A binary geothermal power operates on the simple Rankine cycle with isobutane as the working fluid. The isentropic efficiency of the turbine, the net power output, and the thermal efficiency of the cycle are to be determined
Assumptions :
1. Steady operating conditions exist.
2. Kinetic and potential energy changes are negligible.
Properties: The specific heat of geothermal water (
[) is taken to be 4.18 kJ/kg.ºC.
Analysis (a) We need properties of isobutane, we can obtain the properties from EES.
a. Turbine
P![P_{3} = 3.25mPa = (3.25*1000) kPa\\= 3250kPa\\from the EES TABLE\\h_{3} = 761.54 kJ/kg\\s_{3} = 2.5457 kJ/kg\\P_{4} = 410kPa\\\\s_{4} = s_{3} \\h_{4s} = 470.40kJ/kg\\\\T_{4} = 179.5^{0} C\\\\h_{4} = 689.74 kJ/KG\\\\ The isentropic efficiency, n_{T} = \frac{h_{3}-h_{4} }{h_{3}- h_{4s} }](https://tex.z-dn.net/?f=P_%7B3%7D%20%3D%203.25mPa%20%3D%20%283.25%2A1000%29%20kPa%5C%5C%3D%203250kPa%5C%5Cfrom%20the%20EES%20TABLE%5C%5Ch_%7B3%7D%20%3D%20761.54%20kJ%2Fkg%5C%5Cs_%7B3%7D%20%3D%202.5457%20kJ%2Fkg%5C%5CP_%7B4%7D%20%3D%20410kPa%5C%5C%5C%5Cs_%7B4%7D%20%3D%20s_%7B3%7D%20%5C%5Ch_%7B4s%7D%20%3D%20470.40kJ%2Fkg%5C%5C%5C%5CT_%7B4%7D%20%3D%20179.5%5E%7B0%7D%20C%5C%5C%5C%5Ch_%7B4%7D%20%3D%20689.74%20kJ%2FKG%5C%5C%5C%5C%20The%20%20isentropic%20%20efficiency%2C%20n_%7BT%7D%20%3D%20%5Cfrac%7Bh_%7B3%7D-h_%7B4%7D%20%20%7D%7Bh_%7B3%7D-%20h_%7B4s%7D%20%7D)
=![=\frac{761.54-689.74}{761.54-670.40} \\=\frac{71.8}{91.14} \\=0.788](https://tex.z-dn.net/?f=%3D%5Cfrac%7B761.54-689.74%7D%7B761.54-670.40%7D%20%5C%5C%3D%5Cfrac%7B71.8%7D%7B91.14%7D%20%5C%5C%3D0.788)
b. Pump
![h_{1} = h_{f} @ 410kPa = 273.01kJ/kg\\v_{1} = v_{f} @ 410kPa = 0.001842 m^{3}/kgw_{p,in} = \frac{v_{1}(P_{2}-P_{1}) }{n_{p} } \\\\= \frac{0.01842(3250-410)}{0.9} \\\\ =5.81kJ/kg\\h_{2} =h_{1} + w_{p,in}\\ = 273.01+5.81\\ = 278.82 kJ/kg\\\\w_{T,out} = m^{.} (h_{3} -h_{4} )\\=(305.6)(761.54-689.74)\\=305.6(71.8)\\=21,942kW\\\\](https://tex.z-dn.net/?f=h_%7B1%7D%20%3D%20h_%7Bf%7D%20%40%20410kPa%20%3D%20273.01kJ%2Fkg%5C%5Cv_%7B1%7D%20%3D%20v_%7Bf%7D%20%40%20410kPa%20%3D%200.001842%20m%5E%7B3%7D%2Fkgw_%7Bp%2Cin%7D%20%3D%20%20%5Cfrac%7Bv_%7B1%7D%28P_%7B2%7D-P_%7B1%7D%29%20%20%20%7D%7Bn_%7Bp%7D%20%7D%20%5C%5C%5C%5C%3D%20%5Cfrac%7B0.01842%283250-410%29%7D%7B0.9%7D%20%5C%5C%5C%5C%20%3D5.81kJ%2Fkg%5C%5Ch_%7B2%7D%20%3Dh_%7B1%7D%20%2B%20w_%7Bp%2Cin%7D%5C%5C%20%20%20%20%20%20%20%20%20%20%3D%20273.01%2B5.81%5C%5C%20%20%20%20%20%20%20%20%20%20%20%3D%20278.82%20kJ%2Fkg%5C%5C%5C%5Cw_%7BT%2Cout%7D%20%3D%20m%5E%7B.%7D%20%20%28h_%7B3%7D%20-h_%7B4%7D%20%29%5C%5C%3D%28305.6%29%28761.54-689.74%29%5C%5C%3D305.6%2871.8%29%5C%5C%3D21%2C942kW%5C%5C%5C%5C)
![W^{.} _ {P,in} = m^{.} (h_{2} -h_{1}) \\=m^{.} w_{p,in \\=305.6(5.81)\\\\=1,777kW\\W^{.} _{net} = W^{.} _{T, out} - W^{.} _{P,in} \\= 21,942-1,777\\=20,166 kW\\\\HEAT EXCHANGER\\\\Q_{in} = m^{.} _{geo} c_{geo} (T_{in-T_{out} } )\\=555.9(4.18)(160-90)\\=162.656kW\\](https://tex.z-dn.net/?f=W%5E%7B.%7D%20_%20%7BP%2Cin%7D%20%3D%20m%5E%7B.%7D%20%28h_%7B2%7D%20-h_%7B1%7D%29%20%5C%5C%3Dm%5E%7B.%7D%20%20w_%7Bp%2Cin%20%5C%5C%3D305.6%285.81%29%5C%5C%5C%5C%3D1%2C777kW%5C%5CW%5E%7B.%7D%20%20_%7Bnet%7D%20%3D%20W%5E%7B.%7D%20_%7BT%2C%20out%7D%20-%20W%5E%7B.%7D%20%20_%7BP%2Cin%7D%20%5C%5C%3D%2021%2C942-1%2C777%5C%5C%3D20%2C166%20kW%5C%5C%5C%5CHEAT%20EXCHANGER%5C%5C%5C%5CQ_%7Bin%7D%20%3D%20m%5E%7B.%7D%20_%7Bgeo%7D%20c_%7Bgeo%7D%20%28T_%7Bin-T_%7Bout%7D%20%7D%20%29%5C%5C%3D555.9%284.18%29%28160-90%29%5C%5C%3D162.656kW%5C%5C)
