Mcconkey Chapter 10 solution

1. Assignment of Power Plant-I Designed by Sir Engr. Masood Khan
SOLVED PROBLEMS OF CHAPTER # 10
TITLE: NOZZLES & JET PROPULSION

2. Nozzles & Jet Propulsion Designed by Sir Engr. Masood Khan
PROBLEM # 10.1
Calculate the throat & exit areas of a nozzle to expand air at
the rate of 4.5 kg/s from 8.3 bar, 327 C into a space at 1.38
bar. Neglect the inlet velocity & assume isentropic flow.
GIVEN DATA:
Working Fluid = Air
Mass flow rate, mo
= 4.5
P1=8.3 bar P2 = 1.38 bar
T1 = 327 = 600 K C1 = 0
Flow is Isentropic.
REQUIRED:
At =? A2 =?
SOLUTION:
Since we have (Pc / P1) = (2/γ+1) γ / γ-1
γ = 1.4
(Pc / P1) = (2 / 2.4) 1.4 / 0.4
= 0.528
Pc = P1*0.528 = 4.385 bar
Since Pc > P2 so flow will achieve critical condition at throat
and nozzle is choked.
Tc/T1 = 2 / γ+1
Tc= T1*2 / γ+1 = 600 (2/2.4) = 500 K
Since Critical Velocity would be
Cc = a = √(γRTc) = √1.4*287*500 = 448.218 m/s
And PcVc = RTc
Vc = RTc/ Pc = (287*500)/4.385*105
= 0.327 m3
/kg
m = CA / V => Ac = mVc/Cc
Ac = 4.50*0.327/448.218 = 3285.5 mm2
For exit condition we have
T2 / T1 = (P2 / P1)γ-1 / γ
T2 = 600 (1.38/8.3)0.4/1.4
= 360 K
V2 = RT2 / P2 = 287*360/1.38*105
= 0.474 m3
/kg
C2 = √{2Cp(T1-T2)} = √2*1.005*103
(600-360) = 695 m/s
A2 = mV1/C2 = 4.5*0.474/695 = 4837 mm2
PROBLEM # 10.2:
It is required to produce a steam of helium at the rate of 0.1
kg/s traveling at sonic velocity at a temperature of 15 C.
Assume negligible inlet velocity, isentropic flow, & a back
pressure of 1.013 bar, calculate:
(i) The required inlet pressure & temperature;
(ii) The exit area of the nozzle.
For helium take the molar mass as 4 kg/kmol & γ = 1.66
GIVEN DATA:
m = 0.1kg/s Tc = 15 C Pc = P2 = 1.013 bar
C = 0 n = γ = 1.66
Molecular weight of helium, M = 4
Rhe = 8314/4 = 2078.5 J/kg
REQUIRED:
P1 =? T1 =? A2 =?
For Sonic flow, Pc = P1
SOLUTION:
Pc / P1 = P2 / P1 = (2/n+1) n/n+1
= (2/1.66+1) 1.66/0.66
= 0.488
T2/T1 = Tc / T1 = 2/n+1 = 2/1.66+1 = 0.752
P1 = P2 / 0.488 = 1.013/0.488 = 2.076 bar
T1 = Tc / 0.752 = 288/0.752 = 383 k
A2 = mV2/C2 = mVc/Cc ------------------------------ (1)
V2 = RT2/P2 = (2078.5*288)/(1.013*105
) = 5.91 m3
/s
Cc = C2 = √γRTc = √(1.66*2078.5*288) = 996.84 m/s

