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Sloan [31]
3 years ago
7

Air is to be compressed so that its volume is reduced to half of its original volume if the initial pressure is 200 kPa and the

temperature is 20°C. Assume the ratio of specific heats is 1.4. Of the compression is done isentropically determine the required pressure.
Engineering
1 answer:
Alex_Xolod [135]3 years ago
7 0

Answer:

P_{2} = 527.803\,kPa

Explanation:

The politropic relationship for a isentropic process is:

\frac{P_{2}}{P_{1}} = \left(\frac{V_{1}}{V_{2}}  \right)^{\gamma}

Where \gamma is the ratio of specific heats

The final pressure is:

P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}}\right)^{\gamma}

P_{2} = (200\,kPa)\cdot (2)^{1.4}

P_{2} = 527.803\,kPa

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An insulated rigid tank is divided into two compartments of different volumes. Initially, each compartment contains the same ide
storchak [24]

Answer:

Explanation:

Given that

Mass of 1 = m_1

Mass of 2 = m_2

Temperature in  1 = T_1

Temperature in 2 = T_2

Pressure remains i  the group apartment

The closed system and energy balance is

E_{in}-E_{out}=\Delta E_{system}

The kinetic energy and potential energy are negligible

since it is insulated tank ,there wont be eat transfer from the system

And there is no work involved

\Delta U = 0

Let the final temperature be final temperature

m_1+c_v(T_3-T_1)+m_2c_v(T_3+T_2)=0\\\\m_1+c_v(T_3-T_1)=m_2c_v(T_2-T_3)---(i)

Using mass balance

m_3+m_2+m_1

from eqn i

m_1+c_v(T_3-T_1)=m_2c_v(T_2-T_3)m_1T_3-m_1T_1=m_2T_2-m_2T_3\\\\m_1T_3+m_2T_3=m_2T_2+m_1T_1\\\\(m_1+m_2)T_3=m_1T_1+m_2T_2\\\\(m_1)T_3=m_1T_1+m_2T_2\\\\T_3=\frac{m_1T_1_m_2T_2}{m_3}

Therefore the final temperature can be express as

\large \boxed {T_3=\frac{m_1}{m_3} T_1+\frac{m_2}{m_3}T_2 }

8 0
3 years ago
PLEASE HELP, THANK YOU
Elena L [17]

Answer:

6,3,2,5,1,4 because they jst are

Explanation:

3 0
3 years ago
The diffusion coefficients for iron in nickel are given at two temperatures:T (K)D (m2/s)12739.4 × 10–1614732.4 × 10–14(a) Deter
hram777 [196]

Answer:

The diffusion coefficients for iron in nickel are given at two temperatures:

T (K)        1273          1473

D (m^{2}/s) 9.4 × 10^{-16}    2.4 × 10^{-14}

(a) Determine the values of Do and the activation energy Qd.

(b) What is the magnitude of D at 1100°C (1373 K)?

<em>A </em>

<em>The pre-exponential factor Do = 2.1 x </em>10^{-16}<em></em>

<em>The activation energy Qd = 252,609 J/mol</em>

<em>B</em>

<em>The diffusion coefficient D= 5.14 x </em>10^{-15}<em></em>

Explanation:

The full explanation is contained in the attached images;

5 0
4 years ago
A data bus can be visualized as a multilane highway
iris [78.8K]

Answer:

B. with each component having an individual address

Explanation:

Data bus is a system within a computer or device, consisting of a connector or set of wires, that provides transportation for data. Data bus needs an address unique to each component in order to deliver the right data to the right place. Every memory location has a unique binary address. A microprocessor architecture is mainly composed of two main buses: The data bus and the address bus.

3 0
3 years ago
a cubical box 20-cm on a side is contructed from 1.2 cm thick concrete panels. A 100-W light bulb is sealed inside the box. What
Flura [38]

Answer:

Temperature on the inside ofthe box

Explanation:

The power of the light bulb is the rate of heat conduction of the bulb, dq/dt = 100 W

The thickness of the wall, L = 1.2 cm = 0.012m

Length of the cube's side, x = 20cm = 0.2 m

The area of the cubical box, A = 6x²

A = 6 * 0.2² = 6 * 0.04

A = 0.24 m²

Temperature of the surrounding, T_0 = 20^0 C = 273 + 20 = 293 K

Temperature of the inside of the box, T_{in} = ?

Coefficient of thermal conductivity, k = 0.8 W/m-K

The formula for the rate of heat conduction is given by:

dq/dt = \frac{kA(T_{in} - T_0)}{L} \\\\100 = \frac{0.8*0.24(T_{in} - 293)}{0.012}\\\\T_{in} - 293 = \frac{100 * 0.012}{0.8*0.24} \\\\T_{in} - 293 = 6.25\\\\T_{in} = 293 + 6.25\\\\T_{in} = 299.25 K\\\\T_{in} = 299.25 - 273\\\\T_{in} = 26.25^0 C

5 0
4 years ago
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