Roof insulation: Insulation of the bare roof with 50 mm thick of polystyrene foam, roof area = 99m?, unit cost = $ 5.7 m'Einstallation cost is included} Overall heat transfer coefficient: for uninsulated roof Uou = (1.93 *1.83) W/m?.K, for insulated roof Unew = (0.53* 1.8) Wrm'K Winter heating degree days = I16l HOO, Summer coling degree days = Heat source : diesel boiler', Energy content of diesel = 97 kwhLiter Cooling Source : electric chiller è Power Plant uses Diesel fuel Boiler efficiency = 0.87, chiller COP = 185 #1.83 electric grid efficiency Cost of diesel = 0.46 $Liter, Cost of electricity = 0.0% $/kwh Discount "interest" rate = I %, analysis period'= \0 years Per I kwh of Diesel : cor emission = 0.271 kgrkwh, NO, emission = 0.57 grkwh, SOn emission= 0.86 grkwh = 1317 COD %3D = 33 % %3D a = 999 B = 1161 6 = 1317 o = 1.83 8 = || Item Value Unit Old site energy losses "Diese, heating Old site energy losses "Electric", cooling New site energy losses "Diesel", heating New site energy losses "Electric", cooling Annual site kWhryr kwhryr kwhyr kwhryr energy savings "Dieser"- kwhryr
Roof insulation: Insulation of the bare roof with 50 mm thick of polystyrene foam, roof area = 99m?, unit cost = $ 5.7 m'Einstallation cost is included} Overall heat transfer coefficient: for uninsulated roof Uou = (1.93 *1.83) W/m?.K, for insulated roof Unew = (0.53* 1.8) Wrm'K Winter heating degree days = I16l HOO, Summer coling degree days = Heat source : diesel boiler', Energy content of diesel = 97 kwhLiter Cooling Source : electric chiller è Power Plant uses Diesel fuel Boiler efficiency = 0.87, chiller COP = 185 #1.83 electric grid efficiency Cost of diesel = 0.46 $Liter, Cost of electricity = 0.0% $/kwh Discount "interest" rate = I %, analysis period'= \0 years Per I kwh of Diesel : cor emission = 0.271 kgrkwh, NO, emission = 0.57 grkwh, SOn emission= 0.86 grkwh = 1317 COD %3D = 33 % %3D a = 999 B = 1161 6 = 1317 o = 1.83 8 = || Item Value Unit Old site energy losses "Diese, heating Old site energy losses "Electric", cooling New site energy losses "Diesel", heating New site energy losses "Electric", cooling Annual site kWhryr kwhryr kwhyr kwhryr energy savings "Dieser"- kwhryr
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
Related questions
Question
the question is in heat conservation

Transcribed Image Text:Annual site energy
savings "electric
Annual total site
kwhyr
$yr
energy cost
savings
Annual Source
kwh/yr
energy savings
"heating"
Annual Source
kwhryr
energy savings
"Cooling"
Annual Source
energy savings
"Tota"
kwhryr
Reduction in COn
Tonsyr
Reduction in NOx
kgryr
Reduction in SOK
kgryr
Initial
investment cost
Net annual cost
$
savings
Present value
Investment
Present value
Savings
NPV
$
SIR
Area (m') I kw 24 hour
Energy Losss in heating season =
R (m.
*\000 w
day

Transcribed Image Text:Roof insulation: Insulation of the bare roof with 50 mm thick of polystyrene
foam,
roof area = 99m?, unit cost = $ 5.7 /m'{installation cost is included}
Overall heat transfer coefficient:
for uninsulated roof Uou = (1.93 *1.83) W/m?.K, for insulated roof Unew = (0.53* 1.83)
Winter heating degree days = l161 HDD, Summer cooling degree days = 1317 COD
Heat Source : diesel boiler', Energy content of diesel = 97 kwhiLiter
Cooling Source : electric chiller è Power Plant uses Diesel fuel
Boiler efficiency = 0.87, chiller CoP = 1,85 *1.83 electric grid efficiency
Cost of diesel = 0.46 $/Liter, Cost of electricity = 0.09 $1kwh
Discount "interest" rate = ll %, analysis period'= 10 years
Per I kwh of Diesel:
COr emission = 0.271 karkwh, NO, emission = 0.57 grkwh, SOx emission= 0,86 grkwh
33 %
%3D
%3D
a = 999
B = 1161
6 = 1317
o = 1.83
ɛ = ||
%3D
エtem
Value
Unit
Old site energy
losses "Diesel,
heating
Old site energ
losses "Electric",
cooling
New site energy
losses "Diesel"
heating
New site energy
losses "Electric",
cooling
kwhyr
kwhyr
kwhyr
kwhryr
Annual site
energy
Savings "Dieser"-
kwhryr
Expert Solution

Step 1
The thickness and thermal conductivity of the mediums through which heat is transmitted affect the overall heat transfer coefficient. The bigger the coefficient, the easier it is to transmit heat from the source to the heated product.
Multiply the amount of kilowatt hours (kWh) per year (the figure on the Energy Rating Label) by your power tariff to estimate how much an electrical item with a star rating will cost to run each year. Consumers are often offered electrical energy in kilowatt-hours.
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