# Chapter 1 : Dimensions and Units¶

### Example 1.1 Page 12¶

In :
# solution

# Variables
# Using conversion factors from table 1.3 (Pg 9)
q1 = 75.                     # [gallon/min] (volumetric flow rate)
q2 = 75./(60*.219969)        # [dm**3/s]
row = 0.8                    # [kg/dm**3]

# Calculation
q3 = q2*row                  # [kg/s] (mass flow rate)

# Result
print "mass flow rate = ",q3," [kg/s] "

mass flow rate =  4.54609513159  [kg/s]


### Example 1.2 Page 12¶

In :
import math
# solution

# Variables
qm = 2000.                   # [kg/h] (mass flow rate)
d1 = 3.068                   # [in] (internal dia of pipe)

# Calculation
# Using conversion factors from table 1.3 (Pg 9)
d2 = 3.068/.0393701          # [mm]
A = ((math.pi/4)*d2**2)/10**6       # [m**2] (cross section area)

# Using steam tables; Appendix IV.3
v = 0.46166                  # [m**3/kg] (sp. vol. of steam at 440 kPa)
qv = (qm*v)/3600.            # [m**3/s]
vs = qv/A                    # [m/s]

# Result
print "velocity of the steam in the pipeline is ",vs," m/s"

velocity of the steam in the pipeline is  53.7752187869  m/s


### Example 1.3 Page 13¶

In :
# solution

# Variables
m = 2000.            # [lb] (mass flow rate)
t = 24.              #[hr]
lf = 144.            # [Btu/lb] (latent heat of fusion)

# Calculation
# Using conversion factors from table 1.3 (Pg 9)
TR = (m*lf*.251996*4.184)/(3600*24.)

# Result
print "1 TR = ",TR," kW"

1 TR =  3.51450421333  kW


### Example 1.4 Page 13¶

In :
# solution

# Variables
# C = 89.2*A*(T/M**).5   [ft**3/s]
k = 89.2
C1 = 1.              # [ft**3/s]

# Calculation
# Using conversion factors from table 1.3 (Pg 9)
C2 = 35.31467*C1
T1 = 1.              #[dgree R]
T2 = 1.8*T1 # [K]
A1 = 1.              # [ft**2]
A2 = 10.76391
k2 = (k*A2*(1.8)**.5)/35.34167

# Result
print "eq in SI becomes  C = ",k2,"*T/M**.5  [m**3/s]"

eq in SI becomes  C =  36.4488724075 *T/M**.5  [m**3/s]