Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
8th Edition
ISBN: 9780073398174
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
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Chapter 17.7, Problem 117P

a)

To determine

The exit velocity, mass flow rate, and exit Mach number if the nozzle is isentropic.

a)

Expert Solution
Check Mark

Answer to Problem 117P

The exit velocity of the stream is 1487ft/s.

The mass flow rate is 18.71lbm/s.

The Mach number at the exit of nozzle is 0.9.

Explanation of Solution

For isentropic,

The flow of steam through the nozzle is steady and isentropic.

Write the expression of energy balance equation for the converging-diverging nozzle.

h1+V12/2=h2+V22/2

Inlet velocity is equal to zero V1=0.

V2=2(h1h2) (I)

Here, enthalpy at exit is h2, enthalpy at inlet is h1, velocity of steam at the inlet of nozzle is V1, and velocity of steam at exit of nozzle is V2 .

Write the expression to calculate the exit area of the nozzle.

A2=m˙v2V2m˙=A2V2v2 (II)

Here, mass flow rate of steam is m˙ , specific volume of the steam at the exit is v2 .

Write the expression to calculate the velocity of sound through the steam at the exit of nozzle.

c2=(Pρ)s1/2=(ΔPΔ(1/v))s1/2 (III)

Here, pressure drop in the nozzle is ΔP, and drop in the specific volume of the steam is Δ(1/v).

Write the expression to calculate the Mach number for the steam at the exit of nozzle.

Ma2=V2c2 (IV)

Here, Mach number of the steam at the exit is Ma2 .

Conclusion:

Refer Table A-6, “Superheated water”, obtain the values of h01, and s2s at inlet pressure of 450psia and temperature of 900°F as 1468.6Btu/lbm, and 1.7117Btu/lbmR respectively.

Here, at superheated condition the entropy of saturated steam is s2s.

Refer Table A-6, “Superheated water”, obtain the isentropic final enthalpy value h2s and final specific volume at an entropy of 1.7117Btu/lbR and a pressure of 275psia as 1400.5Btu/lbm and 2.5732ft3/lbm.

The stagnation enthalpy of steam at the inlet is equal to the actual enthalpy

at the inlet (h01=h1).

Substitute 1468.6Btu/lbm for h1 and 1400.5Btu/lbm for h2 in Equation (II).

V2=2(1468.6Btu/lbm1400.5Btu/lbm)=2(1468.6Btu/lbm1400.5Btu/lbm)(25037ft2/s21Btu/lbm)=1487ft/s

Thus, the exit velocity of the stream is 1487ft/s.

Substitute 3.75in2. for A, 2.5732ft3/lbm for v2, and 1487ft/s for V2 in Equation (III).

m˙=3.75in.2×1847ft/s2.5732ft3/lbm=3.75in.2(1ft2144 in.2)×1847ft/s2.5732ft3/lbm=18.71lbm/s

Thus, the mass flow rate is 18.71lbm/s.

Refer Table A-6, “Superheated water”, obtain the value of specific volume of steam at the entropy of 1.7117Btu/lbR at pressure just below and above the specified pressure of 250psia and 300psia as 2.7709ft3/lbm and 2.4048ft3/lbm respectively.

Substitute (300250)psia for ΔP, and (12.404812.7709)lbm/ft3 for Δ(1/v) in Equation (IV).

c2=((300250)psia(12.404812.7709)lbm/ft3)1/2=((300250)psia(12.404812.7709)lbm/ft3(25037ft2/s21Btu/lbm)(1Btu/lbm5.4039ft3psia))1/2=2053ft/s

Substitute 2053ft/s for c2, and 1487ft/s for V2 in Equation (V).

Ma2=1487ft/s2053ft/s=0.9

Hence, the Mach number at the exit of nozzle is 0.9.

b)

To determine

The exit velocity, mass flow rate, and exit Mach number if the has an efficiency of 94 percent.

b)

Expert Solution
Check Mark

Answer to Problem 117P

The exit velocity of the stream is 1752ft/s.

The mass flow rate is 17.531lbm/s.

