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2 years ago

Elements of parallel computing

Engineering

1 answer:

Leya [2.2K]2 years ago

6 0

$\huge{\orange}\fcolorbox{purple}{cyan}{\bf{\underline{\green{\color{pink}Answer}}}}$

<h3><u>E</u><u>lements of parallel </u><u>computing:</u></h3>

Computer systems organization.
Computing methodologies.
General and reference.
Networks.
Software and its engineering.
Theory of computation.

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Atmospheric pressure is measured to be 14.769 psia. a. What would be the equivalent reading of a water barometer (inches of H20)

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Answer:

(a) water height =408.66 in.

(b) mercury height=30.04 in.

Explanation:

Given: P=14.769 psi ( 1 psi= 6894.76 $\frac{N}{m^2}$ )

we know that $P=\rho\times g\timesh$

where $\rho =Density,g=9.81\frac{m}{s^2}$

h=height.

Given that P=14.769 psi ⇒P= 101828.6 7 $\dfrac{N}{m^2}$

(a) $P=\rho_{w}\times g\times h_{w}$

$\rho_{w}=1000\frac{Kg}{m^3}$

⇒101828.67= $1000\times 9.81\times h_{w}$

$h_{w}$ =10.38 m

So water barometer will read 408.66 in. (1 m=39.37 in)

(b) $P=\rho_{hg}\times g\times h_{hg}$

$\rho_{hg}$ =13600

So 101828.67= $13600\times 9.81\times h_{hg}$

$h_{hg}$ =0.763 m

So mercury barometer will read 30.04 in.

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3 years ago

A 16-lb solid square wooden panel is suspended from a pin support at A and is initially at rest. A 4-lb metal sphere is shot at

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Explanation:

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3 years ago

A four-cylinder, four-stroke internal combustion engine operates at 2800 RPM. The processes within each cylinder are modeled as

Ulleksa [173]

Answer:

1) 287760.4 Hp

2) 18410899.5 kPa

Explanation:

The parameters given are;

p₁ = 14.7 lbf/in² = 101325.9 Pa

v₁ = 0.0196 ft³ = 0.00055501 m³

T₁ = 80°F = 299.8167 K

k = 1.4

Assumptions;

1) Air standard conditions are appropriate

2) There are negligible potential and kinetic energy changes

3) The air behaves as an ideal gas and has constant specific heat capacities of temperature and pressure

1) Process 1 to 2

Isentropic compression

T₂/T₁ = (v₁/v₂)^(1.4 - 1) = 10^0.4

p₂/p₁ = (v₁/v₂)^(1.4)

p₂ = p₁×10^0.4 = 101325.9*10^0.4 = 254519.153 Pa

T₂ = 299.8167*10^0.4 = 753.106 K

p₃ = 1080 lbf/in² = 7,446,338 Pa

Stage 2 to 3 is a constant volume process

p₃/T₃ = p₂/T₂

7,446,338/T₃ = 254519.153/753.106

T₃ = 7,446,338/(254519.153/753.106) = 22033.24 K

T₃/T₄ = (v₁/v₂)^(1.4 - 1) = 10^0.4

T₄ = 22033.24/(10^0.4) = 8771.59 K

The heat supplied, Q₁ = cv(T₃ - T₂) = 0.718*(22033.24 -753.106) = 15279.14 kJ

The heat rejected = cv(T₄ - T₁) = 0.718*(8771.59 - 299.8167) = 6082.73 kJ

W(net) = The heat supplied - The heat rejected = (15279.14 - 6082.73) = 9196.41 kJ

The power = W(net) × RPM/2*1/60 = 9196.41 * 2800/2*1/60 = 214582.9 kW

The power by the engine = 214582.9 kW = 287760.4 Hp

2) The mean effective pressure, MEP = W(net)/(v₁ - v₂)

v₁ = 0.00055501 m³

v₁/v₂ = 10

v₂ = v₁/10 = 0.00055501/10 = 0.000055501

MEP = 9196.41/(0.00055501 - 0.000055501) = 18410899.5 kPa

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3 years ago