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3 years ago

7

. Bơm kiểu piston tác dụng đơn có áp suất p=0,64 Mpa và lưu lượng Q=3,5 l/s. Xác định tốc độ quay của trục bơm và công suất của

bơm nếu biết đường kính piston D=150 mm; bán kính tay quay R=60 mm; hiệu suất thể tích của bơm là 0=0,94; hiệu suất chung của bơm b=0,80.

1 answer:

Viktor [21]3 years ago

4 0

Answer:

not understand language

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An ideal gas undergoes two processes: one frictionless and the other not. In both the cases, the gas is initially at 200 ℉ and 1

Zarrin [17]

Answer:

The process which has friction

Explanation:

The entropy is simply the change in the state of the things or the molecules in the system. It is simply the change in the energy of the system with a focus on the atoms in the system. This is also known as the internal energy of the system and is given the symbol, G. The friction contributes to the change in the energy of the system. This is because friction generates another form of energy - that is heat energy. This energy causes the internal temperature id the system to increase. Hence the greater change in the temperature.

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3 years ago

Read 2 more answers

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

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3 years ago

on the same scale for stress, the tensile true stress-true strain curve is higher than the engineeringstress-engineering strain

Bess [88]

Answer:

The condition does not hold for a compression test

Explanation:

For a compression test the engineering stress - strain curve is higher than the actual stress-strain curve and this is because the force needed in compression is higher than the force needed during Tension. The higher the force in compression leads to increase in the area therefore for the same scale of stress the there is more stress on the Engineering curve making it higher than the actual curve.

<em>Hence the condition of : on the same scale for stress, the tensile true stress-true strain curve is higher than the engineering stress-engineering strain curve.</em><em> </em>does not hold for compression test

5 0

2 years ago

For a steel alloy it has been determined that a carburizing heat treatment of 15 h duration will raise the carbon concentration

Free_Kalibri [48]

Answer:

135 hour

Explanation:

It is given that a carburizing heat treatment of 15 hour will raise the carbon concentration by 0.35 wt% at a point of 2 mm from the surface.

We have to find the time necessary to achieve the same concentration at a 6 mm position.

we know that $\frac{x_1^2}{Dt}=constant$ where x is distance and t is time .As the temperature is constant so D will be also constant

So $\frac{x_1^2}{t}=constant$

then $\frac{x_1^2}{t_1}=\frac{x_2^2}{t_2}$ we have given $x_1=2 mm\ ,t_1=15 hour\ ,x_2=6\ mm$ and we have to find $t_2$ putting all these value in equation

$\frac{2^2}{15}=\frac{6^2}{t_2}$

so $t_2=135\ hour$

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3 years ago

Thin film deposition is a process where: a)-elemental, alloy, or compound thin films are deposited onto a bulk substrate! b)-Pho

marshall27 [118]

Answer:

(A) elemental, alloy, or compound thin films are deposited on to a bulk substrate

Explanation:

In film deposition there is process of depositing of material in form of thin films whose size varies between the nano meters to micrometers onto a surface. The material can be a single element a alloy or a compound.

This technology is very useful in semiconductor industries, in solar panels in CD drives etc

so from above discussion it is clear that option (a) will be the correct answer

8 0

3 years ago