# Chapter 4:WORK AND HEAT¶

## Ex4.1:PG-96¶

In [16]:
#example 1
#work done during different processes
import math

P1=200 #initial pressure inside cylinder in kPa
V2=0.1 #in m^3
V1=0.04 #initial volume of gas in m^3

W1=P1*(V2-V1) #work done in isobaric process in kJ
print"\n hence,the work done during the isobaric process is",round(W1,2),"kJ. \n"

W2=P1*V1*math.log(V2/V1) #work done in isothermal process in kJ
print"\n hence,the work done in isothermal process is",round(W2,2),"kJ. \n"

P2=P1*(V1/V2)**(1.3) #final pressure according to the given process
W3=(P2*V2-P1*V1)/(1-1.3)
print"\n hence,the work done during the described process is",round(W3,2),"kJ. \n"

W4=0 #work done in isochoric process
print"\n hence,the work done in the isochoric process is",round(W4,3),"kJ. \n"

 hence,the work done during the isobaric process is 12.0 kJ.

hence,the work done in isothermal process is 7.33 kJ.

hence,the work done during the described process is 6.41 kJ.

hence,the work done in the isochoric process is 0.0 kJ.



## Ex4.3:PG-99¶

In [10]:
#example 3

#work produced

Psat=190.2 #in kPa
P1=Psat #saturation pressure in state 1
vf=0.001504 #in m^3/kg
vfg=0.62184 #in m^3/kg
x1=0.25 #quality
v1=vf+x1*vfg #specific volume at state 1 in m^3/kg
v2=1.41*v1 #specific volume at state 2 in m^3/kg
P2=600 #pressure in state 2 in kPa from Table B.2.2
m=0.5 #mass of ammonia in kg
W=m*(P1+P2)*(v2-v1)/2 #woork produced by ammonia in kJ
print "The final pressure is",round(P2),"kPa\n"
print "hence,work produced by ammonia is",round(W,2),"kJ"

The final pressure is 600.0 kPa

hence,work produced by ammonia is 12.71 kJ


## Ex4.4:PG-100¶

In [3]:
#example 4

#calculating work done

v1=0.35411 #specific volume at state 1 in m^3/kg
v2=v1/2
m=0.1 #mass of water in kg
P1=1000 #pressure inside cylinder in kPa
W=m*P1*(v2-v1) #in kJ
print "the work in the overall process is",round(W,1),"kJ"

the work in the overall process is -17.7 kJ


## Ex4.7:PG-108¶

In [1]:
#example 7
#heat transfer
k=1.4 #conductivity of glass pane in W/m-K
A=0.5 #total surface area of glass pane
dx=0.005 #thickness of glasspane in m
dT1=20-12.1 #temperature difference between room air and outer glass surface temperature in celsius
Q=-k*A*dT1/dx #conduction through glass slab in W
h=100 #convective heat transfer coefficient in W/m^2-K
dT=12.1-(-10)  #temperature difference between warm room and colder ambient in celsius
Q2=h*A*dT #heat transfer in convective layer in W
#result
print "the rate of heat transfer in the glass and convective layer is",round(Q2),"kW."

the rate of heat transfer in the glass and convective layer is 1105.0 kW.