This notebook contains material from chemeng316_fluids; content is available on Github.

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B.1 Homework 1

ChE 316 – Chem Eng. Fluid Mechanics – Spring 2024

DUE 29 Jan 2024

Problem 1. In the production of fishmeal, the oil is first extracted, leaving a wet fish cake containing 88 wt.% water. This cake is partially dried to reduce the moisture content to 27 wt.%, after which the cake is ground.

• Construct a block flow diagram of this process.
• How many kg/h of wet cake would be necessary to produce 1200 kg/h of "dry" cake?
• How much water is extracted from the wet cake (in kg/h)?

See Supplements 2.2 and 2.3 to get some ideas about how to work in python with units and fluid properties if you would like to give it a try.

See Supplement 1.3 for a direct example of how to solve a system of equations with sympy.

Getting Started: Here is some code to at least get the UnitRegistry and CoolProp modules loaded for you to get started.

try:
    import pint
except:
    !pip install pint
    import pint
finally:
    ur = pint.UnitRegistry()

try:
    import CoolProp.CoolProp as CP
except:
    !pip install CoolProp
    import CoolProp.CoolProp as CP

# we need numpy for constants like pi
# https://numpy.org/doc/stable/reference/constants.html
import numpy as np

# YOUR SOLUTION HERE

Problem 2. Air is flowing steadily through a horizontal heated tube. The air enters at 45 $^\circ$F at a velocity of 55 ft/s. The air leaves the tube at 145 $^\circ$F and 66 ft/s. HINT: See Supplement 2.3 for CoolProp properties examples.

• What is the mass flowrate (in lb/s) of the air within the heated tube if the tube diameter is 3 inch?
• If the average specific heat of air is 0.24 BTU/lb $\cdot^\circ$F, how many BTU/lb of energy is transferred through the wall of the tube?
• Does the mass flowrate within the tube increase, decrease, or remain the same? Explain your reasoning.

# In order to get tempeature and fluid properties you do something like this ...

# GIVEN information ...
T1_degF = ur.Quantity(45, ur.degF);
u1 = 55 * ur.feet / ur.sec;
cp_air = 0.24 * ur.BTU/(ur.lb*ur.degR);

T2_degF = ur.Quantity(145, ur.degF);
u2 = 66 * ur.feet / ur.sec;

d = 3 * ur.inch;

# ASSUMPTION: I'm not told anything about the pressure ... so I have to assume something ...
P1 = 1.0 * ur.atm;

# GET PROPERTIES
# Note ... I'm assuming a simple "dry"-air model for "AIR" ... if humidity needs to be taken
# into account we would need to use HAPropsSI ... see
# http://coolprop.org/fluid_properties/HumidAir.html#haprops-sample
T1_K = T1_degF.to(ur.degK);
rho1_kgPerM3 = CP.PropsSI("D","T",T1_K.magnitude,"P",P1.to(ur.Pa).magnitude,"Air") * ur.kilogram / ur.meter **3;

T2_K = T2_degF.to(ur.degK);
P2 = P1;
rho2_kgPerM3 = CP.PropsSI("D","T",T2_K.magnitude,"P",P2.to(ur.Pa).magnitude,"Air") * ur.kilogram / ur.meter **3;

print(f"cp_air = {cp_air:.4f~P} = {cp_air.to(ur.J/ur.kg/ur.K):.4f~P}")
print(f"rho_1 = {rho1_kgPerM3:.3f~P}")
print(f"rho_2 = {rho2_kgPerM3:.3f~P}")

# YOUR SOLUTION HERE

Problem 3. An evaporator is fed 32 tonnes/hr of a solution consisting of 12% NaOH, 17% NaCl, and 71% water. During evaporation, water evaporates, and salt precipitates as crystals, which settles and is removed from the remaining concentrated liquor. The liquor leaving the evaporator contains 48% NaOH, 2% NaCl, and 50% water.

• Calculate the kilograms of water evaporated per hour.
• Calculate the kilograms of salt precipitated per hour.
• Calculate the kilograms of concentrated liquor produced per hour.

# Some GIVEN information ...
mDot1 = 32000 * ur.kg/ur.hr; # note that tonne is metric ton = 1000 kg

# YOUR SOLUTION HERE

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