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nadya68 [22]
3 years ago
14

At 1 atm, nitrogen (N2) molecules become a liquid below 77 K. Thus the vapor pressure of liquid nitrogen is 1 atm at 77 K. Use t

his information to estimate the binding energy of a nitrogen molecule within the liquid. Note: You may assume that the equilibrium condition for this system is NG/V = nre-A/KT', with ny =1.93 x 1031 m-3, and that the nitrogen gas is ideal.
a. 0.081 eV
b. 81 eV
c. 0.0081 eV
d. 0.81 eV
e. 8.1 eV
Engineering
1 answer:
Bogdan [553]3 years ago
6 0
What did he sayyyyyyy
You might be interested in
Steam heated at constant pressure in a steam generator enters the first stage of a supercritical reheat cycle at 28 MPa, 5208C.
algol [13]

This question is incomplete, the complete question is;

Steam heated at constant pressure in a steam generator enters the first stage of a supercritical reheat cycle at 28 MPa, 520°C. Steam exiting the first-stage turbine at 6 MPa is reheated at constant pressure to 500°C. Each turbine stage has an isentropic efficiency of 78% while the pump has an isentropic efficiency of 82%. Saturated liquid exits the condenser that operates at constant pressure of 6 kPa.

Determine the quality of the steam exiting the second stage of the turbine and the thermal efficiency.

Answer:

- the quality of the steam exiting the second stage of the turbine is 0.9329  

- the thermal efficiency is 36.05%  

Explanation:

get the properties of steam at pressure p1 = 28 MPa and temperature T2 = 520°C .

Specific enthalpy h1= 3192.3 kJ/kg

Specific entropy s1 = 5.9566 kJ/kg.K  

Process 1 to 2s is isentropic expansion process in the turbine

S1 = S2s

get the enthalpy at state 2s at pressure p2 = 6 MPa and S2s = 5.9566 kJ/kg.K

h2s = 2822.2 kJ/kg

get the enthalpy at state 2 using isentropic turbine efficiency of the turbine. nT1 = (h1 - h2) / (h1 - h2s)

0.78 = (3192.3 - h2) / (3192.3 - 2822.2)

h2 = 2903.6 kJ/kg

get the enthalpy at state 3 at pressure p2 = p3 = 6 MPa and T3 = 500°C

h3 = 3422.2 kJ/kg

s3 = 6.8803 kJ/kg.K

Process 3 to 4s is isentropic expansion process in the turbine

S3 = S4s

get the enthalpy at state 4s at pressure p4s = p4 = 6 kPa and S4s = 6.8803 kJ/kg.K

h4s = 2118.8 kJ/kg

get the enthalpy at state 4 using isentropic turbine efficiency of the turbine. nT2 = (h3 - h4) / (h3 - h4s)

0.78 = (3422.2 - h4) / ( 3422.2 - 2118.8 )

h4 = 2405.5 kJ/kg

get the properties at pressure, p5 = 6 kPa

h5 = hf

= 151.53 kJ/kg

v5 = Vf  

= 0.0010064 m³/kg  

get the enthalpy at state 6 using isentropic pump efficiency of the turbine, at

p6 = p1 = 28 MPa

np = v5( p6 - p5) / (h6 - h5)

0.82 =  ((0.0010064)( 28000 - 6)) / (h6 - 151.53)

h6 = 185.89 kJ/kg  

Now to find the quality of the steam at the exit of the second stage of the turbine

At stat4, p4 = 6kPa  

h4f = 151.53 kJ/kg

h4fg = 2415.9 kJ/kg  

h4 = h4f + x4h4fg

2405.5 = 151.53 + (x4 (2415.9))

x4 = 0.9329  

the quality of the steam exiting the second stage of the turbine is 0.9329  

Also to find the efficiency of the power plant, we use the following equation;

n = Wnet / Qin  

= (Wt1 + Wt2 - Wp) / (Q61 + Q23)

=  [(h1 - h2) + (h3 - h4) - (h6 - h5)] / [(h1 - h6) + (h3 - h2)]

[(3192.3 - 2903.6) + (3422.2 - 2405.5) - (185.89 - 151.53)] / [(3192.3 - 185.89) + (3422.2 - 2903.6)]

= 0.3605

n = 36.05%  

therefore the thermal efficiency is 36.05%  

3 0
3 years ago
Calculate the amount of current flowing through a 75-watt light bulb that is connected to a 120-volt circuit in your home.
vodomira [7]

Answer:

I = 0.625 A

Explanation:

Given that,

Power of the light bulb, P = 75 W

Voltage of the circuit, V = 120 V

We need to find the current flowing through it. We know that, Power is given by :

P=V\times I

I is the electric current

I=\dfrac{P}{V}\\\\I=\dfrac{75\ W}{120\ V}\\\\I=0.625\ A

So, the current is 0.625 A.

5 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
Kim is working on the cost estimate and feasible design options for a building. Which stage of a construction plan is Kim workin
melisa1 [442]
Aye she suenebaiaksj
8 0
4 years ago
in verification of ohms law the mass is 100g, initial length is 31, final length is 31.3 what is the extension?​
MissTica

Answer:

24

Explanation:

6 0
3 years ago
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