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Alenkasestr [34]
3 years ago
14

An ideal Diesel Cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the maximum air temperature and the rat

e of heat addition to this cycle when it produces 200 hp of power; the cycle is repeated 1200 times per minute; and the state of the air at the beginning of the compression is 95 kPa and 17oC. Use constant specific heats at room temperature.
Engineering
1 answer:
Mariulka [41]3 years ago
5 0

Answer:

A) Q'_in = 304.66 hp

B) T_max = 1109.29 °C

Explanation:

We are given;

compression ratio; r = 18

cutoff ratio; r_c = 1.5

Power; W' = 200 hp

Temperature; T1 = 17°C = 290 K

A) To find the rate of heat addition, we will use the derived formula;

Q'_in = (W' × r^(k - 1) × k(r_c - 1))/(r^(k - 1) × k(r_c - 1)) - ((r_c)^(k) - 1)

Where k is a constant = 1.4

Thus;

Q'_in = (200 × 18^(1.4 - 1) × 1.4(1.5 - 1))/((18^(1.4 - 1) × 1.4(1.5 - 1)) - ((1.5^(1.4)) - 1)

Q'_in = 304.66 hp

B) To find the maximum temperature, we will use the Formula;

T_max = T1 × r^(k - 1) × r_c

T_max = 290 × 18^(1.4 - 1) × 1.5

T_max = 1382.29 K

Converting to °C gives;

T_max = 1109.29 °C

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Explanation:

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At 99% confidence, Z(critical) = 2.58

That gives 99% confidence interval as,

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The value of the lower limit is,

900 - 25.8 = 874.2

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¿Cuál era probablemente el activo más valioso de Persia?
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su suministro de aceite

Explanation:

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The gas expanding in the combustion space of a reciprocating engine has an initial pressure of 5 MPa and an initial temperature
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Answer:

a). Work transfer = 527.2 kJ

b). Heat Transfer = 197.7 kJ

Explanation:

Given:

P_{1} = 5 Mpa

T_{1} = 1623°C

                       = 1896 K

V_{1} = 0.05 m^{3}

Also given \frac{V_{2}}{V_{1}} = 20

Therefore, V_{2} = 1  m^{3}

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C_{V} = 0.8 kJ / kg-K

Also given : P_{1}V_{1}^{1.25}=C

   Therefore, P_{1}V_{1}^{1.25} = P_{2}V_{2}^{1.25}

                     5\times 0.05^{1.25}=P_{2}\times 1^{1.25}

                     P_{2} = 0.1182 MPa

a). Work transfer, δW = \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}

                                  \left [\frac{5\times 0.05-0.1182\times 1}{1.25-1}  \right ]\times 10^{6}

                              = 527200 J

                             = 527.200 kJ

b). From 1st law of thermodynamics,

Heat transfer, δQ = ΔU+δW

   = \frac{mR(T_{2}-T_{1})}{\gamma -1}+ \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}

  =\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W

  =\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200

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6 0
3 years ago
Suppose we have a database for an investment firm, consisting of the following attributes: B (broker), O (office of a broker), I
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Answer:

Given, FDs are:

S -> D

I -> B

IS -> Q

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a)

"I" and "S" must be there in any candidate key because they do not appear on the right side of any functional dependency.

The only candidate key is: IS

IS -> ISBDQO

b)

Decomposition of R into 3NF: (I, B), (S, D), (B, O), (I, S, Q)

c)

Decomposition of R into BCNF:

Decompose R by I → B into R1 = (I, B) and R2 = (I, O, S, Q, D).

R1 is in BCNF

Decompose R2 by S → D into R21 = (S, D) and R22 = (O, I, S, Q).

R21is in BCNF

Decompose R22 by I → O into R221 = (I, O) and R222 = (I, S, Q).

R221 is in BCNF.

R222 is in BCNF.

The decomposition is: (I, B), (S, D), (I, O), (I, S, Q)

We can also write it as: (I, B), (S, D), (B, O), (I, S, Q)

Explanation:

The answer above is rendered in a very explanatory way.

8 0
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