homework 1 thermodynamics (2)

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Thermodynamics - AME 230 Fall 2023 1 Student name : Shyanne Smith Answers are highlighted in blue Date : 09/01/2023 Homework Assignment No.1 I. Indicate whether the following statements are true or false 1. Kilogram, second, foot, and newton are all examples of SI units. (Hint: see Table 1.3) B: False Kilogram, second and newton are SI units. Foot is an English unit. 2. The specific volume is the reciprocal of the density. (Hint: see Sec. 1.3.4) A: True Specific volume= Density= 3. The value of a temperature expressed using the Rankine scale is always higher than its value expressed using the Fahrenheit temperature scale. (Hint: see Sec. 1.7.3 and Eq. 1.18) A: True Since 459.67 needs to be subtracted from the temperature in Rankine, Fahrenheit is a lower value. II. Perform the following unit conversions (tolerance ±2%), do not use problem-solving technique, just the answer with at least one intermediate step: (a) 85.4 in. 3 to L (this is solved as an example, continue with b)

Thermodynamics - AME 230 Fall 2023 2 Answer: ( ) 3 3 3 3 2 3 3 1 cm 1 m 1 L 85.4 in. 1.4 L 1.399 L also acceptable 0.061024 in. 10 cm 10 m − = (b) 544.719 ft·lbf to kJ Answer: 1 ft*lbf = 1.355818 Joules 1 kJ = 1000 Joules 544.719 ft*lbf * 1.355818 J/1ft*lbf * 1kJ/1000J = 0.739kJ (c) 70.0 hp to kW Answer: 1 hp = 746 Watts 1kW = 1000 Watts 70hp * 746 Watts / 1 hp * 1kW/ 1000Watts= 52.22 kW (d) 700.0 lb/h to kg/s Answer: 1 kg = 2.20462 lb 60 s = 1 hr 1 lb/h = 0.000125998 kg/s 700 lb/h * 0.000125998 kg/s / 1 lb/h= 0.0881986 kg/s (e) 20.5744 lbf/in. 2 to bar Answer: 1lb/in^2 = 0.0689475727 bars 20.5744 lbf/in. 2 * 0.0689475727 bars/1lb/in^2 = 1.419 bar (f) 1750.0 ft 3 /min to m 3 /s Answer: 1 ft^3/min = 0.000471947 m^3/s

Thermodynamics - AME 230 Fall 2023 3 1750 ft 3 /min * 0.000471947 m^3/s / 1 ft^3/min = 0.8259 m^3/s (g) 52.5 mile/h to km/h Answer: 1km=0.621371 miles 52.5 miles/hour * 1 km/hour / 0.621371 miles/hour = 84.49 km/h (h) 0.7 ton (=2000 lbf) to N Answer: 0.7 tons = 2000lbf 1 lbf = 4.44822 N 2000lbf * 4.44822 N / 1 lbf = 8896.44 N III. At a certain elevation, the pilot of a balloon has a mass of 120 lb and a weight of 119 lbf. What is the local acceleration of gravity, in ft/s 2 , at that elevation? If the balloon drifts to another elevation where g = 32.05 ft/s 2 , what is her weight, in lbf, and mass, in lb? (Hint: Refer to Section 1.4) For this problem you must use engineering problem- solving techniques: KNOWN, FIND, … Answer KNOWN : We know the mass of the pilot M= 120lb and the pound-force is 119lbf We know the second elevation where g= 32.05 ft/s^2 FIND :

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Thermodynamics - AME 230 Fall 2023 4 We need to find the local acceleration of gravity in ft/s^2 which is found by the equation W=m 𝑔 ?𝑜??? so 𝑔 ?𝑜??? =W/m We also need to find the the weight of the girl at g= 32.05ft/s^2 in lbf and in lbs SCHEMATIC AND GIVEN DATA : Given: M= 120lb and the pound-force is 119lbf We know the second elevation where g= 32.05 ft/s^2 ENGINEERING MODEL : The dashed lines show a control volume system because the balloon changes with elevation which means it interacts with its surroundings. ANALYSIS :

Thermodynamics - AME 230 Fall 2023 5 To calculate the local acceleration of gravity we need the following information: 1lbf = 1lb x 32.174 ft/s^2 1 lb = 1lbf / 32.174 ft/s^2 W=m 𝑔 ?𝑜??? So 𝑔 ?𝑜??? = 𝑊 ? 𝑔 ?𝑜??? = 119??𝑓 120?? = 119lbf / 120(1 lbf / 32.174 ft/s^2) = 31.905883 𝑔 ?𝑜??? =31.906 ft/s^2 Weight of the balloon at elevation where g = 32.05 ft/s^2 W= m g W = 120lb(1lbf / 32.174 lb-ft/s^2) x (32.05ft/s^2) = 119.537515 lbf Mass in lbs: M= W/g M= 119.537515lbf / 32.05ft/s^2 = 19.537515lbf / 32.05ft/s^2(1lbf / 32.174 lb-ft/s^2) = 119.9998lb lb Mass in lbf= 119.538 lbf Mass in lb= 120lb COMMENT : Not sure what we are supposed to write for the comment section? IV. Water flows through a Venturi meter, as shown in the figure. The pressures of the water at locations a and b in the pipe support columns of water that differ in height by L = 18 in. The atmospheric pressure is 14.7 lbf/in 2 , the specific volume of water is 0.01604 ft 3 /lb, and the acceleration of gravity is g = 32.0 ft/s 2 . Determine the difference in pressure between points a and b , in lbf/in 2 . Does the pressure increase or decrease from point a to point b ? (Hint: Refer to Section 1.6.1) For this problem you must use engineering problem- solving techniques: KNOWN, FIND, …

Thermodynamics - AME 230 Fall 2023 6 Answer KNOWN : We know the atmospheric pressure is Patm=14.7lbf/in^2 The acceleration due to gravity is g= 32ft/s^2 The distance in height from a to b is L=18in The specific volume of water is v=0.01604 ft 3 /lb FIND : We are looking for the difference in pressure between points a and b and whether or not that difference is an increase or decrease from point a to point b. 𝛥𝑃 =? SCHEMATIC AND GIVEN DATA : As mentioned before the given data and diagram look like this

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Thermodynamics - AME 230 Fall 2023 7 The dashed lines represent a control volume system because the atmospheric pressure alters values when calculating answers so this means the system is interacting with its surroundings. ENGINEERING MODEL : The dashed lines are for a control volume system because it is affected by its surroundings (atmospheric pressure) If the two tubes were sealed shut then the system would be a closed system because atmospheric pressure would not affect it. ANALYSIS : Density of water = ρ= 1/ν = 1/0.01604ft^3 /lb

Thermodynamics - AME 230 Fall 2023 8 ρ= 62.344 ft^3/lb Pa=- ρg L = 62.344 ft^3/lb * 32.0 ft/s^2 * h = 1995.008 lb/ft *s^2 x h Pb= ρg(h+ L)= 62.344 ft^3/lb * 32.0 ft/s^2* (h +18/12) = 1995.008h + 2992.512 lb/ft*s^2 𝛥𝑃 = Pb – Pa = 1995.008h + 2992.512 – 1995.008h 𝛥𝑃 = Pb – Pa = 2992.512 lb/ft * s^2 * 1/32 ft/s^2 = 93.516 lb/ft^2 93.516 lb/ft^2 * 1 ft^2/ (12in)^2 = 0.649 lb/in^2 Since the difference, 𝛥𝑃 is positive, the pressure is increasing from point a to point b. The pressure is increasing in the direction of flow. COMMENT :

homework 1 thermodynamics (2)

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