(a) Explanation Write the expression for efficiency of compressor. η c = h ¯ 2 s − h ¯ 1 h ¯ 2 − h ¯ 1 h ¯ 2 − h ¯ 1 = h ¯ 2 s − h ¯ 1 η c (I) Here, the molar specific enthalpy at state 1 is h ¯ 1 , the molar specific enthalpy at state 2 is h ¯ 2 and the molar specific enthalpy at superheated state 2 is h ¯ 2 s . Write the expression for the actual change in enthalpy. h ¯ 2 s − h ¯ 1 = y N 2 ( h ¯ 2 s − h ¯ 1 ) N 2 + y O 2 ( h ¯ 2 s − h ¯ 1 ) O 2 (II) Here, the number of mole of nitrogen is y N 2 and the number of mole of oxygen is y O 2 . Substitute y N 2 ( h ¯ 2 s − h ¯ 1 ) N 2 + y O 2 ( h ¯ 2 s − h ¯ 1 ) O 2 for h ¯ 2 s − h ¯ 1 in Equation (I). h ¯ 2 − h ¯ 1 = y N 2 ( h ¯ 2 s − h ¯ 1 ) N 2 + y O 2 ( h ¯ 2 s − h ¯ 1 ) O 2 η c (III) Write the expression for the theoretical enthalpy change. h ¯ 2 − h ¯ 1 = y N 2 ( h ¯ 2 − h ¯ 1 ) N 2 + y O 2 ( h ¯ 2 − h ¯ 1 ) O 2 (IV) Conclusion: Refer to Table A-23, “Ideal gas properties of selected gases”, obtain the below properties at the temperature of 290 K and 298 K using interpolation method of two variables. 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 y and x are molar specific enthalpy of nitrogen and the temperature of nitrogen. Show the molar specific enthalpy with respect to temperature at 290 K and 298 K as in Table (1). Temperature, K specific volume, kJ / kmol 290 K 8432 293 K y 2 298 K 8669 Calculate molar specific enthalpy at temperature of 293 K for nitrogen using interpolation method. Substitute 290 K for x 1 , 293 K for x 2 , 298 K for x 3 , 8432 kJ / kmol for y 1 , and 8669 kJ / kmol for y 3 in Equation (V). y 2 = ( 293 K − 290 K ) ( 8669 kJ / kmol − 8432 kJ / kmol ) ( 298 K − 290 K ) + 8432 kJ / kmol = ( 3 K ) ( 237 kJ / kmol ) ( 8 K ) + 8432 kJ / kmol = 8521 kJ / kmol Refer to Table A-23, “Ideal gas properties of selected gases”, obtain the below properties at the temperature of 290 K and 298 K using interpolation method of two variables. 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 (VI) Here, the variables denote by y and x are molar specific enthalpy of oxygen and the temperature of oxygen. Show the molar specific enthalpy with respect to temperature at 290 K and 298 K as in Table (2). Temperature, K specific volume, kJ / kmol 290 K 8443 293 K y 2 298 K 8682 Calculate molar specific enthalpy at temperature of 293 K for oxygen using interpolation method. Substitute 290 K for x 1 , 293 K for x 2 , 298 K for x 3 , 8443 kJ / kmol for y 1 , and 8682 kJ / kmol for y 3 in Equation (VI) (b) To determine The power required in kW . (c) To determine The rate of entropy production, in kW / K

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Chapter 12.9, Problem 22P

(a)

To determine

The temperature at nozzle exit in °C.

(b)

To determine

The power required in kW.

(c)

To determine

The rate of entropy production, in kW/K

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Chapter 12.9, Problem 21P

Chapter 12.9, Problem 23P

Chapter 12 Solutions

FUNDAMENTALS OF ENGINEERING THERMODYNAM

Ch. 12.9 - Prob. 1E Ch. 12.9 - Prob. 2E Ch. 12.9 - Prob. 3E Ch. 12.9 - Prob. 4E Ch. 12.9 - Prob. 5E Ch. 12.9 - Prob. 6E Ch. 12.9 - Prob. 7E Ch. 12.9 - Prob. 8E Ch. 12.9 - Prob. 9E Ch. 12.9 - Prob. 10E

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The temperature at nozzle exit in °C .

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The temperature at nozzle exit in °C.

The power required in kW.

The rate of entropy production, in kW/K

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Chapter 12 Solutions