Question

A shell-and tube heat exchanger (two shells, four tube passes) is used to heat 10,000 kg/h of pressurized water from 35 to 120 oC with 5000 kg/h pressurized water entering the exchanger at 300 oC. If the overall heat transfer coefficient is 1500 W/m^2-K, determine the required heat exchanger area.

Answer 1

4.75m^2

Step-by-step explanation:

Given:-

- Temperature of hot fluid at inlet:
`T_h_i = 300` °C

- Temperature of cold fluid at outlet:
`T_c_o = 120` °C

- Temperature of cold fluid at inlet:
`T_c_i = 35` °C

- The overall heat transfer coefficient: U = 1500 W / m^2 K

- The flow rate of cold fluid: m_c = 10,00 kg/ h

- The flow rate of hot fluid: m_h = 5,000 kg/h

Solution:-

- We will evaluate water properties at median temperatures of each fluid using table A-4.

Cold fluid: Tci = 35°C , Tco = 35°C

Tcm = 77.5 °C ≈ 350 K --- >
`C_p_c = 4195 (J)/(kg.K)`

Hot fluid: Thi = 300°C , Tho = 150°C ( assumed )

Thm = 225 °C ≈ 500 K --- >
`C_p_h = 4660 (J)/(kg.K)`

- We will use logarithmic - mean temperature rate equation as follows:

`A_s = (q)/(U*dT_l_m)`

Where,

A_s : The surface area of heat exchange

ΔT_lm: the logarithmic differential mean temperature

q: The rate of heat transfer

- Apply the energy balance on cold fluid as follows:

`q = m_c * C_p_c * ( T_c_o - T_c_i )\\\\q = (10,000)/(3600) * 4195 * ( 120 - 35 )\\\\q = 9.905*10^5 W`

- Similarly, apply the heat balance on hot fluid and evaluate the outlet temperature ( Tho ) :

`T_h_o = T_h_i - (q)/(m_h * C_p_h) \\\\T_h_o = 300 - (9.905*10^5)/((5000)/(3600) * 4660) \\\\T_h_o = 147 C`

- We will use the experimental results of counter flow ( unmixed - unmixed ) plotted as figure ( Fig . 11.11 ) of the " The fundamentals to heat transfer" and determine the value of ( P , R , F ).

- So the relations from the figure 11.11 are:

`P = (T_c_o - T_c_i)/(T_h_i - T_c_i) \\\\P = (120 - 35)/(300 - 35) \\\\P = 0.32`

`R = (T_h_i - T_h_o)/(T_c_o - T_c_i) \\\\R = (300 - 147)/(120 - 35) \\\\R = 1.8`

Therefore, P = 0.32 , R = 1.8 ---- > F ≈ 0.97

- The log-mean temperature ( ΔT_lm - cf ) for counter-flow heat exchange can be determined from the relation:

`dT_l_m = (( T_h_i - T_c_o ) - ( T_h_o - T_c_i ) )/(Ln ( (( T_h_i - T_c_o ))/(( T_h_o - T_c_i )) ) ) \\\\dT_l_m = (( 300 - 120 ) - ( 147 - 35 ) )/(Ln ( (( 300-120 ))/(( 147-35)) ) ) \\\\dT_l_m = 143.3 K`

- The log - mean differential temperature for counter flow is multiplied by the factor of ( F ) to get the standardized value of log - mean differential temperature:

`dT_l = F*dT_l_m = 0.97*143.3 = 139 K`

- The required heat exchange area ( A_s ) can now be calculated:

`A_s = (9.905*10^5 )/(1500*139) \\\\A_s = 4.75 m^2`