3. Nozzles & Jet Propulsion Designed by Sir Engr. Masood Khan
PROBLEM # 10.3:
Recalculate problem 10.1 assuming a coefficient of
discharge of 0.96 & a nozzle efficiency of 0.92.
GIVEN DATA:
Use data of problem 10.1
Coefficient of discharge = 0.96
Nozzle efficiency, n = 0.92
mo
= 4.5 kg/s C1 = 0
P1 = 8.3 bar P2 = 1.38 bar
T1 = 327 C = 600 k
REQUIRED:
At =? A2 =?
SOLUTION:
mo
/m = 0.96
m = mo
/ 0.96 = 4.5 / 0.96 = 4.687 kg/s
Pc / P1 = (2/n+1)n / n-1
= (2/γ+1)γ / γ-1
= 0.528
Pc = 0.528P1 = 0.528*8.3 = 4.38 bar
As Pc > P2 => Nozzle is Choked and a Convergent-
Divergent Nozzle is required.
Pc = P2 = 4.38 bar
Tc/T1 = 2/γ+1 = 0.8333
Tc = 0.8333T1 = 500 k = Tt
Vt = RTt / Pt = (287*500)/(4.38*105
) = 0.32778 m3
/kg
Ct = Cc = √γRTc = γ (1.4*287*500) = 448.22 m/kg
At = mVt/Ct=4.087*0.32778/448.22=3.43*10-3
m2
=3430 mm2
T2 = T1 (P2/P1) γ-1 / γ
= 600 (1.38/8.3) 0.4/1.4
= 359 k
ηn = (T1-T/
2) / (T1-T2) = (600 - T/
2) / (600 -354) = 0.92
T/
2 = 378.28 k
V/
2 = RT2 / P2 = 287*378.28/1.38*105
= 0.787 m3
/kg
C/
2= √2Cp(T1-T/
2) =√2*1.005*103
(600-378.28)=667.57 m/s
A2 = m/
V2/C2 = 4.5*0.787/667.57 = 5305 mm2
PROBLEM # 10.4:
A convergent divergent nozzle expands air at 6.89 bar &
427 C into a space at 1 bar. The throat area of the nozzle is
650 mm² & the exit area is 975 mm². The exit velocity is
found to be 680 m/s when the inlet velocity is negligible.
Assuming that friction in the convergent portion is
negligible, calculate:
(i) The mass flow through the nozzle, stating whether
the nozzle is under-expanding or over-expanding;
(ii) The nozzle efficiency & the coefficient of velocity.
GIVEN DATA:
Convergent – divergent nozzle
Working fluid = air where γ = 1.4
P1 = 6.84 bar P2 = 1bar
T1 = 427 C = 700 k
At = 650 mm2
A2 = 975 mm2
C1 = 0 C/
2 = 680 m/s
REQUIRED:
mo
=? ηn =? C/
2/C2 =?
Also Under-Expansion or Over-Expansion =?
Friction in the convergent portion is negligible.
SOLUTION:
(Pc / P1) = (2/γ+1) γ / γ+1
= 0.528
Pc = 0.528(6.89) = 3.64 bar
Pc > P1 => Nozzle is choked
Tc/T1 = 2/γ+1 = 2/1+1.4 = 2/2.4 = 0.8333
Tc = 0.8333T1 = 0.8333*700 = 583.33 k

4. Nozzles & Jet Propulsion Designed by Sir Engr. Masood Khan
Ct = Cc = √γRTc = √1.4*287*583.33 = 484.13 m/s
Vt = Vc = RTc/Pc = 287*583.33/3.64*105
= 0.46 m3
/kg
mo
= CtAt/Vt = 484.13*650*10-6
/0.46 = 0.684 kg/s
T2/T1 = (P2/P1)γ-1 / γ
= (1/6.89) 0.4/1.4
= 0.576
T2 = 0.576T1 = 403 k
C2=√2Cp(T1-T2)=√2*1.005*103
(700-403) =772.6 m/s
ηn = (C/
2/C2) = (680/772.6)2
= 0.775 = 77.5 %
C/
2/C2 = 0.88
PROBLEM # 10.6:
Steam at 20 bar & 240 C expands isentropically to a
pressure of 3 bars in a convergent divergent nozzle.
Calculate the mass flow per unit area:
(i) Assuming equilibrium flow;
(ii) Assuming supersaturated flow.
For a supersaturated flow assume that the process follows
the law, pv¹³ = constant.
GIVEN DATA:
Working fluid = Air
P1 = 20 bar P2 = 3 bar T1 = 240 C = 513 k
Convergent-Divergent Nozzle
REQUIRED:
mo
/A =?
When (1) Equilibrium
(2) Supersaturated, PV1.3
= C
SOLUTION:
Since we have at P1 = 20 bar, Tsat = 212 C
T2 >Tsat, => Initially Super-Heated, so
h1 = 2902.4 kj/kg S1 = 6.545 kj/kg k, V1 = 0.1115 m3
/kg
Since the Process is Isentropic, so
S1 = S2 So at
P2 = 3 bar Sg = 6.991 kj/kg k
Since S2 < Sg, So steam is in a wet region, So at P2 = 3 bar
6.545 = 1.672 + x2 (5.319) => x2 = 0.92
h2 = hf + x2hfg = 561.4 + 0.920(2163.8)
v2 = 0.920*0.60553 = 2552 kj/kg = 0.559 m3
/kg
So m = CA/v => m/A2 = C2/V2 -----------------------------(1)
C2 = √2*103
*(h1-h2) = √2*103
(2902-2552) = 837 m/s
m/A2 = 837/0.559 = 1510 kg/m2
s
When the Steam is Super-Heated.
(VR/V1) = (P1/P2) 1/ k
VR = 0.1115 (20/30) 1/1.3
= 0.48 m3
/kg
CR
2
/2 = k/k-1 (P1V1-P2VR)
=1.3*105
/0.3(20*105
*0.1115-3*0.48) = 827 m/s
We have, m/A = CR/VR = 827/0.48 = 1723 kg/m2
s
PROBLEM # 10.7:
A thermometer inserted into an air stream flowing at 33.5
m/s records a temperature of 15 C, the static pressure in the
duct is found to be 1.01 bar. Assuming that the air round
the thermometer bulb is brought to rest adiabatically,
calculate:
(i) The true static temperature of the air;
(ii) The stagnation pressure of the air.
GIVEN DATA:
C1 = 33.5 m/s T0 = 15 C = 255 k
P = 1.01 bar (Air is brought to rest Adiabatically)
REQUIRED:
Tstatic =? P0 =?

