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schepotkina [342]
2 years ago
12

(25%) A well-insulated compressor operating at steady state takes in air at 70 oF and 15 psi, with a volumetric flow rate of 500

ft3 /min, and compresses it to 60 psi. The power input to the compressor is 80 horsepower. Kinetic and potential energy changes from the inlet to exit can be neglected. Determine the exit temperature, in oR.
Engineering
1 answer:
lubasha [3.4K]2 years ago
6 0

Answer:

You can look it up

Explanation: if you don't know what it is look it up on .

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To find the reactance XLXLX_L of an inductor, imagine that a current I(t)=I0sin(ωt)I(t)=I0sin⁡(ωt) , is flowing through the indu
Sophie [7]

Answer:

V(t) = XLI₀sin(π/2 - ωt)

Explanation:

According to Maxwell's equation which is expressed as;

V(t) = dФ/dt ........(1)

Magnetic flux Ф can also be expressed as;

Ф = LI(t)

Where

L = inductance of the inductor

I = current in Ampere

We can therefore Express Maxwell equation as:

V(t) = dLI(t)/dt ....... (2)

Since the inductance is constant then voltage remains

V(t) = LdI(t)/dt

In an AC circuit, the current is time varying and it is given in the form of

I(t) = I₀sin(ωt)

Substitutes the current I(t) into equation (2)

Then the voltage across inductor will be expressed as

V(t) = Ld(I₀sin(ωt))/dt

V(t) = LI₀ωcos(ωt)

Where cos(ωt) = sin(π/2 - ωt)

Then

V(t) = ωLI₀sin(π/2 - ωt) .....(3)

Because the voltage and current are out of phase with the phase difference of π/2 or 90°

The inductive reactance XL = ωL

Substitute ωL for XL in equation (3)

Therefore, the voltage across inductor is can be expressed as;

V(t) = XLI₀sin(π/2 - ωt)

3 0
3 years ago
Air enters the compressor of a simple gas turbine at 100 kPa, 300 K, with a volumetric flow rate of 5 m3/s. The compressor press
Zepler [3.9K]

Answer:

a) 3581.15067 kw

b) 95.4%

Explanation:

<u>Given data:</u>

compressor efficiency = 85%

compressor pressure ratio = 10

Air enters at:    flow rate of 5m^3/s , pressure = 100kPa, temperature = 300 K

At turbine inlet : pressure = 950 kPa, temperature = 1400k

Turbine efficiency = 88% , exit pressure of turbine = 100 kPa

A) Develop a full accounting of the exergy increase of the air passing through the gas turbine combustor in kW

attached below is a detailed solution to the given question

6 0
2 years ago
The following program includes fictional sets of the top 10 male and female baby names for the current year. Write a program tha
otez555 [7]

Answer:

Define Variables and Use List methods to do the following

Explanation:

#<em>Conjoins two lists together</em>

all_names = male_names.union(female_names)

#<em>Finds the names that appear in both lists, just returns those</em>

neutral_names = male_names.intersection(female_names)

#<em>Returns names that are NOT in both lists</em>

specific_names = male_names.symmetric_difference(female_names)

3 0
2 years ago
Read 2 more answers
What i s the value of a capacitor with 250 V applied and has 500 pC of charge? (a) 200 uF (b) 0.5 pF (c) 500 uF (d) 2 pF
exis [7]

Answer:

(d) 2 pF

Explanation: the charge on capacitor is given by the expression

Q=CV

where Q=charge

           C=capacitance

           V=voltage across the plate of the capacitor

here we have given Q=500 pF, V=250 volt

using this formula C=\frac{Q}{V}

=500×10^{-12}×\frac{1}{250}

=2×10^{-12}

=2 pF

3 0
3 years ago
Air is compressed in a reversible, isothermal, steady- flow process from 15 psia, 100°F to 100 psia. Calculate the work of compr
mixas84 [53]

Answer:

|W|=169.28 KJ/kg

ΔS = -0.544 KJ/Kg.K

Explanation:

Given that

T= 100°F

We know that

1 °F = 255.92 K

100°F = 310 .92 K

P _1= 15 psia

P _1= 100 psia

We know that work for isothermal process  

W=mRT\ln \dfrac{P_1}{P_2}

Lets take mass is 1 kg.

So work per unit mass

W=RT\ln \dfrac{P_1}{P_2}

We know that for air R=0.287KJ/kg.K

W=RT\ln \dfrac{P_1}{P_2}

W=0.287\times 310.92\ln \dfrac{15}{100}

W= - 169.28 KJ/kg

Negative sign indicates compression

|W|=169.28 KJ/kg

We know that change in entropy at constant volume

\Delta S=-R\ln \dfrac{P_2}{P_1}

\Delta S=-0.287\ln \dfrac{100}{15}

ΔS = -0.544 KJ/Kg.K

3 0
3 years ago
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