Propellers: Momentum theory for thrust and power
Use momentum theory to determine the thrust, power requirement and thrust loading coefficient for a 4.2 m
diameter propeller that is driving a ship at 20 kn in sea water. Assume that the ideal efficiency of the
propeller is 88%, and ignore the effect of the ship wake and the propeller boss.
Solution
uA  0.5144  20  10.3 m/s
2u A
2  10.3
i  0.88 

 u j  13.1 m/s
u j  u A u j  10.3


A0   D 2   4.22  13.9 m2
4
4
u  uA
10.3  13.1
u A AA  u0 A0  j
 A0 
 13.9  163 m3/s
2
2
T    (u A  AA )  u j  u A   1.025  163  (13.1  10.3)  468 kN
468  10.3
 5.48 MW
i
0.88
T
468
CTh  1
 1
 0.62
2
2
2    u A  A0
2  1.025  10.3  13.9
P
T  uA

Propellers: Momentum theory for changing propeller arrangements
a) A 2.5 m diameter screw propeller is required to produce a thrust of 700 kN at a speed of 24 kn in sea
water. Use momentum theory to estimate the minimum power which must be supplied to the
propeller.
b) Use momentum theory to estimate the power requirement for the situation of item a), if the single
2.5 m diameter propeller is replaced by two 1.5 m diameter propellers.
c) Use momentum theory to estimate the propeller diameter of a twin-screw option that delivers the
same thrust and absorbs the same power as in the situation of item a).
Solution
a) u A  0.5144  24  12.3 m/s
A0 

4
 D2 

 2.52  4.91 m2
4
T
700
 1
 1.84
2
   u A  A0 2  1.025  12.32  4.91
2
2
i 

 0.745
1  1  CTh 1  1  1.84
CTh 
P
1
2
T  uA
i

700  12.3
 11.6 MW
0.745
b) u A  12.3 m/s (as before)
A0 

4
 D2 

4
 1.52  1.77 m2
(for a single propeller)
T
350
 1
 2.55
(for both propellers)
2
   u A  A0 2  1.025  12.32  1.77
2
2
i 

 0.693
(for both propellers)
1  1  CTh 1  1  2.55
CTh 
P
1
2
T  uA
i

700  12.3
 12.4 MW
0.693
c) u A  12.3 m/s (as before)
CTh  1.84
CTh 
A0 
1
2

4
(as before)
T
   u A2  A0
 D2

 D
A0 
4 A0

1
2

T
350
 1
 2.45 m2
2
   u A  CTh 2  1.025  12.32  1.84
4  2.45

 1.77 m
Propeller: Open-water efficiency
Consider a propeller with 4.5 m diameter and the following open-water characteristics:
J
0.650
0.680
0.700
0.730
0.760
0.790
KT
0.2573
0.2415
0.2284
0.2119
0.1972
0.1800
KQ
0.0438
0.0418
0.0401
0.0379
0.0360
0.0336
0
J
0.820
0.833
0.880
0.920
0.980
1.030
KT
0.1646
0.1594
0.1380
0.1128
0.0807
0.0421
0
KQ
0.0316
0.0309
0.0285
0.0244
0.0198
0.0143
Assume sea water density to be  = 1026 kg/m3. The inflow velocity is VA = 11.245 m/s.
a) Compute the corresponding open-water efficiencies.
b) What should be the propeller rpm for the propeller to be at maximum efficiency at ship design
speed?
c) What should then be the power delivered at the propeller PD  2  n  Q ?
d) What is the open-water efficiency?
e) What is the averaged axial velocity in the propeller downstream wake V  VA  1  CTh ?
Solution
a) The open-water efficiency is:
J
0.650
0.680
0.700
0.730
0.760
0.790
KT
0.2573
0.2415
0.2284
0.2119
0.1972
0.1800
0 
KQ
0.0438
0.0418
0.0401
0.0379
0.0360
0.0336
KT J
0.1594 0.833



 0.684
KQ 2 0.0309 2
0
0.608
0.625
0.635
0.650
0.663
0.674
J
0.820
0.833
0.880
0.920
0.980
1.030
KT
0.1646
0.1594
0.1380
0.1128
0.0807
0.0421
KQ
0.0316
0.0309
0.0285
0.0244
0.0198
0.0143
b) The maximum efficiency is reached for J0 = 0.833. This yields:
n
VA
11.245

 3 Hz (or 180 rpm)
J  D 0.833  4.5
The power delivered at the propeller is expressed by the torque:
Q  KQ    n2  D5  0.0309  1026  32  4.55  0.527 MNm
PD  2  n  Q  2  3  0.527  9.92 MW
c) The thrust loading coefficient is:
T
603.5  103
CTh  1

 0.585
  VA2  D 2  4 12  1026  11.2452  4.52  4
2
The axial velocity in the propeller slipstream is:
V  VA  1  CTh  11.245  1  0.585  14.16 m/s
0
0.680
0.684
0.678
0.677
0.636
0.483
Propellers: Propeller curve
Consider a propeller with 4.5 m diameter and the following open-water characteristics:
J
0.650
0.680
0.700
0.730
0.760
0.790
0.820
0.833
0.880
0.920
0.980
1.030
KT
0.2573
0.2415
0.2284
0.2119
0.1972
0.1800
0.1646
0.1594
0.1380
0.1128
0.0807
0.0421
KQ
0.0438
0.0418
0.0401
0.0379
0.0360
0.0336
0.0316
0.0309
0.0285
0.0244
0.0198
0.0143
0
0.6080
0.6250
0.6350
0.6500
0.6630
0.6740
0.6800
0.6810
0.6780
0.6770
0.6360
0.4830
Assume sea water density to be  = 1026 kg/m3.
The inflow velocity at design speed is VA = 11.245 m/s.
a) What should be the propeller rpm for the propeller to be at maximum efficiency at ship design
speed?
b) What should then be the power delivered at the propeller PD  2  n  Q ?
c) What is the open-water efficiency?
d) What is the averaged axial velocity in the propeller downstream wake V  V A  1  CTh ?
Solution
a) The maximum efficiency is reached for J0 = 0.833. The associated rpm is:
n
VA
11.245

 3 Hz
J  D 0.833  4.5
(= 180 rpm)
b) The power delivered at the propeller is expressed by the torque:
Q  KQ    n2  D5  0.0309  1026  32  4.55  0.527 MNm
PD  2  n  Q  2  3  0.527  9.92 MW
c) The open-water efficiency is:
0 
KT J
0.1594 0.833



 0.684
KQ 2 0.0309 2
d) The thrust loading coefficient is:
CTh 
T
603.5  103

 0.585
  VA2  D 2  ( / 8) 1026  11.2452  4.52  ( / 8)
The axial velocity in the propeller slipstream is:
V  VA  1  CTh  12.245  1  0.585  14.16 m/s