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ludmilkaskok [199]

3 years ago

14

(Specific weight) A 1-ft-diameter cylindrical tank that is 5 ft long weighs 125 lb and is filled with a liquid having a specific

weight of 66.4 lb/ft3. Determine the vertical force required to give the tank an upward acceleration of 10.7 ft/s2.

1 answer:

lina2011 [118]3 years ago

5 0

Answer:

The answer to the question is 514.17 lbf

Explanation:

Volume of cylindrical tank = πr²h = 3.92699 ft³

Weight of tank = 125 lb

Specific weight of content = 66.4 lb/ft³

Mass of content = 66.4×3.92699 = 260.752 lb

Total mass = 260.752 + 125 = 385.75 lb = 174.97 kg

=Weight = mass * acceleration = 174.97 *9.81 = 1716.497 N

To have an acceleration of 10.7 ft/s² = 3.261 m/s²

we have F = m*a = 174.97*(9.81+3.261) = 2287.15 N = 514.17 lbf

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Air enters a compressor steadily at the ambient conditions of 100 kPa and 22°C and leaves at 800 kPa. Heat is lost from the comp

telo118 [61]

Answer:

a) 358.8K

b) 181.1 kJ/kg.K

c) 0.0068 kJ/kg.K

Explanation:

Given:

P1 = 100kPa

P2= 800kPa

T1 = 22°C = 22+273 = 295K

q_out = 120 kJ/kg

∆S_air = 0.40 kJ/kg.k

T2 =??

a) Using the formula for change in entropy of air, we have:

∆S_air = $c_p In \frac{T_2}{T_1} - Rln \frac{P_2}{P_1}$

Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K

Solving, we have:

[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]

$-0.40= 1.005(ln T_2 - 5.68697)- 0.5968$

Solving for T2 we have:

$T_2 = 5.8828$

Taking the exponential on the equation (both sides), we have:

[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]

b) Work input to compressor:

$w_in = c_p(T_2 - T_1)+q_out$

$w_in = 1.005(358.8 - 295)+120$

= 184.1 kJ/kg

c) Entropy genered during this process, we use the expression;

Egen = ∆Eair + ∆Es

Where; Egen = generated entropy

∆Eair = Entropy change of air in compressor

∆Es = Entropy change in surrounding.

We need to first find ∆Es, since it is unknown.

Therefore ∆Es = $\frac{q_out}{T_1}$

$\frac{120kJ/kg.k}{295K}$

∆Es = 0.4068kJ/kg.k

Hence, entropy generated, Egen will be calculated as:

= -0.40 kJ/kg.K + 0.40608kJ/kg.K

= 0.0068kJ/kg.k

3 0

3 years ago

The air in a room has a pressure of 1 atm, a dry-bulb temperature of 24C, and a wet-bulb temperature of 17C. Using the psychrome

TEA [102]

Answer:

(a) Relative Humidity = 48%,

Specific humidity = 0.0095

(b) Enthalpy = 65 KJ/Kg of dry sir

Specific volume = 0.86 m^3/Kg of dry air

(c/d) 12.78 degree C

(e) Specific volume = 0.86 m^3/Kg of dry air

8 0

3 years ago

Should the ship breaking business continue why or why not?

Dmitry [639]

Answer:

Ship-breaking or ship demolition is a type of ship disposal involving the breaking up of ships for either a source of parts, which can be sold for re-use, or for the extraction of raw materials, chiefly scrap. It may also be known as ship dismantling, ship cracking, or ship recycling. Modern ships have a lifespan of 25 to 30 years before corrosion, metal fatigue and a lack of parts render them uneconomical to operate.[1] Ship-breaking allows the materials from the ship, especially steel, to be recycled and made into new products. This lowers the demand for mined iron ore and reduces energy use in the steel making process. Fixtures and other equipment on board the vessels can also be reused. While ship-breaking is sustainable, there are concerns about the use of poorer countries without stringent environmental legislation. It is also labor-intensive, and considered one of the world's most dangerous industries.[2]

In 2012, roughly 1,250 ocean ships were broken down, and their average age was 26 years.[3][4] In 2013, the world total of demolished ships amounted to 29,052,000 tonnes, 92% of which were demolished in Asia. As of January 2020, India has the largest global share at 30%;[5] followed by Bangladesh, China and Pakistan.[6] Alang, India currently has the world's largest ship graveyard,[5] followed by Chittagong Ship Breaking Yard in Bangladesh and Gadani in Pakistan.[6]

The largest sources of ships are states of China, Greece and Germany respectively, although there is a greater variation in the source of carriers versus their disposal.[7] The ship-breaking yards of India, Bangladesh, China and Pakistan employ 225,000 workers as well as providing many indirect jobs. In Bangladesh, the recycled steel covers 20% of the country's needs and in India it is almost 10%.[8]

As an alternative to ship-breaking, ships may be sunk to create artificial reefs after legally-mandated removal of hazardous materials, or sunk in deep ocean waters. Storage is a viable temporary option, whether on land or afloat, though all ships will be eventually scrapped, sunk, or preserved for museums.

