Chapter 5 : First Law of Thermodynamics¶

Example 5.1 Page No : 115¶

In [1]:
import math

# Variables
v1_total = 7;			# in m**3/min
v_s1 = 0.35;			# in m**3/kg
v_s2 = 0.12;			# in m**3/kg
p1 = 1;			# in bar
p1 = p1 * 10**5;			# in N/m**2
p2 = 6;			# in bar
p2 = p2 * 10**5;			# in N/m**2
D1 = 110;			# in mm
D1 = D1 * 10**-3;			# in m
D2 = 65;			# in mm
D2 = D2 * 10**-3;			# in m

# Calculations and Results
Af1 = math.pi/4*D1**2;			# in m**2
Af2 = math.pi/4*D2**2;			# in m**2
# v1_total = m1 * v_s1
m1 = v1_total / v_s1;			#in kg/min
print "The mass flow rate of air in kg/min is :",m1

m2 = m1;			# in kg/min
v2_total = m2 * v_s2;			# in m**3/min
del_W_flow = (p2 * v2_total) - (p1 * v1_total);			# in J/min
print "The change in the work flow in kJ/min is : ",del_W_flow*10**-3

v_f1 = v1_total/Af1;			# in m/min
v_f2 = v2_total /Af2;			#in m/min
del_v = v_f2 - v_f1;			# in m/min
print "Change in velocity of the air flow in m/min is : %.2f"%del_v

The mass flow rate of air in kg/min is : 20.0
The change in the work flow in kJ/min is :  740.0
Change in velocity of the air flow in m/min is : -13.32


Example 5.2 Page No : 118¶

In [1]:
# Variables
m = 2.;			# in kg per min
m = m / 60;			# in kg per sec
W = 20;			# in kW
h1 = 1400;			# in kJ/kg
h2 = 1300;			# in kJ/kg

# Calculations
Q = (m * (h2 - h1)) + W;			# in kJ/s

# Results
print "Rate of heat transfer to the water jacket in kJ/sec %.2f"%Q

Rate of heat transfer to the water jacket in kJ/sec 16.67


Example 5.3 Page No : 127¶

In [5]:
# Variables
g= 9.81;
p1 = 3;			# in Mpa
p2 = 10;			# in kPa
T1 = 350;			# in °C
T1 = T1 + 273;			# in K
m = 1;			# in kg per sec
v1 = 50;			# in m per sec
v2 = 120;			# in m per sec
z1 = 2;			# in m
z2 = 5;			# in m
C_p = 1.005;			# in kJ per sec
Q = 5;			# in kJ per sec

# Calculations and Results
Q = -(Q) * 10**3;			# in J per sec
T2 = (p2 * T1)/p1;			# in K
del_h = C_p * (T2-T1);			# in kJ
del_h = del_h * 10**3;			# in J
t = m * ( del_h +(v2**2-v1**2)/2 + (g * (z2 - z1)));			# t is variable taken  for calculation
W_s = Q - t;			# in J per sec
W_s = W_s * 10**-6;			# in MW
print "The power output of the turbine in MW is : %.3f"%W_s

# If kinetic and potential energy are ignored then
W_s2 = Q -(m * del_h);			# in J per sec
W_s2 = W_s2 * 10**-6;			# in MW
errorIntroduced= (abs(W_s)-abs(W_s2))/abs(W_s)*100;			# in %
print "Total error introduced in %% is : %.1f"%errorIntroduced

The power output of the turbine in MW is : -1.471
Total error introduced in % is : 0.4


Example 5.4 Page No : 128¶

In [6]:
# Variables
h1 = 246.6;			# in kJ/kg
h2 = 198.55;			# in kJ/kg
W = 0;
g= 9.8;
Q= -(105000);			# in kJ per hr

# Calculations
# m * (h1 + ((v1***2)/(2*1000)) + ((g * z1)/1000)) + Q =  m * (h2 + ((v2**2)/(2*1000)) + ((g * z2)/1000)) + W
# v1 and v2 is change in velocity is neglected and z2 = z1 + 10
m = Q/( (h2-h1) + ((g * 10)/1000) );			# kg per hr

# Results
print "Quantity of  water circulated through the pipe in kg/hr is : %.2f"%m

Quantity of  water circulated through the pipe in kg/hr is : 2189.69


Example 5.5 Page No : 128¶

In [7]:
import math

# Variables
#Given data
m=15.;			    # in kg/min
m= m/60;			# in kg/sec
H1= 5;	    		# in kJ/kg
H1= H1*10**3;			# in J/kg
H2= 173;			# in kJ/kg
H2= H2*10**3;			# in J/kg
V1= 5;		    	# in m/s
V2= 7.5;			# in m/s
Q= 760;			    # in kJ/min
Q= Q*10**3/60;			# in J/s

# Calculations and Results
# Formula (H1+V1**2/2)+(-Q)= (H2+V2**2/2)+W
W= (H1+V1**2/2)+(-Q)-(H2+V2**2/2);			# in W/kg
W= W*10**-3;			# in kW/kg
# The work done will be
W= m*W;			# in kW
P= abs(W);			# in kW
print "Power of the motor required to drive the compressor in kW is : %.2f"%P

# Part (b)
v1= 0.5;			# in m**3/kg
v2= 0.15;			# in m**3/kg
# A1/A2= rho2*V2/(rho1*V1) = v1*V2/(v2*V1)
ratioOFA1andA2= v1*V2/(v2*V1);

Power of the motor required to drive the compressor in kW is : 45.17