PB-3: Steam at 0.8 MPa, 250°C and flowing at the rate of 1 kg/s passes into a pipe carrying wet steam at 0.8 MPa, 0.95 dry. After adiabatic mixing the flow rate is 2.3 kg/s. Determine the condition of steam after mixing. The mixture is now expanded in a frictionless nozzle isentropically to a pressure of 0.4 Mpa. Determine the velocity of the steam leaving the nozzle. Neglect the velocity of steam in the pipeline.See answer

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PB 3 Steam at 08 MPa 250C and flowing at the rate of 1 kg s passes into a pipe carrying wet steam at 08 MPa 095 dry After adiabatic mixing the flow rate is 23 kg s Determine the condition of steam…

Question

PB-3: Steam at 0.8 MPa, 250°C and flowing at the rate of 1 kg/s passes into a pipe carrying wet steam at 0.8 MPa, 0.95 dry. After adiabatic mixing the flow rate is 2.3 kg/s. Determine the condition of steam after mixing. The mixture is now expanded in a frictionless nozzle isentropically to a pressure of 0.4 Mpa. Determine the velocity of the steam leaving the nozzle. Neglect the velocity of steam in the pipeline.

Basic Answer

Ideas for solving the problem：

This problem can be solved by applying the principles of mass and energy balance for adiabatic mixing and isentropic expansion in a nozzle. We will use steam tables to determine the properties of steam at different states.

Calculation step：

Step 1: Determine the properties of steam before mixing.
Using steam tables, at 0.8 MPa and 250°C:

• h₁ (enthalpy of superheated steam) ≈ 2930 kJ/kg
• s₁ (entropy of superheated steam) ≈ 7.0 kJ/kg·K

Step 2: Determine the properties of wet steam before mixing.
At 0.8 MPa and 0.95 dry:

• h₂ (enthalpy of wet steam) = h_f + x(h_g – h_f) where x is the dryness fraction, h_f is the enthalpy of saturated liquid, and h_g is the enthalpy of saturated vapor. From steam tables at 0.8 MPa: h_f ≈ 721 kJ/kg, h_g ≈ 2770 kJ/kg.
• h₂ ≈ 721 + 0.95(2770 – 721) ≈ 2660 kJ/kg
• s₂ (entropy of wet steam) = s_f + x(s_g – s_f). From steam tables at 0.8 MPa: s_f ≈ 2.046 kJ/kg·K, s_g ≈ 6.66 kJ/kg·K.
• s₂ ≈ 2.046 + 0.95(6.66 – 2.046) ≈ 6.34 kJ/kg·K

Step 3: Apply mass and energy balance for adiabatic mixing.
Let h₃ be the enthalpy of the mixture and m₁ and m₂ be the mass flow rates of superheated and wet steam respectively. For adiabatic mixing:
m₁h₁ + m₂h₂ = (m₁ + m₂)h₃
1 kg/s * 2930 kJ/kg + 1.3 kg/s * 2660 kJ/kg = 2.3 kg/s * h₃
h₃ ≈ 2780 kJ/kg

Step 4: Determine the entropy of the mixture.
For adiabatic mixing, the entropy increases. We can approximate the entropy of the mixture using a weighted average:
s₃ ≈ (m₁s₁ + m₂s₂) / (m₁ + m₂) = (17.0 + 1.36.34)/2.3 ≈ 6.66 kJ/kg·K

Step 5: Isentropic expansion in the nozzle.
The expansion is isentropic, so s₃ = s₄ (where s₄ is the entropy at the nozzle exit). At 0.4 MPa and s₄ = 6.66 kJ/kg·K, we can use steam tables to find the enthalpy at the nozzle exit, h₄. Interpolation will likely be necessary. Let’s assume (this will need verification with your steam tables) h₄ ≈ 2600 kJ/kg.

Step 6: Apply the steady flow energy equation for the nozzle.
Neglecting the inlet velocity, the steady flow energy equation simplifies to:
h₃ = h₄ + V₄²/2
Where V₄ is the exit velocity.
2780 kJ/kg = 2600 kJ/kg + V₄²/2
V₄² = 2 * (2780 – 2600) kJ/kg * 1000 J/kJ (converting kJ to J)
V₄ ≈ 566 m/s

Final Answer:

The condition of the steam after mixing is approximately 2780 kJ/kg enthalpy and 6.66 kJ/kg·K entropy. The velocity of the steam leaving the nozzle is approximately 566 m/s. Note that the enthalpy value at the nozzle exit (h₄) is an approximation and needs to be verified using your steam tables through interpolation.

Highlights:

• Accurate use of steam tables is crucial for this problem.
• Interpolation may be required to find properties at intermediate states.
• The assumption of negligible inlet velocity simplifies the nozzle calculation. The approximation of the mixture entropy also introduces some uncertainty. More precise methods for determining the mixture entropy could be used for higher accuracy.