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3 years ago

I’m bored let’s text

Engineering

2 answers:

aleksandrvk [35]3 years ago

7 0

Answer:

anyways thnx for the points and have great day

Nuetrik [128]3 years ago

3 0

Answer:

hi same here. hru btw

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The heat input to an Otto cycle is 1000kJ/kg. The compression ratio is 8 and the pressure and temperature at the beginning of th

Naddik [55]

Answer:

a. $T_3=2027.1 K,P_3=5750.22 KPa$

b. $\eta =0.564$

c. Work out put = 564 KJ/kg

d. $P_{mean}=786.61 KPa$

Explanation:

Given that

Heat in put = 1000 KJ/kg

Compression ratio,r = 8

$T_1=15 C$

$P_1=100 KPa$

Process 1-2

$\dfrac{T_2}{T_1}=r^{\gamma -1}$

$\dfrac{T_2}{273+15}=8^{1.4 -1}$

$T_2=661.65 K$

$\dfrac{P_2}{P_1}=r^{\gamma}$

$\dfrac{P_2}{100}=8^{1.4}$

$P_2=1837.9 KPa$

Process 2-3

We know that for air

$C_v=0.71\ \frac{KJ}{kg-K}$

$Q=C_v(T_3-T_2)$

$1000=0.71(T_3-661.5)$

$T_3=2027.1 K$

$\dfrac{P_3}{P_2}=\dfrac{T_3}{T_2}$

$\dfrac{P_3}{1837.9}=\dfrac{2027.1}{661.5}$

$P_3=5750.22 KPa$

We know that efficiency of otto cycle

$\eta =1-\dfrac{1}{r^{\gamma -1}}$

$\eta =1-\dfrac{1}{8^{1.4-1}}$

$\eta =0.564$

$\eta =\dfrac{Work\ out\ put}{Heat\ in\ put}$

$0.564 =\dfrac{Work\ out\ put}{1000}$

Work out put = 564 KJ/kg

$v_1=\dfrac{RT_1}{P_1}$

$v_1=\dfrac{0.287\times 288}{100}$

$v_1=0.82656\ \frac{m^3}{kg}$

$v_2=\dfrac{0.82656}{8}\ \frac{m^3}{kg}$

$v_2= 0.103\frac{m^3}{kg}$

$Work\ out\ put\ = P_{mean}\times (v_1-v_2)$

$564= P_{mean}\times (0.82-0.103)$

$P_{mean}=786.61 KPa$

8 0

3 years ago

Two kg of N2 at 450 K, 7 bar is contained in a rigid tank connected by a valve to another rigid tank holding 1 kg of O2 at 300 K

11111nata11111 [884]

Answer:WHo knows.Explanation:

7 0

3 years ago

What is the air change rate (ACH) for a 100 ft^2 (9.3 m^2) space with a 10 ft (3.0 m) ceiling and an airflow rate of 200 cfm (95

kakasveta [241]

Answer:

The ACH is 12/h

Solution:

As per the question:

Area of the space, $A_{s} = 100 ft^{2}$

Height of the given space, h = 10 ft

Air flow rate, $Q_{a} = 200 cfm$

Now, to find the Air Change Rate (ACH):

We calculate the Volume of the given space:

$V_{s} = A_{s}\times h = 100\times 10 = 1000 ft^{3}$

Now, the ACH per min:

= $\frac{V_{s}}{Q_{a}} = \frac{1000}{200} = 5/min$

Now, ACH per hour:

= $\frac{60}{5} = 12/h$

3 0

3 years ago

which of the following tools is used for measuring small diameter holes which a telescoping gauge cannot fit into? A. telescopin

Alex

Answer:

C. Dial indicator

Explanation:

This meassers small diameters

7 0

2 years ago

Choose the best data type for each of the following so that any reasonable value is accommodated but no memory storage is wasted

stiks02 [169]

Answer:

Explanation:

Part (a):

Statement : The number of siblings you have

Suitable Data type : Byte

Typical Value : From -128 and up to 127

Explanation: Byte data type is the most suitable since it can covers minimum and maximum number of siblings one can have.

Part (b):

Statement : Your final grade in this class

Suitable Data type : Char

Typical Value : 1 byte

Explanation: Grades is in the form of alphabetical letter which is either A, B, C, D, F or E which can be stored in character data type.

Part (c):

Statement : Population of Earth

Suitable Data type : Long

Maximum Value : 9223372036854775807

Explanation: Long Data takes up to 8 bytes and can store up to 9223372036854775807 which can cater for more than 36 billion. The population of earth is only around 7 billion currently making Long data type the most suitable data type to store earth population.

Part (d):

Statement : Population of US Country

Suitable Data type : Integer

Typical Value :2147483647

Explanation: Integer data type takes up to 4 bytes and can store up to 2147483647 making it suitable to store U.S population.

Part (e):

Statement : The number of passengers on bus

Suitable Data type : Byte

Typical Value :From -128 up to 127

Explanation: The typical maximum number of passengers of a bus are only around 72. Byte data type is the most suitable since it can cater the number up to 127.

Part (f):

Statement : Player's score in a Scrabble game

Suitable Data type : Short

Typical Value : 32767

Explanation: The maximum point can be scored in the Scrabble game is only 830 therefore the most suitable data type for this case is the short data type.

Part (g):

Statement : One team's score in a Major League Baseball game

Suitable Data type : Byte

Typical Value : From -128 up to 127

Explanation: The maximum point can be scored in the Base ball game is only 49 therefore the most suitable data type for this case is the Byte data type since it can cater up to 127.

Part (h):

Statement : The year an historical event occurred

Suitable Data type : Short

Maximum Value: 32767

Explanation: The historic event year can be any number from 1 to 2020 therefore the most suitable data type is the short data type.

Part (i):

Statement : The number of legs on an animal

Suitable Data type : Short

Maximum Value: 32767

Explanation: The most number of legs found are 750 legs therefore the most suitable data type is the short data type which can cater up to 32767.

Part (j):

Statement : The Price of an automobile

Suitable Data type : Float

Maximum Value: 340282350

Explanation: The most expensive car is around 15 million therefore the most suitable data type is the float data type which can cater up to 340 million.

3 0

3 years ago