force and distance
W=¶
´F dx = F x
W 30 000 000 J
F = x = 40 000 m = 750 N
L 40 km
km
1000 m
V = t = 0.5 h = 80 h = 80 3600 s = 22.2 ms1
Sonntag, Borgnakke and van Wylen
4.33
A 400-L tank A, see figure P4.33, contains argon gas at 250 kPa, 30oC. Cylinder B,
having a frictionless piston of such mass that a pressure of 150 kPa will float it, is
initially empty. The valve is opened and argon flows into B and eventually reaches a
uniform state of 150 kPa, 30oC throughout. What is the work done by the argon?
Solution:
Take C.V. as all the argon in both A and B. Boundary movement work done in
cylinder B against constant external pressure of 150 kPa. Argon is an ideal gas, so
write out that the mass and temperature at state 1 and 2 are the same
PA1VA = mARTA1 = mART2 = P2( VA + VB2)
=> VB2 =
2
250 u 0.4
- 0.4 = 0.2667 m3
150
3
´ PextdV = Pext(VB2 - VB1) = 150 kPa (0.2667 - 0) m = 40 kJ
1W2 = ¶
1
Sonntag, Borgnakke and van Wylen
4.42
A piston cylinder contains 1 kg of liquid water at 20oC and 300 kPa. There is a
linear spring mounted on the piston such that when the water is heated the
pressure reaches 3 MPa with a volume of 0.1 m3.
a) Find the final temperature
b) Plot the process in a P-v diagram.
c) Find the work in the process.
Solution:
Take CV as the water. This is a constant mass:
m2 = m1 = m ;
State 1: Compressed liquid, take saturated liquid at same temperature.
B.1.1: v1 = vf(20) = 0.001002 m3/kg,
State 2: v2 = V2/m = 0.1/1 = 0.1 m3/kg and P = 3000 kPa from B.1.3
=> Superheated vapor close to T = 400oC
Interpolate: T2 = 404oC
Work is done while piston moves at linearly varying pressure, so we get:
1
1W2 = ³ P dV = area = Pavg (V2 V1) = 2 (P1 + P2)(V2 - V1)
= 0.5 (300 + 3000)(0.1 0.001) = 163.35 kJ
P C.P.
T
2
C.P.
2
300 kPa
1
300
T
20
v
1
v
Sonntag, Borgnakke and van Wylen
4.46
Consider the problem of inflating the helium balloon, as described in problem
3.79. For a control volume that consists of the helium inside the balloon
determine the work done during the filling process when the diameter changes
from 1 m to 4 m.
Solution :
Inflation at constant P = P0 = 100 kPa to D1 = 1 m, then
P = P0 + C ( D* -1 - D* -2 ),
D* = D / D1,
to D2 = 4 m, P2 = 400 kPa, from which we find the constant C as:
400 = 100 + C[ (1/4) - (1/4)2 ] => C = 1600 kPa
S
The volumes are:
V = 6 D3 => V1 = 0.5236 m3; V2 = 33.51 m3
2
WCV = ´
¶ PdV
1
2
* -1 - D* -2)dV
= P0(V2 - V1) + ´
¶ C(D
1
S
V = 6 D3,
S
S
dV = 2 D2 dD = 2 D13 D* 2 dD*
D2*=4
Ÿ WCV = P0(V2 - V1) + 3CV1
´
¶
(D*-1)dD*
D1*=1
4
D2* 2 - D1* 2
*
*
= P0(V2 - V1) + 3CV1[
(D
D
)
]
2
1
2
1
16-1
= 100u33.51 – 0.5236) + 3u1600u0.5236 [ 2 – (4–1)]
= 14 608 kJ
Sonntag, Borgnakke and van Wylen
4.48
The piston/cylinder shown in Fig. P4.48 contains carbon dioxide at 300 kPa,
100qC with a volume of 0.2 m3. Mass is added at such a rate that the gas
compresses according to the relation PV1.2 constant to a final temperature of
200qC. Determine the work done during the process.
Solution:
From Eq. 4.4 for the polytopic process PVn = const ( n =/ 1 )
2
P2V2 - P1V1
W
=
PdV
=
´
1 2
¶
1-n
1
Assuming ideal gas, PV = mRT
mR(T2 - T1)
,
1W2 =
1-n
P1V1 300 u 0.2 kPa m3
But
mR = T = 373.15
K = 0.1608 kJ/K
1
1W2
=
0.1608(473.2 - 373.2) kJ K
1 - 1.2
K = -80.4 kJ
Sonntag, Borgnakke and van Wylen
4.63
A piston/cylinder assembly (Fig. P4.63) has 1 kg of R-134a at state 1 with 110qC,
600 kPa, and is then brought to saturated vapor, state 2, by cooling while the
piston is locked with a pin. Now the piston is balanced with an additional constant
force and the pin is removed. The cooling continues to a state 3 where the R-134a
is saturated liquid. Show the processes in a P-V diagram and find the work in
each of the two steps, 1 to 2 and 2 to 3.
Solution :
CV R-134a This is a control mass.
Properties from table B.5.1 and 5.2
State 1: (T,P) B.5.2 => v = 0.04943 m3/kg
State 2: given by fixed volume v2 = v1 and x2 = 1.0
v2 = v1 = vg = 0.04943 m3/kg
State 3 reached at constant P (F = constant)
so from B.5.1
=> T = 10qC
v3 = vf = 0.000794 m3/kg
P
1
3
cb
2
V
Since no volume change from 1 to 2 => 1W2 = 0
Constant pressure
2W3 = ³P dV = P(V3 -V2) = mP(v3 -v2)
= 415.8 (0.000794 - 0.04943) 1 = -20.22 kJ
Sonntag, Borgnakke and van Wylen
4.73
A film of ethanol at 20qC has a surface tension of 22.3 mN/m and is maintained
on a wire frame as shown in Fig. P4.73. Consider the film with two surfaces as a
control mass and find the work done when the wire is moved 10 mm to make the
film 20 u 40 mm.
Solution :
Assume a free surface on both sides of the frame, i.e., there are two surfaces
20u30 mm
-3
-6
W = ´
¶ S dA = 22.3u10 u2(800 600)u10
= 8.92u10-6 J = -8.92 PJ
Sonntag, Borgnakke and van Wylen
4.94
A torque of 650 Nm rotates a shaft of diameter 0.25 m with Z = 50 rad/s. What
are the shaft surface speed and the transmitted power?
Solution:
V = Zr = ZD/2 = 50 u0.25 / 2 = 6.25 m/s
Power = TZ = 650 u 50 Nm/s = 32 500 W = 32.5 kW
Sonntag, Borgnakke and van Wylen
4.97
A water-heater is covered up with insulation boards over a total surface area of 3
m2. The inside board surface is at 75qC and the outside surface is at 20qC and the
board material has a conductivity of 0.08 W/m K. How thick a board should it be
to limit the heat transfer loss to 200 W ?
Solution :
Steady state conduction through a single layer
board.
.
.
'T
Q cond = k A Ÿ'x k $'7Q
'x
75 20
'x = 0.08 u3u 200 = 0.066m