Saturated steam enters a converging–diverging nozzle at 1.75 MPa, 10 percent moisture, and negligible velocity, and it exits at 1.2 MPa. For a nozzle exit area of 25 cm 2 , determine the throat area, exit velocity, mass flow rate, and exit Mach number if the nozzle ( a ) is isentropic and ( b ) has an efficiency of 92 percent.

Question

Want to see more full solutions like this?

Answer 1

Textbook Question

Chapter 17.7, Problem 137RP

Saturated steam enters a converging–diverging nozzle at 1.75 MPa, 10 percent moisture, and negligible velocity, and it exits at 1.2 MPa. For a nozzle exit area of 25 cm², determine the throat area, exit velocity, mass flow rate, and exit Mach number if the nozzle (a) is isentropic and (b) has an efficiency of 92 percent.

(a)

Expert Solution

To determine

The exit velocity for isentropic converging-diverging nozzle.

The mass flow rate for isentropic converging-diverging nozzle.

The exit Mach number for isentropic converging-diverging nozzle.

The throat area for isentropic converging-diverging nozzle.

Answer to Problem 137RP

The exit velocity for isentropic converging-diverging nozzle is 363.7 m/s_.

The mass flow rate for isentropic converging-diverging nozzle is 6.35 kg/s_.

The exit Mach number for isentropic converging-diverging nozzle is 0.829_.

The throat area for isentropic converging-diverging nozzle is 24.26 cm2_.

Explanation of Solution

Determine the initial specific enthalpy of the isentropic converging-diverging nozzle.

h1=(hf+x1hfg) (I)

Here, the specific enthalpy of saturated liquid is hf, the specific enthalpy change upon vaporization is hfg, and the quality of initial state is x1.

Determine the initial entropy of the isentropic converging-diverging nozzle.

s1=sf+x1sfg (II)

Here, the specific entropy of saturated liquid is sf and the specific entropy change upon vaporization is sfg.

Determine the final quality of the isentropic converging-diverging nozzle.

s2=sf+x2sfgx2=s2−sfsfg (III)

Here, the final entropy of the isentropic converging-diverging nozzle is s2.

Determine the final specific enthalpy of the isentropic converging-diverging nozzle.

h2=(hf+x2hfg) (IV)

Determine the final specific volume of the isentropic converging-diverging nozzle.

v2=(vf+x2vfg)=(vf+x2(vg−vf)) (V)

Determine the exit velocity of the isentropic converging-diverging nozzle.

h1+V122=h2+V2220=ht−h1+V222V2=2(h1−h2) (VI)

Note: the fluid flow was complete stop at the initial exit velocity then it becomes zero.

Determine the mass flow rate of the isentropic converging-diverging nozzle.

m˙=1v2A2V2 (VII)

Determine the velocity of sound at the exit of the isentropic converging-diverging nozzle.

c=(∂P∂r)s1/2≅(ΔPΔ(1/v))s1/2 (VIII)

Determine the exit Mach number for isentropic converging-diverging nozzle.

Ma2=V2c2 (IX)

Determine the critical pressure at the throat.

Pt=P∗=0.567×P01 (X)

Here, the stagnation pressure at the initial state is P01.

Determine the velocity of throat using steady-flow energy balance.

h1+V122=ht+Vt220=ht−h1+Vt22Vt=2(h1−ht) (XI)

Note: the initial velocity is zero.

Determine the throat area for isentropic converging-diverging nozzle.

At=m˙vtVt (XII)

Conclusion:

At the inlet,

Perform unit conversion of pressure at state 1 from MPa to kPa.

P01=1.75 MPa×(1000 kPa1 MPa)=1750 kPa

From the Table A-5, “Saturated water-Pressure table”, obtain value of the specific enthalpy of saturated liquid, the specific enthalpy change upon vaporization, specific entropy of saturated liquid, and the specific entropy change upon vaporization at 1750 kPa of temperature as:

hf=878.16 kJ/kghfg=1917.1 kJ/kgsf=2.3844 kJ/kg⋅Ksfg=4.0033 kJ/kg⋅K

Substitute 878.16 kJ/kg for hf, 1917.1 kJ/kg for hfg, and 0.90 for x in Equation (I).

h1=(878.16 kJ/kg)+(0.90)×(1917.1 kJ/kg)=(878.16 kJ/kg)+(1725.39 kJ/kg)=2603.55 kJ/kg

Substitute 2.3844 kJ/kg⋅K for sf, 4.0033 kJ/kg⋅K for sfg, and 0.90 for x in Equation (II).

s1=(2.3844 kJ/kg⋅K)+(0.90)×(4.0033 kJ/kg)=(2.3844 kJ/kg⋅K)+(3.60297 kJ/kg⋅K)=5.98737 kJ/kg⋅K≅5.9874 kJ/kg⋅K

At the exit,

Perform unit conversion of pressure at state 2 from MPa to kPa.

