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yuradex [85]
3 years ago
14

A submarine submerges by admitting seawater (S = 1.03) into its ballast tanks. The amount of water admitted is controlled by air

pressure, because seawater will cease to flow into the tank when the internal pressure (at the hull penetration) is equal to the hydrostatic pressure at the depth of the submarine. Consider a ballast tank, which can be modeled as a vertical half-cylinder (R = 8 ft, L = 20 ft) for which the air pressure control valve has failed shut. The failure occurred at the beginning of a dive from 60 ft to 1000 ft. The tank was initially filled with seawater to a depth of 2 ft and the air was at a temperature of 40 °F. As the weight of water in the tank is important in maintaining the boat’s attitude, determine the weight of water in the tank as a function of depth during the dive. You may assume that tank internal pressure is always in equilibrium with the ocean’s hydrostatic pressure and that the inlet pipe to the tank is at the bottom of the tank and penetrates the hull at the "depth" of the submarine.
Engineering
1 answer:
ser-zykov [4K]3 years ago
6 0

Answer:

Explanation:

Fw= y looking symbol sea water (2010.62-168171.2727/h+32.935)ft^3

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Before you disconnect the service battery from the discharged battery, it is good practice to place a load across the
Lilit [14]

Answer:

it is true i just did this test

Explanation:

6 0
3 years ago
Read 2 more answers
Consider a mixing tank with a volume of 4 m3. Glycerinflows into a mixing tank through pipe A with an average velocity of 6 m/s,
Svetach [21]

This question is incomplete, the complete question as well as the missing diagram is uploaded below;

Consider a mixing tank with a volume of 4 m³. Glycerin flows into a mixing tank through pipe A with an average velocity of 6 m/s, and oil flow into the tank through pipe B at 3 m/s. Determine the average density of the mixture that flows out through the pipe at C. Assume uniform mixing of the fluids occurs within the 4 m³ tank.

Take p_o = 880 kg/m³ and p_{glycerol = 1260 kg/m³    

 

Answer:

the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³  

Explanation:

Given that;

Inlet velocity of Glycerin, V_A = 6 m/s

Inlet velocity of oil, V_B = 3 m/s  

Density velocity of glycerin, p_{glycerol = 1260 kg/m³

Density velocity of glycerin, Take p_o = 880 kg/m³

Volume of tank V = 4 m

from the diagram;

Diameter of glycerin pipe, d_A = 100 mm = 0.1 m

Diameter of oil pipe, d_B = 80 mm = 0.08 m

Diameter of outlet pipe d_C = 120 mm = 0.12 m

Now, Appling the discharge flow equation;

Q_A + Q_B = Q_C

A_Av_A + A_Bv_B = A_Cv_C

π/4 × (d_A)²v_A + π/4 × (d_B )²v_B = π/4 × (d_C)²v_C

we substitute

π/4 × (0.1 )² × 6 + π/4 × (0.08 )² × 3 = π/4 × (0.12)²v_C

0.04712 + 0.0150796 = 0.0113097v_C

0.0621996 = 0.0113097v_C

v_C = 0.0621996 / 0.0113097

v_C  = 5.5 m/s

Now we apply the mass flow rate condition

m_A + m_B = m_C

p_{glycerin}A_Av_A + p_0A_Bv_B = pA_Cv_C  

so we substitute

1260 × π/4 × (0.1 )² × 6 + 880 × π/4 × (0.08 )² × 3 = p × π/4 × (0.12)² × 5.5

1260 × 0.04712 + 880 × 0.0150796 = p × 0.06220335

59.3712 + 13.27 = 0.06220335p  

72.6412 = 0.06220335p    

p = 72.6412 / 0.06220335

p =  1167.8 kg/m³  

Therefore, the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³  

4 0
3 years ago
Explain the difference between the connection of a cumulative compound and a differential compound motor
Alexandra [31]

Answer:

Explanation:

A motor is a device that directs current in electrical energy form to mechanical energy, which is known as direct current (DC) motors.

DC motors are of three types: (a) The series motor, (b) The shunt motor, and (c) the compound motor. Our main focus here is the Compound motor, which is further sub-divided into:

i) The cumulative compound motors

ii) The differential compound motors

The difference between these two are:

Cumulative compound motors                  Differential compound motors

In cumulative compound motors,              In differential compound motors,

both the series and shunt windings          both series and shunt are

are connected in a way that,                     connected in a way that the

production of fluxes through them           production of fluxes via them

assist each other i.e. they aid each          always opposes each other i.e.

other in the production of magnetism      they oppose each other in the

                                                                    production of magnetism.

