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3 years ago

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A 280 km long pipeline connects two pumping stations. It is desired to pump 0.56 m3/s of oil through a 0.62 m diameter line, the

discharge station is 250 m lower in elevation than the upstream station, and the discharge pressure is to be maintained at 300 kPa. The oil has a dynamic viscosity of 3.645 x 10-3 Pa·s and a density of 810 kg/m3. The pipe is constructed of commercial steel and the inlet pressure may be taken as 1 atmosphere. Considering the mechanical-energy friction loss Ff, determine the power required to pump the oil if the power efficiency is 70%. Develop a sketch of the piping system considered.
Ignore energy losses due to contraction, expansion & 90° bends!!!

1 answer:

Arlecino [84]3 years ago

8 0

Can u answer myn and if u do I’ll do urs

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A 90-hp (shaft output) electric car is powered by an electric motor mounted in the engine compartment. If the motor has an avera

Ostrovityanka [42]

Answer:

The rate of heat supply is 8.901 horse-power.

Explanation:

From Thermodynamics energy efficiency of the electric car ( $\eta$ ), no unit, is the ratio of translational mechanical power ( $\dot E_{out}$ ), measured in horse power, to electric power ( $\dot E_{in}$ ), measured in horse-power. The rate of heat supply ( $\dot E_{l}$ ), measured in horse-power, by the motor to the engine compartment at full load is difference between electric energy and translational mechanical energy. That is:

$\eta = \frac{\dot E_{out}}{\dot E_{in}}$ (1)

$\dot E_{l} = \dot E_{in}-\dot E_{out}$ (2)

$\dot E_{l} = \left(\frac{1}{\eta}-1\right)\cdot \dot E_{out}$ (3)

If we know that $\eta = 0.91$ and $\dot E_{out} = 90\,hp$ , then rate of heat supply is:

$\dot E_{l} = \left(\frac{1}{0.91}-1 \right)\cdot (90\,hp)$

$\dot E_{l} = 8.901\,hp$

The rate of heat supply is 8.901 horse-power.

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3 years ago

Why would a Military Base be considered a Planned Community?

ira [324]

What he said is right ^^^

5 0

3 years ago

Read 2 more answers

What is the multiplicity of the root x = -4?

just olya [345]

Answer:

See explanation

Explanation:

The multiplicity of a root of a polynomial equation is the number of times the root repeats.

Let $x=a$ be the root of $f(x)$ , then;

$(x-a)^1=0$ has a multiplicity of 1.

$(x-a)^2=0$ has a multiplicity of 2.

$(x-a)^3=0$ has a multiplicity of 3.

$(x-a)^m=0$ has a multiplicity of m, where m is a positive integer.

We were given the root $x=-4$ .

If $f(x)=(x+4)(x-6)^3$

Then the multiplicity of the root $x=-4$ is 1.

If $f(x)=(x+4)^4(x-6)^3$

Then the multiplicity of the root $x=-4$ is 4.

Since the polynomial function or equation is not given in the question, we cannot determine the multiplicity of this root.

But I hope with this explanation, you can refer to the original(complete) question and choose the correct answer.

8 0

3 years ago

Air expands through a turbine from 10 bar, 900 K to 1 bar, 500 K. The inlet velocity is small compared to the exit velocity of 1

Elis [28]

Answer:

- the mass flow rate of air is 7.53 kg/s

- the exit area is 0.108 m²

Explanation:

Given the data in the question;

lets take a look at the steady state energy equation;

m" = W" $_{cv$ / [ (h₁ - h₂ ) - $\frac{V_2^2}{2}$ ]

Now at;

T₁ = 900K, h₁ = 932.93 k³/kg

T₂ = 500 K, h₂ = 503.02 k³/kg

so we substitute, in our given values

m" = [ 3200 kW × $\frac{1\frac{k^3}{s} }{1kW}$ ] / [ (932.93 - 503.02 )k³/kg - $\frac{100^2\frac{m^2}{s^2} }{2}$ | $\frac{ln}{kg\frac{m}{s^2} }$ || $\frac{1kJ}{10^3N-m}$ | ]

m" = 7.53 kg/s

Therefore, the mass flow rate of air is 7.53 kg/s

now, Exit area A₂ = v₂m" / V₂

we know that; pv = RT

so

A₂ = RT₂m" / P₂V₂

so we substitute

A₂ = {[ $(\frac{8.314}{28.97}\frac{k^3}{kg.K})$ × $500 K$ × $(7.54 kg/s)$ ] / [(1 bar)(100 m/s )]} | $\frac{1 bar}{10N/m^2}$ || $10^3N.m/1k^3$

A₂ = 0.108 m²

Therefore, the exit area is 0.108 m²

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3 years ago

Select the correct answer.

babunello [35]

Answer:

The information he needs next is;

B. Volume of paint needed per unit area

Explanation:

The operation Andy wants to perform = To apply paint on the walls of the house

The information Andy knows = The area of all walls in the house

The total volume of paint needed = The total area of the walls × The volume of paint needed for each unit area

Therefore, the information required is the volume of paint needed per unit area.

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3 years ago