Answer: The energy system related to your question is missing attached below is the energy system.
answer:
a) Work done = Net heat transfer
Q1 - Q2 + Q + W = 0
b) rate of work input ( W ) = 6.88 kW
Explanation:
Assuming CPair = 1.005 KJ/Kg/K
<u>Write the First law balance around the system and rate of work input to the system</u>
First law balance ( thermodynamics ) :
Work done = Net heat transfer
Q1 - Q2 + Q + W = 0 ---- ( 1 )
rate of work input into the system
W = Q2 - Q1 - Q -------- ( 2 )
where : Q2 = mCp T = 1.65 * 1.005 * 293 = 485.86 Kw
Q2 = mCp T = 1.65 * 1.005 * 308 = 510.74 Kw
Q = 18 Kw
Insert values into equation 2 above
W = 6.88 Kw
Answer:
a) V = 0.354
b) G = 25.34 GPA
Explanation:
Solution:
We first determine Modulus of Elasticity and Modulus of rigidity
Elongation of rod ΔL = 1.4 mm
Normal stress, δ = P/A
Where P = Force acting on the cross-section
A = Area of the cross-section
Using Area, A = π/4 · d²
= π/4 · (0.0020)² = 3.14 × 10⁻⁴m²
δ = 50/3.14 × 10⁻⁴ = 159.155 MPA
E(long) = Δl/l = 1.4/600 = 2.33 × 10⁻³mm/mm
Modulus of Elasticity Е = δ/ε
= 159.155 × 10⁶/2.33 × 10⁻³ = 68.306 GPA
Also final diameter d(f) = 19.9837 mm
Initial diameter d(i) = 20 mm
Poisson said that V = Е(elasticity)/Е(long)
= - <u>( 19.9837 - 20 /20)</u>
2.33 × 10⁻³
= 0.354,
∴ v = 0.354
Also G = Е/2. (1+V)
= 68.306 × 10⁹/ 2.(1+ 0.354)
= 25.34 GPA
⇒ G = 25.34 GPA
Answer: The exit temperature of the gas in deg C is
.
Explanation:
The given data is as follows.
= 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)
= 100 kPa,
We know that for an ideal gas the mass flow rate will be calculated as follows.
or, m =
=
= 10 kg/s
Now, according to the steady flow energy equation:
= 5 K
= 5 K + 300 K
= 305 K
= (305 K - 273 K)
=
Therefore, we can conclude that the exit temperature of the gas in deg C is
.
Answer:
1.2727 stokes
Explanation:
specific gravity of fluid A = 1.65
Dynamic viscosity = 210 centipoise
<u>Calculate the kinematic viscosity of Fluid A </u>
First step : determine the density of fluid A
Pa = Pw * Specific gravity = 1000 * 1.65 = 1650 kg/m^3
next : convert dynamic viscosity to kg/m-s
210 centipoise = 0.21 kg/m-s
Kinetic viscosity of Fluid A = dynamic viscosity / density of fluid A
= 0.21 / 1650 = 1.2727 * 10^-4 m^2/sec
Convert to stokes = 1.2727 stokes