6 0

3 years ago

At 45° latitude, the gravitational acceleration as a function of elevation z above sea level is given by g = a − bz , where a =

Ahat [919]

Answer:

8861.75 m approximately 8862 m

Explanation:

We need to remember Newton's 2nd Law which says that the force experienced by an object is proportional to his acceleration and that the constant of proportionality between those two vectors correspond to the mass of the object.

$F=ma$ for the weight of an object (which is a force) we have that the acceleration experienced by that object is equal to the gravitational acceleration, obtaining that $W = mg$

For simplicity we work with $g =9.807$ $\frac{m}{s^{2}}$ despiting the effect of the height above sea level. In this problem, we've been asked by the height above sea level that makes the weight of an object 0.30% more lighter.

In accord with the formula $g = a-bz$ the "normal" or "standard" weight of an object is given by $W = mg = ma$ when $z = 0$ , so we need to find the value of $z$ that makes $W = m(a-bz) = 0.997ma$ meaning that the original weight decrease by a 0.30%, so now we operate...

$m(a-bz) = 0.997ma$ now we group like terms on the same sides $ma(1-0.997) = mbz$ we cancel equal tems on both sides and obtain that $z = \frac{a}{b} (0.003) = \frac{9.807 \frac{m}{s^{2} } }{3.32*10^{-6} s^{-2} } (0.003) = 8861.75 m$

7 0

3 years ago

Steam enters a turbine steadily at 7 MPa and 600°C with a velocity of 60 m/s and leaves at 25 kPa with a quality of 95 percent.

Rufina [12.5K]

Answer:

a) $\dot m = 16.168\,\frac{kg}{s}$ , b) $v_{out} = 680.590\,\frac{m}{s}$ , c) $\dot W_{out} = 18276.307\,kW$

Explanation:

A turbine is a steady-state devices which transforms fluid energy into mechanical energy and is modelled after the Principle of Mass Conservation and First Law of Thermodynamics, whose expressions are described hereafter:

Mass Balance

$\frac{v_{in}\cdot A_{in}}{\nu_{in}} - \frac{v_{out}\cdot A_{out}}{\nu_{out}} = 0$

Energy Balance

$-q_{loss} - w_{out} + h_{in} - h_{out} = 0$

Specific volumes and enthalpies are obtained from property tables for steam:

Inlet (Superheated Steam)

$\nu_{in} = 0.055665\,\frac{m^{3}}{kg}$

$h_{in} = 3650.6\,\frac{kJ}{kg}$

Outlet (Liquid-Vapor Mix)

$\nu_{out} = 5.89328\,\frac{m^{3}}{kg}$

$h_{out} = 2500.2\,\frac{kJ}{kg}$

a) The mass flow rate of the steam is:

$\dot m = \frac{v_{in}\cdot A_{in}}{\nu_{in}}$

$\dot m = \frac{\left(60\,\frac{m}{s} \right)\cdot (0.015\,m^{2})}{0.055665\,\frac{m^{3}}{kg} }$

$\dot m = 16.168\,\frac{kg}{s}$

b) The exit velocity of steam is:

$\dot m = \frac{v_{out}\cdot A_{out}}{\nu_{out}}$

$v_{out} = \frac{\dot m \cdot \nu_{out}}{A_{out}}$

$v_{out} = \frac{\left(16.168\,\frac{kg}{s} \right)\cdot \left(5.89328\,\frac{m^{3}}{kg} \right)}{0.14\,m^{2}}$

$v_{out} = 680.590\,\frac{m}{s}$

c) The power output of the steam turbine is:

$\dot W_{out} = \dot m \cdot (-q_{loss} + h_{in}-h_{out})$

$\dot W_{out} = \left(16.168\,\frac{kg}{s} \right)\cdot \left(-20\,\frac{kJ}{kg} + 3650.6\,\frac{kJ}{kg} - 2500.2\,\frac{kJ}{kg}\right)$

$\dot W_{out} = 18276.307\,kW$

6 0

2 years ago