Question

The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet temperatures are Th,o = 120°C and Tc,o = 125°C. Determine a.) whether the heat exchanger operating in a parallel flow or counter flow configuration, b.) the heat exchanger effectiveness, and c.) the NTU.

Answer 1

Answer:Counter,

0.799,

1.921

Step-by-step explanation:

Given data

`T_(h_i)=200^(\circ)C`

`T_(h_o)=120^(\circ)C`

`T_(c_i)=100^(\circ)C`

`T_(c_o)=125^(\circ)C`

Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger

Equating Heat exchange

`m_hc_(ph)\left [ T_(h_i)-T_(h_o)\right ]=m_cc_(pc)\left [ T_(c_o)-T_(c_i)\right ]`

`(m_hc_(ph))/(m_cc_(pc))`=
`(125-100)/(200-120)=(25)/(80)=C\left ( capacity rate ratio\right )`

we can see that heat capacity of hot fluid is minimum

Also from energy balance

`Q=UA\Delta T_m=\left ( mc_p\right )_(h)\left ( T_(h_i)-T_(h_o)\right )`

`NTU=(UA)/(\left ( mc_p\right )_(h))`=
`(\left ( T_(h_i)-T_(h_o)\right ))/(T_m)`

`T_m=(\left ( 200-125\right )-\left ( 120-100\right ))/(\ln (75)/(20))`

`T_m=41.63^(\circ)C`

NTU=1.921

`And\ effectiveness \epsilon =(1-exp\left ( -NTU\left ( 1-c\right )\right ))/(1-c\left ( -NTU\left ( 1-c\right )\right ))`

`\epsilon =(1-exp\left ( -1.921\left ( 1-0.3125\right )\right ))/(1-0.3125exp\left ( -1.921\left ( 1-0.3125\right )\right ))`

`\epsilon =(1-exp\left ( -1.32068\right ))/(1-0.3125exp\left ( -1.32068\right ))`

`\epsilon =(1-0.2669)/(1-0.0834)`

`\epsilon =0.799`