5. Nozzles & Jet Propulsion Designed by Sir Engr. Masood Khan
SOLUTION:
Since we know that, T0 = Ts + C2
/2Cp
287 = Ts + (33.5) 2
/2*1.005*103
Ts = 287.44 k = 144.44 C
For stagnation pressure
P0 = P + C2
p/2RT
= 1*105
+(33.5) 2
* 1.01*105
/2*287*288
= 10.1685 Pa = 1.017 bar
PROBLEM # 10.8:
A turbojet aircraft is traveling at 925 km/h in atmospheric
conditions of 0.45 bar & -26 C. The compressor pressure
ratio is 8, the air mass flow rate is 45kg/s, & the maximum
allowable cycle temperature is 800 C. The compressor,
turbine, & the jet pipe stagnation isentropic efficiencies are
0.85, 0.89 & 0.9 respectively, the mechanical efficiency of
the drive is 0.98, & the combustion efficiency is 0.99.
Assuming a convergent propulsion nozzle, a loss of
stagnation pressure in the combustion chamber of 0.2 bar,
& a fuel of calorific value 13300 kj/kg, calculate:
(i) The required nozzle exit area;
(ii) The net thrust developed;
(iii) The air- fuel ratio;
(iv) The specific fuel consumption.
For a gases in the turbine & propulsion nozzle take γ =
1.333 & Cp = 1.15kj/kg.k
For the combustion process assume an equivalent Cp value
of 1.15 kJ/kg.k
GIVEN DATE:
Turbo Jet-Air Craft
Ca = 800 km/h = 222.2 m/s
T0 = -50 C0
= 223 k Pa = 0.24 bar
Compression pressure ratio, P02 / P01 = 10/1
T03 = 820 C =1093 k ηid = 0.9 ηc = 0.9
Loss in Cc = 0.14 bar
Calorific value of fuel = 43320045/kg
ηcc = 0.98 ηm = 98% ηJp = 0.92
For compression γ = 1.4, Cp = 1.005 kj/kg k
For CC γ = 1.33, Cp = 1.15 kj/kg k
Convergent Nozzle.
REQUIRED:
Thrust =? SFC =?
DIAGRAM:
SOLUTION:
K.E at inlet of air = C2
a/2 = (222.2)2
/2 = 247 kj/kg
Since T01 = T0 + C2
/2Cp = 223+24.7/1.005 = 247.6 k
At intake ηI = (T/
01-T0) / (T01-T0)
T/
01 = T0 + η (T01-T0) = 233+0.9(247.6-223) = 245 k
Since intake process is Isentropic.
P01 / P0 = (T/
01/T0)γ / γ-1
= (245/223)1.4/ 0.4
P01 = 0.24*1.39 = 0.334 bar
Since compression process is Isentropic.
(T/
02 / T01) = (P02 / P01) γ / γ-1
= (10)0.4/.14
= 1.931
T/
02 = T01*1.931 = 478 k
ηc = (T/
02 – T1) / (T02 – T1) = 0.9
T/
02 = 247.6 + 256 = 503.6 k
P02 / P01 = 10
P02 = 10*0.334 = 3.34 bar
P03 = P02 – loss in cc = 3.34 – 0.14 = 3.20 bar
Compressor work is
Wc = Cp (T02 – T1) = 1.005(503.6-247.6) = 257.3 kj/kg
Since ηn = Wc / WT

