The point of contact of two pitch circles of mating gears is called?

A heat engine operates between a source at 477°C and a sink at 27°C. If heat is supplied to the heat engine at a steady rate of

lara [203]

Answer:

$T_C = 27+273.15 = 300.15 K$

$T_H = 477+273.15 = 750.15 K$

And replacing in the Carnot efficiency we got:

$e= 1- \frac{300.15}{750.15}= 0.59988 = 59.98 \%$

$W_{max}= e* Q_H = 0.59988 * 65000 \frac{KJ}{min}= 38992.2 \frac{KJ}{min}$

Explanation:

For this case we can use the fact that the maximum thermal efficiency for a heat engine between two temperatures are given by the Carnot efficiency:

$e = 1 -frac{T_C}{T_H}$

We have on this case after convert the temperatures in kelvin this:

$T_C = 27+273.15 = 300.15 K$

$T_H = 477+273.15 = 750.15 K$

And replacing in the Carnot efficiency we got:

$e= 1- \frac{300.15}{750.15}= 0.59988 = 59.98 \%$

And the maximum power output on this case would be defined as:

$W_{max}= e* Q_H = 0.59988 * 65000 \frac{KJ}{min}= 38992.2 \frac{KJ}{min}$

Where $Q_H$ represent the heat associated to the deposit with higher temperature.

4 0

3 years ago

2. One of the many methods used for drying air is to cool the air below the dew point so that condensation or freezing of the mo

REY [17]

Answer:

0.5°c

Explanation:

Humidity ratio by mass can be expressed as

the ratio between the actual mass of water vapor present in moist air - to the mass of the dry air

Humidity ratio is normally expressed in kilograms (or pounds) of water vapor per kilogram (or pound) of dry air.

Humidity ratio expressed by mass:

x = mw / ma (1)

where

x = humidity ratio (kgwater/kgdry_air, lbwater/lbdry_air)

mw = mass of water vapor (kg, lb)

ma = mass of dry air (kg, lb)

It can be as:

x = 0.005 (100) / [(100 - 100)]

x = 0.005 x 100 / (100 - 100)

x = 0.005 x 100 / 0

x = 0.5°c

So the temperature to which atmospheric air must be cooled in order to have humidity ratio of 0.005 lb/lb is 0.5°c

6 0

3 years ago

The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta

tatyana61 [14]

Answer:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

$H_o =r x mv=rxL$

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

$\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1$

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is $I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2$ ".

If we analyze the staritning point we see that the initial velocity can be founded like this:

$v_o =\omega r_{OIC}=\omega (0.15m)$

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

$H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af$

$0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)$

And if we integrate the left part and we simplify the right part we have

$1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega$

And if we solve for $\omega$ we got:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

8 0

3 years ago

Air expands through a turbine operating at steady state. At the inlet p1 = 150 lbf/in^2, T1 = 1400R and at the exit p2 = 14.8 lb

Paraphin [41]

Answer:

The power developed in HP is 2702.7hp

Explanation:

Given details.

P1 = 150 lbf/in^2,

T1 = 1400°R

P2 = 14.8 lbf/in^2,

T2 = 700°R

Mass flow rate m1 = m2 = m = 11 lb/s Q = -65000 Btu/h

Using air table to obtain the values for h1 and h2 at T1 and T2

h1 at T1 = 1400°R = 342.9 Btu/h

h2 at T2 = 700°R = 167.6 Btu/h

Using;

Q - W + m(h1) - m(h2) = 0

W = Q - m (h2 -h1)

W = (-65000 Btu/h ) - 11 lb/s (167.6 - 342.9) Btu/h

W = (-65000 Btu/h ) - (-1928.3) Btu/s

W = (-65000 Btu/h ) * {1hr/(60*60)s} - (-1928.3) Btu/s

W = -18.06Btu/s + 1928.3 Btu/s

W = 1910.24Btu/s

Note; Btu/s = 1.4148532hp

W = 2702.7hp

5 0

3 years ago

How many robots does bailey nursery own

givi [52]

Answer:

The Bailey family has flourished during its business’ 110-year history. But Bailey Nurseries’ leaders still operate with the belief that the family doesn’t always know best. The company has grown from a one-man operation selling fruit trees and ornamental shrubs to one of the largest wholesale nurseries in the United States, thanks to insights from those who are family and those who aren’t.

“For a business to thrive, you have to ask for outside help,” says Terri McEnaney, president of the Newport-based company and a fourth-generation family member. “We get an outside perspective through family business programs, advisors and our board, because you can get a bit ingrained in your own way of thinking.”

When Bailey Nurseries chose its current leader in 2000, it brought in a facilitator who gathered insights from key employees, board members and owners. Third-generation leaders (and brothers) Gordie and Rod Bailey picked Rod’s daughter McEnaney, who had experience both inside and outside the company.

Explanation:

5 0

2 years ago