Assignment__3 - Assignment #3 8/16/04 PROBLEM 3.1 PROBLEM...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Assignment #3 8/16/04 PROBLEM 3.1 PROBLEM STATEMENT: 0.5 lbmol of H 2 O (molecular weight M=18 lbm/lbmol) occupies a volume of 0.145 ft 3 at a location where the local acceleration of gravity is 30.5 ft/s 2 . Determine: a) the weight of water, b) its specific volume, and c) its density. All answers should be expressed in U.S.Conventional units. GIVEN: H 2 0, M = 18 lbm/lbmol V=0.145 ft 3 , g=30.5 ft/s 2 FIND: 3 3 m (lbm), v (ft /lbm), (lbm/ft ) ρ ASSUMPTIONS: None GOVERNING RELATIONS: 1. W = mg/g c 2. V v = m 3. m 1 = = V v QUANTITATIVE SOLUTION: a). The weight of water, [ ] 2 c 2 1 (lbf) W =mg/g = 0.5 (lbmol) 18 (lbm/lbmol) 30.5 (ft/s ) 32.2 (lbm ft/s ) = 8.52 lbf × × × g b). The specific volume, 3 3 V 0.145 (ft ) v = = m 0.5 (lbmol) 18 (lbm/lbmol) = 0.0161 ft /lbm × c). The density, 3 3 1 1 = = v 0.0161 (ft /lbm) = 62.1 lbm/ft DISCUSSION OF RESULTS: In the U.S. Conventional system of units, it is especially important to remember to use the conversion constant 2 c lbm ft /s g 32.2 lbf = to convert mass-length-time units to force. Note that g c is not the gravitational acceleration, g (units of ft/s 2 ), which can vary. It is simply a units conversion constant whose numerical value depends on the unit system being used.
Background image of page 1

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

View Full DocumentRight Arrow Icon
Assignment #3 8/16/04 PROBLEM 3.3 PROBLEM STATEMENT : Find the pressure, P, and the specific volume, v, of water at T= 170 o C and x = 0.45. GIVEN: T= 170 o C, x = 0.45 FIND: Pressure, P, and the specific volume,v . ASSUMPTIONS: Pure substance (H 2 O) GOVERNING RELATIONS: For a saturated mixture, L LV v v x(v ) = + PROPERTY DATA: At T= 170 o C and x = 0.45, P SAT = 791.47 kPa, v L = 0.001114 m 3 /kg v LV = 0.2417 m 3 /kg (from Table 10s) QUANTITATIVE SOLUTION: P = P SAT = 791.47 kPa L LV v v x(v ) = + = 0.001114 (m 3 /kg) + (0.45)(0.2417 m 3 /kg) = 0.11 m 3 /kg (Table 10s) DISCUSSION OF RESULTS: Self explanatory
Background image of page 2
Assignment #3 8/16/04 PROBLEM 3.6 PROBLEM STATEMENT: Find the specific volume of water at temperature 400°C and pressure 2.0 MPa. GIVEN: T=400°C, P=2.0 MPa = 2000 kPa FIND: v, specific volume ASSUMPTIONS: Pure substance (H 2 O) PROPERTY DATA: SAT T (2000 kPa)=212.42°C (Table 11s) QUANTITATIVE SOLUTION: SAT T T water is superheated 3 v(2000 kPa, 400°C) = 0.1512 (m /kg) (Table 12s) DISCUSSION OF RESULTS: This problem illustrates the value of using the saturation table as a "roadmap" to define what region the given state is in. Since T>T SAT for the given pressure, it is clear that the water is superheated and hence its properties can be found in the superheated vapor tables.
Background image of page 3

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

View Full DocumentRight Arrow Icon
Assignment #3 8/16/04 PROBLEM 3.9 PROBLEM STATEMENT: Water in an underground reservoir is at a pressure of 7,000 psia and a temperature of 120 °F. How much error would be incurred in estimating the holding capacity (lbm) of a reservoir whose volume is 10 6 ft 3 if the water is assumed to be an incompressible liquid versus using the compressed liquid tables? GIVEN:
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/20/2010 for the course M E 320 taught by Professor Deinert during the Spring '08 term at University of Texas at Austin.

Page1 / 14

Assignment__3 - Assignment #3 8/16/04 PROBLEM 3.1 PROBLEM...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online