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arlik [135]
3 years ago
10

A piston cylinder assembly fitted with a slowly rotating paddle wheel contains 0.13 kg of air at 300K. The air undergoes a const

ant pressure process to a final temp of 400K. During the process heat is transfered to the air by Q=12kJ. Assuming the ideal gas model with k=1.4 and negligible changes in kinetic and potential energy for the air, determine the work done by the paddle on the air and the work done by the air to displace the piston. Gas constant for air is 0.287 kJ/kg*K
note ideal gas model is pV=mRT, and with k=1.4 the total internal energy change of air canbe calculated by U2-U1=(mR/k-1)(delta T)

Engineering
1 answer:
vova2212 [387]3 years ago
7 0

The answer & explanation for this question is given in the attachment below.

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1.technology is more advanced
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6 0
3 years ago
A car hits a tree at an estimated speed of 10 mi/hr on a 2% downgrade. If skid marks of 100 ft. are observed on dry pavement (F=
Dafna11 [192]

Answer:

v_{o} = 22.703\,\frac{m}{s} \left(50.795\,\frac{m}{s}\right)

Explanation:

The deceleration of the car on the dry pavement is found by the Newton's Law:

\Sigma F = -\mu_{k,1}\cdot m\cdot g \cdot \cos \theta + m\cdot g \cdot \sin \theta = m\cdot a_{1}

Where:

a_{1} = (-\mu_{k,1}\cdot \cos \theta + \sin \theta)\cdot g

a_{1} = (-0.33\cdot \cos 1.146^{\textdegree}+\sin 1.146^{\textdegree})\cdot \left(9.807\,\frac{m}{s^{2}} \right)

a_{1} = -3.040\,\frac{m}{s^{2}}

Likewise, the deceleration of the car on the unpaved shoulder is:

a_{2} = (-\mu_{k,2}\cdot \cos \theta + \sin \theta)\cdot g

a_{2} = (-0.28\cdot \cos 1.146^{\textdegree}+\sin 1.146^{\textdegree})\cdot \left(9.807\,\frac{m}{s^{2}} \right)

a_{2} = -2.549\,\frac{m}{s^{2}}

The speed just before the car entered the unpaved shoulder is:

v_{o} = \sqrt{\left(4.469\,\frac{m}{s} \right)^{2}-2\cdot \left(-2.549\,\frac{m}{s^{2}} \right)\cdot (60.88\,m)}

v_{o} = 18.175\,\frac{m}{s}

And, the speed just before the pavement skid was begun is:

v_{o} = \sqrt{\left(18.175\,\frac{m}{s} \right)^{2}-2\cdot \left(-3.040\,\frac{m}{s^{2}} \right)\cdot (30.44\,m)}

v_{o} = 22.703\,\frac{m}{s} \left(50.795\,\frac{m}{s}\right)

8 0
3 years ago
In a full duplex communication by using UTP cable, Pt =140 w, Pr =10 w and NEXTdb =14,47. According to this information which an
Paraphin [41]
The answer that is correct is c
4 0
3 years ago
For a steel alloy it has been determined that a carburizing heat treatment of 14 h duration at 809°C will raise the carbon conce
Yakvenalex [24]

Answer:

t_2 = 27.7 hr

Explanation:

Given data:

carbon concentration  =  0.54%

from the relation given below calculate the time required to achieve concentration at 6.00 mm from surface

\frac{x^2}{Dt} = constant

D considered constant

\frac{x^2}{t} =  constant

here, x POSITION FROM SURFACE, t is time required to achieve concentration

\frac{x_1^2}{t_1} = \frac{x_2^2}{t_2}

\frac{3.6^2}{14} = \frac{6^2}{t_2}

t_2 = 27.7 hr

3 0
4 years ago
A square isothermal chip is of width w 5 mm on a side and is mounted in a substrate such that its side and back surfaces are wel
Lynna [10]

Answer:

a) 0.35 W

b) 5.25 W

Explanation:

Given that

Width of the chip, W = 5 mm

The environment temperature, T(∞) = 15° C

Surface temperature, T(s) = 85° C

The initial convection heat coefficient, h1 = 200 W/m².K

The final convection heat coefficient, h2 = 3000 W/m².K

See attachment for solution

6 0
3 years ago
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