Car A starts from rest at t = 0 and travels along a straight road with a constant acceleration of 6 ft/s2 until it reaches a spe
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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:
The composition of Composite of Volume basis is 2.154% Copper and 97.83% Rubber.
Explanation:
Let us assume that the total mass of composite is 100 lbm So as per the given conditions
- 15 lbm is copper and 85 lbm is rubber.
- Density of rubber is 70 lbm/ft3
- Specific gravity of Copper is 9
So As per the formula of specific gravity
here density of water is 62.4 lbm/ft3
Solving for Density of Copper gives
For composition on volume basis, volume of individual components and composite are calculated as
The composition is given as
So the composition of Composite of Volume basis is 2.154% Copper and 97.83% Rubber.