Question

Water is to be heated from 10°C to 80°C as it flows through a 2-cm-internal-diameter, 13-m-long tube. The tube is equipped with an electric resistance heater, which provides uniform heating throughout the surface of the tube. The outer surface of the heater is well insulated, so that in steady operation all the heat generated in the heater is transferred to the water in the tube. If the system is to provide hot water at a rate of 5 L/min, determine the power rating of the resistance heater. Also, estimate the inner surface temperature of the pipe at the exit.

Answer 1

Answer:

- the power rating of the resistance heater is 24139.5 W

- the inner surface temperature of the pipe at the exit is 96.34°C

Explanation:

Given the data in the question;

Flow rate of water in the tube V" = 5L/min = 8.333 × 10⁻⁵ m³/s        

The water is to be heated from 10°C to 80°C;

so Average or mean temperature $T_{avg$ will be;

$T_{avg$ = (T₁ + T₂) / 2 = (10 + 80) / 2 = 90/2 = 45°C

Now, from the Table " Properties of Water " at average temperature;

at $T_{avg$ = 45°C

density p = 990.1 kg/m³

specific heat $C_p$ = 4180 J/kg-k

thermal conductivity k = 0.637 W/m-°C

Now, we determine the mass flow;

m" = pV"

we substitute

m" = 990.1 × 8.333 × 10⁻⁵

m" = 0.08250 kg/s

we know that the power rating of the resistance heater is equal to the heat transfer rate to the water;

Q' = m"$C_p$( T₂ - T₁ )

we substitute

Q' = (0.08250 × 4180 ) ( 80 - 10 )

Q' =  344.85 × 70

Q' = 24139.5 W

Hence, the power rating of the resistance heater is 24139.5 W

Next, we determine the average velocity of water in the tube;

$V_{avg$ = V" / $A_c$

$V_{avg$ = V" / ( $\frac{1}{4}$πD² )

given that;  flows through a 2-cm-internal-diameter; D = 0.02 m

we substitute

$V_{avg$ = (8.333 × 10⁻⁵) / ( $\frac{1}{4}$π × (0.02)² )

$V_{avg$ = (8.333 × 10⁻⁵) / ( 3.14159 × 10⁻⁴ )

$V_{avg$ =  0.265 m/s

Also, from table " saturated water property table "

At 45°C

viscosity μ = 0.596 × 10⁻³ kg/m-s

Prandtl number Pr = 3.91

Now, we determine the kinematic viscosity

v = μ / p

we substitute

v = ( 0.596 × 10⁻³ ) / 990.1

v = 6.01959 × 10⁻⁷ m²/s

so, Reynolds number in the flow region will be;

Re = ($V_{avg$ × D) / v

we substitute

Re = ( 0.265 × 0.02) / (6.01959 × 10⁻⁷)

Re = 8804.586

we can see that our Reynolds number (  8804.586 ) more than 2300 and less than 10,000.

Hydraulic and thermal entry length are equal in this flow region,

such that;

$L_h$ = $L_t$

⇒ 10 × D = 10 × 0.02 = 0.2 m

we can see that the entry length ( 0.2 m ) is smaller than the given length ( 13 m ) in the question; the flow is a turbulent flow.

So we the Nuddelt number

Nu = $0.023Re^{0.8} Pr^{0.4$

Nu = 0.023 × $8804.586^{0.8$ × $3.91^{0.4$

Nu = 56.8

Hence, the heat transfer coefficient h will be;

h = $\frac{k}{D}$ × Nu

we substitute

h = $\frac{0.637}{0.02}$ × 56.8

h = 31.85 × 56.8

h = 1809.1 W/m²-°C

Now, area of the heat transfer will be

A$_s$ = πDL

we substitute

A$_s$ = π × 0.02 × 13

A$_s$ = 0.8168 m²

Finally we determine the inner temperature of the pipe at exit. using the relation;

Q' = hA$_s$( T₃ - T₂ )

we substitute

24139.5 = 1809.1 × 0.8168( T₃ - 10 )

24139.5 = 1477.67288( T₃ - 80 )

24139.5 = 1477.67288T₃ - 118213.8304

24139.5 + 118213.8304 = 1477.67288T₃

1477.67288T₃ = 142353.3304

T₃ = 142353.3304 / 1477.67288T

T₃ = 96.34°C

Therefore, the inner surface temperature of the pipe at the exit is 96.34°C