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.