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Oksi-84 [34.3K]
3 years ago
11

Tech A says that a torsion bar is used to operate the control valve in a hydraulically controlled power steering system. Tech B

says that in
the neutral position, the pressure in the power steering system is very high. Who is correct?

Select one:

A. Tech A

B.Tech B

C. Both A and B

D. Neither A nor B
Engineering
1 answer:
Margaret [11]3 years ago
5 0

Answer:B

Explanation:

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PROMPT: Decide which 8 oz. serving of drink (Orange Juice or Lo-Carb Monster Energy®) will provide more energy to your brain cel
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A well-insulated rigid tank contains 5 kg of a saturated liquid–vapor mixture of water at 200 kPa. Initially, three-quarters of
vlabodo [156]

Answer:

S_{gen} = 18.519\,\frac{kJ}{K}

Explanation:

Given that rigid tank is a closed system, the following model is constructed after the First Law of Thermodynamics:

W_{heater} + U_{sys,1} - U_{sys,2} = 0

W_{heater} = U_{sys,2} - U_{sys,1}

W_{heater} = m\cdot (u_{2}-u_{1})

The entropy generation inside the rigid tank is determined by appropriate application of the Second Law of Thermodynamics:

S_{sys,1} - S_{sys,2} + S_{gen} = 0

S_{gen} = S_{sys,2} - S_{sys,1}

S_{gen} = m\cdot (s_{2}-s_{1})

The properties of the steam are obtained from steam tables:

Intial State

P = 200\,kPa

T = 120.21\,^{\textdegree}C

\nu = 0.2222\,\frac{m^{3}}{kg}

u = 1010.7\,\frac{kJ}{kg}

s = 2.9294\,\frac{kJ}{kg\cdot K}

x = 0.25

Final State

P = 869.567\,kPa

T = 173.88\,^{\textdegree} C

\nu = 0.2222\,\frac{m^{3}}{kg}

u = 2578.6\,\frac{kJ}{kg}

s = 6.6332\,\frac{kJ}{kg\cdot K}

x = 1.00

The entropy change of the steam during the process is:

S_{gen} = (5\,kg)\cdot \left(6.6332\,\frac{kJ}{kg\cdot K} - 2.9294\,\frac{kJ}{kg\cdot K} \right)

S_{gen} = 18.519\,\frac{kJ}{K}

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