In a heat exchanger- the hot liquid enters with a temperature of 200 - C and leaves at 180 - C- The cooling fluid enters at 50 - C and leaves at 130 - C- The capacity ratio of the heat exchanger is

C_h(T_h_1 T_h_c)=C_e(T_c_e T_c_1) ⇒ C_b(20)=C_c(80) C_minC_max = C_cC_h=2080=0.25=Capacity ratio ...

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Design of Heat Exchanger - LMTD Method

In a heat exc...

Question

In a heat exchanger, the hot liquid enters with a temperature of $200^{\circ} C$ and leaves at $180^{\circ} C$ . The cooling fluid enters at $50^{\circ} C$ and leaves at $130^{\circ} C$ . The capacity ratio of the heat exchanger is

0.35

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0.45

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0.55

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0.25

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Solution

The correct option is D 0.25
$C_{h} (T_{h_{1}} - T_{h_{c}}) = C_{e} (T_{c_{e}} - T_{c_{1}})$

$\Rightarrow C_{b} (20) = C_{c} (80)$

$\frac{C_{min}}{C_{max}} = \frac{C_{c}}{C_{h}} = \frac{20}{80} = 0.25 = Capacity ratio$

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In a heat exchanger, the hot liquid enters with a temperature of 200∘C and leaves at 180 ∘C. The cooling fluid enters at 50∘C and leaves at 130 ∘C. The capacity ratio of the heat exchanger is

The correct option is D 0.25Ch(Th1−Thc)=Ce(Tce−Tc1) ⇒Cb(20)=Cc(80) CminCmax=CcCh=2080=0.25=Capacity ratio

In a heat exchanger, the hot liquid enters with a temperature of $200^{\circ} C$ and leaves at $180^{\circ} C$ . The cooling fluid enters at $50^{\circ} C$ and leaves at $130^{\circ} C$ . The capacity ratio of the heat exchanger is

The correct option is D 0.25
$C_{h} (T_{h_{1}} - T_{h_{c}}) = C_{e} (T_{c_{e}} - T_{c_{1}})$

$\Rightarrow C_{b} (20) = C_{c} (80)$

$\frac{C_{min}}{C_{max}} = \frac{C_{c}}{C_{h}} = \frac{20}{80} = 0.25 = Capacity ratio$