1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
raketka [301]
3 years ago
5

A rigid tank contains 1 kg of oxygen (O2) at p1 = 35 bar, T1 = 180 K. The gas is cooled until the temperature drops to 150 K. De

termine the volume of the tank, in m3, and the final pressure, in bar, using the:
(a) ideal gas equation of state.
(b) Redlich–Kwong equation.
Engineering
1 answer:
andreyandreev [35.5K]3 years ago
3 0

Answer:

a. Volume = 13.36 x 10^-3 m³ Pressure = 29.17 bar  b. Volume = 14.06 x 10^-3 m³ Pressure = 22.5 bar

Explanation:

Mass of O₂ = 1kg, Pressure (P1) = 35bar, T1= 180K, T2= 150k Molecular weight of O₂ = 32kg/Kmol

Volume of tank and final pressure using a)Ideal Gas Equation and b) Redlich - Kwong Equation

a. PV=mRT

V = {1 x (8314/32) x 180}/(35 x 10⁵) = 13.36 x 10^-3

Since it is a rigid tank the volume of the tank must remain constant and hnece we can say

T2/T1 = P2/P1, solving for P2

P2 = (150/180) x 35 = 29.17bar

b. P1 = {RT1/(v1-b)} - {a/v1(v1+b)(√T1)}

where R, a and b are constants with the values of, R = 0.08314bar.m³/kmol.K, a = 17.22(m³/kmol)√k, b = 0.02197m³/kmol

solving for v1

35 = {(0.08314 x 180)/(v1 - 0.02197)} - {17.22/(v1)(v1 + 0.02197)(√180)}

35 = {14.96542/(v1-0.02197)} - {1.2835/v1(v1 + 0.02197)}

Using Trial method to find v1

for v1 = 0.5

Right hand side becomes =  {14.96542/(0.5-0.02197)} - {1.2835/0.5(0.5 + 0.02197)} = 31.30 ≠ Left hand side

for v1 = 0.4

Right hand side becomes =  {14.96542/(0.4-0.02197)} - {1.2835/0.4(0.4 + 0.02197)} = 39.58 ≠ Left hand side

for v1 = 0.45

Right hand side becomes =  {14.96542/(0.45-0.02197)} - {1.2835/0.45(0.45 + 0.02197)} = 34.96 ≅ 35

Specific Volume = 35 m³/kmol

V = m x Vspecific/M = (1 x 0.45)/32 = 14.06 x 10^-3 m³

For Pressure P2, we know that v2= v1

P2 = {RT2/(v2-b)} - {a/v2(v2+b)(√T2)} = {(0.08314 x 150)/(0.45 - 0.02197)} - {17.22/(0.45)(0.45 + 0.02197)(√150)} = 22.5 bar

You might be interested in
An 80-percent-efficient pump with a power input of 20 hp is pumping water from a lake to a nearby pool at a rate of 1.5 ft3/s th
photoshop1234 [79]

Answer:

mechanical power used to overcome frictional effects in piping is 2.37 hp

Explanation:

given data

efficient pump = 80%

power input = 20 hp

rate = 1.5 ft³/s

free surface = 80 ft

solution

we use mechanical pumping power delivered to water is

{W_{u}}= \eta  {W_{pump}}  .............1

put here value

{W_{u}}  = (0.80)(20)

{W_{u}} = 16 hp

and

now we get change in the total mechanical energy of water is equal to the change in its potential energy

\Delta{E_{mech}} = {m} \Delta pe   ..............2

\Delta {E_{mech}} = {m} g \Delta z  

and that can be express as

\Delta {E_{mech}} = \rho Q g \Delta z     ..................3

so

\Delta {E_{mech}} = (62.4lbm/ft^3)(1.5ft^3/s)(32.2ft/s^2)(80ft)[\frac{1lbf}{32.2lbm\cdot ft/s^2}][\frac{1hp}{550lbf \cdot ft/s}]      ......4

solve it we get

\Delta {E_{mech}} = 13.614 hp

so here

due to frictional effects, mechanical power lost in piping

we get here

{W_{frict}} = {W_{u}}-\Delta {E_{mech}}  

put here value

{W_{frict}} = 16 -13.614

{W_{frict}} = 2.37  hp

so mechanical power used to overcome frictional effects in piping is 2.37 hp

4 0
3 years ago
Refrigerant-134a enters the expansion valve of a refrigeration system at 120 psia as a saturated liquid and leaves at 20 psia. D
Shkiper50 [21]

Solution :

