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2 years ago

6) A deep underground cavern Contains 980 cuft

Engineering

1 answer:

Elza [17]2 years ago

3 0

Answer:

15625 moles of methane is present in this gas deposit

Explanation:

As we know,

PV = nRT

P = Pressure = 230 psia = 1585.79 kPA

V = Volume = 980 cuft = 27750.5 Liters

n = number of moles

R = ideal gas constant = 8.315

T = Temperature = 150°F = 338.706 Kelvin

Substituting the given values, we get -

1585.79 kPA * 27750.5 Liters = n * 8.315 * 338.706 Kelvin

n = (1585.79*27750.5)/(8.315 * 338.706) = 15625

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3. Air at 1 atm and 20 0 C flows tangentially on both sides of a smooth flat plate of width W=10 ft and length L=10 ft in the di

8_murik_8 [283]

Answer:

ExplanationAir at 1 atm and 20°C flows tangentially on both sides of a thin, smooth flat plate of ... ... Smooth Flat Plate Of Width W = 10 Ft, And Of Length L 3 Ft In The Direction Of The Flow. The Velocity Outside The Boundary Layer Is Constant At 20 Ft/s.

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3 years ago

) Assuming different AM regulations; the receiver is using mixer with subtracting format. The frequency selectivity ratio is app

Zarrin [17]

Answer:

$F=710KHZ$

Explanation:

From the question we are told that:

Frequency selectivity ratio $R=10$

AM range 750 kHz to 2600 kHz

Therefore Bandwidth is

$B=2100-680$

$B=1420KHZ$

Generally the equation for The intermediate frequency is mathematically given by

Intermediate frequency=\frac{Bandwidth}{2}

$F=\frac{B}{2}$

$F=\frac{1420}{2}$

$F=710KHZ$

8 0

3 years ago

Consider a very long, cylindrical fin. The temperature of the fin at the tip and base are 25 °C and 50 °C, respectively. The dia

Mrrafil [7]

Answer:

The fin temperature in °C at a distance of 10 cm from the base = 33.78°C

Explanation:

The following assumptions will be made to solve this problem

- The heat transfer coefficient does not change with the time or distance.

- The temperature of the fins varies just in only one direction.

The temperature of the fin at x = 10 cm = 0.10 m from the base can be calculated from the temperature variation with distance formula for a very long fin.

(T - T∞) = (T₀ - T∞)e⁻ᵐˣ

T = T(x) = temperature at any point along the fin

T∞ = temperature at the tip of the fin = ambient temperature = 25°C

T₀ = temperature at the base of thw fin = 50°C

x = any distance along the length of the fin from the base of the fin = 0.1 m

m = √(hP/KA)

h = Heat transfer coefficient = 123 W/m².K

P = perimeter in contact with the base = πD = π × 0.03 = 0.0943 m

K = thermal conductivity = 150 W/m.K

A = surface area in contact with the base = πD²/4 = π(0.03)²/4 = 0.0007071 m²

m = √(123 × 0.0943)/(150 × 0.0007071)

m = 10.46

mx = 10.46 × 0.1 = 1.046

(T - 25) = (50 - 25) e⁻¹•⁰⁴⁶

T = 25 + 25 e⁻¹•⁰⁴⁶ = 25 + 8.78 = 33.78°C

8 0

3 years ago

Careful planning will save time, __________, and energy while ensuring the production of a high quality product.

Svet_ta [14]

Juicers nb 345676 at that rate it will be amazing

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2 years ago

If the old radiator is replaced with a new one that has longer tubes made of the same material and same thickness as those in th

Nookie1986 [14]

Answer: hello some parts of your question is missing attached below is the missing information

The radiator of a car is a type of heat exchanger. Hot fluid coming from the car engine, called the coolant, flows through aluminum radiator tubes of thickness d that release heat to the outside air by conduction. The average temperature gradient between the coolant and the outside air is about 130 K/mm . The term ΔT/d is called the temperature gradient which is the temperature difference ΔT between coolant inside and the air outside per unit thickness of tube

answer : Total surface area = 3/2 * area of old radiator

Explanation:

we will use this relation

K = $\frac{Qd }{A* change in T }$

change in T = ΔT

therefore New Area ( A ) = 3/2 * area of old radiator

Given that the thermal conductivity is the same in the new and old radiators

3 0

3 years ago