Cryogenic Information Library
Heat Transfer Problems Solved

Exemplar 5-liter Laboratory Dewar

Assume the following:

Neck Tube: Material NEMA Grade G-10
K for G-10 = 2.03 BTU/Hr/Ft² °F/In
Effective length 4 inches
O.D. 2.00 inches
I.D. 1.96 inches
Area = ((2.00"/2)²) - ((1.96"/2)²) = .1244 in² / 144 in²/ft² = .0009 Ft²

Annular space 24-hour settle pressure of 1 X 10^-4 Torr cold

Outer vessel and outer layer of the MLI effective surface area 236 square inches (1.63 square feet) or (.151 meter²)

Dewar contains LN2 at 77K (-320°F) at atmospheric pressure.  (Latent heat of vaporization of 85.6 BTU / Lb and density of 1.77 Lbs / Liter)

Dewar is in a room temperature environment of 295K (72°F)

There are 60 layers of MLI

Calculate heat conducted into the inner vessel through the neck tube

Q = k * area in square feet * temperature gradient in °F = kA*((T"-T')/l)2

Q = 2.03 BTU/Hr/Ft²°F/In * .0009 Ft² ((-320°F - 72°F)/ 4 in) = .18 BTU / Hr

Calculate heat radiated into inner vessel through the MLI

From the Technifab Cryogenic Insulation Intermediate paper, MLI at a cold settle pressure of 1 * 10^-4 Torr renders a thermal resistance "R" value of approximately 1440 degree F - hour - square foot / BTU

Q = A * (Th-Tc)/R3

Q = Area in square feet * (Temperature of hot surface in °F - Temperature of cold surface in °F) / R in °F - hour - square foot / BTU

Q = 1.63 Ft² * (72°F + 320°F) / 1440°F - hour - Ft² / BTU

Q = .44 BTU / Hour

Calculate expected NER for Dewar filled with LN2

Total heat transfer into inner vessel = Q through neck tube + Q through annular space

Total Q = .18 BTU / Hour + .44 BTU / Hour = .62 BTU / Hour

.62 BTU / Hour (Total heat transferred) / 85.6 BTU / Lb (Latent heat of vaporization) = .007 Lbs / Hour (vaporized)4

.007 Lbs / Hour * 1.78 Lbs / Liter = .01 Liters / hour vaporized

.01 Liters / Hour * 24 Hours / Day = .31 Liters / day natural evaporation rate

The Classic Radiation Theory for Multi-Layered Reflecting Parallel Plates

For reference, another method to calculate the heat transferred into the inner vessel through the annular space is by use of the Stefan-Boltzman constant.  This method is called the classic radiation theory for multi-layered reflecting parallel plates.  It assumes no convective heat transfer in the annular space of the vessels.

Q = ( A e eff (T2^4-T1^4)1

Effective emissivity is calculated as follows:

eeff = [ 2 (1/e0 + 1/e1-1) + ( N-1) ( 2/e1-1)]^-1

e0 = emissivity of the hot and cold surfaces = .02
e1 = emissivity of the layers of MLI = .029
N = number of layers of MLI = 50

eeff = [ 2 (1/.02 + 1/.029-1) + ( 60-1) ( 2/.029-1)]^-1

e eff = .0019

Q = .000000056703 * .151 meter^2 * .0019 * (295 °K^4-77 °K^4)

Q = .12 Watts

Q = .12 Watts / .293 = .41 BTU / Hour

Calculate Heat Transferred through a Wall made up of Various Materials

The problem:

The cross section of a wall of a cold storage room consists of 16 inches of brick (k=3.6 BTU/hr (Ft^2/In), an air space equivalent to the resistance of 8 inches of brick, 8 more inches of brick, 4 inches of corkboard (k=.3 BTU/hr (FFt^2/In), and 1.5 inches of cement plaster brick (k=2.3 BTU/hr (FFt^2/In).  Calculate the amount of heat energy transmitted per day through the wall by conduction per square foot of wall area.  The surface temperature of the outer wall is 70°F and the inner wall is 30°F.