P02=1.2 MPa×(1000 kPa1 MPa)=1200 kPa

From the Table A-5, “Saturated water-Pressure table”, obtain value of the specific volume of saturated liquid, the specific volume change upon vaporization, specific enthalpy of saturated liquid, the specific enthalpy change upon vaporization, specific entropy of saturated liquid, and the specific entropy change upon vaporization 1200 kPa of temperature as:

vf=0.001138 m3/kgvg=0.16326 m3/kghf=798.33 kJ/kghfg=1985.4 kJ/kgsf=2.2159 kJ/kg⋅Ksfg=4.3058 kJ/kg⋅K

Substitute 5.9874 kJ/kg⋅K for s2, 2.2159 kJ/kg⋅K for sf, and 4.3058 kJ/kg⋅K for sfg in Equation (III).

x2=(5.9874 kJ/kg⋅K)−(2.2159 kJ/kg⋅K)(4.3058 kJ/kg⋅K)=(5.9874 kJ/kg⋅K)−(2.2159 kJ/kg⋅K)(4.3058 kJ/kg⋅K)=0.8759

Substitute 798.33 kJ/kg for hf, 1985.4 kJ/kg for hfg, and 0.8759 for x2 in Equation (IV).

h2=(798.33 kJ/kg)+(0.8759)×(1985.4 kJ/kg)=(798.33 kJ/kg)+(1739.01 kJ/kg)=2537.3 kJ/kg

Substitute 0.001138 m3/kg for vf, 0.16326 m3/kg for vg, and 0.8759 for x2 in Equation (V).

v2=[(0.001138 m3/kg)+(0.8759)×(0.16326 m3/kg−0.001138 m3/kg)]=(0.001138 m3/kg)+(0.142999 m3/kg)=0.14314 m3/kg

Substitute 2603.5 kJ/kg for h1 and 2537.4 kJ/kg for h2 in Equation (VI).

V2=2(2603.5 kJ/kg−2537.4 kJ/kg)=2(66.1 kJ/kg)(1000 m2/s21 kJ/kg)=363.59 m/s

Thus, the exit velocity for isentropic converging-diverging nozzle is 363.7 m/s_.

Substitute 0.14314 m3/kg for v2, 25 cm2 for A2, and 363.59 m/s for V2 in Equation (VII).

m˙=1(0.14314 m3/kg)×(25 cm2)×(363.59 m/s)=1(0.14314 m3/kg)×(25 cm2×(1 m210000 cm2))×(363.59 m/s)=1(0.14314 m3/kg)×(25×10−4m2)×(363.59 m/s)=6.35 kg/s

Thus, the mass flow rate for isentropic converging-diverging nozzle is 6.35 kg/s_.

At state 2 entropy value for the specific volume of steam and at pressure just below and just above the specified pressure 1.1 and 1.3 MPa are determined to be 0.1547 m3/kg and 0.1333 m3/kg.

Substitute 1.1 MPa for Pabove, 1.3 MPa for Pbelow, 0.1333 kg/m3 for vabove, and 0.1547 kg/m3 for vbelow in Equation (VIII).

c2=((1.3−1.1)MPa(10.1333−10.1547)kg/m3)1/2=(0.2 MPa×(1000 kPa1 MPa)(7.501875−6.464124)kg/m3)×(1000 m2/s21 kPa⋅m3)1/2=(0.2 MPa×(1000 kPa1 MPa)1.03775kg/m3)×(1000 m2/s21 kPa⋅m3)1/2=438.9 m/s

Substitute 363.59 m/s for V2 and 438.9 m/s for c2 in Equation (IX).

Ma2=363.59 m/s439.0 m/s=0.828

Thus, the exit Mach number for isentropic converging-diverging nozzle is 0.829_.

Substitute 1.75 MPa for P01 in Equation (X).

Pt=0.576×(1.75 MPa)=1.008 MPa

From the above throat pressure the value of enthalpy and specific volume as:

ht=2507.7 kJ/kgvt=0.1672 m3/kg

Substitute 2603.5 kJ/kg for h1 and 2507.7 kJ/kg for ht in Equation (XI).

