African solar systems part 7: heat transfer and steam power

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Introduction

The creation of heat is only part of the story in a CSP - you also need to move the heat to where it can be useful. Many CSP systems use specialized heat transfer fluids (HTFs) to move heat from one spot to another. These HTFs are often oil or a molten salt, both of which have issues in a rural environment. They are not easily replaced in the case of an accident or leak. Oil breaks down at high temperatures and is flammable, whereas molten salt will solidify if the system is off for a significant length of time. Hence let's assume that the heat transfer fluid will be water in both liquid and vapor form (steam), which is not flammable, easily replaced, and remains a liquid even if the system is not operational for a time.

What kind of steam should one use? We have a choice between saturated steam or superheated steam. Saturated steam is more efficient for heat transfer, whereas superheated steam is necessary for running turbines to produce electricity. Whether steam is saturated or superheated depends on both the pressure and temperature of the steam. In either case, there must be a certain amount of steam flow out of the solar receiver to carry the heat to either a storage tank or a electrical generator.

The question

Let us model our solar energy receiver as a long, horizontal, hollow metal tube of length 10 m and inner diameter 2 cm. Water is pumped into one end of the tube at a velocity of $v_i$ and temperature of 25 $^\circ$ C. The total energy input to the tube from a solar parabolic trough around it is 13,700 W. What is the maximum velocity $v_i$ in $m/s$ such that steam and not water exits the receiver from the right end?

Assume that all the received energy gets transferred to the flowing water.

The answer is 0.0169.

1 solution

Beakal Tiliksew
Apr 13, 2014

$P\Delta t=m(c\Delta T+{ L }_{ v })\quad \longrightarrow \quad C=4.186J/g\circ c\quad \quad \quad L_{ v }=2.25\times { 10 }^{ 6 }J/kg\quad \quad \quad (1)\quad \\ \\ \quad \quad m=\rho V=\rho \pi { r }^{ 2 }L\quad \quad ......\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (2)\\ \\ \Delta t=\quad \frac { \rho \pi { r }^{ 2 }L(c\Delta T+{ L }_{ v }\} }{ P } \quad \quad \\ \\ v=\frac { L }{ \Delta t } \Longrightarrow \quad \frac { P }{ \rho \pi { r }^{ 2 }(c\Delta T+{ L }_{ v }\} } \approx 0.0169\\ \\ \\ \\ \\ \\ \\ \\$

African solar systems part 7: heat transfer and steam power

The answer is 0.0169.

1 solution

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