Answer:
The final pressure is 3.16 torr
Solution:
As per the question:
The reduced pressure after drop in it, P' = 3 torr =
At the end of pumping, temperature of air,
After the rise in the air temperature,
Now, we know the ideal gas eqn:
PV = mRT
So
(1)
where
P = Pressure
V = Volume
R = Rydberg's constant
T = Temperature
Using eqn (1):
Now, at constant volume the final pressure, P' is given by:
Answer:
a. 149.74 KJ/KG
b. 97.9%
c. 0.81 kJ/kg K
Explanation:
Answer:
The final specific internal energy of the system is 1509.91 kJ/kg
Explanation:
The parameters given are;
Mass of steam = 1 kg
Initial pressure of saturated steam p₁ = 1000 kPa
Initial volume of steam, = V₁
Final volume of steam = 5 × V₁
Where condition of steam = saturated at 1000 kPa
Initial temperature, T₁ = 179.866 °C = 453.016 K
External pressure = Atmospheric = 60 kPa
Thermodynamic process = Adiabatic expansion
The specific heat ratio for steam = 1.33
Therefore, we have;
Adding the effect of the atmospheric pressure, we have;
p = 1000 + 60 = 1060
We therefore have;
T₁/T₂ = 1.70083
T₁ = 1.70083·T₂
T₂ - T₁ = T₂ - 1.70083·T₂
Whereby the temperature of saturation T₁ = 179.866 °C = 453.016 K, we have;
T₂ = 453.016/1.70083 = 266.35 K
ΔU = 3××(T₂ - T₁)
= cv for steam at 453.016 K = 1.926 + (453.016 -450)/(500-450)*(1.954-1.926) = 1.93 kJ/(kg·K)
cv for steam at 266.35 K = 1.86 kJ/(kg·K)
We use cv given by (1.93 + 1.86)/2 = 1.895 kJ/(kg·K)
ΔU = 3××(T₂ - T₁) = 3*1.895 *(266.35 -453.016) = -1061.2 kJ/kg
The internal energy for steam =
= 2777.12 kJ/kg
= 0.194349 m³/kg
p = 1000 kPa
= 2777.12 - 0.194349 * 1060 = 2571.11 kJ/kg
The final specific internal energy of the system is therefore, + ΔU = 2571.11 - 1061.2 = 1509.91 kJ/kg.
Answer:
Z = 3 + 0.23t
The water level is rising
Explanation:
Please see attachment for the equation