Vt=2(2603.5−2507.7)kJ/kg=2×(95.8 kJ/kg)×(1000 m2/s21 kJ/kg)=191600 m2/s2=437.7 m/s

Substitute 6.35 kg/s for m˙, 0.1672 m3/kg for vt, and 437.7 m/s for Vt in Equation (XII).

At=(6.35 kg/s)(0.1672 m3/kg)(437.7 m/s)=1.06172 m3/s437.7 m/s=0.002426 m2=0.002426 m2×(10000 cm21 m2)

=24.256 cm2≅24.26 cm2

Thus, the throat area for isentropic converging-diverging nozzle is 24.26 cm2_.

(b)

Expert Solution

To determine

The exit velocity for isentropic converging-diverging nozzle.

The mass flow rate for isentropic converging-diverging nozzle.

The exit Mach number for isentropic converging-diverging nozzle.

The throat area for isentropic converging-diverging nozzle.

Answer to Problem 137RP

The exit velocity for isentropic converging-diverging nozzle is 348.7 m/s_.

The mass flow rate for isentropic converging-diverging nozzle is 6.35 kg/s_.

The exit Mach number for isentropic converging-diverging nozzle is 0.829_.

The throat area for isentropic converging-diverging nozzle is 24.26 cm2_.

Explanation of Solution

Determine the initial specific enthalpy of the isentropic converging-diverging nozzle.

h1=(hf+x1hfg) (XIII)

Here, the specific enthalpy of saturated liquid is hf, the specific enthalpy change upon vaporization is hfg, and the quality of initial state is x1.

Determine the initial entropy of the isentropic converging-diverging nozzle.

s1=sf+x1sfg (XIV)

Here, the specific entropy of saturated liquid is sf and the specific entropy change upon vaporization is sfg.

Determine the final quality of saturated steam for the isentropic converging-diverging nozzle.

s2s=sf+x2ssfgx2s=s2−sfsfg (XV)

Here, the final entropy of the isentropic converging-diverging nozzle is s2.

Determine the final specific enthalpy of saturated steam for the isentropic converging-diverging nozzle.

h2s=(hf+x2hfg) (XVI)

Determine the enthalpy of steam at the actual exit state for the isentropic converging-diverging nozzle.

ηN=h01−h2h01−h2s . (XVII)

Determine the final specific enthalpy for the isentropic converging-diverging nozzle.

h2=(hf+x2hfg) (XVIII)

Determine the exit entropy of the isentropic converging-diverging nozzle.

s2=sf+x2sfg (XIX)

Determine the final specific volume of the isentropic converging-diverging nozzle.

v2=(vf+x2vfg)=(vf+x2(vg−vf)) (XX)

Determine the exit velocity of the isentropic converging-diverging nozzle.

h1+V122=h2+V2220=ht−h1+V222V2=2(h1−h2) (XXI)

Note: the fluid flow was complete stop at the initial exit velocity then it becomes zero.

Determine the mass flow rate of the isentropic converging-diverging nozzle.

m˙=1v2A2V2 (XXII)

Determine the velocity of sound at the exit of the isentropic converging-diverging nozzle.

c=(∂P∂r)s1/2≅(ΔPΔ(1/v))s1/2 (XXIII)

Determine the exit Mach number for isentropic converging-diverging nozzle.

Ma2=V2c2 (XXIV)

Determine the critical pressure at the throat.

Pt=P∗=0.567×P01 (XXV)

Here, the stagnation pressure at the initial state is P01.

Determine the actual enthalpy of steam at the throat.

ηN=h01−hth01−hts (XXVI)

Determine the velocity of throat using steady-flow energy balance.

h1+V122=ht+Vt220=ht−h1+Vt22Vt=2(h1−ht) (XXVII)

Note: the initial velocity is zero.

Determine the throat area for isentropic converging-diverging nozzle.

At=m˙vtVt (XXVIII)

Conclusion:

At the inlet,

Perform unit conversion of pressure at state 1 from MPa to kPa.

