Explanation Write the expression for the sonic velocity. C = − v 2 ( ∂ p ∂ V ) s (I) Here, the sonic velocity is C , the specific volume is v , the pressure is p and the volume is V . Write the expression for the heat constant at constant pressure for water using ideal gas model. C p = R [ α + β T + γ T 2 + δ T 3 + ε T 4 ] (II) Here, the temperature of the water is T , the heat constant at constant pressure is C p , the coefficient factor for water are α , β , γ ,and δ and the universal gas constant is R . Write the expression for the specific heat ratio for water. k = C p ( C p − R ) (III) Here, the specific heat ratio for water is k . Write the expression for the sonic velocity using ideal gas model. c = k R T (IV) Here, the specific heat ratio is k , the universal gas constant is R and the temperature is T . Write the formula of interpolation method of two variables. y 2 = ( x 2 − x 1 ) ( y 3 − y 1 ) ( x 3 − x 1 ) + y 1 (V) Here, the variables denote by x and y are specific entropy and specific volume. Write the expression for the partial derivative of volume. ( ∂ p ∂ V ) s = ( p 2 − p 1 ) ( v 2 − v 1 ) (VI) Here, the pressure at inlet is p 1 , the pressure at exit is p 2 , the specific volume at pressure p 1 is v 1 and the specific volume at pressure p 2 is v 2 . Conclusion: Refer to table A-4 “properties of superheated water vapor” to obtain the specific entropy and the specific volume at temperature T = 360 ° C and pressure p = 60 bar as 6.3782 kJ / kg ⋅ K and, 0.04331 m 3 / kg . Refer to table A-4 “properties of superheated water vapor” to obtain the specific volume at specific entropy s = 6.3782 kJ / kg ⋅ K and pressure p 2 = 80 bar as 0.03432 m 3 / kg and, 0.03742 m 3 / kg . Show the pressure at 0.03432 m 3 / kg and 0.03742 m 3 / kg as in below Table. Specific entropy kJ / kg ⋅ K specific volume m 3 / kg 6.3634 kJ / kg ⋅ K 0.03432 m 3 / kg 6.3782 kJ / kg ⋅ K ? 6.5190 kJ / kg ⋅ K 0.03742 m 3 / kg Substitute 6.3634 kJ / kg ⋅ K for x 1 , 6.3782 kJ / kg ⋅ K for x 2 , 6.5190 kJ / kg ⋅ K for x 3 , 0.03432 m 3 / kg for y 1 , and 0.03742 m 3 / kg for y 3 in Equation (V). y 2 = { [ ( 6.3782 kJ / kg ⋅ K − 6.3634 kJ / kg ⋅ K ) ( 0.03742 m 3 / kg − 0.03432 m 3 / kg ) ] ( 6.5190 kJ / kg ⋅ K − 6.3634 kJ / kg ⋅ K ) + 0.03432 m 3 / kg } = ( 4.588 × 10 − 5 ) 0.1556 m 3 / kg + 0.03432 m 3 / kg = 0.002993 m 3 / kg + 0.03432 m 3 / kg ≈ 0.03461 m 3 / kg So the specific volume at p 2 = 80 bar is .03461 m 3 / kg . Refer to table A-4 “properties of superheated water vapor” to obtain the specific volume at specific entropy s = 6.3782 kJ / kg ⋅ K and pressure p 1 = 40 bar as 0.05546 m 3 / kg and, 0.06199 m 3 / kg . Show the pressure at 0.05546 m 3 / kg and 0.06199 m 3 / kg as in below Table. Specific entropy kJ / kg ⋅ K specific volume m 3 / kg 6.2568 kJ / kg ⋅ K 0.05546 m 3 / kg 6.3782 kJ / kg ⋅ K ? 6.4553 kJ / kg ⋅ K 0.06199 m 3 / kg Substitute 6.2568 kJ / kg ⋅ K for x 1 , 6.3782 kJ / kg ⋅ K for x 2 , 6.4553 kJ / kg ⋅ K for x 3 , 0.05546 m 3 / kg for y 1 , and 0.06199 m 3 / kg for y 3 in Equation (V). y 2 = { [ ( 6.3782 kJ / kg ⋅ K − 6.2568 kJ / kg ⋅ K ) ( 0.06199 m 3 / kg − 0.05546 m 3 / kg ) ] ( 6

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ISBN: 9781119391630

Author: MORAN

Publisher: WILEY

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Chapter 9.14, Problem 115P

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The sonic velocity using data from table A-4.

The sonic velocity using ideal gas model.

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Chapter 9.14, Problem 114P

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The sonic velocity using data from table A-4.

The sonic velocity using ideal gas model.