Question

[10 CO3 6. Obtained an expression for the view factor fi Hot oil is cooled by exchanging heat with water in a heat exchanger. The inlet and outlet temperature of oil are 400 K and 350 K, respectively, and oil flows at the rate of 0.5 kg's Water is used at the rate of 0.4 kg/s to cool the hot oil. The inlet temperature of water is 300 K. Calculate the required heat transfer area for the following cases. Assume overall heat transfer coefficient is 200 W/m K. (a) counter flow (b) parallel flow, specific heat of oil , 2 kJ/kg-K.

please help me ASAP

Answer 1

Given data:

The inlet temperature of the oil, T₁ = 400 K

The outlet temperature of the oil, T₂ = 350 K

Oil flow rate, m₁ = 0.5 kg/s

Specific heat of oil, C_po = 2 kJ/kg-K

The inlet temperature of the water, t₁ = 300 K

The outlet temperature of the water, t₂ = ?

Water flow rate, m₂ = 0.4 kg/s

Overall heat transfer coefficient, U = 200 W/m²K

Solution:

Heat transfer rate of oil is

$\\Q = m_{1}C_{po}(T_{2}-T_{1})\\ Q = 0.5\times2\times (400-350)\\ Q = 50 \text{ kJ/s}$

The specific heat of water is $C_{pw}$ = 4.18 kJ/kg-K

Heat transfer rate of water is

$\\Q = m_{2}\times C_{pw}(t_{2}-t_{1})\\$

substitute Q = 50 kJ/s, m₂ = 0.4 kg/s, t₁ = 300 K in the above equation.

$\\50 = 0.4\times4.18\times (t_{2}-300)\\ t_{2} = 330\text{ K}\\$

a) Counter flow

Calculate the Log mean heat temperature for counterflow.

(T1 - t2) - (T2-ti) LMTD 11-t2 T2-ti (400 – 330) - (350 – 300) LMTD In 400-330 350-300 LMTD= 59.44 K

The heat transfer area formula is,

$\\A =\frac{Q}{U\times (LMDT)}\\ \\$

Substitute the values of Q, U and LMTD in the above equation.

$\\A =\frac{50\times 1000}{200\times (59.44)}\\ \\ A = 4.206 \text{ m}^{2}$

b) Parallel flow

Calculate the Log mean heat temperature for parallelflow.

$\\LMTD= \frac{(T_{1}-t_{1})-(T_{2}-t_{2})}{In\frac{T_{1}-t_{1}}{T_{2}-t_{2}}}\\ \\ \\ LMTD= \frac{(400-300)-(350-330)}{In\frac{400-300}{350-330}}\\ \\ \\ LMTD=49.7 \text{ K}$

The heat transfer area formula is,

$\\A =\frac{Q}{U\times (LMDT)}\\ \\$

Substitute the values of Q, U and LMTD in the above equation.

$\\A =\frac{50\times 1000}{200\times (49.7)}\\ \\ A = 5.03 \text{ m}^{2}$