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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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Minchanka [31]

Answer:

Opened Push-button Switch (i.e. a PTM Switch)

Explanation:

Tha's just another symbol for a switch, but this one specifies that the switch is a push-button type of switch.

Since it's not touching and completing the line, the state of the switch is initially open.

6 0
3 years ago
I = 48 mA, R = 1125 2. What is Vs ?
german
54 volts

Ohms law. E= I x R
6 0
3 years ago
An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no
Stells [14]

Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

Analysis:

a. This is reversible adiabatic(i.e isentropic) process and thus s_{1} =s_{2}

From the refrigerant table A11-A13

P_{1} =0.8MPa   \left \{ {{ {{v_{1}=v_{g}  @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g}  @0.8MPa =246.82 kJ/kg } -   also  {{s_{1}=s_{g}  @0.8MPa =0.91853 kJ/kgK } } \right.

sat vapor

m=\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa  \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339}   = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) =  38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}

T_{2} =T_{sat @ 0.2MPa} = -10.09^{o}  C

b.) We take the content of the cylinder as the sysytem.

This is a closed system since no mass leaves or enters.

Hence, the energy balance for adiabatic closed system can be expressed as:

E_{in} - E_{out}  =ΔE

w_{b, out}  =ΔU

w_{b, out} =m([tex]u_{1} -u_{2)

w_{b, out}  = workdone during the isentropic process

=5.8491(246.82-219.9)

=5.8491(26.91)

=157.3993

=157.4kJ

4 0
3 years ago
Steep safety ramps are built beside mountain highway to enable vehichles with fedective brakes to stop safely. a truck enters a
Veronika [31]

Answer:

a. 6 seconds

b. 180 feet

Explanation:

Images attached to show working.

a. You have the position of the truck so you integrate twice. Use the formula and plug in the time t = 7 sec. Check out uniform acceleration. The time at which the truck's velocity is zero  is when it stops.

b. Determine the initial speed. Plug in the time calculated in the previous step. From this we can observe that the truck comes to a stop before the end of the ramp.

7 0
3 years ago
The cubic capacity of a four-stroke over-square spark-ignition engine is 245cc. The over-square ratio is (1.1) The clearance vol
pashok25 [27]

Answer:

Bore = 7 cm

stroke = 6.36 cm

compression ratio = 10.007

Explanation:

Given data:

Cubic capacity of the engine, V = 245 cc

Clearance volume, v = 27.2 cc

over square-ratio = 1.1

thus,

D/L = 1.1

where,

D is the bore

L is the stroke

Now,

V = \frac{\pi}{4}D^2L

or

V = \frac{\pi}{4}\frac{D^3}{1.1}

on substituting the values, we have

245 =  \frac{\pi}{4}\frac{D^3}{1.1}

or

D = 7.00 cm

Now,

we have

D/L = 1.1

thus,

L = D/1.1

L = 7/1.1

or

L= 6.36 cm

Now,

the compression ratio is given as:

\textup{compression ratio}=\frac{V+v}{v}

on substituting the values, we get

\textup{compression ratio}=\frac{245+27.2}{27.2}

or

Compression ratio = 10.007

4 0
3 years ago
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