Answer:
use the dimensions shown in the figure
Answer:
a) isentropic efficiency = 84.905%
b) rate of entropy generation = .341 kj/(kg.k)
Please kindly see explaination and attachment.
Explanation:
a) isentropic efficiency = 84.905%
b) rate of entropy generation = .341 kj/(kg.k)
The Isentropic efficiency of a turbine is a comparison of the actual power output with the Isentropic case.
Entropy can be defined as the thermodynamic quantity representing the unavailability of a system's thermal energy for conversion into mechanical work, often interpreted as the degree of disorder or randomness in the system.
Please refer to attachment for step by step solution of the question.
Answer:
Um...
Explanation:
This is what I like to see teachers giving out.
Answer:
q = 1.73 W
Explanation:
given data
small end = 5 cm
large end = 10 cm
high = 15 cm
small end is held = 600 K
large end at = 300 K
thermal conductivity of asbestos = 0.173 W/mK
solution
first we will get here side of cross section that is express as
...............1
here x is distance from small end and S1 is side of square at small end
and S2 is side of square of large end and L is length
put here value and we get
S = 5 +
S =
m
and
now we get here Area of section at distance x is
area A = S² ...............2
area A =
m²
and
now we take here small length dx and temperature difference is dt
so as per fourier law
heat conduction is express as
heat conduction q =
...............3
put here value and we get
heat conduction q =
it will be express as
now we intergrate it with limit 0 to 0.15 and take temp 600 to 300 K
solve it and we get
q (30) = (0.173) × (600 - 300)
q = 1.73 W
Answer:
$916
Explanation:
To solve this, we use the formula
FV = P/i * [(1+i)^n - 1], where
FV = future value of the all the money invested, $5 million
n = time span, = 500 months
P = payment per month
I = interest rate, 9% by 12 months, = 0.0075
Considering that we have been given all in our question, then we substitute directly and solve. So we have,
5000000 = P/0.0075 * [(1+0.0075)^500 -1]
5000000 * 0.0075 = P * [1.0075^500 - 1]
37500 = P * [41.93 - 1]
37500 = P * 40.93
P = 37500/40.93
P = $916.20
Therefore, the engineer needs to save $916 in a month which is the accrued