$P_1 = 120 \ psia$

$P_2 = 20 \ psia$

Using the data table for refrigerant-134a at P = 120 psia

$h_1=h_f=40.8365 \ Btu/lbm$

$u_1=u_f=40.5485 \ Btu/lbm$

$T_{sat}=87.745^\circ  F$

∴ $h_2=h_1=40.8365 \ Btu/lbm$

For pressure, P = 20 psia

$h_{2f} = 11.445 \ Btu/lbm$

$h_{2g} = 102.73 \ Btu/lbm$

$u_{2f} = 11.401 \ Btu/lbm$

$u_{2g} = 94.3 \ Btu/lbm$

$T_2=T_{sat}=-2.43^\circ  F$

Change in temperature, $\Delta T = T_2-T_1$

                                         $\Delta T = -2.43-87.745$

                                           $\Delta T=-90.175^\circ  F$

Now we find the quality,

$h_2=h_f+x_2(h_g-h_f)$

$40.8365=11.445+x_2(91.282)$

$x_2=0.32198$

The final energy,

$u_2=u_f+x_2.u_{fg}$

   $=11.401+0.32198(82.898)$

   $=38.09297 \ Btu/lbm$

Change in internal energy  

$\Delta u= u_2-u_1$

   = 38.09297-40.5485

  = -2.4556        

5 0
2 years ago
A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h
hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

D_{1} = 150 mm

A_{1}= \frac{\pi }{4}\times 150^{2}

              = 17671.45 mm^{2}

D_{2} = 250 mm

A_{2}= \frac{\pi }{4}\times 250^{2}

              = 49087.78 mm^{2}

The centrifugal force acting on the flywheel is fiven by

F = M ( R_{2} - R_{1} ) x w^{2} ------------(1)

Here F = ( -UTS x A_{1} + UCS x A_{2} )

Since density, \rho = \frac{M}{V}

                        \rho = \frac{M}{A\times t}

                        M = \rho \times A\times tM = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t

                        M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37

                        M = 8252963901

∴ R_{2} - R_{1} = 50 mm

∴ F = 763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}

  F = 33618968.38 N --------(2)

Now comparing (1) and (2)

33618968.38 = 8252963901\times 50\times \omega ^{2}

∴ ω = 4036.61

We know

\omega = \frac{2\pi N}{60}

4036.61 = \frac{2\pi N}{60}

∴ N = 38546.82 rpm

7 0
3 years ago
The __________________ refers to the main screen of the computer.
Sedbober [7]

Answer:

<em>D</em><em>e</em><em>s</em><em>k</em><em>t</em><em>o</em><em>p</em>

Explain:

<em>Desktop refers to the main screen of the computer. It is the first screen you see after logging in. The desktop’s appearance can vary widely because it is highly customizable, but generally desktops will feature a large image, icons, and a taskbar(covered later on this page).</em>

4 0
2 years ago
A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28
Delicious77 [7]

Answer:

5984.67N

Explanation:

A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?

from continuity equation

v1A1=v2A2

equation of continuity

v1=4ft /s=1.21m/s

d1=14 inch=.35m

d2=14-2=0.304m

A1=pi*d^2/4

0.096m^2

a2=0.0706m^2

from continuity once again

1.21*0.096=v2(0.07)

v2=1.65

force on the pipe

(p1A1- p2A2) + m(v2 – v1)

from bernoulli

p1 + ρv1^2/2 = p2 + ρv2^2/2

difference in pressure or pressure drop

p1-p2=2psi

13.789N/m^2=rho(1.65^2-1.21^2)/2

rho=21.91kg/m^3

since the pipe is cylindrical

pressure is egh

13.789=21.91*9.81*h

length of the pipe is

0.064m

AH=volume of the pipe(area *h)

the mass =rho*A*H

0.064*0.07*21.91

m=0.098kg

(193053*0.096- 179263.6* 0.07) + 0.098(1.65 – 1.21)

force =5984.67N

4 0
3 years ago
Other questions:
  • What are the advantages of using 3D ink jet printing?
    10·1 answer
  • What causes decay in the amplitudes of vibration?
    11·1 answer
  • A local surf report provides the height of the wave from the trough to the crest of the wave. How does this relate to the wave’s
    11·1 answer
  • Steel riverts in aluminium drain gutters leak after two years. is it galvanic corrosion? ​
    5·1 answer
  • Although many countries have issues with soil erosion due to deforestation, some of the most serious effects are seen
    8·1 answer
  • Which of the following machine parts always require
    12·1 answer
  • Which process made making copies of technical drawings easier?
    8·1 answer
  • How many millimeters are there in a centimeter?
    10·1 answer
  • What is the difference between the elements of design and the principles of design? Define at
    7·1 answer
  • At time t the resultant force on a particle, of mass 250kg is (300ti-400tj)N. Initially, the particle is at the origin and is mo
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!