6 0
3 years ago
A spark ignition engine burns a fuel of calorific value 45MJkg. It compresses the air-ful mixture in accordance with PV^1.3=cons
antoniya [11.8K]

Answer:

i). Compression ratio = 3.678

ii). fuel consumption = 0.4947 kg/hr

Explanation:

Given  :

PV^{1.3}=C

Fuel calorific value = 45 MJ/kg

We know, engine efficiency is given by,

\eta = 1-\left ( \frac{1}{r_{c}} \right )^{1.3-1}

where r_{c} is compression ratio = \frac{v_{c}+v_{s}}{v_{c}}

           r_{c} = 1+\frac{v_{s}}{v_{c}}

where v_{c} is compression volume

           v_{s} is swept volume

Now it is given that swept volume at 30% of compression, 70% of the swept volume remains.

Then, v_{30}=v_{c}+0.7v_{s}

and at 70% compression, 30% of the swept volume remains

    ∴    v_{70}=v_{c}+0.3v_{s}  

We know,

\frac{P_{2}}{P_{1}}=\left ( \frac{V_{1}}{V_{2}} \right )^{n}

\frac{2.75}{1.5}=\left ( \frac{v_{c}+0.7\times v_{s}}{v_{c}+0.3\times v_{s}} \right )^{1.3}

\left ( 1.833 \right )^{\frac{1}{1.3}}=\frac{v_{c}+0.7v_{s}}{v_{c}+0.3v_{s}}\\

1.594=\frac{v_{c}+0.7v_{s}}{v_{c}+0.3v_{s}}

v_{c}+0.7v_{s}=1.594v_{c}+0.4782v_{s}

0.7v_{s}-0.4782v_{s}=1.594v_{c}-v_{c}

0.2218v_{s} = 0.594v_{c}

v_{c}=0.3734 v_{s}

∴   r_{c}= 1+\frac{v_{s}}{0.3734v_{s}}

Therefore, compression ratio is r_{c} = 3.678

Now efficiency, \eta =\left ( 1-\frac{1}{r_{c}} \right )^{0.3}

 \eta =\left ( 1-\frac{1}{3.678} \right )^{0.3}

 \eta =0.32342 , this is the ideal efficiency

Therefore actual efficiency, \eta_{act} =0.5\times \eta _{ideal}

           \eta_{act} =0.5\times 0.32342

           \eta_{act} =0.1617

Therefore total power required = 1 kW x 3600 J

                                                    = 3600 kJ

∴ we know efficiency, \eta=\frac{W_{net}}{Q_{supply}}

Q_{supply}=\frac{W_{net}}{\eta _{act}}

Q_{supply}=\frac{3600}{0.1617}

Q_{supply}=22261.78 kJ

Therefore fuel required = \frac{22261.78}{45000}

                                        = 0.4947 kg/hr      

5 0
3 years ago
1kg of air (R 287 J/kgK) fills a weighted piston-cylinder device at 50kPa and 100°C. The device is cooled until the temperature
Reika [66]

Answer:

the work done during this cooling is −28.7 kJ

Explanation:

Given data

mass (m) = 1 kg

r = 287 J/kg-K

pressure ( p) = 50 kPa

temperature (T) = 100°C = ( 100 +273 ) = 373 K

to find out

the work done during this cooling

Solution

we know the first law of thermodynamics

pv = mRT     ....................1

here put value of p, m R and T and get volume v(a) when it initial stage in equation 1

50 v(a) = 1 × 0.287  × 373

v(a) = 107.051 / 50

v(a) = 2.1410 m³    .......................2

now we find out volume when temperature is  0°C

so put  put value of p, m R and T and get volume v(b) when temperature is cooled in equation 1

50 v(b) = 1 × 0.287  × 273

v(a) = 78.351 / 50

v(a) = 1.5670 m³    .......................3

by equation 2 and 3 we find out work done to integrate the p with respect to v i.e.

work done = \int\limits^a_b {p} \, dv

integrate it and we get

work done = p ( v(b) - v(a)  ) ................4

put the value p and v(a) and v(b) in equation 4 and we get

work done = p ( v(b) - v(a)  )

work done = 50 ( 1.5670 - 2.1410 )

work done = 50 ( 1.5670 - 2.1410 )

work done = 50 (−0.574)

work done = −28.7 kJ

here we can see work done is negative so its mean work done opposite in direction of inside air

'

7 0
3 years ago
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