This document outlines the design of a custom power plant and cooling system for a large campus. Key details include: - The system will cool 963,400 sq ft of building space with 107,044 occupants. - Thermal loads were estimated from lighting, occupants, and outdoor air. - Refrigeration and power cycles were designed using a vapor compression refrigeration cycle paired with a combined Brayton and Rankine gas turbine system. - The final design produces over 1 MW of power with a 52% thermal efficiency using 325.71 lbm/min of fuel and 57.64 lbm/min of water. Only 5.0 lbm/min of compressed natural gas is required as fuel.

1. Custom Power Plant and Cooling System Design Diary
Starting Assumptions
Total Cooled Square Footage: 963,400 sq ft
Avg Room Sq Ft: 752
Avg Bulbs per Room: 36 (9 fixtures of 4 bulbs)
Total Bulbs (estimate): 46,121
Avg Space Occupied by Person: 9 sq ft
Total People: 107,044
People per room: 83
HX min driving force: 8
Isen Efficiency: 91
Indoor Air: 70
Outdoor Air: 96
Outdoor Water: 77
Max Steam Temp: 1050
Max Flame Temp: 2200
Qin (outside air) = 1.5 cooling tons(total number of sq ft./2000 sq ft) = 722.55 cooling tons = 144,510
BTU/Min
Qin (human bodies): 6 BTU/min/person (10,000 people) = 60,000 BTU/min
Heat given off by human body:
http://ergo.human.cornell.edu/studentdownloads/DEA3500notes/Thermal/thcondnotes.html
Qin (fluorescent lights): 32 watt bulb where 1kW/hr = 3,412.14 BTU, so 1 W/hr = 3.412 BTU, so 1 W/min =
0.05687 BTU. 32 watt bulb = 1.82 BTU/min(46,121 bulbs) = 83,938.79 BTU/min
Qin (total) = 288,448.79 BTU/min
Qin (per room) = (752/963,400)(288,448.79) = 225.155 BTU/Min
http://www.engineeringtoolbox.com/heat-gain-lights-d_709.html

2. REFRIGERATION CYCLE
Within the Room
Max Mass Flow Rate for Air: 163.424 lb/min
Airspeed: 700ft/min (~7.95 mph)
Volumetric Flow (exit): 2100ft^3/min
Air Cooling
-Enthalpy of entering air @70degrees F = 127.2
-Enthalpy of exiting air @50degrees F = 122.4
-Change in Enthalpy = -4.8
-Mass flow rate = (225.155 BTU/min)/(4.8 BTU/lbm) = 46.91 lbm/min.
-Airspeed entering Air Handler: Volumetric Flow Rate =(mass flow rate/molecular
weight)RT/P = (46.91/28.97)(10.73)(530)/14.7 = 626.5 cu. ft./min = 208.8 ft/min =
2.373mph
-Airspeed exiting Air Handler: Volumetric Flow Rate =
(mass flow rate/molecular weight)RT/P =
(46.91/28.97)(10.73)(510)/14.7 = 602.795 cu. ft./min =
200.932 ft/min = 2.283mph
Process of choosing values
We know that s must be constant 1-2
Pressure must be constant 2-3
h must be constant 3-4
Pressure must be constant 4-1
Thus, we will choose a temp/pressure at 1 that gives an s (as a vapor) that can be replicated as
a superheated vapor at 2. 3 will have an enthalpy, reached as a sat. liquid at the same pressure
as 2. 4 will have the same enthalpy as 3, which will be replicated by having a mixture with the
same enthalpy as 3 and pressure as 1.
Evaporator
-At Position 4
-Quality = 0.2363
-Temperature = 15.37 degrees F
-Enthalpy = 37.869 (using h = (1-x)hf + xhg)
-Pressure = 30 psi
-At Position 1
-Quality = 1
-Temperature = 15.37 degrees F
-Enthalpy = 105.32
-Pressure = 30 psi
-Entropy = 0.22383
-Change in Enthalpy across Evaporator = (105.32-37.869) = +67.451 BTU/lbm
-Mass Flow Rate of R134a = (225.155 BTU/min)/
(67.451 BTU/lbm) = 3.3381 lbm/min
Compressor
-At Position 1
-Quality = 1
-Temperature = 15.37 degrees F
-Enthalpy = 105.32
-Pressure = 30 psi
-Entropy = 0.22383