Q = (T1-T2) / (L brick/K brick) + (L cork / K cork) + (L cem / K cem)

Q = (70(F-30(F) / (32 In/ 3.6 BTU/hr(FFt^2/Hr) + (4 In/ .3 BTU/hr(FFt^2/Hr) + (1.5 In/
2.3 BTU/hr(FFt^2/Hr)

Q = 1.75 BTU / hr Ft^2

Q = (1.75 BTU / hr / Ft^2) * 24 Hr / Day = 42 BTU / Ft^2 / Day

Calculate Heat Leak into a Typical Vacuum Insulated Pipe System

Assume a 200-foot-long system of 1.0 inch pipe using 10 bayonet connections and (2) valves.  Assume the system is in constant operation flowing LN2.  Assume the LN2 latent heat of vaporization is 85.6 BTU / Lb and the density of the LN2 is 1.77 Lbs / Liter.

Using information available from the Technifab website for performance of VJ pipe, the following calculation is performed.

Q for 200 feet of VJP = 200 Ft * .48 BTU / Hr / Ft = 96 BTU / Hr = Q

Heat leak for bayonets = Q = 12 * 1.24 * .48 BTU/ Hr = 7.14 BTU / Hr

Total system Q = 7.24 BTU / Hr + 96 BTU / Hr = 103.14 BTU / Hr

Calculate Heat Leak into a Typical Foam Insulated Pipe System

Assume a 200-foot-long system of 1.0 inch pipe using 10 bayonet connections and 2 valves.  Assume the system is in constant operation flowing LN2.  Assume the LN2 latent heat of vaporization is 85.6 BTU / Lb and the density of the LN2 is 1.77 Lbs / Liter.

Using information available from the Technifab website for performance of 7-inch-thick foam insulated 1.0 inch pipe, the following calculation is performed.

Q for 200 feet of VJP = 200 Ft * 20 BTU / Hr / Ft = 4000 BTU / Hr = Q
(Fittings and valves are ignored in this calculation)

Calculate the Amount of Heat Leaked into a Vacuum Insulated Pipe System that has Foam Insulated Connections

Assume a 200-foot-long system of 1.0 inch pipe using 10 foam insulated connections each one foot in length.  Total length of vacuum insulated piping is therefore 200 - 10 or 190 feet long.

Q for 190 feet of VJP = 190 Ft * .48 BTU / Hr / Ft = 91.2 BTU / Hr = Q
(Fittings and valves are ignored in this calculation)

Q for 10 feet of connections insulated with 7-inch thick foam is calculated as:
Q for 10 feet of VJP = 10 Ft * 20 BTU / Hr / Ft = 200 BTU / Hr = Q

Total heat leak therefore = 91.2 BTU / Hr + 200 BTU / Hr = 291.2 BTU / Hr

Calculate the Amount of LN2 Vaporized Per Day as a Result of the Heat Leak into the VJ Pipe System

103.1 BTU / Hr / 85.6 BTU / Lb = 1.2 Lb / Hr

1.2 Lbs / Hr / 6.74 Lbs / Gallon = .18 Gallon / Hour

.18 Gallon / Hour * 24 Hr / Day = 4.27 Gallon vaporized per day.
103.1 BTU / Hr / 85.6 BTU / Lb = 1.2 Lb / Hr

1.2 Lbs / Hr / 6.74 Lbs / Gallon = .18 Gallon / Hour

.18 Gallon / Hour * 24 Hr / Day = 4.27 Gallon vaporized per day.

Calculate the Amount of LN2 Vaporized Per Day as a Result of the Heat Leak into the Foam Insulated System

4000 BTU / Hr / 85.6 BTU / Lb = 46.7 Lb / Hr

46.7 Lbs / Hr / 6.74 Lbs / Gallon = 6.9 Gallon / Hour

6.9 Gallon / Hour * 24 Hr / Day = 166.4 Gallon vaporized per day.

Calculate the Amount of LN2 Vaporized Per Day as a Result of the Heat Leak into the VJ Insulated System Using Foam Insulated Connections

291.2 BTU / Hr / 85.6 BTU / Lb = 3.4 Lb / Hr

3.4 Lbs / Hr / 6.74 Lbs / Gallon = .5 Gallon / Hour

.5 Gallon / Hour * 24 Hr / Day = 12.1 Gallon vaporized per day.

