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ipn [44]
3 years ago
7

Air is contained inside a vertical piston-cylinder assembly by a piston of mass 105 kg and face area of 0.05 m2 . The mass of ai

r inside the cylinder is 20 grams, and the air initially occupies a volume of 50 liters. The atmosphere exerts a pressure of 101.3 kPa on the top of the piston. The volume of the air slowly decreases to 0.005 m3 , while maintaining constant pressure, as the specific internal energy of the air decreases by 175 kJ/kg. Neglecting friction between the piston and the cylinder wall, determine the heat between the surroundings and the air inside the piston-cylinder assembly in kJ
Engineering
1 answer:
alexgriva [62]3 years ago
4 0

Answer:

ΔQ = 1.06 KJ

Explanation:

The amount of heat transfer between the piston-cylinder system and the surrounding can easily be found by using the First Law of Thermodynamics. The first law of thermodynamics can be written as follows:

ΔQ = ΔU + W

ΔQ = mΔu + PΔV

where,

ΔQ = Heat transfer between system and surrounding = ?

Δu = specific change in internal energy of the system = - 175 KJ/kg

m = mass of air = 20 g = 0.02 kg

P = Constant Pressure = 101.3 KPa

ΔV = Change in Volume = 0.05 m³ - 0.005 m³ = 0.045 m³

Therefore,

ΔQ = (0.02 kg)(-175 KJ/kg) + (101.3 KPa)(0.045 m³)

ΔQ = -3.5 KJ + 4.56 KJ

<u>ΔQ = 1.06 KJ</u>

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guapka [62]

Answer:

(a) 705 kg

Explanation:

it is given that volume =500m^3

pressure = 200 kPa

it is given that 80% mass is vapor so dryness fraction =x=0.8

from the standard table at 200 kPa

v_f=0.001061 \frac{m^3}{kg}

v_g=0.88578\frac{m^3}{kg}

specific volume v=v_f+x\left ( v_g-v_f \right )

v=0.00106+0.8\left ( 0.88578-0.00106 \right )=0.7088\frac{m^3}{kg}

we know that specific\ volume =\frac{volume}{mass}

0.7088=\frac{500}{mass}

mass=\frac{500}{0.7088}=705.456 \ kg

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3 years ago
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3 years ago
This question allows you to practice proving a language is non-regular via the Pumping Lemma. Using the Pumping Lemma (Theorem 1
Ulleksa [173]

Answer:

<em>L is not a regular language with formal proofs  </em>

Explanation:

<em>(a) To prove that L is not a regular language, we will use a proof by contradiction. the assumption entails  that L is a regular language. Then by the Pumping Lemma for Regular Languages, </em>

<em>there exists a pumping length p for L such that for any string s ∈ L where |s| ≥ p, </em>

<em>s = xyz subject to the following conditions: </em>

<em>(a) |y| > 0 </em>

<em>(b) |xy| ≤ p, and </em>

<em>(c) ∀i > 0, xyi </em>

<em>z ∈ L</em>

<em />

<em>(b) To determine that L is not a regular language, we mke use of proof by contradiction.  lets assume, that L is regular. Then by the Pumping Lemma for Regular Languages, it states also,</em>

<em>The pumping length, p for L such that for any string s ∈ L where |s| ≥ p, s = xyz subject  to the condtions as follows : </em>

<em>(a) |y| > 0 </em>

<em>(b) |xy| ≤ p, and </em>

<em>(c) ∀i > 0, xyi </em>

<em>z ∈ L. </em>

<em>Choose s = 0p10p </em>

<em>. Clearly, |s| ≥ p and s ∈ L. By condition (b) above, it follows is shown. by the first condition x and y are zeros.</em>

<em>for some  k > 0. Per (c), we can take i = 0 and the resulting string will still be in L. Thus,  xy0 </em>

<em>z should be in L. xy0 </em>

<em>z = xz = 0(p−k)10p </em>

<em>It is shown that is is  not in L. This is a  contraption with the pumping lemma.  our assumption that L is regular is  incorrect, and L is not a regular language</em>

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3 years ago
Which type of elevated stand does not need a tree?
SashulF [63]

Answer:

Elevated stands place the hunter above ground level. They can be tree stands placed in or against trees, or freestanding structures. They have become increasingly popular in recent years with both firearm and bow hunters.

Explanation:

4 0
3 years ago
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dimaraw [331]

Answer:

BYEEEEEEEEEEEE3EEEEEEEEEE

Explanation:

dawg

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