1. A survey vessel has a 10 ft diameter,... propeller speed is 200 rpm, the boat speed is 20...

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1. A survey vessel has a 10 ft diameter, B 5-90 propeller with a pitch of 10 ft. The
propeller speed is 200 rpm, the boat speed is 20 knots, and the thrust reduction factor
(t) is 0.12, wake fraction (w) is 0.18, and the relative rotational efficiency ηR is 1.0.
The propeller operates as indicated by the Wageningen (Troost) Series B propeller
charts. Determine:
a.
b.
c.
d.
e.
Thrust
Shaft torque
EHP of the boat
The propeller shaft power (delivered power) PD
The (Quasi) PC or ηD
The propeller is also tested at zero ship speed (bollard pull) and it is found that the
engine limits the torque to 50,000 lbf ft. Determine:
f.
the propeller rpm and thrust at this condition
a. Prop. Thrust.
Given variables
d := 10ft
p := 10ft
p_over_d :=
t := .12
w := .18
η R := 1
d
n_rpm := 200
n :=
n_rpm
60⋅ sec
n = 3.333
Vs := 20knot
ρ := 1.9905lb⋅
sec
ft
VA := Vs ⋅ ( 1 − w)
Velocity of Approach
Advance Ratio
p
2
4
m
VA = 8.437
s
VA
J1 :=
n⋅ d
1
s
Use the B 5-90 prop curve to determine KT and KQ
J1 = 0.83
KT := .12
KQ := .023
2 4
Thrust := KT⋅ ρ ⋅ n ⋅ d
4
Thrust = 2.654 × 10 lb
b. Shaft Torque
2 5
Torque := KQ⋅ ρ ⋅ n ⋅ d
c. Shaft powerr delivered
PD := 2⋅ π ⋅ n ⋅
4
Torque
lb⋅
550⋅
Torque = 5.087 × 10 lb⋅ ft
ft
sec
hp
3
PD = 1.937 × 10 hp
d. EHP
Vs
PE := Thrust ⋅ ( 1 − t ) ⋅
lb⋅
550⋅
3
ft
PE = 1.433 × 10 hp
sec
hp
e. Quasi Efficiency
η D :=
PE
η D = 0.74
PD
f. Propeller rpm and thrust at 50,000.
Advance_velocity := 0
n o :=
Torque max := 50000⋅ lb⋅ ft
Torque max
5
no = 3.305
ρ ⋅ KQ⋅ d ⋅ η R
2 4
Thrust q := KT⋅ ρ ⋅ n q ⋅ d
1
s
n q := n o ⋅ 60⋅ sec
n q = 198.286
7 2
Thrust q = 9.391× 10 s lb
2. A propeller is to be selected for a single-screw container ship with the following
features:
EHP = 80000 HP, ship speed = 25 kts, maximum propeller diameter = 34 ft,
w = 0.249, t = 0.18, ηR = 1.0, centerline depth, h = 25 ft
a. Using the maximum prop diameter, determine the optimum B 5-90 design. Use the
metrics below to confirm your design.
a.
P/D
b.
KT (optimum)
c.
KQ (optimum)
d.
ηo (optimum)
e.
J
f.
Developed HP
g.
The (Quasi) PC or ηD
h.
RPM
From the consideration of cavitation, determine:
i.
The predicted cavitation (%) using the Burrill correlation
j.
The expanded area ratio (EAR) to provide 5% cavitation for a commercial
ship.
Assume the operating conditions are similar to the B 5-90 propeller.
