Answer:
power produced = 3098.52 kW
Explanation:
given data
insulated turbine = 100 bar
temperature = 400°C
pressure = 200 kPa
mass flow rate = 1.99 kg/s
solution
we use here steam table for At 100 bar and 400°C
h1 = 3096.5 KJ/Kg
and
at P2 = 200 Kpa
h2 = hf + 0.47 hg
h2 = 504.7 + 0.47 × 2201.6
h2 = 1539.452 KJ/Kg
so here
power produced is express as
power produced = m × (h1 - h2) .................1
power produced = 1.99 × ( 3096.5 - 1539.452 )
power produced = 3098.52 kW
Answer:
The viscosities of the oils are 0.967 Pa.s and 1.933 Pa.s
Explanation:
Assuming the two oils are Newtonian fluids.
From Newton's law of viscosity for Newtonian fluids, we know that the shear stress is proportional to the velocity gradient with the viscosity serving as the constant of proportionality.
τ = μ (∂v/∂y)
There are oils above and below the plate, so we can write this expression for the both cases.
τ₁ = μ₁ (∂v/∂y)
τ₂ = μ₂ (∂v/∂y)
dv = 0.3 m/s
dy = (0.06/2) = 0.03 m (the plate is centered in a gap of width 0.06 m)
τ₁ = μ₁ (0.3/0.03) = 10μ₁
τ₂ = μ₂ (0.3/0.03) = 10μ₂
But the shear stress on the plate is given as 29 N per square meter.
τ = 29 N/m²
But this stress is a sum of stress due to both shear stress above and below the plate
τ = τ₁ + τ₂ = 10μ₁ + 10μ₂ = 29
But it is also given that one viscosity is twice the other
μ₁ = 2μ₂
10μ₁ + 10μ₂ = 29
10(2μ₂) + 10μ₂ = 29
30μ₂ = 29
μ₂ = (29/30) = 0.967 Pa.s
μ₁ = 2μ₂ = 2 × 0.967 = 1.933 Pa.s
Hope this Helps!!!
Answer: Rc = 400 Ω and Rb = 57.2 kΩ
Explanation:
Given that;
VCE = 5V
VCC = 15 V
iC = 25 mA
β = 100
VD₀ = 0.7 V
taking a look at the image; at loop 1
-VCC + (i × Rc) + VCE = 0
we substitute
-15 + ( 25 × Rc) + 5 = 0
25Rc = 10
Rc = 10 / 25
Rc = 0.4 k
Rc = 0.4 × 1000
Rc = 400 Ω
iC = βib
25mA = 100(ib)
ib = 25 mA / 100
ib = 0.25 mA
ib = 0.25 × 1000
ib = 250 μAmp
Now at Loop 2
-Vcc + (ib×Rb) + VD₀ = 0
-15 (250 × Rb) + 0.7 = 0
250Rb = 15 - 0.7
250Rb = 14.3
Rb = 14.3 / 250
Rb = 0.0572 μ
Rb = 0.0572 × 1000
Rb = 57.2 kΩ
Therefore Rc = 400 Ω and Rb = 57.2 kΩ
Answer:
4mA
Explanation:
For this problem, we will simply apply Ohm's law:
V = IR
V/R = I
I = V / R
I = 12 volt / 3kΩ
I = 4mA
Hence, the current in the circuit is 4mA.
Cheers.