Solutions Manual
Rocket Propulsion Elements 9th Edition
By
George P. Sutton,
Oscar Biblarz
( All Chapters Included - 100% Verified Solutions )
, 2
CHAPTER 2
1. A jet of fluid hits a stationary flat plate in the manner shown [see textbook].
(a) If there is 50 kg of fluid flowing per minute at an absolute velocity of 200 m/sec,
what will be the force on the plate?
(b) What will this force be when the plate moves in the direction of flow at u = 50
km/h? Explain the methodology.
m
m
(c − u x ) = Fx and
m uy − u y = Fy = 0
2 2
=
m ρ A(c − u x )
a) m = 50 kg/min, c = 200 m/sec, u x = 0
F x = (50/60)(200) = 166.7 N on plate
b) ρA = m /c = (50/60)/200 = 0.0042 kg/m
u x = 50 km/h = 13.9 m/sec
F x = 0.0042(200 – 13.9)2 = 144.24 N on plate
In Part b), the mass flow rate on the plate has decreased because of the decreased relative
velocity. The flow is incompressible.
2. The following data are given for a certain rocket unit: thrust, 8896 N; propellant
consumption, 3.867 kg/sec; velocity of vehicle, 400 m/sec; energy content of propellant,
6.911 MJ/kg. Assume 100% combustion efficiency.
Determine (a) the effective velocity; (b) the kinetic jet energy rate per unit flow of
propellant; (c) the internal efficiency; (d) the propulsive efficiency; (e) the overall
efficiency; (f) the specific impulse; (g) the specific propellant consumption.
F = 8896 N, m = 3.867 kg/sec, u = 400 m/sec, Q R = 6.911 MJ/kg, take η comb = 1.0
a) c = F/ m = 8896/3.867 = 2,300 m/sec
E jet Pjet 0.5m v2 2
b) = = = 0.5c 2 = 2.645x106 m2/sec2 or 2.645 MJ/kg
m m m
0.5m c 2 c2
c) ηint = = = 0.383
m QRηcomb 2QR
d) η p 2u / c 2()
= = = 0.3376
1 + (u / c) 1 + ()2
2
e) η = η pηint = 12.9 %
Fu
OR η= = 13.3%
m QR + 0.5m u 2
SOLUTIONS MANUAL to accompany ROCKET PROPULSION ELEMENTS, 9th EDITION 2
1
,3
F 8896
f) I s = = = 234.5 sec
mg 0 3.867 x9.81
g) SFC = 1/I s = 0.00426 sec-1
3. A certain rocket has an effective exhaust velocity of 7000 ft/sec; it consumes 280
lbm/sec of propellant mass, each of which liberates 2400 Btu/lbm. The unit operates
for 65 sec. Construct a set of curves plotting the propulsive, internal, and overall
efficiencies versus the velocity ratio u/c(0 < u/c < 1.0). The rated flight velocity
equals 5000 ft/sec. Calculate (a) the specific impulse; (b) the total impulse; (c) the
mass of propellants required; (d) the volume that the propellants occupy if their average
specific gravity is 0.925. Neglect gravity and drag.
[Assume sea-level operation where 1 lbm weights 1 lbf so there is no numerical distinction
between the two units.]
c = 7000 ft/se, m = 280 lbm/sec, Q R = 2400 Btu/lbm, t p = 65 sec, u RATED = 5000 ft/sec.
a) I s = c/g 0 = 7000/32.174 = 217.5 sec.
b) I t = Ft p = c m t p /g 0 = (7000)(280/32.2)(65) = 3.96 x 106 lbf-sec
c) w = w t p = (280)(65)(32.2/32.2) = 18,200 lbf or 18,200 lbm at sea level
d) SG = (density)/(density of liquid water at standard conditions) = 0.925
ρ = (0.925)(62.4) = 57.72 lbm/ft3 and Volume = 18,200/57.72 = 315.3 ft3
4. For the rocket in Problem 2, calculate the specific power, assuming a propulsion system
dry mass of 80 kg and a duration of 3 min.
DRY MASS = EMPTY MASS (see Fig. 4-1)
SOLUTIONS MANUAL to accompany ROCKET PROPULSION ELEMENTS, 9th EDITION 3
2
, 4
Pjet 0.5Fc
Specific Power = =
m0 m f + mp
m 0 = (3.867)(180) + 80 = 776.06 kg
Pjet 0.5(8896)(2300)
= = 13.18 kW/kg
m0 776.06
5. A Russian rocket engine (RD-110) consists of four thrust chambers supplied by a single turbopump.
The exhaust from the turbine of the turbopump is ducted to four vernier chamber nozzles (which
can be rotated to provide some control of the flight path). Using the information below, determine the
thrust and mass flow rate of the four vernier thrusters.
