Question

A shell-and-tube heat exchanger with two shell passes and eight tube passes exchanges heat between water and hot air. The air flows in the shell, entering with a temperature of 350°C and leaves the heat exchanger at 175°C. Water flows in the tubes at the rate of 45,500 kg/hr., entering with a temperature of 80°C and leaving at 150°C. The total surface area for the heat exchange is 925 m2 . If the specific heat of water is 4,236 J/kg °C, calculate the overall heat transfer coefficient.

Answer 1

overall heat transfer coefficient is U = 29.614 W/m².k

Step-by-step explanation:

We are given;

Mass Flow rate of water;m' = 45,500 kg/h = 45500/3600 kg/s = 12.6389 kg/s

total surface area for the heat exchange; A = 925 m²

Water heated from 80°C to 150°C

Exhaust gases cool from 350°C to 175°C

specific heat of water;c = 4,236 J/kg °C

Let's first calculate net rate of heat transfer;

The formula for net rate of heat transfer is given as;

q = m'•c•Δt

Where;

m' is mass flow rate

c is specific heat

Δt is change in the temperature of water.

Thus;

q = 12.6389•4,236•(150 - 80)

q = 3747683.33 W

Now, the overall heat transfer coefficient is gotten from the formula;

q = U•A•ΔT_lm

Where;

U is the overall heat transfer coefficient

q is the net rate of heat transfer

ΔT_lm is logarithmic mean temperature difference

A is the total surface area for the heat exchange

f is the correction factor

ΔT_lm is calculated as;

ΔT_lm = [(350 - 150) - (175 - 80)]/[In((350 - 150)/(175 - 80))]

ΔT_lm = [(200 - 95)]/[In(200/95)]

ΔT_lm = 141.04°C

Since:q = U•A•f•ΔT_lm

Let's make U the subject;

U = q/(A•f•ΔT_lm)

From the chart i attached, f is estimated to be 0.97

Plugging in the relevant values to obtain;

U = 3747683.33/(925 x 0.97 x 141.04)

U = 29.614 W/m².k