6. Nozzles & Jet Propulsion Designed by Sir Engr. Masood Khan
WT = Wc / ηn = 257.3 / 0.98 = 262 kj/kg
WT = Cp (T03 – T04)
262.57 = 1.15(T03 - T04)
T04 = 1093 - 227.19 = 865 k
ηT = (T03 - T04)/ (T03 -T/
04)
T/
04 = 1093 - 247.82 = 845 k
Since 3-4 is Isentropic process
(T03 /T/
04)γ / γ-1
= (P03 / P04)
P04 = 3.2*0.357 = 1.156 bar
P04 is the inlet pressure of nozzle so
Pc / P04 = (2/1+γ)γ / γ-1
= (2/2.333) 1.333/0.333
= 0.539
Pc = 1.156*0.593 = 0.624 bar
Since Pc > Pa
So Nozzle will be choking and exit velocity will be sonic.
(Tc - T04) = (T03 - T04) = 2/γ+1 = 2/2.33 = 741.3 k
ηJ = (T04 - T05) / (T04 - T/
05) = 0.92
T/
05 = T04 + 865 - 741.3 / 0.92 = 730.6 k
P04 / P5 = (T04 –T/
5)γ / γ-1
P5 = 1.156 / 1.966 = 0.589 bar
Since R = Cp(γ-1) / γ
R = 1.005(1.333-1)/1.333 = 0.2872 kj/kg k
Vs = Ris / Ps = 287.3*741.3 / 0.589*105
= 3.616 m3
/kg
Jet Velocity = Cj = √γRTs
Jet Velocity = √1.3*287.3*741.3 = 532.8 m/s
Mass flow rate, mo
= CjA / Vs
mo
= 532.8*0.08/3.616 = 11.788 kg/s
Momentum Thrust = mo
(Cj – Ca)
Momentum Thrust = 11.78(532.8–222.2) = 3661N
Pressure Thrust = (P5 – Pa) A
Pressure Thrust = 0.08*105
(0.589-0.24) =2792N
Total Thrust will be = 3661 + 2792 = 6453 N
Heat Supplied = mo
Cpg (T03 - T02)
Heat Supplied =11.788*1.15(1093-50306)=7990 kj/kg
Heat Supplied = mf*Calorific Value
mf = 7990 / 0.98*43300 = 0.188 kg/s
SFC = 0. /188 / Total Thrust = 0.188*103
/6543
SFC = 0.0291 kg/kN
PROBLEM # 10.9:
In a turboprop engine the compressor pressure is 6 & the
maximum cycle temperature is 760 C. The stagnation
isentropic efficiencies of the compressor & the turbine are
0.85 & 0.88 respectively; the mechanical efficiency is 99
%. The aircraft is traveling at 725 km/h at an altitude where
the ambient temperature is -7 C. Taking an intake duct
efficiency of 0.9, neglecting the pressure loss in the
combustion chamber & assuming that the gases in the
turbine expands down to atmospheric pressure, leaving the
aircraft at 725 km/h relative to the aircraft, Calculate:
(i) The specific power output;
(ii) The cycle efficiency.
For the gases in the turbine take γ = 1.333 & Cp
=1.15kj/kgk ; for the combustion process assume an
equivalent Cp value of 1.15 kj/kgk.
GIVEN DATA:
Ca = 650 km h = 180.5 m/s T01 = -18 C = 255 k
Pr. Ratio = 9/1
Maximum Cycle Temperature = 850 C =1123 k
ηid = 0.9 ηc = 0.89 ηT = 0.93
Velocity of exhaust gases = 650 km/h
REQUIRED:
Power Output =? ηCycle =? ηM =?

7. Nozzles & Jet Propulsion Designed by Sir Engr. Masood Khan
DIAGRAM:
SOLUTION:
Air K.E at inlet = C2
/2 = (180.5)2
/2 = 16.29 kg/kg
As we have
T01 = T0 + C2
/2Cp = 255 + (16.29)2
/2(1.005) = 271 k
id = (T/
01 -T0) / ( T01 -T0)
T/
01 = 269.4 k
P01 / Po = (T01 / T0)γ / γ-1
= 1.215
Since Compressor process is Isentropic.
T/
02 = T01*(P02 / Po1) γ-1 / γ
=271.2*(9)0.4/1.4
= 508 k
ηc = (T/
01 -T01) / (T01 -T01) = 0.89
T02 = 271.2 +266 = 537.24
Compressor Input Work,
Wc = Cpa (T02 -T01) = 1.005(537.2-271.2) = 267.5 kg/kg
P03 / P4 = P02 / Po1 * P0 / P4 = 9*1.215 = 10.935
Therefore T4 / T03 = P4 / Po3 = 1/10.935
T4 = 617.8 k
T/
04 = T04 + Ca
2
/2 = 617.8+(180.5)2
/2*103
1.15 = 632 k
WT = Cp (T03 -T04) = 525.12 kj/kg
Net Output = (525.12 - 267.5) / 0.98 = 252.5 kj/kg
Heat Supplied = Cp (T03 - T02) = 1.15(1123-537.2) =673.67
ηcycle = 252.5 / 673.67 = 37.4 %

8. Nozzles & Jet Propulsion Designed by S