P01=1.75 MPa×(1000 kPa1 MPa)=1750 kPa

From the Table A-5, “Saturated water-Pressure table”, obtain value of the specific enthalpy of saturated liquid, the specific enthalpy change upon vaporization, specific entropy of saturated liquid, and the specific entropy change upon vaporization at 1750 kPa of temperature as:

hf=878.16 kJ/kghfg=1917.1 kJ/kgsf=2.3844 kJ/kg⋅Ksfg=4.0033 kJ/kg⋅K

Substitute 878.16 kJ/kg for hf, 1917.1 kJ/kg for hfg, and 0.90 for x in Equation (XIII).

h1=(878.16 kJ/kg)+(0.90)×(1917.1 kJ/kg)=(878.16 kJ/kg)+(1725.39 kJ/kg)=2603.55 kJ/kg

Substitute 2.3844 kJ/kg⋅K for sf, 4.0033 kJ/kg⋅K for sfg, and 0.90 for x in Equation (XIV).

s1=(2.3844 kJ/kg⋅K)+(0.90)×(4.0033 kJ/kg)=(2.3844 kJ/kg⋅K)+(3.60297 kJ/kg⋅K)=5.98737 kJ/kg⋅K≅5.9874 kJ/kg⋅K

At the exit,

Perform unit conversion of pressure at state 2 from MPa to kPa.

P02=1.2 MPa×(1000 kPa1 MPa)=1200 kPa

From the Table A-5, “Saturated water-Pressure table”, obtain value of the specific volume of saturated liquid, the specific volume change upon vaporization, specific enthalpy of saturated liquid, the specific enthalpy change upon vaporization, specific entropy of saturated liquid, and the specific entropy change upon vaporization 1200 kPa of temperature as:

vf=0.001138 m3/kgvg=0.16326 m3/kghf=798.33 kJ/kghfg=1985.4 kJ/kgsf=2.2159 kJ/kg⋅Ksfg=4.3058 kJ/kg⋅K

Substitute 5.9874 kJ/kg⋅K for s2, 2.2159 kJ/kg⋅K for sf, and 4.3058 kJ/kg⋅K for sfg in Equation (XV).

x2=(5.9874 kJ/kg⋅K)−(2.2159 kJ/kg⋅K)(4.3058 kJ/kg⋅K)=(5.9874 kJ/kg⋅K)−(2.2159 kJ/kg⋅K)(4.3058 kJ/kg⋅K)=0.8759

Substitute 798.33 kJ/kg for hf, 1985.4 kJ/kg for hfg, and 0.8759 for x2 in Equation (XVI).

h2=(798.33 kJ/kg)+(0.8759)×(1985.4 kJ/kg)=(798.33 kJ/kg)+(1739.01 kJ/kg)=2537.3 kJ/kg

Substitute 2603.5 kJ/kg for h01, 2537.3 kJ/kg for h2s, and 0.92 for ηN in Equation (XVII).

0.92=(2603.5 kJ/kg)−h2(2603.5 kJ/kg−2537.3 kJ/kg)0.92=(2603.5 kJ/kg)−h2(66.2 kJ/kg)h2=2542.7 kJ/kg

Substitute 2542.7 kJ/kg for h2, 798.33 kJ/kg for hf and 1985.4 kJ/kg for hfg, in Equation (XVIII).

2542.7 kJ/kg=(798.33 kJ/kg)+x2×(1985.4 kJ/kg)2542.7 kJ/kg−(798.33 kJ/kg)=x2×(1985.4 kJ/kg)x2=0.8786

Substitute 0.8786 for x2, 2.2159 kJ/kg⋅K for sf, and 4.3058 kJ/kg⋅K for sfg in Equation (XIX).

s2=2.2159 kJ/kg⋅K+(0.8786)×(4.3058 kJ/kg⋅K)=2.2159 kJ/kg⋅K+(3.783 kJ/kg⋅K)=5.9989 kJ/kg⋅K

Substitute 0.001138 m3/kg for vf, 0.16326 m3/kg for vg, and 0.8786 for x2 in Equation (XX).

v2=[(0.001138 m3/kg)+(0.8786)×(0.16326 m3/kg−0.001138 m3/kg)]=(0.001138 m3/kg)+(0.14244 m3/kg)=0.14357 m3/kg≅0.1436 m3/kg

Substitute 2603.5 kJ/kg for h1 and 2542.7 kJ/kg for h2 in Equation (XXI).

V2=2(2603.5 kJ/kg−2542.7 kJ/kg)=2(60.8 kJ/kg)(1000 m2/s21 kJ/kg)=121600 m2/s2=348.7 m/s

Thus, the exit velocity for isentropic converging-diverging nozzle is 348.7 m/s_.

Substitute 0.1436 m3/kg for v2, 25 cm2 for A2, and 348.7 m/s for V2 in Equation (XXII).

m˙=1(0.1436 m3/kg)×(25 cm2)×(348.7 m/s)=1(0.1436 m3/kg)×(25 cm2×(1 m210000 cm2))×(348.7 m/s)=1(0.1436 m3/kg)×(25×10−4m2)×(348.7 m/s)=6.07 kg/s

Thus, the mass flow rate for isentropic converging-diverging nozzle is 6.07 kg/s_.

