Maximum Depth Model Of Uss Sturgeon

Published Date: 02 Nov 2017

Displacement Surfaced 3640 tons

Submerged 4762 tons

Propulsion S5W nuclear reactor 15000 shp

System two steam turbines,

One propeller

Dimensions Beam 31’ 8"

Leangth 292’ 3"

Draft 28’ 8"

Speed surface >15 knots

Submerged >20 knots

Depth limit >400

Crew 107

Thresher’s remains were located on the seafloor, some 8,400 ft (2,690 m) below the surface

ρ, seawater = 1025 kg/m3

What is the pressure that seawater exerts at the site of the lost ship?

Consider the following

P = pressure exerted by seawater at the site of the lost ship = ?

g= acceleration due to gravity = 9.8 m/s2

h= height of the water level = 8400 ft = 2600 m

ρ, seawater = Density of seawater= 1025 kg/m3

Calculation of pressure

P = ρxgxh

P = 1025 x 9.8 x 2600 =2.611 x 107 kg/m2

PROBLEM 2: Maximum Speed

Given:

D = diameter of the submarine = 31’ 8 "= 9.652 m

L = length of the submarine = 292’ 3" = 89.0778 m

Draft = 28’ 8" = 8.7376 m

V= Maximum Speed =?

η = viscosity

Suppose to find the entire force (which must be zero) on one square meter of the thin layer of

Water between heights z and z + Δ z :

First, gravity: if the river is flowing downhill at some low angle θ , this square meter of

the layer (volume , density ) experiences a gravitational force mg g sinθ ≈ ρg Δz.θ pulling it downstream

considering

sinθ =θ

Gravity essentialy balance out the difference between the two viscous forces:

Dividing throughout by η and by Δz,

Taking now the limit and recalling the definite Δ z → 0, and recalling the differential eq.ion of the differential

Hence the differential equation

Solving it, we get

Where C, D is constant

We know that

Pmax = Power = 15000 shp = 11185.5 Kw

Power = R = F.V

R = P.A.V eq. 1

Area A = π R2.L = 6517.694 m2

Using equation 1

V = R/(P.A) = 11185.5X1000/(6517.694X185) = 9.27 m/s

Vmax= 9.27 m/s = 18.01944 Knots

Problem #3; Steam Flow to Main Engine

Given

T1 = 250 0C = 523 K

T2 = 100 0C = 373 K

P1= 2.5 Mpa

P2= 0.05 Mpa

m = mass flow rate = ?

We know that the maximum power

Pmax =

h1= T1. CP

h2= T2. CP

2x11185.5x1000=(523-373).m.Cp

m= 16 kg/s

Mass flow of steam required to achieve the maximum engine power 16 kg/s

Problem #4; Steam Line Pressure Loss

Bernoulli’s equation:

The Bernoulli’s equation is the foremost helpful equations that's applied in a large choice of fluid flow connected issues. This equation are often derived in several ways in which, e.g. by integration Euler’s equation on a contour, by applying 1st and second laws of natural philosophy to steady, irrotational, in adhesive and incompressible flows etc. In easy kind the Bernoulli’s equation relates the pressure, rate and elevation between any 2 points within the flow field. it's a scalar equation and is given by:

Bernoulli’s equation could also be written in terms of pressures (i.e., Pascals in SI units) as:

Between any two points 1 and 2 in the flow field for ir-rotational flows, so Bernoulli’s equation is written as:

Presentation of Bernoulli equation to pipe flow

The Darcy-Weisbach calculation is usually used equations for approximating frictional pressure drops in internal flows. This equivalence is given by:

Where

L= length if the pipe = 84 feet = 25.6032 m

D= length if the pipe = 6 inch = 168 mm = 0.168 m

ρ = density = 0.6 kg/m3

m=Q= mass flow rate velocity = 25 kg/s

V = Velocity of fluid = Q/ ρ A = 25/ ρ (π.D2) = 1879.66 m/s

f= friction factor = 0.009

Q =Discharge Rate = Area xV

V = Q/A = 16X4/(0.6X3.14X0.1682) =1203.59 m/s

Head loss in bend

ΔPb = KV2/(2g)

V= Velocity of fluid

K=confident of bend = 0.16

g= acc. Due to gravity

ΔPb = pressure drop in bend

Head loss at sudden/contraction

ΔPc = 0.5V2/(2g)

Hence total head loss

hp = Δpf + Δph + Δpc1 + Δpc2

hp = F. L .V2/(2g) + KV2/(2g) + 0.5V2/(2g) +0.5V2/(2g)

Putting the values we get,

hp =102661.3316 m

Pressure loss = Pp = ρ x g x hp = 60.4264 N/m2

Problem #5 Thermal Loss and Temperature Drop

1-D radial conduction over a cylinder:

One of times encountered downside is that of warmth flow through the walls of a pipe or through the insulation placed around a pipe. Contemplate the cylinder rdisplayed. The pipe is either insulated on the ends or is of enough length, L, that heat losses through the ends square measure negligible. Assume no heat sources among the wall of the tube. If T1> T2, heat can flow outward, radially, from the within radius, R1, to the surface radius, R2. the method are delineate by the Fourier Law.

The differential equation leading heat diffusion is:

As constant k, the answer is

The heat flow rate through the wall is given by:

Hence, the thermal resistance in that case can be expressed as:

Thermal resistance network:

Overall thermal resistance:

Insulation thickness: ro-ri

Using above relation and given parameter

Rtot = ln(0.168275/0.168)/(2x3.14x0.0451x25.6032) +1/(2x3.14x0.168275x.25.603x088)

Rtot = 2.254 x 10-4 + 0.4197 = 0.4200 m2K/W

Extra Credit

T2 = temperature at the Exit = 2500 C

T1 = temperature at the inlet = ?

Energy loss =?

Temperature at the inlet

W +mh1 = mh2

W = m(h2-h1) = m Cpx (h2-h1)

Putting the mass rate we get

W = 16 X1.006X(523-373) = 2401.44 KJ/s

Pmax =

h1= T1. CP

h2= T2. CP

2x11185.5x1000=(523-373).m.Cp

m= 16 kg/s

As P1/T1 = P2/T2

SO

T1 = T2XP1/P2

T1 = 100 0C