A rigid tank with a volume of 4 m^3 contains argon at 500 kPa and 30 deg C. It is connected to a piston cylinder (initially empt
y) with a line and a closed valve. The pressure in the cylinder should be 300 kPa to float the piston. Now the valvle is slowly opened and heat is transferred so the argon reaches a final state of 30 deg C with the valve open. The heat transfer takes place from a 60 deg C space through a conduction layer 1 cm thick with a conductivity k = 0.1 W/mK and a surface area of 1.5 m^2.
a. Find the final volume of the argon and the total work for the process.
b. Find the total heat transfer needed.
c. How long will the process take?
a. Find the final volume of the argon and the total work for the process.
b. Find the total heat transfer needed.
c. How long will the process take?
1 answer:
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Answer:
b) False
Explanation:
in trapezoidal rule the error is proportional to and the order of accuracy is proportional to
.
Trapezoidal rule is numerical integration method .Trapezoidal rule is used to find the area of curves.In trapezoidal rule we finds the approximate value of integration.But the real value of integration will not differ to much from the value which finds by using trapezoidal rule.
Answer:
a. which is constant therefore, n = constant
b. The temperature at the end of the process is 109.6°C
c. The work done by the balloon boundaries = 10.81 MJ
The work done on the surrounding atmospheric air = 10.6 MJ
Explanation:
p₁ = 100 kPa
T₁ = 27°C
D₁ = 10 m
v₂ = 1.2 × v₁
p ∝ α·D
α = Constant
v₂ = 1.2 × v₁ = 1.2 × 523.6 = 628.32 m³
Therefore, D₂ = 10.63 m
We check the following relation for a polytropic process;
We have;
n = -1/3
Therefore, the relation, pVⁿ = Constant
b. The temperature T₂ is found as follows;
T₂ = 300.15/0.784 = 382.75 K = 109.6°C
c.
p₂ = 100000/0.941 = 106.265 kPa
The work done by the balloon boundaries = 10.81 MJ
Work done against atmospheric pressure, Pₐ, is given by the relation;
Pₐ × (V₂ - V₁) = 1.01×10⁵×(628.32 - 523.6) = 10576695.3 J
The work done on the surrounding atmospheric air = 10.6 MJ
Answer:
Hello your question is incomplete attached below is the missing part
a) p1 = 454.83 kPa, p2 = 283.359 Kpa , p3 = 536.423 kpa , p4 = 860.959kPa
b) W12 = 3.4 kJ, W23 = -3.5875 KJ, W34 = -4.0735 KJ, W41 = 3.5875 KJ
c) 5
Explanation:
Given data:
mass of air ( m ) = 1/10 kg
adiabatic index ( k ) = 1.4
temperature for isothermal expansion = 250K
rate of heat transfer ( Q12 ) = 3.4 KJ
temperature for Isothermal compression ( T4 ) = 300k
final volume ( V4 ) = 0.01m ^3
a) Calculate the pressure, in Kpa, at each of the four principal states
from an ideal gas equation
P4V4 = mRT4 ( input values above )
hence P4 = 860.959kPa
attached below is the detailed solution
b) Calculate work done for each processes
attached below is the detailed solution
C) Calculate the coefficient of performance
attached below is detailed solution
