Answer :
Answer:
The rate of heat transfer is [tex]H = 31.35\ kW[/tex]
The heat transfer surface area is [tex]A_s = 0.4818 m^2[/tex]
Explanation:
From the question we are told that
The specific heat of water is [tex]cp = 4180 \ J/kg \cdot K[/tex]
The temperature of cold water is [tex]T_c = 15^o C[/tex]
The rate of cold the flow is [tex]\r m = 0.25 kg/s[/tex]
The temperature of the heated water [tex]T_h = 45 ^oC[/tex]
The specific heat of hot water is [tex]c_p__{H}} = 4190 J/kg \cdot K[/tex]
The temperature of the hot water is [tex]T_H = 100^oC[/tex]
The rate of hot the flow is [tex]\r m_H = 3 kg/s[/tex]
The heat transfer coefficient is [tex]U = 950 W/m^2 \cdot K,[/tex]
From the [tex]\epsilon -NTU[/tex] method we have that
[tex]C_h = \r m_H c_p__{H}}[/tex]
Where [tex]C_h[/tex] is the heat capacity rate of hot water
Substituting the value
[tex]C_h = 3 * (4190)[/tex]
[tex]= 12,5700\ W/^oC[/tex]
Also
[tex]C_c = \r m c_p[/tex]
Where [tex]C_c[/tex] is the heat capacity rate of cold water
[tex]C_c = 0.25 * 4180[/tex]
[tex]= 1045 \ W / ^oC[/tex]
The maximum heat capacity [tex]C_h[/tex] and the minimum heat capacity is [tex]C_c[/tex]
The maximum heat transfer is
[tex]H_{max} = C_c (T_H - T_c)[/tex]
Substituting values
[tex]H_{max} = (1045)(100- 15)[/tex]
[tex]= 88,825\ W[/tex]
The actual heat transfer is mathematically evaluated as
[tex]H = C_c (T_h - T_c)[/tex]
Substituting values
[tex]H = 1045 (45 - 15 )[/tex]
[tex]H = 31350 \ W[/tex]
[tex]H = 31.35\ kW[/tex]
The effectiveness of the heat exchanger is mathematically evaluated as
[tex]\epsilon = \frac{H}{H_{max}}[/tex]
Substituting values
[tex]\epsilon = \frac{31350}{88,825}[/tex]
[tex]= 0.35[/tex]
The NTU of the heat exchanger is mathematically represented as
[tex]NTU = \frac{1}{C-1} ln [\frac{\epsilon - 1}{\epsilon C -1} ][/tex]
Where C is the ratio of the minimum to the maximum heat capacity which is mathematically represented as
[tex]C = \frac{C_c}{C_h}[/tex]
Substituting values
[tex]C = \frac{1045}{12,570}[/tex]
[tex]= 0.083[/tex]
Substituting values in to the equation for NTU
[tex]NTU = \frac{1}{0.083 -1} ln[\frac{0.35 - 1}{0.35 * (0.083 - 1)} ][/tex]
[tex]= 0.438[/tex]
Generally the heat transfer surface area can be mathematically represented as
[tex]A_s = \frac{NTU C_c}{U}[/tex]
Substituting values
[tex]A_s = \frac{(0.438) (1045)}{950}[/tex]
[tex]A_s = 0.4818 m^2[/tex]