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gayaneshka [121]

4 years ago

9

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 150 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 17°C. Use constant specific heats at room temperature. The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4.

1 answer:

defon4 years ago

8 0

Answer:

A) Rate of heat addition = 228.53 Hp

B) Maximum Air temperature = 1109 °C

Explanation:

We are given;

Cut off ratio; r_c = 1.5

Compression ratio; r = 18

Power produced; W` = 150 Hp

cp = 1.005 kJ/kg·K

cv = 0.718 kJ/kg·K

R = 0.287 kJ/kg·K

k = 1.4

Temperature;T1 = 17°C = 17 + 273 K = 290K

I've attached the rest of the explanation below.

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3 years ago

A vacuum gage connected to a tank reads 30 kPa at a location where the barometric reading is 755 mmHg. Determine the absolute pr

navik [9.2K]

Answer:

Absolute pressure=70.72 KPa

Explanation:

Given that Vacuum gauge pressure= 30 KPa

Barometer reading =755 mm Hg

We know that barometer always reads atmospheric pressure at given situation.So atmospheric pressure is equal to 755 mm Hg.

We know that P= ρ g h

Density of $Hg=13600 \frac{kg}{m^3}$

So P=13600 x 9.81 x 0.755

P=100.72 KPa

We know that

Absolute pressure=atmospheric pressure + gauge pressure

But here given that 30 KPa is a Vacuum pressure ,so we will take it as negative.

Absolute pressure=atmospheric pressure + gauge pressure

Absolute pressure=100.72 - 30 KPa

So

Absolute pressure=70.72 KPa

8 0

3 years ago

16 . You are turning onto a two-lane road divided by a broken yellow line. You know immediately that:

Over [174]

When a person is turning onto a two-lane road divided by a broken yellow line, you know immediately that you are on a two-way road.

<h3>What is the road about?</h3>

Note that a Yellow centerlines can be seen in roads and it is one that is often used to separate traffic moving in different directions.

Note also that Broken lines can be crossed to allow slower-moving traffic and as such, When a person is turning onto a two-lane road divided by a broken yellow line, you know immediately that you are on a two-way road.

See full question below

You are turning onto a two-lane road divided by a broken yellow line. You know immediately that:

Answers

You are on a two-way road.

You are on a one-way road.

The road is under repair.

You must stay to the left of the broken yellow lines.

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5 0

2 years ago

1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil

kakasveta [241]

Answer:

a) $n = 1000\,users$ , b) $\Delta t_{min} = \frac{1}{30}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$ , c) $n = 10000000\,users$ , $\Delta t_{min} = \frac{1}{3000}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

Explanation:

a) The total number of users that can be accomodated in the system is:

$n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )$

$n = 1000\,users$

b) The length of the side of each cell is:

$l = \sqrt{1\,km^{2}}$

$l = 1\,km$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{30}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$

c) The total number of users that can be accomodated in the system is:

$n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )$

$n = 10000000\,users$

The length of each side of the cell is:

$l = \sqrt{100\,m^{2}}$

$l = 10\,m$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{3000}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

8 0

4 years ago

Calculate the viscosity(dynamic) and kinematic viscosity of airwhen

nikitadnepr [17]

Answer:

(a) dynamic viscosity = $1.812\times 10^{-5}Pa-sec$

(b) kinematic viscosity = $1.4732\times 10^{-5}m^2/sec$

Explanation:

We have given temperature T = 288.15 K

Density $d=1.23kg/m^3$

According to Sutherland's Formula dynamic viscosity is given by

${\mu} = {\mu}_0 \frac {T_0+C} {T + C} \left (\frac {T} {T_0} \right )^{3/2}$ , here

μ = dynamic viscosity in (Pa·s) at input temperature T,

$\mu _0$ = reference viscosity in(Pa·s) at reference temperature T0,

T = input temperature in kelvin,

$T_0$ = reference temperature in kelvin,

C = Sutherland's constant for the gaseous material in question here C =120

$\mu _0=4\pi \times 10^{-7}$

$T_0$ = 291.15

$\mu =4\pi \times 10^{-7}\times \frac{291.15+120}{285.15+120}\times \left ( \frac{288.15}{291.15} \right )^{\frac{3}{2}}=1.812\times 10^{-5}Pa-s$ when T = 288.15 K

For kinematic viscosity :

$\nu = \frac {\mu} {\rho}$

$kinemic\ viscosity=\frac{1.812\times 10^{-5}}{1.23}=1.4732\times 10^{-5}m^2/sec$

3 0

3 years ago