c. ![The thermal efficiency of the cycle n_{th} =\frac{W^{.} _{net} }{Q^{._{in} } } \\\\= \frac{20,166}{162,656} \\=0.124\\=12.4%](https://tex.z-dn.net/?f=The%20thermal%20efficiency%20of%20the%20cycle%20%20n_%7Bth%7D%20%20%3D%5Cfrac%7BW%5E%7B.%7D%20_%7Bnet%7D%20%7D%7BQ%5E%7B._%7Bin%7D%20%7D%20%7D%20%5C%5C%5C%5C%3D%20%5Cfrac%7B20%2C166%7D%7B162%2C656%7D%20%5C%5C%3D0.124%5C%5C%3D12.4%25)
Answer:
5.6 mm
Explanation:
Given that:
A cylindrical tank is required to contain a:
Gage Pressure P = 560 kPa
Allowable normal stress
= 150 MPa = 150000 Kpa.
The inner diameter of the tank = 3 m
In a closed cylinder there exist both the circumferential stress and the longitudinal stress.
Circumferential stress ![\sigma = \dfrac{pd}{2t}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Cdfrac%7Bpd%7D%7B2t%7D)
Making thickness t the subject; we have
![t = \dfrac{pd}{2* \sigma}](https://tex.z-dn.net/?f=t%20%3D%20%5Cdfrac%7Bpd%7D%7B2%2A%20%5Csigma%7D)
![t = \dfrac{560000*3}{2*150000000}](https://tex.z-dn.net/?f=t%20%3D%20%5Cdfrac%7B560000%2A3%7D%7B2%2A150000000%7D)
t = 0.0056 m
t = 5.6 mm
For longitudinal stress.
![\sigma = \dfrac{pd}{4t}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Cdfrac%7Bpd%7D%7B4t%7D)
![t= \dfrac{pd}{4*\sigma }](https://tex.z-dn.net/?f=t%3D%20%5Cdfrac%7Bpd%7D%7B4%2A%5Csigma%20%7D)
![t = \dfrac{560000*3}{4*150000000}](https://tex.z-dn.net/?f=t%20%3D%20%5Cdfrac%7B560000%2A3%7D%7B4%2A150000000%7D)
t = 0.0028 mm
t = 2.8 mm
From the above circumferential stress and longitudinal stress; the stress with the higher value will be considered ; which is circumferential stress and it's minimum value with the maximum thickness = 5.6 mm
Answer:
4. A series of steps engineers use to solve problems.
Explanation:
The process of engineering design is a sequence of procedures that engineers pursue to arrive at a solution to a specific problem. Most times the solution includes creating a product such as a computer code, which fulfills certain conditions or performs a function. If the project in-hand includes designing, constructing, and testing it, then engineers probably adopt the design process. Steps of the process include defining the problem, doing background research, specifying requirements, brainstorming solutions, etc.
Answer:
Time taken = 136.32 minutes
Explanation:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.
Answer:
The governing ratio for thin walled cylinders is 10 if you use the radius. So if you divide the cylinder´s radius by its thickness and your result is more than 10, then you can use the thin walled cylinder stress formulas, in other words:
- if
then you have a thin walled cylinder
or using the diameter:
- if
then you have a thin walled cylinder