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nadezda [96]
3 years ago
5

.a. What size vessel holds 2 kg water at 80°C such that 70% is vapor? What are the pressure and internal energy? b. A 1.6 m3 ves

sel holds 2 kg water at 0.2 MPa. What are the quality, temperature, and internal energy?
Engineering
1 answer:
vesna_86 [32]3 years ago
5 0

Answer:

Part a: The volume of vessel is 4.7680m^3 and total internal energy is 3680 kJ.

Part b: The quality of the mixture is 90.3%  or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.

Explanation:

Part a:

As per given data

m=2 kg

T=80 °C =80+273=353 K

Dryness=70% vapour =0.7

<em>From the steam tables at 80 °C</em>

Specific volume of saturated vapours=v_g=3.40527 m^3/kg

Specific volume of saturated liquid=v_f=0.00102 m^3/kg

Now the relation  of total specific volume for a specific dryness value is given as

                                  v=v_f+x(v_g-v_f)

Substituting the values give

v=v_f+x(v_g-v_f)\\v=0.00102+0.7(3.40527-0.00102)\\v_f=2.38399 m^3/kg

Now the volume of vessel is given as

v=\frac{V}{m}\\V=v \times m\\V=2.38399 \times 2\\V=4.7680 m^3

So the volume of vessel is 4.7680m^3.

Similarly for T=80 and dryness ratio of 0.7 from the table of steam

Pressure=P=47.4 kPa

Specific internal energy is given as u=1840 kJ/kg

So the total internal energy is given as

u=\frac{U}{m}\\U=u \times m\\U=1840 \times 2\\U=3680 kJ

The total internal energy is 3680 kJ.

So the volume of vessel is 4.7680m^3 and total internal energy is 3680 kJ.

Part b

Volume of vessel is given as 1.6

mass is given as 2 kg

Pressure is given as 0.2 MPa or 200 kPa

Now the specific volume is given as

v=\frac{V}{m}\\v=\frac{1.6}{2}\\v=0.8 m^3/kg

So from steam tables for Pressure=200 kPa and specific volume as 0.8 gives

Temperature=T=120 °C

Quality=x=0.903 ≈ 90.3%

Specific internal energy =u=2330 kJ/kg

The total internal energy is given as

u=\frac{U}{m}\\U=u \times m\\U=2330 \times 2\\U=4660 kJ

So the quality of the mixture is 90.3%  or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.

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