Answer:
yes
Explanation:
blueprint of the construction is a prediction of project its is slightly auto cad
Answer:
net boiler heat = 301.94 kW
Explanation:
given data
saturated steam = 6.0 bars
temperature = 18°C
flow rate = 115 m³/h = 0.03194 m³/s
heat use by boiler = 90 %
to find out
rate of heat does the boiler output
solution
we can say saturated steam is produce at 6 bar from liquid water 18°C
we know at 6 bar from steam table
hg = 2756 kJ/kg
and
enthalpy of water at 18°C
hf = 75.64 kJ/kg
so heat required for 1 kg is
=hg - hf
= 2680.36 kJ/kg
and
from steam table specific volume of saturated steam at 6 bar is 0.315 m³/kg
so here mass flow rate is
mass flow rate = 
mass flow rate m = 0.10139 kg/s
so heat required is
H = h × m
here h is heat required and m is mass flow rate
H = 2680.36 × 0.10139
H = 271.75 kJ/s = 271.75 kW
now 90 % of boiler heat is used for generate saturated stream
so net boiler heat = 
net boiler heat = 
net boiler heat = 301.94 kW
Answer: what traffic patterns people will follow,
how people will feel when they enter or leave a building,
where to put doors, walls, and windows,
what building materials to use.
Explanation:
so the answers ar 2,3,5, and 6
Answer:
I=0.3636
Explanation:
See the attached picture for explanation.
Answer:
a) 25000 pcs/yr
b) cost per part produced is $ 1.84 per pc
c) Cpc = $ 1.365 per pcs
Explanation:
a)
the relation to calculate the number of parts produced annually by manual process is;
Q = Hw / Tc
Hw is the hourly rate ( 2000hr/yr) and Tc is the cycle time ( 4.8 min)
so we substitute
Q = (2000 × 60) / 4.8
= 120000 / 4.8
= 25000 pcs/yr
b)
cost per part produced
the relation to calculate the cost per part produced is expressed as;
Cpc = $23(Hw) / Q
Cpc is the cost per part produced
so we substitute
Cpc = $23(2000) / 25000
= 46000 / 25000
= $ 1.8 per pc
therefore cost per part produced is $ 1.84 per pc
c)
for the robot cell, at a service life of 4 years and a 10% rate of return , the factor is expressed as;
f = [r(1 + r)^t] / [((1 + r)^t ) - 1 ]
our rate r = 10% = 0.1 and our t = 4
so we substitute
f = [0.1 (1 + 0.1)^4] / [((1 + 0.1)^4 ) - 1 ]
f = 0.14641 / 0.4641
f = 0.3155
now we find the total cost
TC = 120000(0.3155) + 2000(0.3) + 2500
TC = $ 40,960
next we find the parts produced annually
Q = (2000 × 60) / 4
Q = 120,000 / 4
Q = 30000 pcs/yr
finally we find the cost per part produced;
Cpc = TC / Q
we substitute
Cpc = 40,960 / 30000
Cpc = $ 1.365 per pcs
