4.27
The rolling resistance of a car depends on its weight as: F = 0.006 mg. How long will a car of 1400 kg drive for a work input of 25 kJ?
Solution:
Work is force times distance so assuming a constant force we get
W = ⌡⌠ F dx = F x = 0.006 mgx
Solve for x
x = W0.006 mg = 25 kJ0.006 × 1400 kg × 9.807 m/s2 =
303.5 m
4.39
Air in a spring loaded piston/cylinder has a pressure that is linear with volume, P = A + BV. With an initial state of P = 150 kPa, V = 1 L and a final state of 800 kPa and volume
1.5 L it is similar to the setup in Problem 3.113. Find the work done by the air.
Solution:
Knowing the process equation: P = A + BV giving a linear variation of pressure versus volume the straight line in the PV diagram is fixed by the two points as
state 1 and state 2. The work as the integral of PdV equals the area under the process curve in the PV diagram.
State 1: P1 = 150 kPa V1 = 1 L = 0.001 m3
State 2: P2 = 800 kPa V2 = 1.5 L = 0.0015 m3
Process: P = A + BV linear in V
⇒
1W2 = ⌡⌠ 1 2 PdV = (P1 + P22)(V2  V1)
= 12 (150 + 800) kPa (1.5  1) × 0.001 m3
= 0.2375 kJ
4.48
The piston/cylinder shown in Fig. P4.48 contains carbon dioxide at 300 kPa, 100°C 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 200°C. Determine the work done during the process.
Solution:
From Eq. 4.4 for the polytopic process PVn = const ( n =/ 1 )
1W2 = ⌡⌠ 1 2 PdV = P2V2  P1V11  n
Assuming ideal gas, PV = mRT
1W2 = mR(T2  T1)1  n ,
But mR = P1V1T1 = 300 × 0.2373.15 kPa m3K = 0.1608 kJ/K
1W2 = 0.1608(473.2  373.2)1  1.2 kJ KK =
80.4 kJ
4.60
A cylinder containing 1 kg of ammonia has an externally loaded piston. Initially the ammonia is at 2 MPa, 180°C and is now cooled to saturated vapor at 40°C, and then further cooled
to 20°C, at which point the quality is 50%. Find the total work for the process, assuming a piecewise linear variation of P versus V.
Solution:
State 1: (T, P) Table B.2.2
v1 = 0.10571 m3/kg
State 2: (T, x) Table B.2.1 sat. vap.
P2 = 1555 kPa,
v2 = 0.08313 m3/kg
State 3: (T, x) P3 = 857 kPa, v3 = (0.001638 + 0.14922)/2 = 0.07543 m3/kg
Sum the the work as two integrals each evaluated by the area in the Pv diagram.
1W3 = ⌡⌠13 PdV ≈ ( P1 + P22 ) m(v2  v1) + ( P2 + P32 ) m(v3  v2)
= 2000 + 15552 1(0.08313  0.10571) + 1555 + 8572 1(0.07543  0.08313)
=
49.4 kJ
4.63
A piston/cylinder assembly (Fig. P4.63) has 1 kg of R134a at state 1 with 110°C, 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 R134a is saturated liquid.
Show the processes in a PV diagram and find the work in each of the two steps, 1 to 2 and 2 to 3.
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 Spring '07
 MUNTZ
 Thermodynamics, Energy, Heat, Heat Transfer, kPa

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