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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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torisob [31]

Answer:

The correct option is;

c. Leaving the chuck key in the drill chuck

Explanation:

A Common safety issues with a drill press leaving the chuck key in the drill chuck

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It is also required to ensure that the switch is in the OFF position before turning plugging in the power cable

Be sure that the chuck key is removed from the chuck before turning on the power. Using a self-ejecting chuck key is a good way of insuring that the key is not left in the chuck accidentally. Also to avoid accidental starting, make sure the switch is in the OFF position before plugging in the cord. Always disconnect the drill from the power source when making repairs.

5 0
3 years ago
What type of engineer would be most likely to develop a design for cars? chemical civil materials mechanical
Andreyy89
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4 0
3 years ago
Calculate the pressure drop in a duct (measured by a differential oil manometer) if the differential height between the two flui
Burka [1]

Answer:

The pressure drop is 269.7N/m^2

Explanation:

∆P = ∆h × rho × g

∆h = 3.2cm = 3.2/100 = 0.032m, rho = 860kg/m^3, g = 9.8m/s^2

∆P = 0.032×860×9.8 = 269.7N/m^2

6 0
3 years ago
7.4 A pretimed four-timing-stage signal has critical lane group flow rates for the first three timing stages of 200, 187, and 21
Irina18 [472]

Answer:

16 seconds

Explanation:

Given:

C = 60

L = 4 seconds each = 4*4 =16

In this problem, the first 3 timing stages are given as:

200, 187, and 210 veh/h.

We are to find the estimated effective green time of the fourth timing stage. The formula for the estimated effective green time is:

g = (\frac{v}{s}) (\frac{C}{X})

Let's first find the fourth stage critical lane group ratio \frac{v}{s} , using the formula:

C = \frac{1.5L +5}{1 - ( \frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800}) + ( \frac{v}{s})}

60 = \frac{1.5*16 + 5}{1 - ( \frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800}) + ( \frac{v}{s})}

60 = \frac{24+5}{1 - (0.332 + ( \frac{v}{s}))}

Solving for (\frac{v}{s}), we have:

(\frac{v}{s}) = 0.185

Let's also calculate the volume capacity ratio X,

X = (\frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800} + 0.185)(\frac{60}{60-16}

X = 0.704

For the the estimated effective green time of the fourth timing stage, we have:

g_4 = (\frac{v}{s}) (\frac{C}{X})

Substituting figures in the equation, we now have:

g_4 = (0.185) (\frac{60}{0.704})

g_4 = 15.78 seconds

15.78 ≈ 16 seconds

The estimated effective green time of the fourth timing stage is 16 seconds

8 0
2 years ago
Problem 7.16 (GO Tutorial) A cylindrical bar of steel 10.4 mm (0.4094 in.) in diameter is to be deformed elastically by applicat
shepuryov [24]

Answer:

P = 18035.25 N

Explanation:

Given

D = 10.4 mm

ΔD = 3.2 ×10⁻³ mm

E = 207 GPa

ν = 0.30

If

σ = P/A

A = 0.25*π*D²

σ = E*εx

ν = - εz / εx

εz = ΔD / D

We can get εx as follows

εz = ΔD / D = 3.2 ×10⁻³ mm / 10.4 mm = 3.0769*10⁻⁴

Now we find εx

ν = - εz / εx   ⇒   εx = - εz / ν = - 3.0769*10⁻⁴ / 0.30 = - 1.0256*10⁻³

then

σ = E*εz = (207 GPa)*(-1.0256*10⁻³) = - 2.123*10⁸ Pa

we have to obtain A:

A = 0.25*π*D² = 0.25*π*(10.4*10⁻3)² = 8.49*10⁻⁵ m²

Finally we apply the following equation in order o get P

σ = P/A   ⇒  P =  σ*A = (- 2.123*10⁸Pa)*(8.49*10⁻⁵ m²) = 18035.25 N

4 0
3 years ago
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