Explanation Given: Temperature of air at state 1 ( T 1 ) is 565 R . Pressure of air at state 1 ( P 1 ) is 14 psia . Temperature of air at state 3 ( T 3 ) is 2860 R . Number of cylinder ( N cyl ) is 6 . Stroke of the piston ( S ) is 3.9 in . . Bore diameter of the cylinder ( D ) is 3.5 in . . Speed of the engine ( n ) is 2500 rpm . Calculation: Draw the P − v diagram of the cycle as in Figure (1). Refer Table A-2E, “Ideal-gas specific heats of various common gases”, obtain the following properties of the air. R = 0.37047 psia ⋅ ft 3 / lbm ⋅ R c v = 0.240 Btu / lbm ⋅ R c p = 0.171 Btu / lbm ⋅ R k = 1.4 It is given that the minimum enclosed volume is 9.8% of the maximum enclosed volume. v 2 = 0.098 v 1 Calculate the compression ratio ( r ) . r = v 1 v 2 = v 1 0.098 v 1 = 10.20 Calculate the temperature at state 2 ( T 2 ) . T 2 = T 1 ( v 1 v 2 ) k − 1 = T 1 ( r ) k − 1 = ( 565 R ) ( 10.20 ) 1.4 − 1 = 1430.7 R Calculate the temperature at state 4 ( T 4 ) . T 4 = T 3 ( v 3 v 4 ) k − 1 = T 3 ( 1 r ) k − 1 = ( 2860 R ) ( 1 10.20 ) 1.4 − 1 = 1129.4 R Calculate the net-work output of the cycle ( w net,out ) . w net,out = q in − q out = c v ( T 3 − T 4 ) − c v ( T 2 − T 1 ) = { ( 0.171 Btu / lbm ⋅ R ) ( 2860 R − 1129.4 R ) − ( 0.171 Btu / lbm ⋅ R ) ( 1430

5th Edition

ISBN: 9780078027680

Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala

Publisher: McGraw-Hill Education

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Chapter 9, Problem 34P

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Problem (17): water flowing in an open channel of a rectangular cross-section with width (b) transitions from a mild slope to a steep slope (i.e., from subcritical to supercritical flow) with normal water depths of (y₁) and (y2), respectively. Given the values of y₁ [m], y₂ [m], and b [m], calculate the discharge in the channel (Q) in [Lit/s]. Givens: y1 = 4.112 m y2 = 0.387 m b = 0.942 m Answers: ( 1 ) 1880.186 lit/s ( 2 ) 4042.945 lit/s ( 3 ) 2553.11 lit/s ( 4 ) 3130.448 lit/s

Problem (14): A pump is being used to lift water from an underground tank through a pipe of diameter (d) at discharge (Q). The total head loss until the pump entrance can be calculated as (h₁ = K[V²/2g]), h where (V) is the flow velocity in the pipe. The elevation difference between the pump and tank surface is (h). Given the values of h [cm], d [cm], and K [-], calculate the maximum discharge Q [Lit/s] beyond which cavitation would take place at the pump entrance. Assume Turbulent flow conditions. Givens: h = 120.31 cm d = 14.455 cm K = 8.976 Q Answers: (1) 94.917 lit/s (2) 49.048 lit/s ( 3 ) 80.722 lit/s 68.588 lit/s 4

Problem (13): A pump is being used to lift water from the bottom tank to the top tank in a galvanized iron pipe at a discharge (Q). The length and diameter of the pipe section from the bottom tank to the pump are (L₁) and (d₁), respectively. The length and diameter of the pipe section from the pump to the top tank are (L2) and (d2), respectively. Given the values of Q [L/s], L₁ [m], d₁ [m], L₂ [m], d₂ [m], calculate total head loss due to friction (i.e., major loss) in the pipe (hmajor-loss) in [cm]. Givens: L₁,d₁ Pump L₂,d2 오 0.533 lit/s L1 = 6920.729 m d1 = 1.065 m L2 = 70.946 m d2 0.072 m Answers: (1) 3.069 cm (2) 3.914 cm ( 3 ) 2.519 cm ( 4 ) 1.855 cm TABLE 8.1 Equivalent Roughness for New Pipes Pipe Riveted steel Concrete Wood stave Cast iron Galvanized iron Equivalent Roughness, & Feet Millimeters 0.003-0.03 0.9-9.0 0.001-0.01 0.3-3.0 0.0006-0.003 0.18-0.9 0.00085 0.26 0.0005 0.15 0.045 0.000005 0.0015 0.0 (smooth) 0.0 (smooth) Commercial steel or wrought iron 0.00015 Drawn…

Chapter 9 Solutions

Fundamentals of Thermal-Fluid Sciences

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