Thermodynamics Thermodynamics The following table list temperatures and specific

Question

Thermodynamics Thermodynamics The following table list temperatures and specific volumes of water vapor at two pressures: 3 P-1.0 MPa P-1.5 MPa 200 240 280 200 240 280 Data encountering in solving problems often do not fall exactly on 0.1325 0.1627 the grid of values provided by property tables, and linear 0.2060 0.2275 0.2480 0.148 interpolation between adjacent table entries becomes necessary. Using the data provided here estimate a- The specific volume at 240°c, p- 1.25 MPa in m2/kg. b- The temperature at p-1.5 MPa, v-0.1555m2/kg in °c. 4 As gas contained within a piston cylinder assembly undergoes a thermodynamic cycle consisting of three processes: Process 1-2: compression with p-V constant from p1-1 bar, V1-1.0m3 and V2-0.2m3. Process 2-3: Constant pressure expansion to V-1.0m3 Process 3-1 Constant volume Sketch the cycle on a p-V diagram labeled with pressure and volume values at each numbered state.

Explanation / Answer

a) specific volume at given temperature at 1.0 MPa is 0.2275 and at 1.5 MPa is 0.1483.

Hence, using linear interpolation, specific volume at this temerature at 1.25 MPa (mid value of 1 and 1.5) = (0.2275+0.1483)/2 = 0.1879 m3/kg

b) At given pressure, Temperature at specific gravity 0.1483 m3/kg is 240 degree C and specific gravity 0.1627 m3/kg is 280 degree C

Hence, using linear interpolation, temperature at this pressure at specific gravity 0.1555 m3/kg (mid value of 0.1483 and 0.1627) = (240+280)/2= 260 Degree C

c) Specific volume at 220 degree C (at p=1)= (.2060+0.2275)/2= 0.2167

  Specific volume at 220 degree C (at p=1.5)= (.1325+0.1483)/2= 0.1404

Hence, at 220 degree C, the specific volume at pressure 1.4= 0.2167-(0.2167-0.1404)*4/5

=0.1557 m3/kg

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