Water flows at the rate of 0.5 kg/s in a 2.5-cm-diameter tube having a length of 4 m. A constant heat flux is imposed at the tube wall so that the tube wall temperature is always 50°C higher than the water temperature. Calculate the heat transfer rate and estimate the temperature rise in the water as it exits the pipe if the water enters at 5°C. The water is pressurized so that boiling cannot occur. (Assume h 3000 W/m2 oC). Note the interplay between heat transfer and thermo!

Q = 47.1 kW

Texit = 25.93ºC

Explanation:

The heat transfer is occurring by convection, which means that the heat is flowing from different materials (tube for water). So, by Newton Cooling Law:

Q = h x A x ΔT

The area of the tube will be:

A = πdL

Where d is the diameter and L is the length.

d = 2.5 cm = 0.025 m

A = 3.14x0.025x4 = 0.314 m²

Then

Q = 3000x0.314x50

Q = 47100 W = 47.1 kW

By the heat equation for water, knowing that the specific heat (c) of water is 4.5 kJ/ºC

Q = mcΔT

47.11 = 0.5x4.5x (Texits - Tenters)

47.1 = 2.25x (Texits - 5ºC)

Texits - 5ºC = 20.93 ºC

Texits = 25.93ºC