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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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Air is saturated with water vapor at 35.0 oC and a total pressure of 1.50 atmospheres. If the molar flow rate of the dry air in
Marizza181 [45]

Answer:

11.541 mol/min

Explanation:

temperature = 35°C

Total pressure = 1.5 * 1.013 * 10^5 = 151.95 kPa

note : partial pressure of water in mixture = saturation pressure of water at T = 35°c )

from steam table it is = 5.6291 Kpa

calculate the mole fraction of H_{2}o ( YH_{2}o )

= 5.6291 / 151.95

= 0.03704

calculate the mole fraction of air ( Yair )

 = 1 - mole fraction of water

= 1 - 0.03704 = 0.9629

Now to determine the molar flow rate of water vapor in the stream

lets assume N = Total molar flow rate

NH_{2}o = molar flow rate of water

Nair = molar  flow rate of air = 300 moles /min

note : Yair * n = Nair

therefore n = 300 / 0.9629 = 311.541  moles /min

Molar flowrate of water

=  n -  Nair

= 311.541 - 300 = 11.541 mol/min

4 0
3 years ago
What are the factors that influence the power input to the compressor?
Lena [83]

Answer:

option e is correct answer

5 0
2 years ago
(a)Compute the electrical conductivity of a cylindrical silicon specimen 7.0 mm (0.28 in.) diameter and 57 mm (2.25 in.) in leng
igor_vitrenko [27]

Answer:

a) \sigma = 12.2 (Ω-m)^{-1}

b) Resistance = 121.4 Ω

Explanation:

given data:

diameter is 7.0 mm

length 57 mm

current I = 0.25 A

voltage v = 24 v

distance between the probes is 45 mm

electrical conductivity is given as

\sigma = \frac{I l}{V \pi r^2}

\sigma  = \frac{0.25 \times 45\times 10^{-3}}{24 \pi [\frac{7 \times 10^{-3}}{2}]^2}

\sigma = 12.2(Ω-m)^{-1}[/tex]

b)

Resistance = \frac{l}{\sigma A}

                  = \frac{l}{ \sigma \pi r^2}

= \frac{57  \times 10^{-3}}{12.2 \times \pi [\frac{7 \times 10^{-3}}{2}]^2}

Resistance = 121.4 Ω

8 0
3 years ago
A rigid tank with a total volume of 0.05 m3 initially contains a two-phase liquid-vapor mixture of water at a pressure of 15 bar
Westkost [7]

Answer:

a) m_{2} = 0.753\,kg, b) Q_{in} = 2122.963\,kJ

Explanation:

A rigid tank means a storage whose volume is constant. Process is entirely isobaric. Initial and final properties of water are included below:

State 1 - Gas-Vapor Mixture

P = 1500\,kPa

T = 198.29^{\textdegree}C

\nu = 0.02726\,\frac{m^{3}}{kg}

u = 1192.94\,\frac{kJ}{kg}

h_{g} = 2791.0\,\frac{kJ}{kg}

x = 0.2

State 2 - Gas-Vapor Mixture

P = 1500\,kPa

T = 198.29^{\textdegree}C

\nu = 0.06643\,\frac{m^{3}}{kg}

u = 1718.12\,\frac{kJ}{kg}

h_{g} = 2791.0\,\frac{kJ}{kg}

x = 0.5

The model for the rigid tank is created by using the First Law of Thermodynamics:

Q_{in} - (m_{1}-m_{2})\cdot h_{g} = m_{2}\cdot u_{2}-m_{1}\cdot u_{1}

Initial and final masses are:

m_{1} = \frac{V_{1}}{\nu_{1}}

m_{1} = \frac{0.05\,m^{3}}{0.02726\,\frac{m^{3}}{kg} }

m_{1} = 1.834\,kg

m_{2} = \frac{V_{2}}{\nu_{2}}

m_{2} = \frac{0.05\,m^{3}}{0.06643\,\frac{m^{3}}{kg} }

m_{2} = 0.753\,kg

a) The final mass within the tank is:

m_{2} = 0.753\,kg

b) The total amount of heat transfer is:

Q_{in} = m_{2}\cdot u_{2}-m_{1}\cdot u_{1}+ (m_{1}-m_{2})\cdot h_{g}

Q_{in} = (0.753\,kg)\cdot (1718.12\,\frac{kJ}{kg} )- (1.834\,kg)\cdot (1192.94\,\frac{kJ}{kg} ) + (1.081\,kg)\cdot (2791.0\,\frac{kJ}{kg} )

Q_{in} = 2122.963\,kJ

5 0
3 years ago
Explain how the ideal vapour compression refrigerator cycle is an improvement upon the Carnot cycle. Use any two of the problems
Grace [21]
  • answer explanation:
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  • <u>Also, the Carnot cycle is an idealized theoretical model, not a practically achievable engineering goal. It illustrates the essential mechanism by which heat differentials can be turned into mechanical energy and vice versa, it has to ignore the fact of inefficiency in conversion between mechanical energy and heat differential for the sake of illustrating a set of ideal processes. A real refrigeration system has to deal with the question of whether the system is pumping away more heat than it generates itself through inefficiency.</u>
8 0
2 years ago
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