A glycerin pump is powered by a 5-kW electric motor. The pressure differential between the outlet and the inlet of the pump at f
1 answer:
Answer:
Explanation:
Given that:
The pressure differential between the outlet and the inlet of the pump at full load is measured to be ΔP is 211 kPa.
the flow rate through the pump is 18 L/s
The Formula of flow rate.
Rearrange
Substitute
The formula for the overall efficiency of the pump.
input the values 3.8kW for and 5kW for
The overall efficiency of the glycerin pump is 76% .
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Answer: Inherent width in the emission line: 9.20 × 10⁻¹⁵ m or 9.20 fm
length of the photon emitted: 6.0 m
Explanation:
The emitted wavelength is 589 nm and the transition time is ∆t = 20 ns.
Recall the Heisenberg's uncertainty principle:-
∆t∆E ≈ h ( Planck's Constant)
The transition time ∆t corresponds to the energy that is ∆E
.
The corresponding uncertainty in the emitted frequency ∆v is:
∆v= ∆E/h = (5.273*10^-27 J)/(6.626*10^ J.s)= 7.958 × 10^6 s^-1
To find the corresponding spread in wavelength and hence the line width ∆λ, we can differentiate
λ = c/v
dλ/dv = -c/v² = -λ²/c
Therefore,
∆λ = (λ²/c)*(∆v) = {(589*10⁻⁹ m)²/(3.0*10⁸ m/s)} * (7.958*10⁶ s⁻¹)
= 9.20 × 10⁻¹⁵ m or 9.20 fm
The length of the photon (<em>l)</em> is
l = (light velocity) × (emission duration)
= (3.0 × 10⁸ m/s)(20 × 10⁻⁹ s) = 6.0 m
To resolve this problem we have,
is unknown.
With these dates we can calculate the Flexural strenght of the specimen,
After that, we can calculate the flexural strenght for the square cross section using the previously value.
Answer:
A) condenser pressure = 9.73 kPa,
B) 10242 kw
C) 36.9%
Explanation:
given data
entrance pressure of steam = 12.5 MPa
temperature of steam = 550⁰c
flow rate of steam = 7.7 kg/s
outer pressure = 2 MPa
reheated steam temperature = 450⁰c
isentropic efficiency of turbine( nt ) = 85% = 0.85
isentropic efficiency of pump = 90% = 0.90
From steam tables
at 12.5 MPa and 550⁰c ; h3 = 3476.5 kJ/kg, S3 = 6.6317 kJ/kgK
also for an Isentropic expansion
S4s = S3 .
therefore when S4s = 6.6317 kJ/kg and P4 = 2 MPa
h4s = 2948.1 kJ/kg
nt = 0.85
nt (0.85) = =
making h4 subject of the equation
h4 = 3476.5 - 0.85 (3476.5 - 2948.1)
h4 = 3027.3 kj/kg
at P5 = 2 MPa and T5 = 450⁰c
h5 = 3358.2 kj/kg, s5 = 7.2815 kj/kgk
at P6 , x6 = 0.95 and s5 = s6
using nt = 0.85 we can calculate for h6 and h6s
from the chart attached below we can see that
p6 = 9.73 kPa, h6 = 2463.3 kj/kg
B) the net power output
solution is attached below
c) thermal efficiency
thermal efficiency = 1 - = 1 - ( 2273.7/ 3603.8) = 36.9% ≈ 37%
