Ask question

Ask question

All categories

tankabanditka [31]

2 years ago

10

An operating gear box (transmission) has 350 hp at its input shaft while 250. hp are delivered to the output shaft. The gear box

has a steady state surface temperature of 180. °F. Determine the rate of entropy production by the gear box.

1 answer:

True [87]2 years ago

8 0

Answer:

Rate of Entropy =210.14 J/K-s

Explanation:

given data:

power delivered to input = 350 hp

power delivered to output = 250 hp

temperature of surface = 180°F

rate of entropy is given as

$Rate\ of\ entropy = \frac{Rate\ of \ heat\ released}{Temperature}$

T = 180°F = 82°C = 355 K

Rate of heat = (350 - 250) hp = 100 hp = 74600 W

Rate of Entropy $= \frac{74600}{355} = 210.14 J/K-s$

You might be interested in

How can you evaluate whether the slope of the dependent variable with an independent variable is the same for each level of the

s2008m [1.1K]

Answer:

By running multiple regression with dummy variables

Explanation:

A dummy variable is a variable that takes on the value 1 or 0. Dummy variables are also called binary

variables. Multiple regression expresses a dependent, or response, variable as a linear

function of two or more independent variables. The slope is the change in the response variable. Therefore, we have to run a multiple regression analysis when the variables are measured in the same measurement.The number of dummy variables you will need to capture a categorical variable

will be one less than the number of categories. When there is no obvious order to the categories or when there are three or more categories and differences between them are not all assumed to be equal, such variables need to be coded as dummy variables for inclusion into a regression model.

3 0

2 years ago

A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po

Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: $4.3\cdot 10^{-15} J$

Explanation:

a)

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

and is located at the origin (x=0, y=0)

Charge 2 is

$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

$r_{1A}=0.40 m$

So the potential due to charge 2 is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V$

The distance of point A from charge 2 is

$r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m$

So the potential due to charge 1 is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V$

Therefore, the net potential at point A is

$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

b)

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

$r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m$

So the potential due to charge 1 at point B is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V$

The distance of charge 2 from point B is

$r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m$

So the potential due to charge 2 at point B is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V$

Therefore, the net potential at point B is

$V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V$

So the potential difference is

$V_B-V_A=-29,500 V-(-2700 V)=-26,800 V$

c)

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

$W=q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

In this problem, we have:

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=-26,800 V$ is the potential difference

Therefore, the work required on the electron is

$W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J$

4 0

3 years ago

we wish to send at a rate of 10Mbits/s over a passband channel. Assuming that an excess bandwidth of 50% is used, how much bandw

gayaneshka [121]

Answer:

QPSK: 7.5 MHz

64-QAM:2.5 MHz

64-Walsh-Hadamard: 160 MHz

Explanation:

See attached picture.

6 0

3 years ago

Q10: Technician A says that nearly all brands of scan tools will pull DTCs from the ABS

Llana [10]

Hey! How are you? My name is Maria, 19 years old. Yesterday broke up with a guy, looking for casual sex.

Write me here and I will give you my phone number - *pofsex.com*

My nickname - Lovely

7 0

3 years ago

What is the average linear (seepage) velocity of water in an aquifer with a hydraulic conductivity of 6.9 x 10-4 m/s and porosit

jeka94

Answer:

a. 0.28

Explanation:

Given that

porosity =30%

hydraulic gradient = 0.0014

hydraulic conductivity = 6.9 x 10⁻4 m/s

We know that average linear velocity given as

$v=\dfrac{K}{n_e}\dfrac{dh}{dl}$

$v=\dfrac{6.9\times 10^{-4}}{0.3}\times0.0014\ m/s$

$v=3.22\times 10^{-6}\ m/s$

The velocity in m/d ( 1 m/s =86400 m/d)

v= 0.27 m/d

So the nearest answer is 'a'.

a. 0.28

4 0

2 years ago