Answer:
0.5 kW
Explanation:
The given parameters are;
Volume of tank = 1 m³
Pressure of air entering tank = 1 bar
Temperature of air = 27°C = 300.15 K
Temperature after heating = 477 °C = 750.15 K
V₂ = 1 m³
P₁V₁/T₁ = P₂V₂/T₂
P₁ = P₂
V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³
For ideal gas, 
= 5/2×R = 5/2*0.287 = 0.7175 kJ
PV = NKT
N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)
N = 9.66×10²⁴
Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles
The average mass of one mole of air = 28.8 g
Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg
∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ
The power input required = The rate of heat transfer = 149.211/(60*5)
The power input required = 0.49737 kW ≈ 0.5 kW.
Answer:
hello your question lacks the required question attached below is the missing diagram
Forces in GJ = -4.4444 i.e. 4.4444 tons
Forces in IG = 15.382 tons ( T )
Explanation:
Forces in GJ = -4.4444 i.e. 4.4444 tons
Forces in IG = 15.382 tons ( T )
attached below is the detailed solution
Answer:
Simple G Code Example Mill - This is a very simple G code example for beginner level ... Run the program on your cnc machine (Safety first, keep a professional around). ... This G code program example don't use Tool radius compensation ...
Explanation:
Answer:
Step 1: State your null and alternate hypothesis. ...
Step 2: Collect data. ...
Step 3: Perform a statistical test. ...
Step 4: Decide whether the null hypothesis is supported or refuted. ...
Step 5: Present your findings.
Answer: Incoherent question
Explanation: This is an act of plagiarism at subjecting the tutor to unnecessary stress at answering the purported question.
