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Schach [20]
2 years ago
12

A spherical balloon with a diameter of 9 m is filled with helium at 20°C and 200 kPa. Determine the mole number and the mass of

the helium in the balloon.
Engineering
1 answer:
SIZIF [17.4K]2 years ago
3 0

Answer:

<em>number of mole is 31342.36 moles</em>

<em>mass is 125.369 kg</em>

<em></em>

Explanation:

Diameter of the spherical balloon d = 9 m

radius r = d/2 = 9/2 = 4.5 m

The volume pf the sphere balloon ca be calculated from

V = \frac{4}{3} \pi r^3

V = \frac{4}{3}* 3.142* 4.5^3 = 381.75 m^3

Temperature of the gas T = 20 °C = 20 + 273 = 293 K

Pressure of the helium gas = 200 kPa = 200 x 10^3 Pa

number of moles n = ?

Using

PV = nRT

where

P is the pressure of the gas

V is the volume of the gas

n is the mole number of the gas

R is the gas constant = 8.314 m^3⋅Pa⋅K^−1⋅mol^−1

T is the temperature of the gas (must be converted to kelvin K)

substituting values, we have

200 x 10^3 x 381.75 = n x 8.314 x 293

number of moles n  = 76350000/2436 = <em>31342.36 moles</em>

We recall that n = m/MM

or m = n x MM

where

n is the number of moles

m is the mass of the gas

MM is the molar mass of the gas

For helium, the molar mass = 4 g/mol

substituting values, we have

m = 31342.36 x 4

m = 125369.44 g

m =<em> 125.369 kg</em>

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Outline an algorithm in **pseudo code** for checking whether an array H[1..n] is a heap and determine its time efficiency.
svlad2 [7]

Answer:

Condition to break: H[j] \geq max {H[2j] , H[2j+1]}

Efficiency: O(n).

Explanation:

Previous concepts

Heap algorithm is used to create all the possible permutations with K possible objects. Was created by B. R Heap in 1963.

Parental dominance condition represent a condition that is satisfied when the parent element is greater than his children.

Solution to the problem

We assume that we have an array H of size n for the algorithm.

It's important on this case analyze the parental dominance condition in order to the algorithm can work and construc a heap.

For this case we can set a counter j =1,2,... [n/2] (We just check until n/2 since in order to create a heap we need to satisfy minimum n/2 possible comparisionsand we need to check this:Break condition: [tex]H[j] \geq max {H[2j] , H[2j+1]}

And we just need to check on the array the last condition and if is not satisfied for any value of the counter j we need to stop the algorithm and the array would not a heap. Otherwise if we satisfy the condition for each j =1,2,.....,[n/2]p then we will have a heap.

On this case this algorithm needs to compare 2*(n/2) times the values and the efficiency is given by O(n).

3 0
2 years ago
In a heat-treating process, a 1-kg metal part, initially at 1075 K, is quenched in a closed tank containing 100 kg of water, ini
storchak [24]

Answer:

attached below

Explanation:

8 0
2 years ago
Bananas are to be cooled from 28°C to 12°C at a rate of 1140 kg/h by a refrigerator that operates on a vapor-compression refrige
Lera25 [3.4K]

Answer:

A) COP = \frac{16.97}{9.8} = 1.731

B) P_{IN} = 0.4763

C) Second law efficiency 4.85%

exergy destruction for the cycle = 9.3237 kW

Explanation:

Given data:

T_1 = 28 degree celcius

T_2 = 12 degree celcius

\dot m = 1140 kg/h

Power to refrigerator = 9.8 kW

Cp = 3.35 kJ/kg degree C

A) Q = \dot m Cp \Delta T

        = 1140 \times 3.35\times (28-12) = 61,104 kJ/h

Q_{abs} = 61,104 kJ/h = 16.97 kJ/sec

COP = \frac{16.97}{9.8} = 1.731

b)

COP ∝ \frac{1}{P_{in}}

P_{in} wil be max when COP maximum

taking surrounding temperature T_H = 20 degree celcius

COP_{max} = \frac{T_L}{T_H- T_L} = \frac{285}{293 - 285} = 35.625

we know that

COP = \frac{heat\ obsorbed}{P_{in}}

P_{IN} = \frac{16.97}{35.62} = 0.4763

c) second law efficiency

\eta_{11} = \frac{COP_R}{(COP)_max} = \frac{1.731}{35.625} = 4.85\%

exergy destruction os given as X = W_{IN} - X_{Q2}

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8 0
2 years ago
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andriy [413]

Answer:

Apalancamiento.

Explanation:

El apalancamiento es el uso de dinero prestado (deuda) para aumentar el rendimiento esperado del capital. El apalancamiento se mide como la relación entre la deuda que devenga intereses y los activos totales. Cuanto mayor sea la deuda que devenga intereses, mayor será el apalancamiento financiero o "aceleración". Esto puede tener un efecto positivo o negativo.

Los costos por intereses de este capital de préstamo suelen ser fijos y se deducen de los ingresos. Un préstamo permite que una organización genere más ingresos sin un aumento necesario en el capital. Como no es necesario recaudar ni mantener capital social adicional, no se requieren pagos de dividendos adicionales (que no se pueden deducir de las ganancias). Sin embargo, un alto apalancamiento puede ser beneficioso durante los tiempos de auge, pero puede conducir a serios problemas de flujo de efectivo durante una recesión, ya que es posible que no haya suficientes retornos para cubrir mayores costos de intereses y obligaciones de reembolso.

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1 year ago
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