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tatiyna
3 years ago
11

6) A deep underground cavern Contains 980 cuft

Engineering
1 answer:
Elza [17]3 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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What is the maximum thermal efficiency possible for a power cycle operating between 600P'c and 110°C? a). 47% b). 56% c). 63% d)
Anastasy [175]

Answer:

(b) 56%

Explanation:

the maximum thermal efficiency is possible only when power cycle is reversible in nature and when power cycle is reversible in nature the thermal efficiency depends on the temperature

here we have given T₁ (Higher temperature)= 600+273=873

lower temperature T₂=110+273=383

Efficiency of power cycle is given by =1-\frac{T2}{T1}

=1-\frac{383}{873}

=1-0.43871

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=56%

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Why do giant stars become planetary nebulas while supergiant stars become supernovas when their nuclear fusion slows and is over
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2 years ago
A gas mixture containing 3 moles CO2, 5 moles H2 and 1 mole water is undergoing the following reactions CO2+3H2 →cH3OH + H2O Dev
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3 years ago
A thin-walled cylinder of average radius 50 mm, and wall thickness 1.0 mm is of length 500 mm, and is built into a wall at one e
kotykmax [81]

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Explanation:

7 0
3 years ago
The wall shear stress in a fully developed flow portion of a 12-in.-diameter pipe carrying water is 2.00 lb/ft^2. Determine the
soldier1979 [14.2K]

Answer:

a) -8 lb / ft^3

b) -70.4 lb / ft^3

c) 54.4 lb / ft^3

Explanation:

Given:

- Diameter of pipe D = 12 in

- Shear stress t = 2.0 lb/ft^2

- y = 62.4 lb / ft^3

Find pressure gradient dP / dx when:

a) x is in horizontal flow direction

b) Vertical flow up

c) vertical flow down

Solution:

- dP / dx as function of shear stress and radial distance r:

                      (dP - y*L*sin(Q))/ L = 2*t / r

                      dP / L - y*sin(Q) = 2*t / r

Where            dP / L = - dP/dx,

                      dP / dx = -2*t / r - y*sin(Q)

Where            r = D /2 ,

                      dP / dx = -4*t / D - y*sin(Q)

a) Horizontal Pipe Q = 0

Hence,           dP / dx = -4*2 / 1 - 62.4*sin(0)

                      dP / dx = -8 + 0

                      dP/dx   = -8 lb / ft^3

b) Vertical pipe flow up Q = pi/2

Hence,           dP / dx = -4*2 / 1 - 62.4*sin(pi/2)

                      dP / dx = 8 - 62.4

                      dP/dx   = -70.4 lb / ft^3

c) Vertical flow down Q = -pi/2

Hence,           dP / dx = -4*2 / 1 - 62.4*sin(-pi/2)

                      dP / dx = -8 + 62.4

                      dP/dx   = 54.4 lb / ft^3                      

7 0
3 years ago
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