3. -At Position 2
-Quality = SH
-Temperature = 88.3 degrees F (found using linear interpolation to
Produce entropy equal to that found at Position 1)
-Enthalpy (ideal)= 116.326 (found using linear interpolation to
discover the value at 88.3 degrees)
-Pressure = 100 psi
-Mass Flow Rate = 3.3381 lbm/min
-Entropy (ideal) = 0.22383
Condenser
-At Position 2
-Quality = SH
-Temperature = 88.3 degrees F
-Enthalpy (ideal) = 116.326
-Enthalpy (real) = 117.415
This is derived from the facts that W-ideal = (mass flow rate)
(h2-ideal - h1), the isentropic efficiency is 91%, Isentropic
Efficiency = W-ide, Efficiency = W-ideal/W-real that W-real =
(mass flow rate)(h2-real - h1). W-ideal = 3.3381(116.326 - 105.32)
= 36.739. Isentropic Efficiency = 0.91 = W-i/W-r = 0.91 =
37.212/W-r, which gives
W-real = 40.373
W-real = (mass flow rate)(h2-real - h1) 40.373 =
3.3381(h2-real - 105.32)
H2-real = 117.415
-Pressure = 100 psi
-Mass Flow Rate = 3.3381 lbm/min
-Entropy (ideal) = 0.22383
-At Position 3
-Quality = 0
--Enthalpy = 37.869
-Pressure = 100 psi
-Temperature = 79.12
-At Position 4
-Quality = 0.2363
-Temperature = 15.37 degrees F
-Enthalpy = 37.869 (using h = (1-x)hf + xhg)
-Pressure = 30 psi
NOTES:
Across 4-1, 67.451 BTU/lbm are gained
Across 2-3, 79.546 BTU/lbm are lost
Qin = QL = 225.155 BTU/min
Qout = QH = 265.533 BTU/min
Air across 7-8
Has same mass flow rate of 49.91 lbm/min as indoors
265.533 BTU/min = (mass flow rate)(Cp)(change in temp) →
(46.91)0.24(change in temp) = 265.533 → Change in temp =
23.58532296. Thus, temp @ 8 = 119.585323 degrees F
COP: 225.155/40.373 = 5.576
Work required to power one 752 ft^2 room: 40.373 BTU/min = 0.952 hp = 0.7099 kW
Work required to power campus: 963,400/752 = 1281.12 rooms. Round up to 1282
Rooms. 40.373(1282)= 51,758.186 BTU/min = 1,220.424 hp = 910.1 kW

4. POWER GENERATION: BRAYTON-RANKINE CYCLE
Brayton Cycle
Given Facts
-Across 5-6, Isentropic Compression
-Across 6-7, Constant-pressure Heat Addition
-7-8,Isentropic Expansion
-8-5, Constant-pressure Heat Rejection
-Pressure Ratio = rp =P2/P1; the higher this is, the greater the efficiency. Generally
ranges from 11 to 16
-Thermal Efficiency = Wnet/Qin; shoot for above 45% (preferably 50%)
-Wnet must = 51,758.186 BTU/min
Compressor (14:1 Compression Ratio)
At Position 5 (Air)
-Pressure = 14.7 psi
-Temperature = 556 degrees Rankine (96 degrees F)
At Position 6 (14:1 Compression Ratio)
-Pressure = 205.8 psi
-Temperature (ideal) = found using the formula T6/T5 = (P6/P5)^(k-1)/k,
with k = 1.4. T6 = T5(P6/P5)^(k-1)/k → T6 = 556(205.8/14.7)^
(1.4-1)/1.4 = 1181.789 degrees Rankine
-Change in Enthalpy (Work-ideal): 0.24(1181.789-556) = 150.1894
-Temperature (real) = found knowing that isentropic efficiency is
capped at 0.91, and thus Work-real = Work-ideal/0.91 =
165.0433, and since Work-real = Cp(T6r-T5),
165.0433 = 0.24(T6r-556) ---> T6r = 1243.68
Combustion Chamber
At Position 6
-Pressure = 205.8 psi
-Temperature = 1243.68
At Position 7
-Pressure = 205.8 psi
-Temperature = 2660 degrees Rankine
-Change in Enthalpy = 0.24 (2660-1243.68) = 339.92 BTU/lbm
= Qin = QH
Gas Turbine
At Position 7
-Pressure = 205.8 psi
-Temperature = 2660 degrees Rankine
At Position 8
-Pressure = 14.7 psi
-Temperature (ideal) → T8 = T7(P8/P7)^((1.4-1)/1.4) = 2660
(14.7/205)^(.4/1.4) = 1251.4586 degrees Rankine
-Change in Enthalpy (Work-ideal) = 0.24(2660-1251.4586) = 338.05
-Temperature (real) = found knowing that isentropic efficiency is
capped at 0.91, and thus Work-real = Work-ideal(0.91) =
307.625, and since Work-real = Cp(T7-T8r),
307.625 = 0.24(2660-T8r) ---> T8-real = 1378.22723
Total System-Work-net = Work-out-real - Work-in-real = 307.625 - 165.0433 =
142.5817 BTU/lbm
-Thermal Efficiency = Work-net/Qin = 142.5817/339.92 = 0.41946
Rankine Cycle

