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2 years ago

How to comment other people

Engineering

2 answers:

andrew-mc [135]2 years ago

7 0

You go to answer on the bottom then press on whichever one you want to answer

Mekhanik [1.2K]2 years ago

4 0

What you mean?? .!.!.!.!!.

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The specific gravity of a substance that has mass of 10 kg and occupies a volume of 0.02 m^3 is a) 0.5 b) 1.5 c) 2.5 d) 3.5 e) n

zhuklara [117]

Answer:

Specific gravity is 0.5 so option (a) is correct option

Explanation:

We have given mass of substance m =10 kg

Volume $V=0.02m^3$

Density of substance is given by $Density\ d=\frac{mass}{volume}=\frac{10}{0.02}=500kg/m^3$

Specific gravity is given by $specific\ gravity=\frac{density\ of\ substance}{density\ f\ water}=\frac{500}{1000}=0.5$

So option (a) is correct option

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3 years ago

Describe the make-up of an internal combustion engine.<br> Pls answer quickly.

Whitepunk [10]

Answer:

The engine consists of a fixed cylinder and a moving piston. The expanding combustion gases push the piston, which in turn rotates the crankshaft. Ultimately, through a system of gears in the power-train, this motion drives the vehicle's wheels.

Explanation:

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2 years ago

A piston-cylinder device contains 0.8 kg of steam at 300°C and 1 MPa. Steam is cooled at constant pressure until one-half of the

liberstina [14]

The answer & explanation for this question is given in the attachment below.

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3 years ago

Additional scals apply to the

REY [17]

Location of the class depends on satiation

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1 year 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