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3 years ago

Why did fprtmiu78t7ty87uhyu

Engineering

1 answer:

alekssr [168]3 years ago

6 0

Yea why did fprtmiu78t7ty87uhyu

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QUESTION:

svet-max [94.6K]

OA bloom is smaller than a bar

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3 years ago

Build a 32-bit accumulator circuit. The circuit features a control signal inc and enable input en. If en is 1 and inc is 1, the

Alex_Xolod [135]

Sorry need.points I'm new

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3 years ago

Oil with a density of 800 kg/m3 is pumped from a pressure of 0.6 bar to a pressure of 1.4 bar, and the outlet is 3 m above the i

Naddik [55]

Answer:

23.3808 kW

20.7088 kW

Explanation:

ρ = Density of oil = 800 kg/m³

P₁ = Initial Pressure = 0.6 bar

P₂ = Final Pressure = 1.4 bar

Q = Volumetric flow rate = 0.2 m³/s

A₁ = Area of inlet = 0.06 m²

A₂ = Area of outlet = 0.03 m²

Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s

Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s

Height between inlet and outlet = z₂ - z₁ = 3m

Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation

$\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m$

Work done by pump

$W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 14.896\\\Rightarrow W_{p}=23380.8\ W$

∴ Power input to the pump 23.3808 kW

Now neglecting kinetic energy

$h=\frac{P_2-P_1}{\rho g}+z_2-z_1\\\Righarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+3\\\Righarrow h=13.19\ m\\$

Work done by pump

$W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 13.193\\\Rightarrow W_{p}=20708.8\ W$

∴ Power input to the pump 20.7088 kW

6 0

3 years ago

The cube measures 3.0-ft on all sides and has a density of 3.1 slugs/ft3. How much does it weigh?

kodGreya [7K]

Answer:

W = 2695.14 lb

Explanation:

given,

side of cube = 3 ft

density of the cube = 3.1 slugs/ft³

we know,

$density = \dfrac{mass}{volume}$

mass = density x volume

volume = 3³ = 27 ft³

mass = 3.1 x 27

m = 83.7 slugs.

weight calculation

converting mass from slug to pound

weight of 1 slug is equal to 32.2 lb

now,

weight of the cube is equal to

W = 83.7 slugs x 32.2 lb/slug

W = 2695.14 lb

hence, weight is equal to W = 2695.14 lb

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