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arlik [135]
2 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]2 years ago
7 0

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

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Water flows around a 6-ft diameter bridge pier with a velocity of 12 ft/s. Estimate the force (per unit length) that the water e
jolli1 [7]

Answer: hello the diagram related to your question is missing please the third image is the missing part of the question

Fx = 977.76 Ib/ft

Explanation:

<u>Estimate the force that water exerts on the pier </u>

V = 12 ft/s

D( diameter ) = 6 ft

first express the force  on the first half of the cylinder  as

Fx1 =  - -2\int\limits^\pi _\frac{\pi }{2}   {Ps*cos\beta *a} \, d\beta   ---------------- ( 1 )

where ; Fy = 0

Ps = Po + 1/2 Pv^2 ( 1 - 4 sin^2β )  ------------- ( 2 )

Input equation (2)  into equation ( 1 )         (note :  assuming Po = 0 )

attached below is the remaining part of the solution

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Explain what a margin of safety is in driving as well as how it can help minimize risk.
Yakvenalex [24]

Answer:

A safety margin is the space left between your vehicle and the next to provide room, time and visibility at every instant

Explanation:

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Safety margins help minimize risks in the following way

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6 0
2 years ago
The aluminum rod (E1 = 68 GPa) is reinforced with the firmly bonded steel tube (E2 = 201 GPa). The diameter of the aluminum rod
Vsevolod [243]

Answer:

Explanation:

From the information given:

E_1 = 68 \ GPa \\ \\ E_2 = 201 \ GPa  \\ \\ d = 25 \ mm \  \\ \\ D = 45 \ mm \ \\ \\ L   = 761 \ mm  \\ \\ P = -88 kN

The total load is distributed across both the rod and tube:

P = P_1+P_2 --- (1)

Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.

\delta_1=\delta_2

\dfrac{P_1L}{A_1E_1}= \dfrac{P_2L}{A_2E_2}

\dfrac{P_1 \times 0.761}{(\dfrac{\pi}{4}\times .0025^2 ) \times 68\times 10^4}= \dfrac{P_2\times 0.761}{(\dfrac{\pi}{4}\times (0.045^2-0.025^2))\times 201 \times 10^9}

P_1(2.27984775\times 10^{-8}) = P_2(3.44326686\times 10^{-9})

P_2 = \dfrac{ (2.27984775\times 10^{-8}) P_1}{(3.44326686\times 10^{-9})}

P_2 = 6.6212 \ P_1

Replace P_2 into equation (1)

P= P_1 + 6.6212 \ P_1\\ \\ P= 7.6212\ P_1 \\ \\  -88 = 7.6212 \ P_1  \\ \\ P_1 = \dfrac{-88}{7.6212} \\ \\  P_1 = -11.547 \ kN

Finally, to determine the normal stress in aluminum rod:

\sigma _1 = \dfrac{P_1}{A_1} \\ \\  \sigma _1 = \dfrac{-11.547 \times 10^3}{\dfrac{\pi}{4} \times 25^2}

\sigma_1 = - 23.523 \ MPa}

Thus, the normal stress = 23.523 MPa in compression.

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Explanation:

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Only put ciilant into ur radiator when the engine is cool (D)

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