Question

Saturated water vapor leaves a steam turbine at a flow rate of 1.5 kg/s and a pressure of 0.51 bar. The vapor is to be completely condensed to a saturated liquid in a shell-and-tube heat exchanger that uses city water as the cold fluid. The water enters the thin-walled tubes at 17°C and is to leave at 57°C.
1. Assuming an overall heat transfer coefficient of 2000W/m^2*K, determine the required heat exchanger surface area and the water flow rate.
After extended operation, fouling causes the overall heat transfer coefficient to decrease to 1000W/m^2*K, and to completely condense the vapor, there must be an attendant reduction in the vapor flow rate.
2. For the same water inlet temperature and flow rate, what is the new vapor flow rate required for complete condensation?

Answer 1

The new vapor flow rate required for complete condensation is 0.94 Kg/s.

Explanation:

The conditions :

• The cold water is cold fluid and hot water is hot fluid.
• Rate of flow $m_{h}$ = 1.5 Kg/s
• Hot fluid pressure P = 0.51 bar
• The inlet temperature and outlet temperature of hot fluid
• $T_{c,i}$ = 17 + 273 = 290 K
• $T_{c,o}$ = 57 + 273 = 330 K
• U = 2000 W/$m^{2}$ K
• $U_{f}$ = 1000 W/$m^{2}$ K

a) The conditions diagram is attached below.

The loss of heat by cold fluid is similar to gain of heat by hot fluid in energy balance performance.

we have

$q=\dot{m}_{h} h_{f g}=C_{c}\left(T_{c, o}-T_{c, i}\right)$

For hot fluid, $T_{Sat}$ = 355 K at 0.51 bar.

$h_{fg}$ = 2304 KJ/Kg

q = 1.5 × 2304

= 3456 KW

Cold fluids heat capacity is

$C_{c}=\dot{m}_{c} c_{p, c}$

For cold fluid, T = $\frac{330+290}{2}$ = 310 K

Specific cold fluid heat is $C_{p,c}$ = 4.178 KJ/Kg . K

$q=\dot{m}_{c} c_{p, c}\left(T_{c, o}-T_{c, i}\right)$

3456 = $m_{c}$ × 4.178 (330-290)

$m_{c}$ = 20.679 Kg/s

$C_{c}$ = 20.679 × 4178

$C_{c}$ = 86396.862 W/K

$C_{min}$ = $C_{c}$

$C_{max}$ = $C_{h}$ = ∞

$C_{r}$ = $\frac{C_{min} }{C_{max} }$

$C_{r}$ = 86396.862 W/K/∞

Effectiveness of heat exchanger is

ε = $\frac{q}{q_{max} }$

$\varepsilon=\frac{q}{C_{\min }\left(T_{h, i}-T_{c, i}\right)}$

$\varepsilon=\frac{q}{C_{c}\left(T_{h, i}-T_{c, i}\right)}$

Therefore, $C_{min}$ = $C_{c}$

$\varepsilon=\frac{3456000 \mathrm{W}}{86396.862 \mathrm{W} / \mathrm{K} \times(355-290) \mathrm{K}}$

ε = 0.6154

For the heat exchanger, $C_{r}$ = 0

NTU relation is expressed,

NTU = - ㏑ ( 1 - ε )

NTU = - ㏑ ( 1 - 0.6154 )

NTU = 0.96

$\mathrm{NTU}=\frac{U A}{C_{\min }}$

0.96 = $\frac{(2000)A}{86396.862}$

A = 41.5 $m^{2}$

b)  The new vapor flow rate required for complete condensation is estimated.

$\mathrm{NTU}=\frac{1000 \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K} \times 41.5 \mathrm{m}^{2}}{86396.862 \mathrm{W} / \mathrm{K}}$

NTU = 0.49

For the heat exchanger, $C_{r}$ = 0

ε = 1 - exp (-NTU)

ε = 1 - exp (-0.49)

ε = 0.387

Rate of heat exchanger is expressed as

q = ε $q_{max}$

$q=\varepsilon C_{c}\left(T_{h, i}-T_{c, i}\right)$ Therefore, $C_{min}$ = $C_{c}$,

q = 0.387 × 86396.862×(355-290)

q = 2173.313 kW

From the equation, $q=\dot{m}_{h, n} h_{f g}$

2173.313 kW = $m_{h,n}$ × 2304

$m_{h,n}$ = 0.94 Kg/s