Answer:
R=1923Ω
Explanation:
Resistivity(R) of copper wire at 20 degrees Celsius is 1.72x10^-8Ωm.
Coil length(L) of the wire=37.0m
Cross-sectional area of the conductor or wire (A) = πr^2
A= π * (2.053/1000)/2=3.31*10^-6
To calculate for the resistance (R):
R=ρ*L/A
R=(1.72*10^8)*(37.0)/(3.31*10^-6)
R=1922.65Ω
Approximately, R=1923Ω
Answer:
3A
Explanation:
Using Ohms law U=I×R solve for I by I=U/R
Answer:
The mass flow rate of the mixture in the manifold is 6.654 kg/min
Explanation;
In this question, we are asked to calculate mass flow rate of the mixture in the manifold
Please check attachment for complete solution and step by step explanation.
Answer:
a) Tբ = 151.8°C
b) ΔV = - 0.194 m³
c) The T-V diagram is sketched in the image attached.
Explanation:
Using steam tables,
At the given pressure of 0.5 MPa, the saturation temperature is the final temperature.
Right from the steam tables (A-5) with a little interpolation, Tբ = 151.793°C
b) The volume change
Using data from A-5 and A-6 of the steam tables,
The volume change will be calculated from the mass (0.58 kg), the initial specific volume (αᵢ) and the final specific volume
(αբ) (which is calculated from the final quality and the consituents of the specific volumes).
ΔV = m(αբ - αᵢ)
αբ = αₗ + q(αₗᵥ) = αₗ + q (αᵥ - αₗ)
q = 0.5, αₗ = 0.00109 m³/kg, αᵥ = 0.3748 m³/kg
αբ = 0.00109 + 0.5(0.3748 - 0.00109)
αբ = 0.187945 m³/kg
αᵢ = 0.5226 m³/kg
ΔV = 0.58 (0.187945 - 0.5226) = - 0.194 m³
c) The T-V diagram is sketched in the image attached