# HW 3 - 3.20 PROBLEM Prob 3.1 PROBLEM STATEMENT 0.5 lbmol of...

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3.20 8/16/04 PROBLEM Prob 3.1 PROBLEM STATEMENT: 0.5 lbmol of H2O (molecular weight M=18 lbm/lbmol) occupies a volume of 0.145 ft3 at a location where the local acceleration of gravity is 30.5 ft/s2. 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: H20, M = 18 lbm/lbmol V=0.145 ft3, g=30.5 ft/s2 FIND: m (lbm), v (ft 3 /lbm), (lbm/ft 3 ) ASSUMPTIONS: None GOVERNING RELATIONS: 1. W = mg/gc 2. v = 3. = V m m 1 = V v QUANTITATIVE SOLUTION: a). The weight of water, 1 (lbf) W = mg/gc = [0.5 (lbmol) 18 (lbm/lbmol)] 30.5 (ft/s2 ) 2 32.2 (lbm ft/s ) = 8.52 lbf b). The specific volume, V 0.145 (ft 3 ) v= = m 0.5 (lbmol) 18 (lbm/lbmol) = 0.0161 ft 3 /lbm c). The density, 1 1 = = v 0.0161 (ft 3 /lbm) = 62.1 lbm/ft 3 DISCUSSION OF RESULTS: In the U.S. Conventional system of units, it is especially important to remember to use lbm ft / s2 to convert mass-length-time units to lbf force. Note that gc is not the gravitational acceleration, g (units of ft/s2), which can vary. It is simply a units conversion constant whose numerical value depends on the unit system being used. the conversion constant gc = 32.2 Prob 3.20 8/16/04 PROBLEM 3.2 PROBLEM STATEMENT: Find the pressure, P, and specific volume, v, of water at T=200oC and h = 852.38 kJ/kg. GIVEN: H2O, T = 200oC, h = 852.38 kJ/kg FIND: Pressure, P, and specific volume, v. ASSUMPTIONS: Pure substance (H2O) PROPERTY DATA: At T= 200oC, hL = 852.38 kJ/kg, and PSAT = 1.5536 MPa (from Table 10s) QUANTITATIVE SOLUTION: Since h=hL, the water is a saturated liquid. P=PSAT = 1.5536 MPa v=vL= 1.156 x 10-3 m3 /kg DISCUSSION OF RESULTS: Self-explanatory PROBLEM 3.6 PROBLEM STATEMENT: Find the specific volume of water at temperature 400C and pressure 2.0 MPa. GIVEN: T = 400C, P =2.0 MPa =2000 kPa FIND: v, specific volume ASSUMPTIONS: Pure substance (H2O) PROPERTY DATA: TSAT (2000 kPa)=212.42C (Table 11s) QUANTITATIVE SOLUTION: T > TSAT water is superheated v (2000 kPa, 400C) = 0.1512 (m3 / 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>TSAT for the given pressure, it is clear that the water is superheated and hence its properties can be found in the superheated vapor tables. Prob 3.20 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 106 ft3 if the water is assumed to be an incompressible liquid versus using the compressed liquid tables? GIVEN: Water, P = 7000 psia, T = 120F, V = 106 ft3 FIND: Holding capacity (lbm) using ICL assumption and compressed liquid table GOVERNING RELATIONS: Holding capacity, m = V/v ASSUMPTIONS: Water is a pure substance QUANTITATIVE SOLUTION: From Table 10e, at 120F, the corresponding saturation pressure is 1.6939

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HW 3 - 3.20 PROBLEM Prob 3.1 PROBLEM STATEMENT 0.5 lbmol of...

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