Question 6.8: A 1/30 model of a ship with 900 m^2 wetted area, towed in wa...

First of all we should identify the pertinent dimensionless parameters that describe the ship resistance problem. For this we have to physically define the problem as follows:

The total drag force F , on a ship depends on ship velocity V , its characteristic geometrical length l , acceleration due to gravity g, density \rho , and viscosity \mu , of the fluid. Therefore, the total number of variables which describe the problem = 6 and the number of fundamental dimensions involved with the variables = 3.

Hence, according to the \pi theorem, number of independent

V, l \text { and } \rho are chosen as the repeating variables.

Then,  \pi_{1}=V^{a} l^{b} \rho^{c} F

Expressing the \pi terms by the dimensional formula of the variables involved we can write

\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}=\left(\mathrm{LT}^{-1}\right)^{a}(\mathrm{~L})^{b}\left(\mathrm{ML}^{-3}\right)^{c}\left(\mathrm{MLT}^{-2}\right)                   (6.36)

\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}=\left(\mathrm{LT}^{-1}\right)^{a}(\mathrm{~L})^{b}\left(\mathrm{ML}^{-3}\right)^{c}\left(\mathrm{LT}^{-2}\right)                      (6.37)

\mathbf{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}=\left(\mathrm{LT}^{-1}\right)^{a}(\mathrm{~L})^{b}\left(\mathrm{ML}^{-3}\right)^{c}\left(\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right)          (6.38)

Equating the exponents of the fundamental dimensions on both sides of the above equations we have

From Eq. (6.36):

c + 1 = 0

a+b-3 c+1=0

-a-2=0

Which give  a = –2, b = –2 and c = –1

Therefore,  \pi_{1}=\frac{F}{\rho V^{2} l^{2}}

From Eq. (6.37)

c=0

a+b-3 c+1=0

-a-2=0

which give  a = – 2, b = 1 and c = 0

therefore,  \pi_{2}=\frac{\lg }{V^{2}}

\pi_{2} is the reciprocal of the square of the Froude number, Fr.

From Eq. (6.38)

c+1=0

a+b-3 c-1=0

-a-1=0

Which give  a = –1, b = –1 and c = –1.

Hence,

which is the reciprocal of the Reynolds number, Re. Hence the problem of ship resistance can be expressed as

Or   F=\rho V^{2} l^{2} \phi\left(\frac{V^{2}}{l g}, \frac{V l \rho}{\mu}\right)                            (6.39)

Therefore, it is found from Eq. (6.39) that the total resistance depends on both the Reynolds number and the Froude number. For complete similarity between a prototype and its model, the Reynolds number must be the same, i.e.,

\frac{V_{p} l_{p} \rho_{p}}{\mu_{p}}=\frac{V_{m} l_{m} \rho_{m}}{\mu_{m}}                                  (6.40)

and also the Froude number must be the same, that i.e.,

\frac{V_{p}}{\left(l_{p} g_{p}\right)^{1 / 2}}=\frac{V_{m}}{\left(l_{m} g_{m}\right)^{1 / 2}}                                     (6.41)

Equation (6.40) gives V_{m} / V_{p}=\left(l_{p} / l_{m}\right)\left(v_{m} / v_{p}\right) where, V (kinematic viscosity) =\mu / \rho . On the other hand, Eq. (6.41) gives V_{m} / V_{p}=\left(l_{m} / l_{p}\right)^{1 / 2} since, in practice, g_{m} cannot be different from g_{p} . For testing small models these conditions are incompatible. The two conditions together require \left(l_{m} / l_{p}\right)^{3 / 2}=v_{m} / v_{p}  and since both the model and prototype usually operate in water, this condition for the scale factor cannot be satisfied. There is, in fact, no practicable liquid which would enable V_{m}  to be less than V_{p} . Therefore, it concludes that the similarity of viscous forces (represented by the Reynolds number) and similarity of gravity forces (represented by the Froude number) cannot be achieved simultaneously between the model and the prototype.

The way out of the difficulty was suggested by Froude. The assumption is made that the total resistance is the sum of three distinct parts: (a) the wave-making resistance; (b) skin friction; and (c) the eddy-making resistance. The part (a) is usually uninfluenced by viscosity but depends on gravity and is therefore independent of the Reynolds number, Re. Part (c), in most cases, is a small portion of the total resistance and varies little with the Reynolds number. Part (b) depends only on the Reynolds number. Therefore, it is usual to lump (c) together with (a). These assumptions allow us to express the function of Re and Fr in Eq. (6.39) as the sum of two separate functions, \phi_{1}(\mathrm{Re})+\phi_{2} (\mathrm{Fr}) . Now the skin friction part may be estimated by assuming that it has the same
value as that for a flat plate, with the same length and wetted surface area, which moves end on through the water at same velocity. Hence, the function \phi_{1}(\operatorname{Re}) is provided by the empirical information of drag resistance on such surfaces. Since the part of the resistance which depends on the Reynolds number is separately determined, the test on the model is conducted at the corresponding velocity which gives equality of the Froude number between the model and the prototype; thus dynamic similarity for the wave-making resistance is obtained. Therefore, the solution of present problem (Example 6.8) is made as follows:

From the equality of the Froude number,

(i) The corresponding speed of the ship V_{p}=\sqrt{\frac{l_{p}}{l_{m}}} \cdot V_{m}

= 10.95 m/s

Area ratio,       =\frac{A_{m}}{A_{p}}=\left(\frac{1}{30}\right)^{2}=\frac{1}{900}

Therefore A_{m} (area of the model) =\frac{900 \mathrm{~m}^{2}}{900}=1 \mathrm{~m}^{2}

(ii) If F_{w} and F_{s} represent the wave-making resistance and skin friction resistance of the ship respectively, then from the definition of the drag coefficient C_{D}  , we can write

= 8N

Now the total resistance on the model F_{m}=F_{s_{m}}+F_{w_{m}}

Hence,  F_{w_{\mathrm{m}}}=F_{m}-F_{s_{m}}=20-8=12 \mathrm{~N}

Now from dynamic similarity for wave making resistance

Or          F_{w_{p}}=F_{w m} \frac{\rho_{p}}{\rho_{m}}\left(\frac{V_{p}}{V_{m}}\right)^{2}\left(\frac{l_{p}}{l_{m}}\right)^{2}

(iii) F_{s_{p}} (skin friction of the prototype)

= 809.34 kN

(iv) Therefore, F_{p} (total drag resistance of the prototype)

(v) Propulsive power required =1133.34 \times 10.95=12410 kW

= 12.41 MW