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2 years ago

6

What is a robot’s work envelope?

1 answer:

Dmitry_Shevchenko [17]2 years ago

3 0

Answer:

B

Explanation:

A robot's work envelope is its range of movement. It is the shape created when a manipulator reaches forward, backward, up and down. These distances are determined by the length of a robot's arm and the design of its axes. ... A robot can only perform within the confines of this work envelope.

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What is the difference between a refrigeration cycle and a heat pump cycle?

sukhopar [10]

Answer:

In refrigeration cycle heat transfer from inside refrigeration

In heat pump cycle heat transfer from environment

Explanation:

heat cycle is mechanical process use for cool the temperature but

In refrigeration heat transfer from inside of refrigeration that decrease temperature of refrigerator and in heat pump it decrease temperature negligible as compare to refrigerator

5 0

3 years ago

A rigid tank contains 3 kg of water initially at 43.97% quality and at a temperature of 120°C. The water is heated until it reac

makkiz [27]

Explanation: see attachment below

6 0

2 years ago

The constant A in Equation 17.2 is 12π4 R/5(θD) 3 where R is the gas constant and θD is the Debye temperature (K). Estimate θD f

kirill115 [55]

Answer:

The Debye temperature for aluminum is 375.2361 K

Explanation:

Molecular weight of aluminum=26.98 g/mol

T=15 K

The mathematical equation for the specific heat and the absolute temperature is:

$C_{v} =AT^{3}$

Substituting in the expression of the question:

$C_{v} =(\frac{12\pi ^{4}R }{5\theta _{D}^{3} } )T^{3}$

$\theta _{D} =(\frac{12\pi ^{4}RT^{3} }{5C_{v} } ) ^{1/3}$

Here

$C_{v} =4.6\frac{J}{kg-K} *\frac{1kg}{1000g} *\frac{26.98g}{1mol} =0.1241J/mol-K$

Replacing:

$\theta _{D} =(\frac{12\pi ^{4}*8.31*15^{3} }{5*0.1241} )^{1/3} =375.2361K$

3 0

2 years ago

A common procedure for measuring the velocity of an air stream involves insertion of an electrically heated wire (called a hot-w

timurjin [86]

Answer:

V = 6.33 m/s

Explanation:

Given:

- The length of the wire L = 0.02 m

- The diameter of the wire D = 0.0005 m

- The calibration expression V = 0.0000625*h^2

- Environment temperature T_inf = 298 K

- Surface temperature T_s = 348 K

- The voltage drop dV = 5 V

- The electric current I = 0.1 A

Find:

- the velocity of Air

Solution:

- Calculate the surface area of the wire:

A = pi*D*L

A = pi*(0.0005)*(0.02) = 0.00003142 m^2

- The rate of energy in the wire P:

P = I*dV = 0.1*5 = 0.5 W

- Apply Newton's Law of Cooling:

P = h*A*(T_s - T_inf)

h = P /A*(T_s - T_inf)

Plug in the values:

h= 0.5/ 0.00003142*(348 - 298)

h = 318.27 W /m^2K

- Using the calibration relationship given, compute the velocity of air:

V = 6.25*10^-5 * h^2

V = 6.25*10^-5 * (318.27)^2

V = 6.33 m/s

5 0

2 years ago

A steam turbine has isentropic efficiency of 0.8. Isentropically, it is supposed to deliver work of 100 kW. What is the actual w

trapecia [35]

Answer:

80 kW; 11 kW; 8 kW; 0.6

Explanation:

Part 1

Isentropic turbine efficiency:

$\eta_t = \frac{\text{Real turbine work}}{\text{isentropic turbine work}} = \frac{W_{real}}{W_s}$

$W_{real} = \eta_t*W_s$

$W_{real} = 0.8*100 kW$

$W_{real} = 80 kW$

Part 2

Coefficient of performance COP is defined by:

$COP = \frac{Q_{out}}{W}$

$Q_{out} = W*COP$

$Q_{out} = 5 kW*2.2$

$Q_{out} = 11 kW$

Part 3

(a)

Energy balance for a refrigeration cycle gives:

$Q_{in} + W = Q_{out}$

$3 kW + 5 kW = Q_{out}$

$8 kW = Q_{out}$

(b)

$COP = \frac{Q_{in}}{W}$

$COP = \frac{3 kW}{5 kW}$

$COP = 0.6$

3 0

3 years ago