A cylinder fitted with a frictionless piston contains 2 kg of R-134a at 3.5 bar and 100 C. The cylinder is now cooled so that the R-134a is kept at constant pressure until a final state is reached with a quality of 25%. Calculate the heat transfer in the process.

The answer to the question is

The heat transferred in the process is -274.645 kJ

Explanation:

To solve the question, we list out the variables thus

R-134a = Tetrafluoroethane

Intitial Temperaturte t₁ = 100 °C

Initial pressure = 3.5 bar = 350 kPa

For closed system we have m₁ = m₂ = m

ΔU = m×(u₂ - u₁) = ₁Q₂ -₁W₂

For constant pressure process we have

Work done = W =  = P×ΔV = P × (V₂ - V₁) = P×m×(v₂ - v₁)

From the tables we have

State 1 we have h₁ = (490.48 +489.52)/2 = 490 kJ/kg

State 2 gives h₂ = 206.75 + 0.75 × 194.57= 352.6775 kJ/kg

Therefore Q₁₂ = m×(u₂ - u₁) + W₁₂ = m × (u₂ - u₁) + P×m×(v₂ - v₁)

= m×(h₂ - h₁) = 2.0 kg × (352.6775 kJ/kg - 490 kJ/kg) =-274.645 kJ

oil flow is a highly convective flow compared to water flow at normal temperature and pressure.

answer:

a

explanation: