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alexdok [17]
3 years ago
15

1. A compressed air system consists of a compressor and receiver, 1500 ft of 4-in pipe, two gate valves, six standard elbows, an

d a manifold. Four rock drills requiring 200 cu-ft/min each are connected to the manifold by 1.25-in hoses 100 ft long. Pressure drop in the manifold is 3 psig and line leakage is 5%. Determine the pressure at the drill when all four drills are operating simultaneously and receiver pressure is 100 psig.
Engineering
1 answer:
Pani-rosa [81]3 years ago
4 0

Answer:

Answer for the question:

1. A compressed air system consists of a compressor and receiver, 1500 ft of 4-in pipe, two gate valves, six standard elbows, and a manifold. Four rock drills requiring 200 cu-ft/min each are connected to the manifold by 1.25-in hoses 100 ft long. Pressure drop in the manifold is 3 psig and line leakage is 5%. Determine the pressure at the drill when all four drills are operating simultaneously and receiver pressure is 100 psig.

is explained in the attachment.

Explanation:

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Assume a function requires 20 lines of machine code and will be called 10 times in the main program. You can choose to implement
Volgvan

Answer:

"Macro Instruction"

Explanation:

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Which option distinguishes the type of software the team should use in the following scenario?
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A well insulated rigid tank contains 4 kg of argon gas at 450 kPa and 30 C. A valve is opened, allowing the argon to escape unti
natima [27]

Answer:

Final mass of Argon=  2.46 kg

Explanation:

Initial mass of Argon gas ( M1 ) = 4 kg

P1 = 450 kPa

T1 = 30°C = 303 K

P2 = 200 kPa

k ( specific heat ratio of Argon ) = 1.667

assuming a reversible adiabatic process

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Applying ideal gas equation ( PV = mRT )

P₁V / P₂V = m₁ RT₁ / m₂ RT₂

hence : m2 = P₂T₁ / P₁T₂ * m₁

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<em>Note: Calculation for T2 is attached below</em>

5 0
2 years ago
Calculate the viscosity(dynamic) and kinematic viscosity of airwhen
nikitadnepr [17]

Answer:

(a) dynamic viscosity = 1.812\times 10^{-5}Pa-sec

(b) kinematic viscosity = 1.4732\times 10^{-5}m^2/sec

Explanation:

We have given temperature T = 288.15 K

Density d=1.23kg/m^3

According to Sutherland's Formula  dynamic viscosity is given by

{\mu} = {\mu}_0 \frac {T_0+C} {T + C} \left (\frac {T} {T_0} \right )^{3/2}, here

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\mu _0= reference viscosity in(Pa·s) at reference temperature T0,

T = input temperature in kelvin,

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C = Sutherland's constant for the gaseous material in question here C =120

\mu _0=4\pi \times 10^{-7}

T_0 = 291.15

\mu =4\pi \times 10^{-7}\times \frac{291.15+120}{285.15+120}\times \left ( \frac{288.15}{291.15} \right )^{\frac{3}{2}}=1.812\times 10^{-5}Pa-swhen T = 288.15 K

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3 0
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