Calculation to Determine Pressure Rise in a Closed Cryogenic Vessel

If we know the contents and the condition of the contents of a closed vessel containing a cryogenic fluid and know the heat leak rate, then we can calculate the pressure rise for a particular time period as follows:

Given: Vessel contains LN2

Initial saturated pressure = PI = 53 PSIG = 68 PSIA

Initial Enthalpy at 68 PSIA = h1 = -38.1 BTU / LB²

Initial Liquid Density at 68 PSIA = (1 = 45.5 Lb / Ft^3

Initial Ullage space = U = 10% = .10

Vessel heat leak rate = Q = 10 BTU / Hr

Total tank volume = Vt = 8.92 Ft^3 = 66.7 Gallon

Initial gas volume = Uvt = .892 Ft^3 = 6.7 Gallon

Polytropic Process Constant = n = 2.51

Period of time = 72 Hours

Total heat leaked = Ht = 72 Hours * 10 BTU / Hr = 720 BTU

Final liquid enthalpy = hf = h1 + [ (h1 * Period) / [Vt * (1(1-U)]

Hf = -38.1 BTU / Lb + [ (10 BTU / Hr * 72 Hr) / (8.92 Ft^3 * 45.5 Lbs / Ft^3 *(1-.10))] = -36.12 BTU / Lb

Final liquid density = r2 = 44.8 Lbs / Ft³

Equation:

P2 = P1 [Uvt / [ Uvt - (Vt - (Uvt * (r1 /r2 - 1))]] ^n

P2 = 53 PSIG [.892 Ft^3 / [.892 Ft^3 - (8.92 Ft^3 * .02) ]]^2.5

P2 = 92.58 PSIG

Calculation to Determine Heat Leak into a Closed Cryogenic Vessel

It is also possible to calculate the heat leak of the vessel if we know the initial pressure, final pressure, the starting temperature of the contents, the initial vapor and liquid densities, and the initial volume of the liquid and vapor phases.  The following equation models the process:

V = Total vessel volume

Pf = Final pressure = 250 PSIA = 230 PSIG

Tf = Venting contents temperature =
M liquid = V total * (1-% ullage) *r liquid

M vapor = V total * % ullage * vapor

X = M vapor / M liquid

r average = [((1-X)/ r liquid) + (X/r vapor) ] ^-1

D U = (M liquid + M vapor ) ( Uf - U1)

time = D U / Q

Q = D U / time

The cryogenic fluid stored in a vessel is LNG.

Given that P1 = 35 PSIG (20 PSIA)

Given that P2 = 250 PSIG ( 235 PSIA)

V vapor = 1.4 Ft^3; (10.5 Gallon)

V liquid = 12.4 ft^3 (92.8 Gallon)

The saturation temperature of the LNG = -238°F

The density of the liquid phase is 25.3 Lbs / Ft^3

The density of the gas phase is .25 Lbs / Ft^3

Initial Internal Energy of the LNG = 197.9 BTU / Lb

Final Internal energy of the LNG = 212.9 BTU / Lb

NG mass = .35 pounds

LNG mass = 31 pounds

X = M vapor / M liquid

X = .3 5/313 = .001

r average = [((1-X)/ r liquid) + (X/r vapor) ] ^-1

r average = 22.8 Lbs / Ft^3

D U = (M liquid + M vapor ) ( Uf - U1)

D U =(25.3 Lb / Ft^3 + .25 Lbs / Ft^3 (212.9 BTU / Lb - 197.9 BTU / Lb)

D U = 383. 25 BTU

Q = D U / time

Q = 383.35 BTU / 72 hours = 5.32 BTU / Hour

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