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SOVA2 [1]
3 years ago
10

Air expands through a turbine operating at steady state. At the inlet p1 = 150 lbf/in^2, T1 = 1400R and at the exit p2 = 14.8 lb

f/in^2, T2 = 700R the mass flow rate of air entering the turbine is 11 lb/s, and 65000 Btu/h of energy is rejected by heat transfer.
a. Neglecting kinetic and potential energy effects, determine the power developed in hp.
Engineering
1 answer:
Paraphin [41]3 years ago
5 0

Answer:

The power developed in HP is 2702.7hp

Explanation:

Given details.

P1 = 150 lbf/in^2,

T1 = 1400°R

P2 = 14.8 lbf/in^2,

T2 = 700°R

Mass flow rate m1 = m2 = m = 11 lb/s Q = -65000 Btu/h

Using air table to obtain the values for h1 and h2 at T1 and T2

h1 at T1 = 1400°R = 342.9 Btu/h

h2 at T2 = 700°R = 167.6 Btu/h

Using;

Q - W + m(h1) - m(h2) = 0

W = Q - m (h2 -h1)

W = (-65000 Btu/h ) - 11 lb/s (167.6 - 342.9) Btu/h

W = (-65000 Btu/h ) - (-1928.3) Btu/s

W = (-65000 Btu/h ) * {1hr/(60*60)s} - (-1928.3) Btu/s

W = -18.06Btu/s + 1928.3 Btu/s

W = 1910.24Btu/s

Note; Btu/s = 1.4148532hp

W = 2702.7hp

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A flame ionization detector, which is often used in gas chromatography, responds to a change in
Lena [83]

Answer:

Option A

Explanation:

We know that ions are present in hydrogen-air flame and when the burning of an organic compound takes place in this flame more ions are produced in the flame.  

Thus when we apply a voltage across this flame, the ion collector plate attracts the all the ions in the flame.

The presence of organic compounds increases the voltage across the hydrogen ion flame produced at the ion collector increases and as the voltage increases, the detection of the organic compound can be made in turn.

Thus flame ionization detector clearly responds to the variation in the collection of ions or electrons in a flame.

3 0
3 years ago
On July 23, 1983, Air Canada Flight 143 required 22,300 kg of jet fuel to fly from Montreal to Edmonton. The density of jet fuel
Natasha2012 [34]

Answer:

20, 083 L

Explanation:

The mistake was the result of not using units when converting the 7862 l to Kg. They used the density in pounds hence they multiplied by 1.77 Lb/L and obtained 13597 Lb not Kg as they assumed.

To obtain the amount needed to refuel they subtracted this quantity from the 22,300 Kg required for the trip again obtaining the wrong quantity of 8703 Kg and they converted this to liters by dividing the density to get 4916 L and then placed then 5000 L of fuel

The quantity required was

7862 L * 1.77 Lb/L = 13915.74 Lb (pounds not kilos)

then converting this pounds to Kg by multiplying by  0.454 Kg/L one gets

6173 Kg on board

Amount Required

( 22,300 -6173)  :  16127 Kg

16127 Kg/ 0.803 Kg/L =  20083 L

5 0
3 years ago
Two Carnot engines operate in series such that the heat rejected from one is the heat input to the other. The heat transfer from
kykrilka [37]

Answer:

Given:

high temperature reservoir T_{H} =1000k

low temperature reservoir T_{L} =400k

thermal efficiency n_{1}= n_{2}

The engines are said to  operate on Carnot cycle which is totally reversible.

To find the intermediate temperature between the two engines, The thermal efficiency of the first heat engine can be defined as

n_{1} =1-\frac{T}{T_{H} }

The thermal efficiency of second heat engine can be written as

n_{2} =1-\frac{T_{L} }{T}

The temperature of intermediate reservoir can be defined as  

1-\frac{T}{T_{H} } =1-\frac{T_{L} }{T} \\T^2=T_{L} T_{H} \\T=\sqrt{T_{L} T_{H} }\\T=\sqrt{400*1000} =632k

8 0
3 years ago
one number is 11 more than another number. find the two number if three times the larger number exceeds four times the smaller n
vaieri [72.5K]

Answer:

a = 40

b = 29

Explanation:

Give a place holder for the numbers that we don't know.

