The molecular weight of a 10g rubber band
1 answer:
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Answer:
new boiler total cost = $229706.825
new boiler total cost = $127512
Explanation:
given data
power p1 = 80 kW
cost C = $160000
cost index CI 1 = 187
cost index CI 2= 194
cost capacity factor f = 0.6
power p2 = 120 kW
current cost = $18000
to find out
total cost and cost of 40 kW
solution
we consider here CN cost for new boiler and CO cost for old boiler
and x is capacity of new boiler
first we find old boiler current cost that is
current cost CO = C × .............1
put here value
current cost = 160000 ×
new current cost = $165989.304
and
use here power sizing technique for 124 kW
CN/CO = ...............2
put here value and find CN
CN/CO =
CN / 165989.304 =
CN = 211706.825
so new cost = $211706.825
so
total cost for new boiler is
total cost = new cost + current cost
total cost = 211706.825 + 18000
new boiler total cost = $229706.825
and
for 40 kW new cost will be
use equation 2
CN/CO =
CN / 165989.304 =
CN = 109512
so new cost is $109512
so
total cost for new boiler is
total cost = new cost + current cost
total cost = 109512 + 18000
new boiler total cost = $127512
Answer:
a. 4.279 MPa
b. 3.198 MPa to 4.279 MPa
c. 0.939 MPa
d. Below 3.198 MPa
Explanation:
From the given parameters
M = 1.75 MPa
M at 1.6 MPa gives A/A* = 1.2502
M at 1.8 MPa gives A/A* = 1.4390
Therefore, by interpolation, we have M = 1.75 MPa gives A
However, we shall use M = 1.75 MPa and A
Similarly,
P/P₀ = 0.1878
a) Where the nozzle is choked at the throat there is subsonic flow in the following diverging part of the nozzle. From tables, we have
A/A* = 1.387. by interpolation M
Therefore P = P₀ × P
Which shows that the nozzle is choked for back pressures lower than 4.279 MPa
b) Where there is a normal shock at the exit of the nozzle, we have;
M₁ = 1.75 MPa, P₁ = 0.1878 × 5 = 0.939 MPa
Where the normal shock is at M₁ = 1.75 MPa, P₂/P₁ = 3.406
Where the normal shock occurs at the nozzle exit, we have
Where the shock occurs t the section prior to the nozzle exit from the throat, the back pressure was derived as = 4.279 MPa
Therefore the back pressure value ranges from 3.198 MPa to 4.279 MPa
c) At M = 1.75 MPa and P
d) Where the back pressure is less than 3.198 MPa according to isentropic flow relations supersonic flow will exist at the exit plane
Answer:
1113kN
Explanation:
The ouside diameter OD of the pipe is 61cm and the thickness T is 0.9cm, so the inside diameter ID will be:
Inside Diameter = Outside Diameter - Thickness
Inside Diameter = 61cm - 0.9cm = 60.1cm
Converting this diameter to meters, we have:
This inside diameter is useful to calculate the volume V of water inside the pipe, that is the volume of a cylinder:
The problem gives you the water density d as 1.0kg/L, but we need to convert it to proper units, so:
Now, water density is given by the equation , where m is the water mass and V is the water volume. Solving the equation for water mass and replacing the values we have:
With the water mass we can find the weight of water:
Finally we find the total weight add up the weight of the water and the weight of the pipe,so:
Converting this total weight to kN, we have:
