# Chapter01:Fundamental of Energy - Science and Technology¶

## Ex1.1:pg-36¶

In [7]:
# Given data:
T1=500+273.0 #source temp in kelvin
T2=100+273.0 #sink temperature in kelvin
W=1 # output power in kW

nth=1-(T2/T1) # thermal efficiency

Q1=1/nth # heat supplied in kW

Q2=Q1-W # heat rejected in kW

print " The heat rejected is ",Q2,"kW"

# the answer in book is wrong due to incorrect  value of T1

 The heat rejected is  0.9325 kW


## Ex1.2:pg-37¶

In [14]:
# Given data:
T1=40+273.0 # ambient temp in kelvin
T2=-10+273.0 # freezer temp in kelvin

Q1=T1*(Q2/T2) # heat transfer rate in kJ/s

W=Q1-Q2 # work in kW

print "The least power required is ",round(W,2),"kW"

The least power required is  0.38 kW


## Ex1.3:pg-37¶

In [16]:
# Given data:
Q1=3e4 # heat required in kJ/h
W=2e3 # work required in kJ/h

Q2=Q1-W # heat abstracted from outside in kJ/h

COP=Q1/(Q1-Q2) # COP of heat pump

print "Heat abstracted from outside air is ",Q2,"kJ/h"
print "COP heat pump is ",COP

Heat abstracted from outside air is  28000.0 kJ/h
COP het pump is  15.0


## Ex1.4:pg-38¶

In [17]:
# Given data:
T11=320+273.0 # temp at source 1 in kelvin
Q1=10000.0 # heat transfer rate at source 1 in kJ/min
T12=65+273.0 # temp at source 2 in kelvin
Q2=120000.0 #  heat transfer rate at source 2 in kJ/min
T2=35+273.0 # temp of surrounding in kelvin

n1=1-(T2/T11) # efficiency by source 1
n2=1-(T2/T12) # efficiency by source 2

W1=Q1*n1 # work at source 1 in kJ/min
W2=Q2*n2 # work at source 2 in kJ/min

print "The work done at two sources is W1=",W1," kJ/min W2=",W2,"kJ/min"
print "The larger power is provided by source 2"

The work done at two sources is W1= 4806.07082631  kJ/min W2= 10650.887574 kJ/min
The larger power is provided by source 2