5. Given Facts
-Across 1-2, Isentropic Compression
-Across 2-3, Constant-pressure Heat Addition
-3-4,Isentropic Expansion
-4-1, Constant-pressure Heat Rejection
Max steam temp: 1050
From Position 8 to Position 9 (which has the same temp as 5), the change in enthalpy
is 0.24(1378.22723-556) = 197.3345 BTU/lbm, which is the Qin for the Rankine
Cycle. 197.3345 BTU/lbm(111.2lbm/min), we get 21,943.5964 BTU/min for Qin.
Compressor
At Position 1
-Pressure = 14.709 psi
-Temperature = 212 degrees F
-Specific Volume = 0.01671
-Entropy = 0.31222
-Enthalpy = 180.21
-Quality = 0
At Position 2
-Pressure = 200 psi
-Temperature = 212 degrees
-Work-ideal, pump = Work-ideal = (spec. vol @1)
(P2-P1)(conversion factor) = 0.01671(200-14.709)(.185)
= 0.5725 BTU/lbm
-Quality = Compressed Liquid
-Work-real (pump) = 0.5725/0.91 = 0.62912 BTU/lbm
-Enthalpy (real) → Work-real = (h2-real - h1)
→ 0.62912 = h2-real-180.21 → h2-real = 180.839 BTU/lbm
Heat Exchanger
At Position 2
-Pressure = 200 psi
-Temperature = 212 degrees
-Enthalpy = 180.839 BTU/lbm
At Position 3 (superheated)
-Pressure = 200 psi
-Temperature = 550 degrees F
-Enthalpy = 1296 BTU/lbm
-Entropy = 1.6516
-Change in Enthalpy (h2 to h3) = 1115.161 BTU/lbm = Qh
-Quality = SH vapor
-Determine Mass Flow Rate: 21,943.5964 BTU/min for Qh / 1115.161 =
19.6775 lbm/min
Steam Turbine
At Position 3 (superheated)
-Pressure = 200 psi
-Temperature = 550 degrees F
-Enthalpy = 1296 BTU/lbm
-Entropy = 1.6516
At Position 4
-Pressure = 14.709 psi
-Temperature = 212

6. -Enthalpy = Found using quality --> h = (1-x)hf + xhg = (1-0.927)180.21
+ 0.927(1150.3) = 1079.48343 BTU/lbm
-Entropy = 1.6516
-Quality = x = (entropy - entropy of saturated liquid @ psi/
(entropy of mixture @ psi) = (1.6516 -0.31222)/1.44427 = 0.927
-Work (ideal) Across Turbine = h3-h4 = 1296 - 1079.48343 = 216.517
BTU/lbm
-Work (real) Across Turbine = 0.91 (216.517) = 197.03047 BTU/lbm
-Work (net) = 197.03047 - 0.62912 = 196.40135 BTU/lbm
-Enthalpy (real) = 196.40135 = 1296 - h4-real → 1099.59865
Condenser
At Position 4
-Pressure = 14.709
-Temperature = 212
-Enthalpy = 1099.59865 BTU/lbm
At Position 1
-Pressure = 14.709 psi
-Temperature = 212 degrees F
-Enthalpy = 180.21
-Change in Enthalpy (Ql) = 919.38865
-Thermal Efficiency = 1- Ql/Qh = 1-(919.38865/115.161) =0.17556
TOTAL SYSTEM NET OUTPUT
For the combined cycle we add the Work-net-Brayton to the product of
(Rankine mfr/Brayton mfr)(Work-net-Rankine) →
142.5817 + (19.6775/111.2)(196.40135) = ​177.3361 BTU/lbm
Thermal Efficiency = Wnet/Qin = 177.3361/339.92 = ​0.52
Now, these BTU/lbm figures are in terms of fuel lbm. To get the necessary
output, we will divide the total energy needs, rounded up to 57,760
BTU/min, by this Wnet for the Brayton-Rankine cycle. 57,760/177.3361
= 325.71 lbm/min fuel. Now, we need to maintain the ratio of (water mass
flow rate)/(fuel mass flow rate), which was originally 19.6775/111.2 =
0.176956. New water mass flow rate = 0.176956(325.71) = 57.64 lbm/min
FINAL SPECS
-Fuel Mass Flow Rate: 325.71 lbm/min
-Total Output: 57,760 BTU/min = 1015.601 kW = 1.0156 MW
-Thermal Efficiency: 52%
-Coefficient of Performance: 5.576
Cost Analysis:
If the Brayton were fueled by Compressed Natural Gas, which has an energy density of
22,268 BTU/lbm, and this design only requires an energy density of 339.92 BTU/lbm, only a
ratio of 339.92/22,268 = 0.0153 CNG/Air is needed. Thus, only 0.0153 (325.71 lbm/min) =
4.983363 lbm/min of CNG is needed. Rounding up to 5.0 lbm/min CNG to account for system
inefficiencies, and converting to ft^3 → (5.0 lbm/min)(1 ft^3/0.05 lbm) = 100 ft^3/min [drawn
from EngineeringToolbox.com, ​http://www.engineeringtoolbox.com/gas-density-d_158.html​].
During one hour of this powerplant’s operation, 1,015.601 kW/hr are produced, consuming
6,000 ft^3 of CNG at an estimated cost of $3.54/1,000 ft^3 [drawn from the US Energy
Information Administration, ​https://www.eia.gov/dnav/ng/hist/n3035us3m.htm​], for a total cost of
$21.24. This translates to a cost of approximately $0.021 per kW/hr. This is significantly less
that the $0.044 per kW/hr that is charged, on average, for industrial uses in Lake Charles
[drawn from ElectricityLocal.com ​http://www.electricitylocal.com/states/louisiana/lake-charles/​]
Per kW/hr, this represents a cost savings of over 52.27%