Given
V2 := 25⋅ knot
EHP := 80000⋅ hp
d 2 := 34⋅ ft
w2 := .249
t2 := .18
η R := 1
h := 25
First we must combine a couple of equations in order to get all the information we know in terms of KT and
J.
ft
⎛
lb⋅
⎜
sec
R2 := ⎜ 550⋅
hp
⎝
⎞
⎟ EHP
⎟⋅
⎠ V2
Kt
→
2
T2 :=
R2
1 − t2
ft
⎛
lb⋅
⎜
sec
⎜ 550⋅
hp
⎝
3
2
Kt :=
⎞
⎟
⎟ ⋅ ( EHP)
⎠
(
)(
)
ρ ⋅ V2 ⋅ d 2 ⋅ 1 − t2 ⋅ 1 − w2
J2
2
T2
2
4
ρ ⋅ n2 ⋅ d2
V2
J2 :=
n 2⋅ d 2
= 0.55
Now we can plot the function KT = 0.55 * J2 on the B 5-90 curve graph. Drawing a verticle line
where the function plot and each KT - P/D intersect will provide a value for KT and ηo. Starting
with a logical P/D (.5 for example), step though P/D values, recording KT and ηo. Take note at
the peak value for ηo, That will determine optimal values.
found:
P/D = 1.2
KT=.29
ηo = .6
Using the curves posted on the web, I
a. P/D = 1.2
J2 :=
b. KT(opt) = .29
Kt
Kq := .055
.55
c. KQ(opt) = .055
η o2 := .6
2 5
d. ηo = .6
Kt := .29
Q2 := Kq ⋅ ρ ⋅ n ⋅ d
J2 = 0.726
e. J = 0.726
1−t ⎞
PC := η o2⋅ ⎛⎜
⎟ ⋅η
⎝ 1 − w⎠ R
EHP
PD2 :=
PC
f. HP = 124200 HP
1 − w2
n 2 := V2⋅
J2⋅ d 2
g. PC = .644
h. RPM = 77.012
n2 = 1.284
1
s
N2 := n 2⋅ 60⋅ s
Cavitation Calculations
EAR := 90
2
A E := EAR ⋅
π ⋅ d2
4
assume AD ~ AE
P_over_D_ans := 1.2
5
PD2 = 1.242 × 10 hp
PC = 0.644
N2 = 77.012
h := 25ft
1
A P := A E⋅ [ 1.067 − 0.229( P_over_D_ans ) ]
2
2
VR := ⎡⎡V2⋅ ( 1 − w2)⎤ + ( 0.7π ⋅ n ⋅ d 2) ⎤
⎣⎣
⎦
⎦
2
T
τC :=
AP
1
2
3
2
⋅ ρ ⋅ VR
2026
σ0.7R :=
ft
4
sec
4
+ 64.4
ft
τC = 5.421× 10
1
A⋅ s
2
3
sec
4
⋅h
⋅
2
sec
1
2 ft
VR + 4.836⋅ ⎛⎜ N2⋅ ⎞⎟ ⋅ d 2
⎝ s⎠
2
2
−4
2
σ0.7R = 1.095 × 10
0.2
2
C :=
− 10
2
A E = 7.591× 10 m
τC⋅ A ⋅ s + 0.3064 − 0.523⋅ σ0.7R
0.2
⋅ 0.7R
0.0305σ
C = −17.791
− 0.0174
% cavitation
Negative cavitation indicates that it is not a problem with at this speed
i. Cavitation = - 17.8%
τCn := C⋅ ⎡.0305⋅ ⎛ σ0.7R
⎣
A pn :=
⎝
0.2⎞
0.2
⎤
⎠ − . × 0174⎦ − .3064 + .523⋅ σ0.7R
T
⎛ .5⋅ ρ ⋅ τ ⋅ V 2⎞
Cn R ⎠
⎝
EARn :=
A pn
( )
2 ⎤
⎡
⎢1.067 − .229⋅ 1.2⋅ π 34 ⎥
4 ⎦
⎣
j. = EAR is much less than one, Changing to meet these requirements would
not be necessary. (This will be considered extra credit)
3. List the advantages and disadvantages of the fixed pitch propeller, controllable
pitch propeller, and waterjet propulsion systems. List the best applications (or
platform(s)) for each propulsor and supporting reasons considering the mission of the
platform. (expectation: half a page of concise thought).
For full credit - A brief discussion similar to that in chapter 6 of the text, At least 2 advantages and 2
disadvantages of each and an example of where each has been used sucessfully.
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