From the calculated heat transfer rate of the evaporating

Question

Solar Thermal Energy Engineering Homework-

Question 1- Finish the thermodynamic analysis of the hot air power system we looked at in Class, given by

Cp - Cv = R, the ideal gas constant = 8.3 J/mol K. To = temperature of the surroundings = 293 K. Assume work done by a system is positive. Use absolute temperatures where needed and proceed as follows:

- Find Wc, the work of compression. Need T2 - the temperature at point 2. For R&A compression we can find T2 from

ΔS = Cp In(T2 /T1) + R In(P1 /P2) = 0

Then we can find for compression Wideal = - (ΔH) (for constant S) = - Cp (T2 - T1)

Now find Wc from the efficiency n that is given in the diagram for R&A compression, in J/mol of air.

- Next we do a similar analysis for the turbine. Find T4 for R&A expansion. Then find Wideal as above, then W of the turbine. Find the net output work Ws from the difference between compression and turbine work in J/mol of air.

Next we compute thermodynamic losses.

- Need T2 for adiabatic (not R&A) compression. From 1st law ΔH = Q - Wc = - Wc = - Cp (T2 - T1) for adiabatic compression.

Using Wc from above, find T2, then find ΔS = Cp In(T2 /T1) + R In(P1 /P2) for adiabatic compression. Then find Wlost = To ΔS for adiabatic compression

- Find T4 for adiabatic expansion in the turbine. Then make a similar analysis as for the compression to find ΔS for adiabatic expansion in the turbine. Then find Wlost = To ΔS for adiabatic expansion

- Find Wlost in the storage unit. Heat transfer to the air q = - Cp (T3 - T2) using T2 for adiabatic compression. Then we have ΔS for the storage unit = - q/T of the storage unit. ΔS for the air in the storage unit = Cp In(T3 /T2)

Total ΔS = ΔS for the storage unit + ΔS for the air

Then Wlost in the storage unit = To*(total ΔS)

Make a table that summarizes your results.

	J/mol of air	% of Wideal
Ws net
Wlost in Compressor
Wlost in Storage Unit
Total Ideal Work
		100%

So in this manner, we evaluate the thermodynamic efficiency of the net turbine work Ws as a percent of Wideal.

Question: What happens to the thermodynamic efficiency Ws/(Total Ideal Work) when the temperature of the air out of the storage bed increases? Why is that?

Question 2- For the open Brayton cycle power system of Question 1, sketch the power cycle on a PV-diagram and on a TS-diagram. On these diagrams, sketch the lines of the ideal compression and expansion of the compressor and turbine, and sketch the lines of the actual compression and expansion. We can get the shape of the curves correctly by noting that for adiabatic and reversible (ΔS= 0) expansion and compression of an ideal gas we have

PV^t = constant (these are hyperbolas)

or equivalently

(T2/T1) = (P2/P1)(γ-1)γ

where γ = Cp/Cv is the ratio of (constant) ideal gas heat capacities.

The cycle isn't closed, but we can draw the line from turbine exit to air inlet for closed cycles on the diagram. What does the area enclosed by these curves represent?

Question 3- For the solar heated basin type of seawater distillation system, experiments have shown that the convection heat transfer coefficient in the system is

Hc = 0.884[(Tb - Tg) + (pwb-pwg/2016-pwb)Tb]^1/3

and the maximum evaporation rate is

me = 9.15E-07hc (pwb - pwg)

where hc (Wm² K) is the convection eat transfer coefficient, Tb is the water temp in the basin (deg.K), Tg is the temp of the glass (deg.K), pwb and pwg are the equilibrium vapor pressures of the water in millimeters of mercury (mm Hg) at temperatures Tb and Tg and me is the evaporation rate (kg/m² s).

Suppose that the temperature of the water in the basin is 80 deg.C and the temperature of the glass is 50 deg.C. Then if hfg is the enthalpy of vaporization of water at 80deg.C, calculate the heat transfer rate of the evaporating water from me and hfg in W/m².

From the calculated heat transfer rate of the evaporating water, calculate the maximum efficiency of the solar distillation system on a 1 m² piece of glass for a solar flux of 800 W/m². This is a maximum, because the vapor pressure of sea water is less than the vapor pressure of pure water. Why is that?

Question 4- A linear parabolic concentrator of length 10 m has an aperture a=1.36 m and focal length f =1.26 m. The azimuth of alignment is 0 degrees. This concentrator is located in Albuquerque, NM.

a. For a parabola, the rim angle ψ = tan-1[(f/a)/(2(f/a))²  - 1/8]. Find the receiver diameter to intercept 0.95 of the solar disk for a reflector with angular error δ = 0.15 deg. The sun's diameter is 5 deg.

b. For this concentrator, if U_L = 14 Win² deg.C, FR = 0.9, (τα) = 0.78, reflectance = 0.90, GB at normal incidence = 2.14 MJ/m², Ti = 230 deg.C, Ta = 10 deg.C at solar noon on March 21, what is the useful heat gain Qu in MJ of the concentrator for the tube of diameter of part a? What is the efficiency?

Question 5- Pictured below is a CSP power tower operation with molten nitrate salt storage.

a. Suppose the hot storage is 56 deg.C, cold storage 290 deg.C, water inlet temp 25 deg.C, steam outlet 550 deg.C, what is the log mean temperature difference LMTD in a counter-cured salt/water heat exchanger? Lock up what LMTD is.

b. The steam at 550 deg.C is at 60 bar pressure is expanded in a turbine with the condenser at 0.1 bar. All the steam is condensed. Pump and turbine efficiencies of a simple Rankine cycle are 90%, estimate the cycle efficiency.

d. Will the actual Rankine cycle efficiency increase or decrease if the LMTD increases?

e. If the CSP system is designed to generate 17 MWe and the storage capacity is 6 hours, what net annual electricity production GW-hr can be expected from 300 days of operation?

Question 6- The table describes some characteristics CSP:

Collector & Power Cycle Technology	Solar- Electric Aperture Related Efficiency	Land Use Factor	Land Use Efficiency
Parabolic Trough Steam Cycle
Central Receiver Steam Cycle
Linear Fresnel Steam Cycle

Look up information related to several commercial or planned CSP and fill in the table with ranges of values. "Aperture related efficiency" is the expected annual solar heat to electricity conversion percent. "Land use factor" is the total solar field reflector aperture area divided by the total field area. "Land use efficiency" is the product of the aperture efficiency and land use factor.