Thermodynamics: Addition of saturated steam to closed tank

In summary: I is looking for a new intern.In summary, the tank initially contains 25000 kg of water distributed between liquid and vapor phases at 30 deg C. Saturated steam at 11 bar is admitted to the tank until pressure reaches 7 bar. The mass of steam added is 50000 kg. The tank's final mass is 2779.7 kg. The addition of the steam adds 3146 Joules of energy to the system.
  • #1
staceybiomed
12
0

Homework Statement


A well-insulated closed tank is of volume 70 m^3. Initially, it contains 25000 kg of water distributed between liquid and vapor phases at 30 deg C. Saturated steam at 11 bar is admitted to the tank until pressure reaches 7 bar. What mass of steam is added?


Homework Equations


m' = m2 - m1 (where m' is mass of steam, m2 is final mass in tank and m1 is initial mass in tank, m1 = 25000 kg)
m2H2 - (m1H1 + m'H') = 0
m2/m1 = (U1 - H')/(U2 - H') = (H' -U1)/(H' - U2)


The Attempt at a Solution


Q = 0 if well insulated
Assume W = 0
Kinetic and potential energies can be ignored
V(tank) = constant and is taken to be the control volume
H' = 2779.7 kJ/kg (from steam tables)

I know m1 and H'. I thought if volume of tank is constant than i can calculate P1 from P1V1=P2V2 but then that just gives me that the pressure is constant too. I need to figure out the enthalpies of the contents in the tank before and after the addition of the sat. steam and then I should be okay. Can anyone help? Thanks!
 
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  • #2
I have since worked further on this problem and ended up with 5 equations, 5 unknowns but it just seems too complicated for this problem. Can anybody help?

m2 = m' + m1 = m' + 25000 = m(Liquid)2 + m(Vapor)2 (at 7 bar)
m(L)2*V(L)2 + m(V)2*V(V)2 = 70 m^3, where V(L)2=0.1108 & V(V)2=0.2727
H2 = m(V)2*H(V)2 + m(L)2*H(L)2, where H(V)2=2762 & H(L)2=697.061
m1H1 +m'H' = m2H2, where m1=25000, H'=2779.7
H1=m(V)1*H(V)1 + m(L)1*H(L)1 = 3.146 x 10^6 kJ
 
  • #3
I think you went down the wrong road with this problem. I would approach this problem by simply adding up the enthalpy H. You know the enthalpy of the original contents and the incoming enthalpy. You also know the enthalpy of the final state. Since the system is closed, it should be a simple relation of;

Hf = Hi + Hadded

H = h*m

Since you know the condition of enthalpy added, you just have to reverse calculate to get your mass.
 
  • #4
Topher925 said:
I think you went down the wrong road with this problem. I would approach this problem by simply adding up the enthalpy H. You know the enthalpy of the original contents and the incoming enthalpy. You also know the enthalpy of the final state. Since the system is closed, it should be a simple relation of;

Hf = Hi + Hadded

H = h*m

Since you know the condition of enthalpy added, you just have to reverse calculate to get your mass.

The system is not closed (initially it is, but then steam is allow to enter it). It is a control volume since there is a mass flow coming into the tank. This also makes it a transient system so the RHS of the energy balance is not zero as it is for a steady flow.

CS
 

1. What is the purpose of adding saturated steam to a closed tank in thermodynamics?

The addition of saturated steam to a closed tank is often used in thermodynamics to increase the pressure and temperature of the system. This can be useful in various industrial processes, such as sterilization or heating in power plants.

2. How does the addition of saturated steam affect the pressure and temperature of the closed tank?

The addition of saturated steam to a closed tank increases the pressure and temperature of the system, as the steam condenses and releases its latent heat. This leads to an increase in the overall energy of the system.

3. Is the addition of saturated steam reversible in a closed tank?

The addition of saturated steam to a closed tank is not reversible, as the steam cannot be easily removed from the system once it has been condensed. This irreversible process is known as a constant volume process.

4. What is the significance of using a closed tank in this thermodynamic process?

The use of a closed tank allows for the addition of saturated steam to be contained within the system, preventing any external factors from affecting the process. This ensures that the pressure and temperature changes are solely due to the addition of steam.

5. Are there any limitations to using the addition of saturated steam in a closed tank for thermodynamic processes?

While the addition of saturated steam can be useful in certain industrial processes, it is not always the most efficient method. It requires a significant amount of energy to produce the steam, and the constant volume process may not be suitable for all applications. Additionally, the steam may contain impurities that can affect the overall performance of the system.

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