Explanation Given information: Temperature is T = 200 ° C Initial pressure is P 1 = 350 kPa Final pressure is P 2 = 700 kPa Stage 1: Pressure: 3 bar and 4 bar Temperature: 150 ° C and 250 ° C Write the expression of interpolation method. y 2 = ( x 2 − x 1 ) ( y 3 − y 1 ) ( x 3 − x 1 ) + y 1 (1) Refer table A-3, “Superheated steam”, obtain the specific enthalpy at 3 bar and 4 bar at 150 ° C using interpolation method from Equation (1). 1 2 3 y (enthalpy) in kJ / kg 2761.2 y 2 = ? 2752.8 x (pressure) in bar 3 3.5 4 Substitute 3 bar for x 1 , 3.5 bar for x 2 , 4 bar for x 3 , 2761.2 kJ / kg for y 1 , and 2752.8 kJ / kg for y 3 in Equation (1). y 2 = ( 3.5 bar − 3 bar ) ( 2 , 752.8 kJ / kg − 2 , 761.2 kJ / kg ) ( 4 bar − 3 bar ) + 2 , 761.2 kJ / kg = 2 , 757.0 kJ / kg Thus, the specific enthalpy at pressure and temperature of 3.5 bar and 150 ° C ( H 3 .5 bar @ 150 ° C ) is 2 , 757.0 kJ / kg . Refer table A-3, “Superheated steam”, obtain the specific enthalpy at 3 bar and 4 bar at 250 ° C using interpolation method from Equation (1). 1 2 3 y (enthalpy) in kJ / kg 2967.9 y 2 = ? 2964.5 x (pressure) in bar 3 3.5 4 From Equation (1), substitute 3 bar for x 1 , 3.5 bar for x 2 , 4 bar for x 3 , 2 , 967.9 kJ / kg for y 1 , and 2 , 964.5 kJ / kg for y 3 . y 2 = ( 3.5 bar − 3 bar ) ( 2 , 967.9 kJ / kg − 2 , 964.5 kJ / kg ) ( 4 bar − 3 bar ) + 2 , 964.5 kJ / kg = 2 , 966.2 kJ / kg Thus, the specific enthalpy at pressure and temperature of 3.5 bar and 250 ° C ( H 3 .5 bar @2 50 ° C ) is 2 , 966.2 kJ / kg . Write the expression to obtain the constant pressure heat capacity ( C P ) . C P = H 3 .5 bar @2 50 ° C − H 3.5 bar @ 1 50 ° C T 2 − T 1 (2) Substitute 2 , 966.2 kJ / kg for H 3.5 bar @ 250 ° C , 2 , 757.0 kJ / kg for H 3 .5 bar @ 150 ° C , 250 ° C for T 2 , and 150 ° C for T 1 in Equation (2)

The constant pressure heat capacity at 350 kPa, 200 ° C and at 700 kPa, 200 ° C . Concept Introduction: The expression of interpolation method. y 2 = ( x 2 − x 1 ) ( y 3 − y 1 ) ( x 3 − x 1 ) + y 1 Here, y 1 , y 2 , y 3 refers to specific enthalpy and x 1 , x 2 , x 3 refers to pressures. The expression to obtain the constant pressure heat capacity ( C P ) . C P = Δ H Δ T = H 3 .5 bar @2 50 ° C − H 3.5 bar @ 1 50 ° C T 2 − T 1 Here, the change in specific enthalpy is Δ H and change in temperature is Δ T . The expression to obtain the constant pressure heat capacity. C P = Δ H Δ T = H 7 bar @ 2 50 ° C − H 7 bar @ 1 50 ° C T 2 − T 1 Here, the change in specific enthalpy is Δ H and change in temperature is Δ T .

Fundamentals of Chemical Engineering Thermodynamics (MindTap Course List)

1st Edition

ISBN: 9781111580704

Author: Kevin D. Dahm, Donald P. Visco

Publisher: Cengage Learning

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Question

Chapter 2.6, Problem 12P

Interpretation Introduction

Interpretation:

The constant pressure heat capacity at 350 kPa, 200°C and at 700 kPa, 200°C.

Concept Introduction:

The expression of interpolation method.

y2=(x2−x1)(y3−y1)(x3−x1)+y1

Here, y1,y2,y3 refers to specific enthalpy and x1,x2,x3 refers to pressures.

The expression to obtain the constant pressure heat capacity (CP).

CP=ΔHΔT=H3.5 bar @250°C−H3.5 bar @ 150°CT2−T1

Here, the change in specific enthalpy is ΔH and change in temperature is ΔT.

The expression to obtain the constant pressure heat capacity.

CP=ΔHΔT=H7 bar @ 250°C−H7 bar @ 150°CT2−T1

Here, the change in specific enthalpy is ΔH and change in temperature is ΔT.

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Chapter 2.6, Problem 11P

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