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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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Answer:

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The cars of a roller-coaster ride have a speed of 19.0 km/h as they pass over the top of the circular track. Neglect any frictio
Dmitrij [34]

Complete Question

The cars of a roller-coaster ride have a speed of 19.0 km/h as they pass over the top of the circular track. Neglect any friction and calculate their speed v when they reach the horizontal bottom position. At the top position, the radius of the circular path of their mass centers is 21 m, and all six cars have the same mass.V = -18 m What is v?X km/h

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v=23.6m/s

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initial velocity uu=19=>5.27778

Generally the equation for Angle is mathematically given by

\theta=\frac{v_c}{2r}

\theta=\frac{18}{2*21}

\theta=0.45

\theta=25.7831 \textdegree

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Height of mass

h=\frac{rsin\theta}{\theta}

h=\frac{21sin25.78}{0.45}

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Generally the equation for Work Energy is mathematically given by

0.5mv_0^2+mgh=0.5mv^2

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Answer:

The answer is "\$29.7072".

Explanation:

Please find the complete question in the attachment file.

As the original prices are 24 euros, \$ 1.2378 must be multiplied by 24.  

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4 0
3 years ago
A solid shaft and a hollow shaft of the same material have same length and outer radius R. The inner radius of the hollow shaft
alexandr402 [8]

Answer with Explanation:

By the equation or Torque we have

\frac{T}{I_{p}}=\frac{\tau }{r}=\frac{G\theta }{L}

where

T is the torque applied on the shaft

I_{p} is the polar moment of inertia of the shaft

\tau is the shear stress developed at a distance 'r' from the center of the shaft

\theta is the angle of twist of the shaft

'G' is the modulus of rigidity of the shaft

We know that for solid shaft I_{p}=\frac{\pi R^4}{2}

For a hollow shaft I_{p}=\frac{\pi (R_o^4-R_i^4)}{2}

Since the two shafts are subjected to same torque from the relation of Torque we have

1) For solid shaft

\frac{2T}{\pi R^4}\times r=\tau _{solid}

2) For hollow shaft we have

\tau _{hollow}=\frac{2T}{\pi (R^4-0.7R^4)}\times r=\frac{2T}{\pi 0.76R^4}

Comparing the above 2 relations we see

\frac{\tau _{solid}}{\tau _{hollow}}=0.76

Similarly for angle of twist we can see

\frac{\theta _{solid}}{\theta _{hollow}}=\frac{\frac{LT}{I_{solid}}}{\frac{LT}{I_{hollow}}}=\frac{I_{hollow}}{I_{solid}}=1.316

Part b)

Strength of solid shaft = \tau _{max}=\frac{T\times R}{I_{solid}}

Weight of solid shaft =\rho \times \pi R^2\times L

Strength per unit weight of solid shaft = \frac{\tau _{max}}{W}=\frac{T\times R}{I_{solid}}\times \frac{1}{\rho \times \pi R^2\times L}=\frac{2T}{\rho \pi ^2R^5L}

Strength of hollow shaft = \tau '_{max}=\frac{T\times R}{I_{hollow}}

Weight of hollow shaft =\rho \times \pi (R^2-0.7R^2)\times L

Strength per unit weight of hollow shaft = \frac{\tau _{max}}{W}=\frac{T\times R}{I_{hollow}}\times \frac{1}{\rho \times \pi (R^2-0.7^2)\times L}=\frac{5.16T}{\rho \pi ^2R^5L}

Thus \frac{Strength/Weight _{hollow}}{Strength/Weight _{Solid}}=5.16

3 0
4 years ago
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