At state 2 entropy value for the specific volume of steam and at pressure just below and just above the specified pressure 1.1 and 1.3 MPa are determined to be 0.1551 m3/kg and 0.1337 m3/kg.

Substitute 1.1 MPa for Pabove, 1.3 MPa for Pbelow, 0.1337 kg/m3 for vabove, and 0.1551 kg/m3 for vbelow in Equation (XXIII).

c2=((1.3−1.1)MPa(10.1337−10.1551)kg/m3)1/2=(0.2 MPa×(1000 kPa1 MPa)(7.479432−6.447453)kg/m3)×(1000 m2/s21 kPa⋅m3)1/2=(0.2 MPa×(1000 kPa1 MPa)1.031978kg/m3)×(1000 m2/s21 kPa⋅m3)1/2=439.7 m/s

Substitute 348.9 m/s for V2 and 439.7 m/s for c2 in Equation (XXIV).

Ma2=348.9 m/s439.7 m/s=0.793

Thus, the exit Mach number for isentropic converging-diverging nozzle is 0.793_.

Substitute 1.75 MPa for P01 in Equation (XXV).

Pt=0.576×(1.75 MPa)=1.008 MPa

From the above throat pressure and entropy at the state 2ts the value of enthalpy as:

hts=2507.7 kJ/kg

Substitute 2603.5 kJ/kg for h01, 2507.7 kJ/kg for hts, and 0.92 for ηN in Equation (XXVI).

0.92=(2603.5 kJ/kg)−ht(2603.5 kJ/kg)−(2507.7 kJ/kg)0.92=(2603.5 kJ/kg)−ht95.8 kJ/kght=2515.4 kJ/kg

From the above throat pressure and enthalpy at the state 2 the value of specific volume as:

vt=0.1679 m3/kg

Substitute 2603.5 kJ/kg for h1 and 2515.4 kJ/kg for ht in Equation (XXVII).

Vt=2(2603.5−2515.4)kJ/kg=2×(88.1 kJ/kg)×(1000 m2/s21 kJ/kg)=176200 m2/s2=419.76 m/s

Substitute 6.07 kg/s for m˙, 0.1679 m3/kg for vt, and 419.76 m/s for Vt in Equation (XXVIII).

At=(6.07 kg/s)(0.1679 m3/kg)(419.76 m/s)=1.019153 m3/s419.76 m/s=0.002428 m2=0.002428 m2×(10000 cm21 m2)

=24.28 cm2

Thus, the throat area for isentropic converging-diverging nozzle is 24.28 cm2_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Q Derive (continuity equation)? I want to derive clear mathematics.

motor supplies 200 kW at 6 Hz to flange A of the shaft shown in Figure. Gear B transfers 125 W of power to operating machinery in the factory, and the remaining power in the shaft is mansferred by gear D. Shafts (1) and (2) are solid aluminum (G = 28 GPa) shafts that have the same diameter and an allowable shear stress of t= 40 MPa. Shaft (3) is a solid steel (G = 80 GPa) shaft with an allowable shear stress of t = 55 MPa. Determine: a) the minimum permissible diameter for aluminum shafts (1) and (2) b) the minimum permissible diameter for steel shaft (3). c) the rotation angle of gear D with respect to flange A if the shafts have the minimum permissible diameters as determined in (a) and (b).

First monthly exam Gas dynamics Third stage Q1/Water at 15° C flow through a 300 mm diameter riveted steel pipe, E-3 mm with a head loss of 6 m in 300 m length. Determine the flow rate in pipe. Use moody chart. Q2/ Assume a car's exhaust system can be approximated as 14 ft long and 0.125 ft-diameter cast-iron pipe ( = 0.00085 ft) with the equivalent of (6) regular 90° flanged elbows (KL = 0.3) and a muffler. The muffler acts as a resistor with a loss coefficient of KL= 8.5. Determine the pressure at the beginning of the exhaust system (pl) if the flowrate is 0.10 cfs, and the exhaust has the same properties as air.(p = 1.74 × 10-3 slug/ft³, u= 4.7 x 10-7 lb.s/ft²) Use moody chart (1) MIDAS Kel=0.3 Q3/Liquid ammonia at -20°C is flowing through a 30 m long section of a 5 mm diameter copper tube(e = 1.5 × 10-6 m) at a rate of 0.15 kg/s. Determine the pressure drop and the head losses. .μ= 2.36 × 10-4 kg/m.s)p = 665.1 kg/m³

Answer 2