Area "A"

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SMath Current saved version of program 3 appVersion 0.99.6654

SMath Last Saved version of program 3 - appVersion 0.99.6654

Fluid Temperature
Ref 2 T.sA 140 °F * :

Min allowable winter concrete wall temperature
Ref 2 T.1 100 °F * :

Surface  Temperature
Ref 2 T.sC 61 °F * :

Spray foam "R"/inch
Ref. 4 R.ins 6.2 hr ft 2^* Δ°F * BTU in * / * :

k.ins 1 R.ins / : BTU in * hr ft 2^* Δ°F * / 0.1613

Winter design temperature
Project specified T.ambw 32 °F * :

Assumptions

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Concrete and shotconcrete (assumed homogeneous)
Thermal Conductivity
Ref 1, 26.11 Table 1
"cement/lime, mortar and stucco 120 lb/ft" k.C 9.7 BTU in * hr ft 2^* Δ°F * / * : BTU hr ft * Δ°F * / 0.8083

~10 in concrete.  Considered homogeneous with shotcrete
Ref 2 X.A 10 in * :

~2 in shotcrete. Considered homogeneous with the concrete
Ref 2 X.B 2 in * :

Estimated surface coefficient for concrete and shotconcrete (assumed homogeneous)
per example
Ref 1; 23.19 h.scC 1.0 BTU hr ft 2^* Δ°F * / * :

Surface Film Resistance winter 15 mph wind
Ref 1; 26.2 Table 10 R.airw .17 ft 2^hr * Δ°F * BTU / * :

Surface Film Resistance summer 7.5 mph wind
Ref 1; 26.2 Table 10 R.airs .25 ft 2^hr * Δ°F * BTU / * :

Margin margin 1.3 :

The following assumptions apply

Calculation is for steady state

There is no heat generated within the volume

Heat flow through a composite layer is the same. When there is more than one layer {in the composite}, the total resistance of the circuit must be calculated as one homogeneous piece.

Areas as the same

Negligible contact resistance at the interfaces

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Solve

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Sign convention: right direction is + -->

First, we need to find the maximum transfer rate or heat flux (air side)

Maximum limit of heat transfer rate
allowed by the air R.airw T.sC T.ambw Q.airmax 1 R.airw / T.sC T.ambw - (* :

Max allowed flux as a function of the variables R.airw T.sC T.ambw Q.airmax BTU hr ft 2^* / 170.5882

Area "A"

Heat balance for the concrete at the "A" interface gives the heat flux on the "A" side

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Heat transfer rate through the wall and insulation
Ref 3 Q.rateA k.C X.A / T.sA T.1 - (* : BTU hr ft 2^* / 38.8

Q.rateA is less than Flux .max; therefore, use Q.rateA

Area "B"

Heat balance for the shotcrete at the "B" interface gives the heat flux on the "B" side

Reminder that concrete and shotecrete is homogeneous, so we are solving for T₂

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Q.rateB Q.rateA :

Q.rateB k.C X.B / T.1 T.2 - (* ≡ T.2 80 °F * ≡ FindRoot °F 92

Temperature of outside of shotcrete towards ambient T.2 °F 92

Minimum insulation thickness
Ref 1; 23.19 Eq 14 X.C k.ins h.scC / T.sA T.2 - T.2 T.sC - / (* margin * : in 0.3247

Sanity check

R.t X.A k.C / X.B k.C / + X.C k.ins / + : hr Δ°F * ft 2^* BTU / 3.25

Homogeneous steady state heat transfer for the assembly per sq ft Q.rate T.sA T.ambw - (R.t / : BTU hr ft 2^* / 33.2306

T.1 T.sA Q.rate X.A k.C / * (- : °F 105.7417

T.2 which you are solving for, already has a value.
 Thus, we need to clear the value to solve for the variable T.2 Clear 1

Need to work out the units.
This should work Q.rateB k.C X.B / T.1 T.2 - (* ≡ T.2 80 °F * ≡ FindRoot °F 97.7417

Summer conditions

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Surface  Temperature
Ref 2 T.sCs 90 °F * :

R.t X.A k.C / X.B k.C / + X.C k.ins / + R.airs + : hr Δ°F * ft 2^* BTU / 3.5

R.t hr Δ°F * ft 2^* BTU / 3.5

Homogeneous steady state heat transfer per sq ft Q.rate T.sA T.sCs - (R.t / : BTU hr ft 2^* / 14.2856

T.1s T.sA Q.rate X.B k.C / * - : °F 137.0545

Conclusion

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Tank Fluid
Surface Temp T.sA °F 140

Temp interface
between A and B
WINTER T.1 °F 105.7417

Temp interface
between B and C T.2 °F 97.7417

Insulation
Surface Temp T.sC °F 61

X.A in 10

X.B in 2

Minimum insulation thickness X.C in 0.3247

Temp interface
between A and B
SUMMER T.1s °F 137.0545

Insulation
Surface Temp
SUMMER T.sCs °F 90

X.C k.ins h.scC / T.sA T.2 - T.2 T.sC - / (* (1.5 in * (k.ins h.scC / T.sA T.2 - T.2 T.sC - / (* (.1 in * + (range :

MKRASKA Temperature gauges/units are special in that they may differ by a factor and an offset from a reference gauge. When using gauges without offset, just like K or °Rankine you can obtain the unitless numeric value by dividing the quantity by the unit. When you have an offset, you have to remove this first, hence the -0°F term. Then you divide by the relative unit matching your gauge.

X.C T.1 T.sA T.sA T.ambw - (X.A k.C / X.B k.C / + X.C k.ins / + / X.A k.C / * (- vectorize :

X.C T.2 X.C T.1 T.sA T.ambw - (X.A k.C / X.B k.C / + X.C k.ins / + / X.B k.C / * (- vectorize :

X.C in / X.C T.1 0 °F * - Δ°F / augment X.C in / X.C T.2 0 °F * - Δ°F / augment 2 1 sys