{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

shpora

# shpora - 4.27 The rolling resistance of a car depends on...

This preview shows pages 1–2. Sign up to view the full content.

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 P-V 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 P-V 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 P-v 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 R-134a 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 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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 4

shpora - 4.27 The rolling resistance of a car depends on...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online