import numpy as np
import matplotlib.pylab as plt
%matplotlib inline
The ammonia will be distributed as liquid at ambient temperatures which may range up to 50 $^\circ$C. What would be the minimum operating pressure of the piping network in psig?
The solution to this problem is found using Antoine's equation to predict ammonia vapor pressure as a function of temperature.
A = 7.36050
B = 926.132
C = 240.17
def Psat(T):
return 10.0**(A - B/(T + C))
T = np.linspace(-83,60)
plt.plot(T,Psat(T))
plt.xlabel('Temperature [deg C]')
plt.ylabel('Pressure [mmHg]')
plt.grid()
At the operating conditions
print "Absolute Pressure at 50 deg C = ", Psat(50), "mmHg"
print " Gauge Pressure at 50 deg C = ", (Psat(50)-760)*14.696/760.0, "psig"
Absolute Pressure at 50 deg C = 14750.6911081 mmHg Gauge Pressure at 50 deg C = 270.5357849 psig
print 14.696*34.47/1.01325
499.946824574
So the minimum operating pressure would be 271 psig in order to keep ammonia in the liquid phase.
The liquid ammonia is delivered to the solar collector at \unit[50]{$^\circ$C} and \unit[500]{psig}. The solar collector contains a pumping unit to bring the pressure to \unit[15]{MPa}. What is the pump requirement in watts for a flowrate of 1 kilogram/min of ammonia?
rho = 0.60 #kg/liter
rho = 600.0 # kg/m3
Vspecific = 1.0/rho
Pin = (500+14.696)*101325/14.696
Pout = 15.0*1.0e6
W = Vspecific*(Pout-Pin)/60.0
print W
318.091899156
dGf = -16600.0
dHf = -46150.0
Grxn = -2.0*dGf
Hrxn = -2.0*dHf
T = 600.0 + 273.15
lnK = -((Grxn-Hrxn)/298.15 + Hrxn/T)/8.314
K = np.exp(lnK)
rho = 600.0
MW =
Hf_NH3 = -46150.0
Hf_N2 = 0.0
Hf_H2 = 0.
Cp_NH3 = 35.6
Cp_N2 = 29.1
Cp_H2 = 29.1
dH_0 =
dH_1 = 2.0*Cp_NH3*(25-50)
dH_2 = 2.0*Hf_NH3 - Hf_N2 - 3.0*Hf_H2
dH_3 = 1.0*Cp_N2*(600-25)
dH_4 = 3.0*Cp_H2*(600-25)
Hrxn = dH_1 + dH_2 + dH_3 + dH_4
print dH_1, dH_2, dH_3, dH_4
print "Net heat of reaction at operating conditions = ", Hrxn, "Joules/gmole"
-1780.0 -92300.0 16732.5 50197.5 -27150.0
15.0*14.696
2175.5736491487787