The Mach number at the exit of nozzle is 0.849.

Explanation of Solution

Nozzle has an efficiency of 90 percent:

Write the expression for the efficiency of nozzle.

ηN=h01h2h01h2s (V)

Here, efficiency of nozzle is ηN, stagnation enthalpy at the inlet is h01, enthalpy at exit is h2, and superheated enthalpy at exit is h2s.

Write the expression of energy balance equation for the converging-diverging nozzle.

h1+V12/2=h2+V22/2

Inlet velocity is equal to zero V1=0.

V2=2(h1h2) (VI)

Here, velocity of steam at the inlet of nozzle is V1 and velocity of steam at exit of nozzle is V2 .

Write the expression to calculate the exit area of the nozzle.

A2=m˙v2V2 (VII)

Here, mass flow rate of steam is m˙ , specific volume of the steam at the exit is v2 .

Write the expression to calculate the velocity of sound through the steam at the exit of nozzle.

c2=(Pρ)s1/2=(ΔPΔ(1/v))s1/2 (VIII)

Here, pressure drop in the nozzle is ΔP, and drop in the specific volume of the steam is Δ(1/v).

Write the expression to calculate the Mach number for the steam at the exit of nozzle.

Ma2=V2c2 (IX)

Here, Mach number of the steam at the exit is Ma2.

Conclusion:

Refer Table A-6, “Superheated water”, obtain the values of h01, and s2s at inlet pressure of 450psia and temperature of 900°F as 1468.6Btu/lbm, and 1.7117Btu/lbmR respectively.

Here, at superheated condition the entropy of saturated steam is s2s.

Refer Table A-6, “Superheated water”, obtain the isentropic final entropy value h2s and final specific volume at an enthalpy of 1400.5Btu/lbm and a pressure of 275psia as 1.7117Btu/lbR and 2.6034ft3/lbm.

Substitute 1400.5Btu/lbm for h2s, 1468.6Btu/lbm for h01, and 0.90 for ηN in Equation (V).

90%=1468.6Btu/lbmh21468.6Btu/lbm1400.5Btu/lbm90(1100)=1400.5Btu/lbmh21468.6Btu/lbm1400.5Btu/lbmh2=1407.3Btu/lbm

The stagnation enthalpy of steam at the inlet is equal to the actual enthalpy

at the inlet (h01=h1).

Substitute 1468.6Btu/lbm for h1 and 1407.3Btu/lbm for h2 in Equation (VI).

V2=2(1468.6Btu/lbm1407.3Btu/lbm)=2(1468.6Btu/lbm1407.3Btu/lbm)(25037ft2/s21Btu/lbm)=1752ft/s

Thus, the exit velocity of the stream is 1752ft/s.

Substitute 3.75in2. for A, 2.6034ft3/lbm for v2, and 1752ft/s for V2 in Equation (VII).

m˙=3.75in.2×1752ft/s2.6034ft3/lbm=3.75in.2(1ft2144 in.2)×1752ft/s2.6034ft3/lbm=17.531lbm/s

Thus, the mass flow rate is 17.531lbm/s.

Refer Table A-6, “Superheated water”, obtain the value of specific volume of steam Δ(1/v) at the entropy of 1.7173Btu/lbR at pressure just below and above the specified pressure of 250Psia and 300Psia as 2.8036ft3/lbm and 2.4329ft3/lbm respectively..

Substitute (300250)psia for ΔP, and (12.432912.8036)lbm/ft3 for Δ(1/v) in Equation (VIII).

c2=((300250)psia(12.432912.8036)lbm/ft3)1/2=((300250)psia(12.432912.8036)lbm/ft3(25037ft2/s21Btu/lbm)(1Btu/lbm5.4039ft3psia))1/2=2065ft/s

Substitute 2065ft/s for c2, and 1752ft/s for V2 in Equation (IX).

Ma2=1752ft/s2065ft/s=0.849

Thus, the Mach number at the exit of nozzle is 0.849.

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Chapter 17 Solutions

Thermodynamics: An Engineering Approach

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