For the individual thrust chambers (vacuum):
F c = 73.14 kN, c c = 2857 m/sec
For the overall engine with verniers (vacuum):
F oa = 297.93 kN, c oa = 2845 m/sec
(a) vernier thrust ; F v = 297.93 – 4x73.14 = 5.37 kN
(b) vernier mass flow rate: m oa = m c + m v
so m v = F oa /c oa – F c /c c = 297930/2845 - 4x73140/2857 = 2.32 kg/sec
(c) vernier effective exhaust velocity: c v = F v / m v = 5370/2.32 = 2315 m/sec
6. A certain rocket engine has a specific impulse of 250 sec. What range of vehicle
velocities (u, in units of ft/sec) would keep the propulsive efficiencies at or greater
than 80%. Also, how could rocket–vehicle staging be used to maintain these high
propulsive efficiencies for the range of vehicle velocities encountered during launch?
I s = 250 sec, c = (32.17)(250) = 8,042.5 ft/sec
a) By inspection of Fig. 2-3, η p ≥ 0.8 for ½ ≤ u/c ≤ 2
So, 4,021 ≤ u ≤ 16,085 ft/sec
b) Design upper stages with increasing I s to keep u/c ≤ 2.0 within atmospheric flight.
7. For a solid propellant rocket motor with a sea-level thrust of 207,000 lbf, determine: (a) the (constant) propellant mass flow
rate m and the specific impulse I s at sea level, (b) the altitude for optimum nozzle expansion as well as the thrust and specific
impulse at this optimum condition and (c) at vacuum conditions. The initial total mass of the rocket motor is 50,000 lbm and its
propellant mass fraction is 0.90. The residual propellant (called slivers, combustion stops when the chamber pressure falls below
a deflagration limit) amounts to 3% of the burnt. The burn time is 50 seconds; the nozzle throat area (A t ) is 164.2 in.2 and its area
ratio (A 2 /A t ) is 10. The chamber pressure (p 1 ) is 780 psia and the pressure ratio (p 1 /p 2 ) across the nozzle may be taken as 90.0.
Neglect any start/stop transients and use the information in Appendix 2.
m p = m m ζ = 50,00x0.9 = 45,000 lbm (propellant loading); usable part = 45,00x0.97 = 43,650 lbm
m = m p /t = 43650/50 = 873.0 lbm/sec
I t = Ft = 207000x50 = 10,350,000 lb-sec
SOLUTIONS MANUAL to accompany ROCKET PROPULSION ELEMENTS, 9th EDITION 4
3
Rocket Propulsion Elements 9th Edition
By
George P. Sutton,
Oscar Biblarz
( All Chapters Included - 100% Verified Solutions )
, 2
CHAPTER 2
1. A jet of fluid hits a stationary flat plate in the manner shown [see textbook].
(a) If there is 50 kg of fluid flowing per minute at an absolute velocity of 200 m/sec,
what will be the force on the plate?
(b) What will this force be when the plate moves in the direction of flow at u = 50
km/h? Explain the methodology.
m
m
(c − u x ) = Fx and
m uy − u y = Fy = 0
2 2
=
m ρ A(c − u x )
a) m = 50 kg/min, c = 200 m/sec, u x = 0
F x = (50/60)(200) = 166.7 N on plate
b) ρA = m /c = (50/60)/200 = 0.0042 kg/m
u x = 50 km/h = 13.9 m/sec
F x = 0.0042(200 – 13.9)2 = 144.24 N on plate
In Part b), the mass flow rate on the plate has decreased because of the decreased relative
velocity. The flow is incompressible.
2. The following data are given for a certain rocket unit: thrust, 8896 N; propellant
consumption, 3.867 kg/sec; velocity of vehicle, 400 m/sec; energy content of propellant,
6.911 MJ/kg. Assume 100% combustion efficiency.
Determine (a) the effective velocity; (b) the kinetic jet energy rate per unit flow of
propellant; (c) the internal efficiency; (d) the propulsive efficiency; (e) the overall
efficiency; (f) the specific impulse; (g) the specific propellant consumption.
F = 8896 N, m = 3.867 kg/sec, u = 400 m/sec, Q R = 6.911 MJ/kg, take η comb = 1.0
a) c = F/ m = 8896/3.867 = 2,300 m/sec
E jet Pjet 0.5m v2 2
b) = = = 0.5c 2 = 2.645x106 m2/sec2 or 2.645 MJ/kg
m m m
0.5m c 2 c2
c) ηint = = = 0.383
m QRηcomb 2QR
d) η p 2u / c 2()
= = = 0.3376
1 + (u / c) 1 + ()2
2
e) η = η pηint = 12.9 %
Fu
OR η= = 13.3%
m QR + 0.5m u 2
SOLUTIONS MANUAL to accompany ROCKET PROPULSION ELEMENTS, 9th EDITION 2
1
,3
F 8896
f) I s = = = 234.5 sec
mg 0 3.867 x9.81
g) SFC = 1/I s = 0.00426 sec-1
3. A certain rocket has an effective exhaust velocity of 7000 ft/sec; it consumes 280
lbm/sec of propellant mass, each of which liberates 2400 Btu/lbm. The unit operates
for 65 sec. Construct a set of curves plotting the propulsive, internal, and overall
efficiencies versus the velocity ratio u/c(0 < u/c < 1.0). The rated flight velocity
equals 5000 ft/sec. Calculate (a) the specific impulse; (b) the total impulse; (c) the
mass of propellants required; (d) the volume that the propellants occupy if their average
specific gravity is 0.925. Neglect gravity and drag.
