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4 years ago

Air enters a compressor operating at steady state at 1 atm with a specific enthalpy of

Engineering

1 answer:

Savatey [412]4 years ago

8 0

Answer:

option C

option A

Explanation:

Enthalpy gained by air= 1023-290

= 733 kJ/kg

Rate of energy gain= mass flow rate × Enthalpy gained by air

= 0.1 × 733

= 73.3 kJ/s

rate of heat transfer between compressor and air= 77kW

Heat loss by air to surroundings= 77-73.3

=3.7kW

Enthalpy lost by steam in turbine= 1407.6-1236.4

= 171.2 Btu/lb

Rate of energy transfer to turbine= Enthalpy lost by steam× mass flow rate

= 171.2×5

= 856 Btu/s

Net rate of energy transfer to turbine=rate of Energy transfer to turbine- rate of heat transfer to turbine

= 856-40

= 816 Btu/s

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Answer:ahahahahahah

Explanation:boyboy

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3 years ago

Technician A says that a lack of lubrication on the back of the disc brake pads can cause brake noise. Technician B says that pa

sweet-ann [11.9K]

Answer:

Technician A and B both are correct.

Explanation:

According to Technician A, we can stop noise by applying lubricant (Break friction).

According to Technician A, we can stop noise by set-up right position for break pad.

7 0

3 years ago

What is polarized electrical receptacle used for

lidiya [134]

Answer:

they are used for electrical currents so that they can flow along the appropriate wires in the circuit

Explanation:

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4 years ago

An electronic toy is powered by three 1.58-V alkaline cells, each with an internal resistance of 0.0205 Ω, and a 1.53-V carbon-z

mina [271]

Answer:

(a) The current in amperes that flows through the toy's circuit is 0.923A

(b) The power supplied to the toy is 5.78721W

(c) The internal resistance r2 of the failed dry cell is 72Ω

Explanation:

From the circuit diagram attached. We have the electric component:

B1 = 3* 1.58 = 4.74V

B2 = 1.53V

r1 = 3*0.0205 = 0.0615Ω

r2 = 0.105Ω

R = 6.625Ω

Since the internal resistances and the resistor R are connected in series, we can calculate the total resistance RT as

RT = r1 + r2 +R = 0.0615 + 0.105 + 6.625

= 6.7915Ω

Total Voltage supplied to the circuit by both batteries V = B1 + B2 = 4.74 + 1.53 = 6.27V

(a) CIRCUIT CURRENT

The current I, flowing through the circuit is i = $\frac{V}{R_{T}}$ = $\frac{6.27}{6.7915}$ =0.923A

The current in amperes that flows through the toy's circuit is 0.923A

(b) THE POWER SUPPLIED TO THE TOY

Power P = I*V =0.923*6.27 = 5.78721W

The power supplied to the toy is 5.78721W

Due to dry cell failure, the power supplied to the toy is reduced to 0.5W

Now Power P = $\frac{V^{2} }{R}$ . To calculate the new total resistance of the circuit we will make R the subject of the formular

R= $\frac{V^{2} }{P}$ = $\frac{6.27^{2} }{0.5}$ = $\frac{39.3129}{0.5}$ = 78.6258Ω

Remember that RT = r1 + r2 + R

r1 =RT- (R +r2)

r1 = 78.6258 - 6.6865 =71.9393Ω

The internal resistance r2 of the failed dry cell is 72Ω

6 0

3 years ago

Read 2 more answers

A shaft 70 mm in diameter is being pushed at a speed of 400 mm/s through a bearing sleeve 70.2 mm in diameter and 250 mm long. T

Allushta [10]

Answer:

F=989.6 N

Explanation:

Given that

Diameter of shaft = 70 mm

Diameter of bearing sleeve =70.2 mm

So clearance h=0.1 mm

Speed V= 400 mm/s

Length of shaft = 250 mm

$\nu =0.005\ \frac{m^2}{s}\ , \rho=900\ \frac{kg}{m^3}$

We know that

$\mu =\rho\times\nu$

μ= 900 x 0.005 Pa-s

μ= 4.5 Pa-s

As we know that

From Newton's law of viscosity ,the shear stress given as follows

$\tau =\mu \dfrac{dU}{dy}$

We also know that

Force = shear stress x area

Now by putting the values

$\tau =4.5\times \dfrac{400}{0.1}$

$\tau=18,000 Pa$

So force

F= 18,000 x π x 0.07 x 0.25

F=988.6 N

6 0

3 years ago