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slavikrds [6]
3 years ago
15

Carbon dioxide (CO2) is used in industrial and research application, and is sometimes stored at very high pressure in rigid meta

l tanks. Consider a tank with a volume of 0.8 m^3. (a) When the tank is first deliverd to a customer on a hot summer day, the pressure is measure to be 18,000 kPa at a temperature of 35 Celsius. What the mass of CO2 in the tank? (b) If the temperature remains constant, when enough CO2 is used such that the tank pressure is reduced in half to 9,000 kPa, what is the mass of CO2 in the tank?
Engineering
1 answer:
12345 [234]3 years ago
3 0

Answer:

a)m =  247.43 kg

b) m = 123.71 kg

Explanation:

a)

Given data:

volume  =0.8 m^3

P =  18,000 kPa

T =35 degree C = 308 K

By ideal gas equation we have following relation

PV = mRT

where R is gas constant

R = \frac{8.314}{44} = 0.18895 kJ/ kg K

m =\frac{PV}{RT}

m = \frac{18000\times 0.8}{0.18895 \times 308}

m = 247.43 kg

b)

when pressure = 9000 kPa

from ideal gas equation

PV = mRT

where R is gas constant

R = \frac{8.314}{44} = 0.18895 kJ/ kg K

m =\frac{PV}{RT}

m = \frac{9000\times 0.8}{0.18895 \times 308}

m = 123.71 kg

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Explanation:

8 0
3 years ago
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Explain combined normal and shear stresses with sketch. Write the general expression for (a) Normal and shear stresses on inclin
Alborosie

Answer:

a) Normal stress :

бn =[ ( бx + бy ) / 2  + ( бx - бy ) / 2  ] cos2∅ + Txysin2∅

shear stress

Tn = ( - бx - бy ) / 2  sin2∅ + Txy cos2∅

b) principal stress :

б1 = ( бx + бy ) / 2  - \sqrt{}( ( бx - бy ) / 2 )^2 + T^2xy

maximum shear stress:

Tmax  = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy

Explanation:

Combined normal stress and shear stress  sketches attached below

The terms in the sketch are :

бx = tensile stress in x direction

бy =  tensile stress in y direction

Txy = y component of shear stress acting on the perpendicular plane to x axis

бn = Normal stress acting on the inclined plane EF

Tn = shear stress acting on the inclined plane EF

A) Normal and shear stresses on inclined plane

Normal stress :

бn =[ ( бx + бy ) / 2  + ( бx - бy ) / 2  ] cos2∅ + Txysin2∅

shear stress

Tn = ( - бx - бy ) / 2  sin2∅ + Txy cos2∅

B) principal and maximum shear stresses

principal stress :

б1 = ( бx + бy ) / 2  - \sqrt{}( ( бx - бy ) / 2 )^2 + T^2xy

maximum shear stress:

Tmax  = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy

6 0
3 years ago
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Answer:

COP = 3.828

W' = 39.18 Kw

Explanation:

From the table A-11 i attached, we can find the entropy for the state 1 at -20°C.

h1 = 238.43 KJ/Kg

s1 = 0.94575 KJ/Kg.K

From table A-12 attached we can do the same for states 3 and 4 but just enthalpy at 800 KPa.

h3 = h4 = hf = 95.47 KJ/Kg

For state 2, we can calculate the enthalpy from table A-13 attached using interpolation at 800 KPa and the condition s2 = s1. We have;

h2 = 275.75 KJ/Kg

The power would be determined from the energy balance in state 1-2 where the mass flow rate will be expressed through the energy balance in state 4-1.

W' = m'(h2 - h1)

W' = Q'_L((h2 - h1)/(h1 - h4))

Where Q'_L = 150 kW

Plugging in the relevant values, we have;

W' = 150((275.75 - 238.43)/(238.43 - 95.47))

W' = 39.18 Kw

Formula foe COP is;

COP = Q'_L/W'

COP = 150/39.18

COP = 3.828

4 0
3 years ago
A particle moves along a straight line with a velocity V=(200s) mm/s, where s is in millimeters. Determine acceleration of the p
iragen [17]

Answer:

200 mm/s²

Explanation:

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Answer:

Chandigarh

Explanation:

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