Question

Cold water (cp = 4180 J/kg·K) leading to a shower enters a thin-walled double-pipe counterflow heat exchanger at 15°C at a rate of 0.25 kg/s and is heated to 45°C by hot water (cp = 4190 J/kg·K) that enters at 100°C at a rate of 3 kg/s. If the overall heat transfer coefficient is 950 W/m2·K, determine the rate of heat transfer and the heat transfer surface area of the heat exchanger using the ε–NTU method. Answers: 31.35 kW, 0.482 m2

Answer 1

Answer:

The rate of heat transfer is  $H = 31.35\ kW$

The heat transfer surface area is  $A_s = 0.4818 m^2$

Explanation:

From the question we are told that  

          The  specific heat of water is $cp = 4180 \ J/kg \cdot K$

           The temperature of cold water is $T_c = 15^o C$

            The rate of cold the flow is $\r m = 0.25 kg/s$

           The temperature of the heated water $T_h = 45 ^oC$

            The specific heat of hot water is  $c_p__{H}} = 4190 J/kg \cdot K$

              The temperature of the hot water is $T_H = 100^oC$

               The rate of hot the flow is $\r m_H = 3 kg/s$

               The heat transfer coefficient is $U = 950 W/m^2 \cdot K,$

From the $\epsilon -NTU$ method we have that

       $C_h = \r m_H c_p__{H}}$

Where $C_h$ is the heat capacity rate of hot  water

      Substituting the value

            $C_h = 3 * (4190)$

                 $= 12,5700\ W/^oC$

Also

      $C_c = \r m c_p$

Where $C_c$ is the heat capacity rate of cold  water

        $C_c = 0.25 * 4180$

              $= 1045 \ W / ^oC$

The maximum heat capacity $C_h$ and the minimum  heat capacity is $C_c$

       The maximum heat transfer is

                $H_{max} = C_c (T_H - T_c)$

Substituting values  

               $H_{max} = (1045)(100- 15)$

                          $= 88,825\ W$

The actual heat transfer is mathematically evaluated as

               $H = C_c (T_h - T_c)$

Substituting values

               $H = 1045 (45 - 15 )$

                    $H = 31350 \ W$

                    $H = 31.35\ kW$

The effectiveness of the heat exchanger is mathematically evaluated as

             $\epsilon = \frac{H}{H_{max}}$

  Substituting values  

           $\epsilon = \frac{31350}{88,825}$

              $= 0.35$

The NTU of the heat exchanger is mathematically represented as

          $NTU = \frac{1}{C-1} ln [\frac{\epsilon - 1}{\epsilon C -1} ]$

Where C is the ratio of the minimum to the maximum heat capacity which is mathematically represented as

             $C = \frac{C_c}{C_h}$

Substituting values

             $C = \frac{1045}{12,570}$

                 $= 0.083$

Substituting values in to the equation for NTU

         $NTU = \frac{1}{0.083 -1} ln[\frac{0.35 - 1}{0.35 * (0.083 - 1)} ]$

                   $= 0.438$

Generally the heat transfer surface area can be mathematically represented as

         $A_s = \frac{NTU C_c}{U}$

Substituting values

          $A_s = \frac{(0.438) (1045)}{950}$

              $A_s = 0.4818 m^2$