Answer:
a) 0.39795 kJ/K
b) 79.589.37 kJ
Explanation:
m = Mass of air = 2 kg
Temperature = 200 K
P₁ = Initial pressure = 300 kPa
P₂ = Final pressure = 600 kPa
R = mass-specific gas constant for air = 287.058 J/kgK
a) For isentropic process
∴ Entropy is generated in the process is 0.39795 kJ/K
b)
∴ Amount of lost work is 79.589.37 kJ
D. Kelsi kanals Alabama’s Italian Oslo Goldie talisman Jacoby Allah gamma Ireland Hannah Lucy UCD tv RSA was was ex
Answer:
No, series parallel, as it implies has components of the circuit configured in both series and parallel. This is typically done to achieve a desired resistance in the circuit. A parallel circuit is a circuit that only has the components hooked in parallel, which would result in a lower total resistance in the circuit than if the components were hooked up in a series parallel configuration.
Explanation:
Answer:
Work done = 125π J
Explanation:
Given:
P = P_i * ( 1 - (x/d)^2 / 25)
d = 5.0 cm
x = 5 * d cm = 25d
Pi = 12 bar
Work done = integral ( F . dx )
F (x) = P(x) * A
F (x) = (πd^2 / 4) * P_i * (1 - (x/d)^2 / 25)
Work done = integral ((πd^2 / 4) * P_i * (1 - (x/d)^2 / 25) ) . dx
For Limits 0 < x < 5d
Work done = (πd^2 / 4) * P_i integral ( (1 - (x/d)^2) / 25)) . dx
Integrate the function wrt x
Work done = (πd^2 / 4) * P_i * ( x - d*(x/d)^3 / 75 )
Evaluate Limits 0 < x < 5d :
Work done = (πd^2 / 4) * P_i * (5d - 5d / 3)
Work done = (πd^2 / 4) * P_i * (10*d / 3)
Work done = (5 π / 6)d^3 * P_i
Input the values:
Work done = (5 π / 6)(0.05)^3 * (1.2*10^6)
Work done = 125π J
Answer:
surface temperature = 128.74⁰c
Explanation:
Given data
diameter of cable = 5 mm = 0.005 m
length of cable = 4 m
T∞ ( surrounding temperature ) = 20⁰c
voltage drop across cable ( dv )= 60 V
current across cable = 1.5 A
attached to this answer is the comprehensive analysis and solution to the problem.
The assumption made is not a good one since the calculated Ts ( surface temperature ) is very much different from the assumed Ts
