Answer:
Q = 65.388 KJ
Explanation:
To calculate the heat required for the given process Q, we recall the energy balance equation.
Therefore, : Q = Δ U = m (u₂ - u₁) ..................equation (1)
We should note that there are no kinetic or potential energy change so the heat input in the system is converted only to internal energy.
Therefore, we will start the equation with the mass of the water (m) using given the initial percentage quality as x₁ = 0.123 and initial temperature t₁ = 100⁰c , we can them determine the initial specific volume v₁ of the mixture. For the calculation, we will also need the specific volume of liquid vₙ = 0.001043m³/kg and water vapour (vₐ) = 1.6720m³/kg
Therefore, u₁ = vₙ + x₁ . ( vₐ - vₙ)
u₁ = 0.001043m³/kg + 0.123 . ( 1.6720m³/kg - 0.001043m³/kg)
u₁ = 0.2066m³/kg
Moving forward, the mass of the vapor can then be calculated using the given volume of tank V = 14 L but before the calculation, we need to convert the volume to from liters to m³.
Therefore, V = 14L . 1m² / 1000L = 0.014 m³
Hence, m = V / u₁
0.014m³ / 0.2066 m³/kg
m = 0. 0677 kg
Also, the initial specific internal energy u₁ can be calculated using the given the initial given quality of x₁ , the specific internal energy of liquid water vₐ = 419.06 kj / kg and the specific internal energy of evaporation vₐₙ = 2087.0 kj/kg.
Therefore, u₁ = vₐ + x₁ . vₐₙ
u₁ = 419.06 kj / kg + 0.123 . 2087.0 kj/kg
u₁ = 675.76 kj/kg
For the final specific internal energy u₂, we first need to calculate the final quality of the mixture x₂ . The tank is rigid meaning the volume does not change and it is also closed meaning the mass does not change.from this, we can conclude the the specific volume also does not change during the process u₁ = u₂. This allows us to use the given final temperature T₂ = 180⁰c to determine the final quality x₂ of the mixture. for the calculation, we will also need the specific volume of liquid vₙ=0.001091m³/kg and vapor vₐ = 0.39248m³/kg
Hence, x₂ = u₂ - vₙ / uₐ
x₂ = 0.2066 m³/kg - 0.001091m³/kg / 0.39248m³/kg
x₂ = 0.524
Moving forward to calculate the final internal energy u₂, we have :
u₂ = vₙ + x₂ . vₙₐ
u₂ = 631.66 kj/kg + 0.524 . 1927.4 kj/kg
u₂ = 1641.62 kj/kg
We now return to equation (1) to plug in the values generated thus far
Q = m (u₂ - u₁)
0. 0677 kg ( 1641.62 kj/kg - 675.76 kj/kg)
Q = 65.388KJ
= 1527.7 rev/min
