in a straight tube of 60mm diameter water is flowing at a velocity of 12m/s the tube surface temp is maintained at 70 degrees c and flowing water is heated from 15 degrees c upto 45 degrees c. take thermophysical properties at bulk mean temp of 30 degrees c find the following quantities the heat transfer coefficient to water. rewrite the complete questionn and give detailed solution

 To find the heat transfer coefficient for water flowing in a straight tube with a diameter of 60 mm, a velocity of 12 m/s, and a surface temperature maintained at 70°C, we use the Dittus-Boelter equation for turbulent flow. The thermophysical properties of water are taken at the bulk mean temperature of 30°C.

Given Data:

• Diameter, $D=0.06\text{m}$

• Velocity, $u=12\text{m/s}$

• Bulk mean temperature, $T_{\text{bulk}}=30^{\circ }\text{C}$

• Surface temperature, $T_{\text{surface}}=70^{\circ }\text{C}$

• Inlet temperature, $T_{\text{in}}=15^{\circ }\text{C}$

• Outlet temperature, $T_{\text{out}}=45^{\circ }\text{C}$

Thermophysical Properties at 30°C:

• Density, $\rho =995.7{\text{kg/m}}^3$

• Dynamic viscosity,

• Thermal conductivity,

• Specific heat capacity,

• Prandtl number, $\text{Pr}=\frac{C_p\mu }{k}=\frac{4178\times 0.0007973}{0.615}\approx 5.42$

Reynolds Number Calculation:

$$\text{Re}=\frac{\rho uD}{\mu }=\frac{995.7\times 12\times 0.06}{0.0007973}\approx 899,350$$

Since $\text{Re}>4000$, the flow is turbulent.

Nusselt Number using Dittus-Boelter Equation:

$$\text{Nu}=0.023\times {\text{Re}}^{0.8}\times {\text{Pr}}^{0.4}$$

Calculating ${\text{Re}}^{0.8}$:

$${\text{Re}}^{0.8}\approx 57,800$$

Calculating ${\text{Pr}}^{0.4}$:

$${\text{Pr}}^{0.4}\approx 1.965$$$$\text{Nu}=0.023\times 57,800\times 1.965\approx 2,613$$

Heat Transfer Coefficient Calculation:

$$h=\frac{\text{Nu}\times k}{D}=\frac{2,613\times 0.615}{0.06}\approx 26,770{\text{W/(m}}^2\cdot \text{K)}$$

Final Answer:
The heat transfer coefficient to water is \boxed{26700} W/(m²·K).