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3 years ago

Help Please!!!!!!! Whatever3443 Please help!

Engineering

2 answers:

KIM [24]3 years ago

8 0

Answer:

Thanks for the pointssssssssssssss

Explanation:

Olin [163]3 years ago

7 0

Answer:

I need some help understanding this question what is it you need to know about doc. whatever 3443?

744 with 3040 1230 0 215 141 184 247 353 whatever 3443 1230 0 215 140 246 337 400 492 581 673 737 for 4232 1230 0 215 50 154 218 which 4499 1230

Explanation:

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A simple ideal Rankine cycle with water as the working fluid operates between the pressure limits of 4 MPa in the boiler and 20

viktelen [127]

Answer:

a) 69,630KW

b) 203 KW

Explanation:

The data obtained from Tables A-4, A-5 and A-6 is as follows:

$h_{1} = h_{f,@20KPa} = 251.42 KJ/kg\\v_{1} = v_{f,@20KPa} = 0.001017 KJ/kgK\\\\w_{p,in} = v_{1} * (P_{2} - P_{1})\\w_{p,in} = (0.001017)*(4000-20)\\\\w_{p,in} = 4.05 KJ/kg\\\\h_{2} = h_{1} - w_{p,in} \\h_{2} = 251.42 + 4.05\\\\h_{2} = 255.47KJ/kg\\\\P_{3} = 4000KPa\\T_{3} = 700 C\\s_{3} = 7.6214 KJ/kgK\\\\h_{3} = 3906.3 KJ/kg\\\\P_{4} = 20 KPa\\s_{3} = s_{4} = 7.6214KJ/kgK\\s_{f} = 0.8320 KJ/kgK\\s_{fg} = 7.0752 KJ/kgK\\\\$

$x_{4} = \frac{s_{4} - s_{f} }{s_{fg} } \\\\x_{4} = \frac{7.6214-0.8320}{7.0752} = 0.9596\\\\h_{f} = 251.42KJ/kg \\h_{fg} = 2357.5KJ/kg \\\\h_{4} = h_{f} + x_{4}*h_{fg} = 251.42 + 0.9596*2357.5 = 2513.7KJ/kg\\\\$

The power produced and consumed by turbine and pump respectively are:

$W_{T,out} = flow(m) *(h_{3} - h_{4}) \\W_{T,out} = 50 *(3906.3-2513.7)\\\\W_{T,out} = 69,630 KW\\\\W_{p,in} = flow(m) *w_{p,in} = 50*4.05 = 203 KW$

7 0

3 years ago

The air conditioner in a house or a car has a cooler that brings atmospheric air from 30C to 10C, with both states at 101KPa. If

torisob [31]

The rate of heat transfer by the air conditioner using constant specific heat of 1.004kj/kg.K is 15.06 kW.

<h3>What is the rate of heat transfer?</h3>

Rate of heat transfer is the power rating of the machine.

Work done and changes in potential and kinetic energy are neglected since it is a steady state process.

The specific heat in terms of specific heat capacity and temperature change is given as:

$q_{out} = Cp(Ti - Te)$

$q_{out} = 1.004(30 - 10) = 20.08 kJ/kg \\$

The rate of heat transfer, is then determined as follows:

Qout = flow rate × specific heat

Qout = 0.75 × 20.08 = 15.06 kW

Therefore, the rate of heat transfer by the air conditioner is 15.06 kW.

Learn more about rate of heat transfer at: brainly.com/question/17152804

#SPJ1

6 0

2 years ago

What is the hardest part of engineering?

Vikki [24]

ANSWER:

Aerospace Engineering. ...

Chemical Engineering. ...

Biomedical Engineering.

EXPLANATION:

This is all i know but ... I hope this helps~

7 0

1 year ago

A transformer has 300,000 windings in its primary coil and uses 12,000V AC input. (4 points) How many windings would be needed t

viva [34]

Answer:

2750

Explanation:

The number of windings and the voltage are proportional.

Let n represent the number of windings to produce 110 Vac. Then the proportion is ...

n/110 = 300,000/12,000

n = 110(300/12) = 2750 . . . . multiply by 110

2750 windings would be needed to produce 110 Vac at the output.

7 0

1 year ago

Determine the following parameters for the water having quality x=0.7 at 200 kPa:

ra1l [238]

Solution :

Given :

Water have quality x = 0.7 (dryness fraction) at around pressure of 200 kPa

The phase diagram is provided below.

a). The phase is a standard mixture.

b). At pressure, p = 200 kPa, T = $T_{saturated}$

Temperature = 120.21°C

c). Specific volume

$v_{f}= 0.001061, \ \ v_g=0.88578 \ m^3/kg$

$v_x=v_f+x(v_g-v_f)$

$=0.001061+0.7(0.88578-0.001061)$

$=0.62036 \ m^3/kg$

d). Specific energy ( $u_x$ )

$u_f=504.5 \ kJ/kg, \ \ u_{fg}=2024.6 \ kJ/kg$

$u_x=504.5 + 0.7(2024.6)$

$=1921.72 \ kJ/kg$

e). Specific enthalpy $(h_x)$

At $h_f = 504.71, \ \ h_{fg} = 2201.6$

$h_x=504.71+(0.7\times 2201.6)$

$= 2045.83 \ kJ/kg$

f). Enthalpy at m = 0.5 kg

$H=mh_x$

$= 0.5 \times 2045.83$

= 1022.91 kJ

7 0

3 years ago