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Lisa [10]
2 years ago
13

Which part of a machine control unit interacts with the machine tools through electric signals?=]

Engineering
1 answer:
FrozenT [24]2 years ago
4 0

Answer:

control loop unit

Explanation:

Edmentum/Plato

You might be interested in
Five kilograms of air at 427°C and 600 kPa are contained in a piston–cylinder device. The air expands adiabatically until the pr
son4ous [18]

Answer:

The entropy change of the air is 0.240kJ/kgK

Explanation:

T_{1} =427+273K,T_{1} =700K\\P_{1} =600kPa\\P_{2} =100kPa

T_{2}  is unknown

we can apply the following expression to find T_{2}

-w_{out} =mc_{v} (T_{2} -T_{1} )

T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }

now substitute

T_{2} =700K-\frac{600kJ}{5kg*0.718kJ/kgK} \\T_{2}=533K

To find entropy change of the air we can apply the ideal gas relationship

Δs_{air}=c_{p} ln\frac{T_{2} }{T_{1} } -Rln\frac{P_{2} }{P_{1} }

Δs_{air} =1.005*ln(\frac{533}{700})-0.287* in(\frac{100}{600} )

Δs_{air} =0.240kJ/kgK

4 0
2 years ago
A cylinder with a frictionless piston contains 0.05 m3 of air at 60kPa. The linear spring holding the piston is in tension. The
AleksAgata [21]

Answer:

18 kJ

Explanation:

Given:

Initial volume of air = 0.05 m³

Initial pressure = 60 kPa

Final volume = 0.2 m³

Final pressure = 180 kPa

Now,

the Work done by air will be calculated as:

Work Done = Average pressure × Change in volume

thus,

Average pressure = \frac{60+180}{2}  = 120 kPa

and,

Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³

Therefore,

the work done = 120 × 0.15 = 18 kJ

4 0
2 years ago
Water vapor at 1.0 MPa, 300°C enters a turbine operating at steady state and expands to 15 kPa. The work developed by the turbin
Andre45 [30]

Answer:

a) isentropic efficiency = 84.905%

b) rate of entropy generation = .341 kj/(kg.k)

Please kindly see explaination and attachment.

Explanation:

a) isentropic efficiency = 84.905%

b) rate of entropy generation = .341 kj/(kg.k)

The Isentropic efficiency of a turbine is a comparison of the actual power output with the Isentropic case.

Entropy can be defined as the thermodynamic quantity representing the unavailability of a system's thermal energy for conversion into mechanical work, often interpreted as the degree of disorder or randomness in the system.

Please refer to attachment for step by step solution of the question.

5 0
3 years ago
Water leaves a penstock (the flow path through a hydroelectric dam) at a velocity of 100 ft/s. How deep is the water behind the
Marysya12 [62]

Answer:

155fts

Explanation:

We apply the bernoulli's equation to get the depth of water.

We have the following information

P1 = pressure at top water surface = 0

V1 = velocity at too water surface = 0

X1 = height of water surface = h

Hf = friction loss = 0

P2 = pressure at exit = 0

V2 = velocity at exit if penstock = 100ft/s

X2 = height of penstock = 0

g = acceleration due to gravity = 32.2ft/s²

Applying these values to the equation

0 + 0 + h = 0 + v2²/2g +0 + 0

= h = 100²/2x32.2

= 10000/64.4

= 155.28ft

= 155

8 0
2 years ago
A rigid, well-insulated tank of volume 0.9 m is initially evacuated. At time t = 0, air from the surroundings at 1 bar, 27°C beg
Eva8 [605]

Answer:

\dot{w}= -0.303 KW

Explanation:

This is the case of unsteady flow process because properties are changing with time.

From first law of thermodynamics for unsteady flow process

\dfrac{dU}{dt}=\dot{m_i}h_i+\dot{Q}-\dot{m_e}h_i+\dot{w}

Given that tank is insulated so\dot{Q}=0 and no mass is leaving so

\dot{m_e}=0

\int dU=\int \dot{m_i}h_i\ dt-\int \dot{w}\ dt

m_2u_2-m_1u_1=(m_2-m_1)h_i- \dot{w}\Delta t

Mass conservation m_2-m_1=m_e-m_i

m_1,m_2 is the initial and final mass in the system respectively.

Initially tank is evacuated so m_1=0

We know that for air u=C_vT ,h=C_p T,P_2v_2=m_2RT_2

m_2=0.42 kg

So now putting values

0.42 \times 0.71 \times 730=0.42\times 1.005\times 300- \dot{w} \times 300

\dot{w}= -0.303 KW

3 0
3 years ago
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