Lets call the two numbers a and b.

From the given info, we can write an expression and solve it:

"one number is 11 more than another number"

a = 11 + b

from this, we know that a > b.

''three times the larger number exceeds four times the smaller number by 4"

3a = 4b + 4

Now we have 2 equations, we can use them to solve using whatever method you want.

a = 11 + b

3a = 4b + 4

I will be using matrices RREF to solve for this.

a - b = 11

3a - 4b = 4

\begin{bmatrix}1 & -1  & 11\\3 & -4 & 4 \end{bmatrix}

\begin{bmatrix}1 & 0  & 40\\0 & 1 & 29 \end{bmatrix}

a = 40

b = 29

6 0
3 years ago
Please answer fast. With full step by step solution.​
lina2011 [118]

Let <em>f(z)</em> = (4<em>z </em>² + 2<em>z</em>) / (2<em>z </em>² - 3<em>z</em> + 1).

First, carry out the division:

<em>f(z)</em> = 2 + (8<em>z</em> - 2) / (2<em>z </em>² - 3<em>z</em> + 1)

Observe that

2<em>z </em>² - 3<em>z</em> + 1 = (2<em>z</em> - 1) (<em>z</em> - 1)

so you can separate the rational part of <em>f(z)</em> into partial fractions. We have

(8<em>z</em> - 2) / (2<em>z </em>² - 3<em>z</em> + 1) = <em>a</em> / (2<em>z</em> - 1) + <em>b</em> / (<em>z</em> - 1)

8<em>z</em> - 2 = <em>a</em> (<em>z</em> - 1) + <em>b</em> (2<em>z</em> - 1)

8<em>z</em> - 2 = (<em>a</em> + 2<em>b</em>) <em>z</em> - (<em>a</em> + <em>b</em>)

so that <em>a</em> + 2<em>b</em> = 8 and <em>a</em> + <em>b</em> = 2, yielding <em>a</em> = -4 and <em>b</em> = 6.

So we have

<em>f(z)</em> = 2 - 4 / (2<em>z</em> - 1) + 6 / (<em>z</em> - 1)

or

<em>f(z)</em> = 2 - (2/<em>z</em>) (1 / (1 - 1/(2<em>z</em>))) + (6/<em>z</em>) (1 / (1 - 1/<em>z</em>))

Recall that for |<em>z</em>| < 1, we have

\displaystyle\frac1{1-z}=\sum_{n=0}^\infty z^n

Replace <em>z</em> with 1/<em>z</em> to get

\displaystyle\frac1{1-\frac1z}=\sum_{n=0}^\infty z^{-n}

so that by substitution, we can write

\displaystyle f(z) = 2 - \frac2z \sum_{n=0}^\infty (2z)^{-n} + \frac6z \sum_{n=0}^\infty z^{-n}

Now condense <em>f(z)</em> into one series:

\displaystyle f(z) = 2 - \sum_{n=0}^\infty 2^{-n+1} z^{-(n+1)} + 6 \sum_{n=0}^\infty z^{-n-1}

\displaystyle f(z) = 2 - \sum_{n=0}^\infty \left(6+2^{-n+1}\right) z^{-(n+1)}

\displaystyle f(z) = 2 - \sum_{n=1}^\infty \left(6+2^{-(n-1)+1}\right) z^{-n}

\displaystyle f(z) = 2 - \sum_{n=1}^\infty \left(6+2^{2-n}\right) z^{-n}

So, the inverse <em>Z</em> transform of <em>f(z)</em> is \boxed{6+2^{2-n}}.

4 0
3 years ago
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