91影视

Step 1: Determine the heat transfer coefficients We'll first find the heat transfer coefficients for heat transfer from the water to the inner surface of the copper tube and from the outer surface of the tube to the air. For heat transfer from water to the inner surface, we can use the Nusselt number definition: \(Nu = \dfrac{hD_i}{k}\), where \(h\) is the heat transfer coefficient between water and the inner tube surface, and \(k\) is the thermal conductivity of water. We can look up the Nusselt number for heat transfer from a fluid in a vertical tube to obtain \(Nu 鈮� 4.36\). For water at \(0^{\circ}C\), the thermal conductivity is \(k 鈮� 0.555 W/m路K\). With the given inner diameter of the tube, \(D_i = 20 mm\), we can find the heat transfer coefficient between water and the inner tube surface: \(h_{wi} = \dfrac{Nu路k}{D_i} 鈮� \dfrac{4.36路0.555}{0.02} 鈮� 120.45 W/m^2路K\). For heat transfer from the outer surface of the tube to the air, we'll use the convection heat transfer equation: \(h_{oa} = C路Re^m路Pr^n\), where \(Re\) is the Reynolds number, \(Pr\) is the Prandtl number, and \(C, m\), and \(n\) are constant values, which depend on the flow regime. Given the air speed over the tube, we can calculate the Reynolds number: \(Re = \dfrac{蟻V(D_o - D_i)}{渭}\), where \(蟻\) is the density of air (1.292 kg/m鲁 at -20潞C), \(V\) is the air speed (3 m/s), \(D_o\) is the outer diameter of the tube (\(D_o = D_i + 2t = 20 + 2 路 2 = 24 mm = 0.024 m\)), and \(渭\) is the dynamic viscosity of air (1.717 x 10^-5 kg/m路s at -20潞C). So, \(Re 鈮� \dfrac{1.292路3路(0.024 - 0.02)}{1.717路10^{-5}} 鈮� 645.98\). Since the Reynolds number is less than 2300, we have laminar flow. The constants that correspond to laminar flow are \(C = 0.664\), \(m = 0.5\), and \(n = 1/3\). We can look up the Prandtl number for air (\(Pr 鈮� 0.728\) at -20潞C) and calculate the heat transfer coefficient between the outer surface and air: \(h_{oa} = 0.664路(645.98)^{0.5}路(0.728)^{(1/3)} 鈮� 36.23 W/m^2路K\). Step 2: Calculate the heat transfer rate Next, we'll calculate the heat transfer rate from the water and tube to the ambient air using the heat transfer coefficients we obtained. The heat flow rate is expressed as: \(Q = area * U * (T_w - T_{\infty})\), where \(U\) is the overall heat transfer coefficient, which can be calculated using the thermal resistance of the copper tube and the convective heat transfer coefficients: \(\dfrac{1}{U} = \dfrac{1}{h_{wi}} + \dfrac{t}{k_t路A_i} + \dfrac{1}{h_{oa}}\), where \(k_t\) is the thermal conductivity of copper (\(k_t 鈮� 391 W/m路K\)) and \(A_i\) is the inner surface area of the tube, which is given by: \(A_i = 蟺D_iL = 蟺路0.02路1 = 0.0628 m^2\). Hence, we have: \(\dfrac{1}{U} = \dfrac{1}{120.45} + \dfrac{0.002}{391路0.0628} + \dfrac{1}{36.23} 鈮� 0.0123\) So, \(U 鈮� 81.30 W/m^2路K\). Now we can calculate the heat transfer rate: \(Q = 0.0628路81.30路(0 - (-20)) 鈮� 102.3 W\). Step 3: Calculate the heat loss per unit mass To determine the heat loss per unit mass, we'll use the specific heat capacity of liquid water (\(c_p = 4,186 J/kg路K\)): \(Q_m = \dfrac{Q}{c_p路(T_w - T_\infty)} = \dfrac{102.3}{4,186路} 鈮� 0.025 W/kg\) So, when the tube is full of water, the heat loss per unit mass is approximately 0.025 W/kg.