[Assume sea-level operation where 1 lbm weights 1 lbf so there is no numerical distinction
between the two units.]
c = 7000 ft/se, m = 280 lbm/sec, Q R = 2400 Btu/lbm, t p = 65 sec, u RATED = 5000 ft/sec.
a) I s = c/g 0 = 7000/32.174 = 217.5 sec.
b) I t = Ft p = c m t p /g 0 = (7000)(280/32.2)(65) = 3.96 x 106 lbf-sec
c) w = w t p = (280)(65)(32.2/32.2) = 18,200 lbf or 18,200 lbm at sea level
d) SG = (density)/(density of liquid water at standard conditions) = 0.925
ρ = (0.925)(62.4) = 57.72 lbm/ft3 and Volume = 18,200/57.72 = 315.3 ft3
4. For the rocket in Problem 2, calculate the specific power, assuming a propulsion system
dry mass of 80 kg and a duration of 3 min.
DRY MASS = EMPTY MASS (see Fig. 4-1)
SOLUTIONS MANUAL to accompany ROCKET PROPULSION ELEMENTS, 9th EDITION 3
2
, 4
Pjet 0.5Fc
Specific Power = =
m0 m f + mp
m 0 = (3.867)(180) + 80 = 776.06 kg
Pjet 0.5(8896)(2300)
= = 13.18 kW/kg
m0 776.06
5. A Russian rocket engine (RD-110) consists of four thrust chambers supplied by a single turbopump.
The exhaust from the turbine of the turbopump is ducted to four vernier chamber nozzles (which
can be rotated to provide some control of the flight path). Using the information below, determine the
thrust and mass flow rate of the four vernier thrusters.
For the individual thrust chambers (vacuum):
F c = 73.14 kN, c c = 2857 m/sec
For the overall engine with verniers (vacuum):
F oa = 297.93 kN, c oa = 2845 m/sec
(a) vernier thrust ; F v = 297.93 – 4x73.14 = 5.37 kN
(b) vernier mass flow rate: m oa = m c + m v
so m v = F oa /c oa – F c /c c = 297930/2845 - 4x73140/2857 = 2.32 kg/sec
(c) vernier effective exhaust velocity: c v = F v / m v = 5370/2.32 = 2315 m/sec
6. A certain rocket engine has a specific impulse of 250 sec. What range of vehicle
velocities (u, in units of ft/sec) would keep the propulsive efficiencies at or greater
than 80%. Also, how could rocket–vehicle staging be used to maintain these high
propulsive efficiencies for the range of vehicle velocities encountered during launch?
I s = 250 sec, c = (32.17)(250) = 8,042.5 ft/sec
a) By inspection of Fig. 2-3, η p ≥ 0.8 for ½ ≤ u/c ≤ 2
So, 4,021 ≤ u ≤ 16,085 ft/sec
b) Design upper stages with increasing I s to keep u/c ≤ 2.0 within atmospheric flight.
7. For a solid propellant rocket motor with a sea-level thrust of 207,000 lbf, determine: (a) the (constant) propellant mass flow
rate m and the specific impulse I s at sea level, (b) the altitude for optimum nozzle expansion as well as the thrust and specific
impulse at this optimum condition and (c) at vacuum conditions. The initial total mass of the rocket motor is 50,000 lbm and its
propellant mass fraction is 0.90. The residual propellant (called slivers, combustion stops when the chamber pressure falls below
a deflagration limit) amounts to 3% of the burnt. The burn time is 50 seconds; the nozzle throat area (A t ) is 164.2 in.2 and its area
ratio (A 2 /A t ) is 10. The chamber pressure (p 1 ) is 780 psia and the pressure ratio (p 1 /p 2 ) across the nozzle may be taken as 90.0.
Neglect any start/stop transients and use the information in Appendix 2.
m p = m m ζ = 50,00x0.9 = 45,000 lbm (propellant loading); usable part = 45,00x0.97 = 43,650 lbm
m = m p /t = 43650/50 = 873.0 lbm/sec
I t = Ft = 207000x50 = 10,350,000 lb-sec
SOLUTIONS MANUAL to accompany ROCKET PROPULSION ELEMENTS, 9th EDITION 4
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