Inline Heater

A1) Rating = (60-x) C_p ρ 12/60000 where x is the starting temperature, C_p the thermal capacity of the oil and ρ its density.

A2) 12 l/m = 2e-4m³/sec. 10mm pipe will be adequate.

A3) Discuss with heat exchanger manufacturers directly.

Dear sir thank you for providing valuable information but sir i wnat to know the KW rating of heater. Please reply.

In order to calculate it, one needs the information expressed algebraically in the formula in response #1. As it hasn't been provided yet, the answer cannot be given yet, no matter how many times the question is asked.

<...sir...> How abstruse.

Dear sir, there is the requirement of heater which is to be selected for following condition.

inlet condition

fluid : FO(furnace oil)

pressure: Atm

Tem: Ambient (35 Deg C)

Flow: 12 lpm

outlet Conditionfluid :

FO(furnace oil)

pressure: Atm

Tem: 60 deg C

Flow: 12 lpm

Can you please share knowledge for following

Kw Rating of heater

size of heater

type of heater (required type is coil type)

please can you provide any drawing.

thanks in advance for your valuable suggestion

OK, so now the forum has the starting temperature of the oil. What about the thermal capacity and the density <sigh>?

Blood and stones come to mind. You've tried your best PWS.

Dear Sir,

Please use the formule below and solve according...it is done for old project....hope use full for u...any doubts kindly reply

A.0 DESIGN DATA :

A.1 Type of Heater : Immersion Type Floor Coil

A.2 HFO Oil Data :

A.2.1 D₁ = Density : 900 kg/m³

A.2.2 C_p1 = Specific heat : 0.48 kcal/kg ⁰C

A.2.3 Heating Range :

Initial Oil Temperature, T₁ (Average Annual Mean Temp.) : 6.6 ⁰C

Final Oil Temperature, T₂ (Oil Temp. to be maintained) : 60 ⁰C

A.3 Aux. Steam Data :

A.3.1 P₁ = Inlet Pressure (Upstream of PRS) : 15 kg/cm²(g)

A.3.2 T₃ = Inlet Temperature (Upstream of PRS) : 220.4 ⁰C

A.3.3 P₂ = Operating Pressure (Downstream of PRS) : 3 kg/cm²(g)

A.3.4 T₄ = Operating Temperature at 3.0 kg/cm²(g) : 193.8 ⁰C

A.3.5 h_fg = Specific Enthalpy of Evaporation at 3.0 kg/cm²(g) : 679 kcal/kg

A.4 Insulation Material

A.4.1 Material : LRB, Mineral Wool

A.4.2 D₂ = Density : 150 kg/m³

A.4.3 Cladding Material : Aluminum, 22G sheet

B.0 PROCESS REQUIREMENT DATA :

B.1 V = Volume of Liquid to be heated : 1100 m³

B.2 H_t = Heating Time : 72 hr

B.3 M₁ = Mass Flow Rate (V x D₁/H_t) : 13750 kg/hr

B.4 D_t = Difference in Temperature (T₂-T₁) : 53.4 ⁰C

B.5 Heat Load to heat oil of Tank , Q₁ : 352440 kcal/hr

(i.e. M₁ C_p1 D_t)

B.6 Heat loss from shell, bottom and roof of tank, Q₂ : 29312.7 kcal/hr

(i.e. q1+q2+q3)

B.6.1 Heat loss through shell, q₁

q₁ = (q_n-q_m) / [ (d/2k x ln d₁/d ) + (d / fd₁) , where

q_n = Final Oil Temperature = T₂ : 60 ⁰C

q_m=Initial Oil Temperature = T₁ : 6.6 ⁰C

d = Outer dia. of Tank shell without insulation : 12.016 m

d₁ = Outer dia. of Tank shell with insulation : 12.096 m

(considering shell insulation thick as 40mm)

h = Height of Tank shell without insulation : 9.5 m

k = Thermal conductivity of Insulation : 0.042 kcal/hr m ⁰C

f : 32.26 kcal/hr m^{2 0}C

a₁ = Surface area of shell (Π x d x h) : 358.14 m²

Therefore,

Heat loss through shell, q₁ : 54.54 kcal/hr m²

: 19535 kcal/hr

B.6.2 Heat loss through roof of tank, q₂

q₂ = (q_n-q_m) / [ l₁/k + 1/f ] , where

l₁ = Roof insulation thickness as considered : 40 mm

a₃ = Surface area of roof (Π x r x s) : 113.097 m²

Therefore,

Heat loss through roof, q₂ : 52.382 kcal/hr m²

: 6142 kcal/hr

B.6.3 Heat loss through bottom of tank, q₃

q₃ = (a₃ x k₂ x D_t) / l₂ , where

l₂ = Thickness of Sand Bed assumed : 500 mm

a₃ = Surface area of bottom : 113.097 m²

k₂ = Thermal Conductivity of Sand (considered) : 0.350 w /m ⁰C

: 0.301 kcal /hr m ⁰C

D_t = Difference in Temperature : 53.4 ⁰C

Therefore,

Heat loss through bottom, q3 : 3635.7 kcal/hr

B.7 Heat required per hour to heat tank material upto T₂ ⁰C, Q₃

Q₃ = (M₂ C_p2 D_t) / H_t , where

M₂ = Mass of Tank material (approx.) :40000 kg

C_p2 = Specific Heat of Tank Material (i.e. M.S.) : 500 J/kg ⁰K

: 0.120 kcal/kg ⁰C

D_t = Difference in Temperature : 53.4 ⁰C

H = Heating Time : 72 hr

Therefore,

Heat required to heat tank material, Q₃ : 3560 kcal/hr

B.8 Total Heat Load per hour, Q (I.e. Q1+Q2+Q3) : 385312.7 kcal/hr

B.9 Heat transfer area of floor coil heater =

Total Heat Load / (Heat Transfer Coefficient x Average Mean Temperature difference)

B.9.1 Average Mean Temperature difference, AMTD : 160.5 ⁰C

AMTD = T_s - (T₂ + T₁)/2 , where

T₂ = Final Oil Temperature : 60 ⁰C

T₁ = Initial Oil Temperature : 6.6 ⁰C

T_s = Super heated steam Temperature at 4.0 bar : 193.8 ⁰C

B.9.2 Heat Transfer Coefficient (Steam to Oil), k : 75 kcal/m² hr ⁰C

Therefore,

Heat transfer area required for floor coil heater , a_h : 32 m²

B.9.3 Actual Length of Floor Coil Pipe required = Heat transfer area / Area of pipe per

meter length

Considering Flooer Coil Pipe as 50NB Seamless Carbon Steel to ASTM A-106, Gr.B,

Sch-40 ;

Surface area of 50NB Pipe per meter, a₃ : 0.189 m²/m

Therefore,

Length of Pipe required (a_h / a₃) :169.36. m

Actual Length of Floor Coil Pipe Provided, L_p : 212 m

(with 25% margin)

B.10 Steam Load, H_s = Heating load per hour / Latent heat of Steam (Q / h_f)

Therefore,

H_s : 567.47 kg/hr

Actual Steam Requirement (with 10% margin) : 625 kg/hr

I suspect that you are asking for the size and duty of an Immersion Heater. Immersion heaters are an efficient way of heating fluids in pipe work.

Do you know the diameter of the pipe where your furnace oil is flowing through?

How does the oil reach the furnace @ atmospheric pressure? Is it gravity fed?

Neil