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An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 27°C, and 75

katen-ka-za [31]

Answer:

Part A) 3899 kPa

Part B) 392.33 kJ/kg

Part C) 0.523

Part D) 495 kPa

Explanation:

Part A

First from the temperature at state 1 the relative specific volume and the internal energy at that state are determined from:

$u_{1}$ = 214.07 kJ/kg

$\alpha$ $r_{1}$ = 621.2

The relative specific volume at state 2 is obtained from the compression ratio:

$\alpha$ $r_{2}$ = $\frac{\alpha r_{1} }{r}$

=621.2/ 8

= 77.65

From this the temperature and internal energy at state 2 can be determined using interpolation with data from A-17(table):

$T_{2}$ = 673 K

$u_{2}$ = 491.2 kJ/kg

The pressure at state 2 can be determined by manipulating the ideal gas relations at state 1 and 2:

$P_{2}$ = $P_{1} r\frac{T_{2} }{T_{1} }$

= 95*8*673/300

= 1705 kPa

Now from the energy balance for stage 2-3 the internal energy at state 3 can be obtained:

$deltau_{2-3} =q_{in}\\ u_{3} -u_{2} =q_{in}\\u_{3}=u_{2}+q_{in}$

= 1241.2 kJ/kg

From this the temperature and relative specific volume at state 3 can be determined by interpolation with data from A-17(table):

$T_{3}$ = 1539 K

$\alpha r_{3}$ = 6.588

The pressure at state 3 can be obtained by manipulating the ideal gas relations for state 2 and 3:

$P_{3} =P_{2} \frac{T_{3} }{T_{2} }$

= 3899 kPa

Part B

The relative specific volume at state 4 is obtained from the compression ratio:

$\alpha r_{4}= r\alpha r_{3}$

= 52.7

From this the temperature and internal energy at state 4 can be determined by interpolation with data from A-17:

$T_{4}$ =775 K

$u_{4}$ = 571.74 kJ/kg

The net work output is the difference of the heat input and heat rejection where the heat rejection is determined from the decrease in internal energy in stage 4-1:

$w=q_{in}-q_{out}\\q_{in}-(u_{4} -u_{1} )\\=392.33 kJ/kg$

Part C

The thermal efficiency is obtained from the work and the heat input:

η= $\frac{w}{q_{in} }$

=0.523

Part D

The mean effective pressure is determined from its standard relation:

MEP= $\frac{w}{\alpha_{1}- \alpha_{2} }$

= $\frac{w}{\alpha_{1}(1- \frac{1}{r} }$

= $\frac{rwP_{1} }{RT_{1} (r-1) }$

=495 kPa

8 0

4 years ago

Which organism is the primary consumer in this food chain?

VashaNatasha [74]

Rabbit hope this u out hollow

3 0

3 years ago

How can I rewrite the equation a - b = d using addition?

Flauer [41]

Explanation:

A=b+d that is the way to rewrite the equation

5 0

3 years ago

Two workers are sliding 290 kg crate across the floor. One worker pushes forward on the crate with a force of 430 N while the ot

frozen [14]

Answer:

0.285

Explanation:

Given two forces of different magnitude, it is important to note that the product of normal force and coefficient of kinetic friction should be equal to the sum of these two forces at equilibrium. Therefore, this can be Mathematically expressed as:

$N/\mu_k=F_1+F_2\\\\$

where N is normal force, $\mu$ is coefficient of static friction, F is force and subscripts 1 and 2 represent larger and smaller magnitude forces respectively. Making $\mu$ the subject of the formula then

$\mu_k=\frac{F_1+F_2}{N}$

Since normal force N is also given by mg where m is mass of object and g is acceleration due to gravity then substituting N with mg we obtain that

$\mu_k=\frac{F_1+F_2}{mg}$ and substituting the figures given in the question, taking g as 9.81 we obtain that

$\mu_k=\frac { 430 N+380 N}{290\times 9.81}=0.285$

Hence,the coefficient of kinetic energy is 0.285 as calculated

4 0

4 years ago

The greatest ocean depths on the earth are found in the marianas trench near the philippines, where the depth of the bottom of t

Doss [256]

To find the pressure with a given data for the height, you are asked to get the hydraulic pressure. Hydraulic pressure has the following formula:

P = density*acceleration due to gravity*height

Assume that the density of seawater is the same as that for pure water,density = 1000 kg/m^3.

P = 1000 kg/m3*9.81m/s2*9100m
P = 89271000 Pascals or 89.271 megapascals

3 0

3 years ago