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Bas_tet [7]
2 years ago
13

2 definiciones de personas en Técnicos en ingeniería eléctrica y agregar link. Por favor❗❗❗

Engineering
1 answer:
Grace [21]2 years ago
3 0

Answer:

no se si te sirva pero

Explanation:

https://www.freelancermap.com/blog/es/que-hace-ingeniero-electrico/

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Consider a simple ideal Rankine cycle and an ideal regenerative Rankine cycle with one open feedwater heater. The two cycles are
borishaifa [10]

Answer:

They both have the same efficiency.

Explanation:

The simple ideal Rankine cycle and an ideal regenerative Rankine cycle with one open feedwater heater would both have the same efficiency because the extraction steam would just create a mini cycle that recirculates. The energy given to the feedwater heater is proportional to the added heat in the boiler to the feedwater in the simple cycle to raise its temperature to the same boiler inlet condition.

Therefore in comparison, the efficiency is the same for both.

4 0
3 years ago
25 points and brainliest is it A, B, C, D
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SIR IT IS D HOPE THIS HELPS (☞゚ヮ゚)☞☜(゚ヮ゚☜)

Explanation:

4 0
2 years ago
Read 2 more answers
A plumbed eyewash station is portable.
Hitman42 [59]
Plumbed stations are permanently connected to a source of potable water, whereas portable stations are self-contained gravity-fed units with their own flushing fluid that must be replaced after each use. ... Eyewash fluid must irrigate and flush both eyes simultaneously.
Hopefully this helped.
8 0
2 years ago
For a bronze alloy, the stress at which plastic deformation begins is 266 MPa and the modulus of elasticity is105 GPa.
pentagon [3]

Answer:

88750 N

Explanation:

given data:

plastic deformation σy=266 MPa=266*10^6 N/m^2

cross-sectional area Ao=333 mm^2=333*10^-6 m^2

solution:

To determine the maximum load that can be applied without

plastic deformation (Fy).

Fy=σy*Ao

   =88750 N

7 0
3 years ago
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500
Rashid [163]

Answer:

Exit temperature = 32 °C

Explanation:

We are given;

Initial Pressure;P1 = 100 KPa

Cp =1000 J/kg.K = 1 KJ/kg.k

R = 500 J/kg.K = 0.5 Kj/Kg.k

Initial temperature;T1 = 27°C = 273 + 27K = 300 K

volume flow rate;V' = 15 m³/s

W = 130 Kw

Q = 80 Kw

Using ideal gas equation,

PV' = m'RT

Where m' is mass flow rate.

Thus;making m' the subject, we have;

m' = PV'/RT

So at inlet,

m' = P1•V1'/(R•T1)

m' = (100 × 15)/(0.5 × 300)

m' = 10 kg/s

From steady flow energy equation, we know that;

m'•h1 + Q = m'h2 + W

Dividing through by m', we have;

h1 + Q/m' = h2 + W/m'

h = Cp•T

Thus,

Cp•T1 + Q/m' = Cp•T2 + W/m'

Plugging in the relevant values, we have;

(1*300) - (80/10) = (1*T2) - (130/10)

Q and M negative because heat is being lost.

300 - 8 + 13 = T2

T2 = 305 K = 305 - 273 °C = 32 °C

13000 + 300 - 8000 = T2

6 0
3 years ago
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