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Liono4ka [1.6K]

4 years ago

5

An irreversible heat pump and a Carnot heat pump operate between the same two thermal reservoirs. Which heat pump has higher COP

?

1 answer:

Shkiper50 [21]4 years ago

6 0

Answer:

Carnot heat pump

Explanation:

Carnot heat pump is an ideal heat pump in which all processes are reversible and that consume minimum amount of work to and produces maximum amount of heating effect compare to all real engine.And that is why COP of Carnot heat pump is more as compare to real heat pump.

All real heat pump are not perfectly reversible heat pump So this is also called irreversible heat pump .Due to irreversibility the COP of irreversible heat pump is always less than the COP of Carnot heat pump.

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Based on the hardness values determined in Part 1, what is the tensile strength (in MPa) for each of the alloys?

tigry1 [53]

Answer:

a. 115

b. 135

c. 220

d. 185

Explanation:

Spheriodite is microscopic constituents in some steels which is composed of spherically shaped cementide particle. It is most ductile and softest type of steel. Pearlite is two phased lamellar compose of alternating layer of ferrite and cementite. It is hard and strong but not tough. It is applied on cutting tools like chopper, blades and knives.

7 0

3 years ago

Water flows in a tube that has a diameter of D= 0.1 m. Determine the Reynolds number if the average velocity is 10 diameters per

Cloud [144]

Answer:

a) $Re_{D} = 111896.745$ , b) $Re_{D} = 1.119\times 10^{-7}$

Explanation:

a) The Reynolds number for the water flowing in a circular tube is:

$Re_{D} = \frac{\rho\cdot v\cdot D}{\mu}$

Let assume that density and dynamic viscosity at 25 °C are $997\,\frac{kg}{m^{3}}$ $0.891\times 10^{-3}\,\frac{kg}{m\cdot s}$ , respectively. Then:

$Re_{D}=\frac{(997\,\frac{kg}{m^{3}} )\cdot (1\,\frac{m}{s} )\cdot (0.1\,m)}{0.891\times 10^{-3}\,\frac{kg}{m\cdot s} }$

$Re_{D} = 111896.745$

b) The result is:

$Re_{D}=\frac{(997\,\frac{kg}{m^{3}} )\cdot (10^{-6}\,\frac{m}{s} )\cdot (10^{-7}\,m)}{0.891\times 10^{-3}\,\frac{kg}{m\cdot s} }$

$Re_{D} = 1.119\times 10^{-7}$

6 0

4 years ago

Please answer fast. With full step by step solution.

lina2011 [118]

Let <em>f(z)</em> = (4<em>z </em>² + 2<em>z</em>) / (2<em>z </em>² - 3<em>z</em> + 1).

First, carry out the division:

<em>f(z)</em> = 2 + (8<em>z</em> - 2) / (2<em>z </em>² - 3<em>z</em> + 1)

Observe that

2<em>z </em>² - 3<em>z</em> + 1 = (2<em>z</em> - 1) (<em>z</em> - 1)

so you can separate the rational part of <em>f(z)</em> into partial fractions. We have

(8<em>z</em> - 2) / (2<em>z </em>² - 3<em>z</em> + 1) = <em>a</em> / (2<em>z</em> - 1) + <em>b</em> / (<em>z</em> - 1)

8<em>z</em> - 2 = <em>a</em> (<em>z</em> - 1) + <em>b</em> (2<em>z</em> - 1)

8<em>z</em> - 2 = (<em>a</em> + 2<em>b</em>) <em>z</em> - (<em>a</em> + <em>b</em>)

so that <em>a</em> + 2<em>b</em> = 8 and <em>a</em> + <em>b</em> = 2, yielding <em>a</em> = -4 and <em>b</em> = 6.

So we have

<em>f(z)</em> = 2 - 4 / (2<em>z</em> - 1) + 6 / (<em>z</em> - 1)

or

<em>f(z)</em> = 2 - (2/<em>z</em>) (1 / (1 - 1/(2<em>z</em>))) + (6/<em>z</em>) (1 / (1 - 1/<em>z</em>))

Recall that for |<em>z</em>| < 1, we have

$\displaystyle\frac1{1-z}=\sum_{n=0}^\infty z^n$

Replace <em>z</em> with 1/<em>z</em> to get

$\displaystyle\frac1{1-\frac1z}=\sum_{n=0}^\infty z^{-n}$

so that by substitution, we can write

$\displaystyle f(z) = 2 - \frac2z \sum_{n=0}^\infty (2z)^{-n} + \frac6z \sum_{n=0}^\infty z^{-n}$

Now condense <em>f(z)</em> into one series:

$\displaystyle f(z) = 2 - \sum_{n=0}^\infty 2^{-n+1} z^{-(n+1)} + 6 \sum_{n=0}^\infty z^{-n-1}$

$\displaystyle f(z) = 2 - \sum_{n=0}^\infty \left(6+2^{-n+1}\right) z^{-(n+1)}$

$\displaystyle f(z) = 2 - \sum_{n=1}^\infty \left(6+2^{-(n-1)+1}\right) z^{-n}$

$\displaystyle f(z) = 2 - \sum_{n=1}^\infty \left(6+2^{2-n}\right) z^{-n}$

So, the inverse <em>Z</em> transform of <em>f(z)</em> is $\boxed{6+2^{2-n}}$ .

4 0

3 years ago

An engine operates on gasoline (LHV=44 MJ/kg) with a brake thermal efficiency of 37.9 % What is the brake specific fuel consumpt

scZoUnD [109]

Answer:

$s =0.21\ kg/Kw.hr$

Explanation:

Given that

Calorific value (CV) = 44 MJ/Kg

CV= 44,000 KJ/kg

Brake thermal efficiency(η) = 37.9 %

We know that

$\eta =\dfrac{BP}{\dot{m_f}\times CV}$

Where BP is the brake power

$\eta =\dfrac{BP}{\dot{m_f}\times CV}$

$0.379 =\dfrac{BP}{\dot{m_f}\times 44000}$

$\dfrac{BP}{\dot{m_f}}=16676$

Brake specific fuel consumption (s)

$s =\dfrac{\dot{m_f}}{BP}$

$s =\dfrac{3600\times \dot{m_f}}{BP}$

$s =\dfrac{3600}{16,676}\ kg/Kw.hr$

$s =0.21\ kg/Kw.hr$

7 0

3 years ago

A loss in value caused by an undesirable or hazardous influence offsite is which type of depreciation?

Lubov Fominskaja [6]

External depreciation may be defined as a loss in value caused by an undesirable or hazardous influence offsite.

<h3>What is depreciation?</h3>

Depreciation may be defined as a situation when the financial value of an acquisition declines over time due to exploitation, fray, and incision, or obsolescence.

External depreciation may also be referred to as "economic obsolescence". It causes a negative influence on the financial value gradually.

Therefore, it is well described above.

To learn more about Depreciation, refer to the link:

brainly.com/question/1203926

#SPJ1

5 0

2 years ago