Question

30 kg of Oxygen gas is contained in a vessel of volume 3.5m3 at 100°C, evaluate 1. Pressure and Specific Volume of gas 2. If the gas is cooled to 30°C, find the final pressure of the gas, change in enthalpy, internal energy per kg of gas. Take gamma = 1.44 and R=0.257kJ/kg-K.​

Answer 1

Answer:

Explanation:

       

0.751 (809 271) 4040

10 (809 271) 5595

v

p

U mC T

H mC T

d) The IRC Calculator provides the internal energies and enthalpies for dry air at both

temperatures. 

U = 10 × (600 – 192) = 4080 kJ

H = 10 × (833 – 269) = 5640 kJ

2.82 Use the enthalpy data for steam from Table C-3.

a) P = 0.2 MPa and T = 400°C. We will use central differencing between 300° and 500° to

estimate the specific heat at 400°:

3487.1 3071.8 2.08 kJ/kg C

200 p

h

C

T

     



17

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b) P = 1.8 MPa and T = 400°C. We will use central differencing between 300° and 500° to

estimate the specific heat at 400°:

3469.8 3029.2 2.20 kJ/kg C

200 p

h

C

T

     



2.83 Using enthalpy data for steam from Table C-3.

b) P = 1 MPa and T = 300°C. We will use central differencing between 250°C and 350°C to get

the specific heat at 300°C.

3157.7 2942.6 2.151 kJ/kg K

100

 

   



p

h

C

T

2.84 40 kg of water is heated at 100 kPa from 16°C to 92°C. The water will exist as a subcooled

liquid. The compressed liquid properties will be approximated by the saturated liquid

properties. Use Table C.1 to get the internal energies and enthalpies:

At 16°C, u1 = uf = 67.18 kJ/kg and h1 = hf = 67.18 kJ/kg.

At 92°C, u2 = uf = 385.2 kJ/kg and h2 = hf = 385.3 kJ/kg

U = 40 × (385.2 – 67.18) = 12 720 kJ

H = 40 × (385.3 – 67.18) = 12 720 kJ

The actual difference is

          h u Pv P v ( ) 100 (0.001038 0.001001)

kJ  0.0037 /kg or 3.7 J/kg

2.86 In this problem, 10 kg of ice at 20°C is heated to become liquid water at 50°C. The phase

change occurs at 0°C since the pressure is presumably atmospheric.

b) Use the equation Cp, ice = 2.1 + 0.0069T:

  

2

ice

1

0 2 2

20

0 20 10 2.1 0.0069 10 2.1(0 20) 0.0069 434 kJ

2

H m C dT p

T dT



 

  

         







As stated on Section 2.2.4, the heat of fusion for ice is 330 kJ/kg, so that

Hmelt = 10 kg × 330 kJ/kg = 3300 kJ

To heat the liquid water from 0°C to 50°C, use Cp from Table B-4:

Hliquid = mCpT =10 × 4.177 × 50 = 2090 kJ



Htotal = 434 kJ + 3300 kJ + 2090 kJ = 5820 kJ