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3 years ago

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A(n) 78-hp compressor in a facility that operates at full load for 2500 h a year is powered by an electric motor that has an eff

iciency of 93 percent. If the unit cost of electricity is $0.11/kWh, what is the annual electricity cost of this compressor

1 answer:

Jet001 [13]3 years ago

5 0

Answer: $17,206.13

Explanation:

Hi, to answer this question we have to apply the next formula:

Annual electricity cost = (P x 0.746 x Ckwh x h) /η

P = compressor power = 78 hp

0.746 kw/hp= constant (conversion to kw)

Ckwh = Cost per kilowatt hour = $0.11/kWh

h = operating hours per year = 2500 h

η = efficiency = 93% = 0.93 (decimal form)

Replacing with the values given :

C = ( 78 hp x 0.746 kw/hp x 0.11 $/kwh x 2500 h ) / 0.93 = $17,206.13

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Answer:

Pressure = 115.6 psia

Explanation:

Given:

v=800ft/s

Air temperature = 10 psia

Air pressure = 20F

Compression pressure ratio = 8

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Conversion:

1 Btu =775.5 ft lbf, $g_{c}$ = 32.2 lbm.ft/lbf.s², 1Btu/lbm=25037ft²/s²

Air standard assumptions:

$c_{p}$ = 0.0240Btu/lbm.°R, R = 53.34ft.lbf/lbm.°R = 1717.5ft²/s².°R 0.0686Btu/lbm.°R

k= 1.4

Energy balance:

$h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} + \frac{v_{a} ^{2} }{2}\\$

As enthalpy exerts more influence than the kinetic energy inside the engine, kinetic energy of the fluid inside the engine is negligible

hence $v_{a} ^{2} = 0$

$h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} \\h_{1} -h_{a} = - \frac{v_{1} ^{2} }{2} \\ c_{p} (T_{1} -T_{a})= - \frac{v_{1} ^{2} }{2} \\(T_{1} -T_{a}) = - \frac{v_{1} ^{2} }{2c_{p} }\\ T_{a}=T_{1} + \frac{v_{1} ^{2} }{2c_{p} }$

$T_{1}$ = 20+460 = 480°R

$T_{a} =480+ \frac{(800)(800}{2(0.240)(25037}$ = 533.25°R

Pressure at the inlet of compressor at isentropic condition

$P_{a } =P_{1}(\frac{T_{a} }{T_{1} }) ^{k/(k-1)}$

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Do heavier cars really use more gasoline? Suppose a car is chosen at random. Let x be the weight of the car (in hundreds of poun

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Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Note: This question is incomplete and lacks necessary data to solve. But I have found the similar question on the internet. So, I will be using the data from that question to solve this question for the sack of concept and understanding.

Data Given:

x = 27 , 44 , 32 , 47, 23 , 40, 34, 52

y = 30, 19, 24, 13 , 29, 19, 21, 14

It is given that,

∑x = 299

∑y = 167

∑ $x^{2}$ = 11887

∑ $y^{2}$ = 3773

We are asked to verify the above values manually in this question.

So,

1. ∑x = 299

Let's verify it:

∑x = 27 + 44 + 32 + 47 + 23 + 40 + 34 + 52

∑x = 299

Yes, it is equal to the given value. Hence, verified.

2. ∑y = 167

Let's verify it:

∑y = 30 + 19 + 24 + 13 + 29 + 19 + 21 + 14

∑y = 169

No, it is not equal to the given value.

3. ∑ $x^{2}$ = 11887

Let's verify it:

For this to find, first we need to square all the value of x individually and then add them together to verify.

∑ $x^{2}$ = $27^{2}$ + $44^{2}$ + $32^{2}$ + $47^{2}$ + $23^{2}$ + $40^{2}$ + $34^{2}$ + $52^{2}$

∑ $x^{2}$ = 11,887

Yes, it is equal to the given value. Hence, verified.

4. ∑ $y^{2}$ = 3773

Let's verify it:

Again, for this we need to find the squares of all the y values and then add them together to verify it.

∑ $y^{2}$ = $30^{2}$ + $19^{2}$ + $24^{2}$ + $13^{2}$ + $29^{2}$ + $19^{2}$ + $21^{2}$ + $14^{2}$

∑ $y^{2}$ = 3,845

No, it is not equal to the given value.

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Answer:

401.3 kg/s

Explanation:

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qw = 0.85 * q2

q2 = 0.64 * q1

p = 0.36 * q1

q1 = p /0.36

q2 = 0.64/0.36 * p

qw = 0.85 *0.64/0.36 * p

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