Hey, can anyone tell me if Igneous rock is good to build on? Cheers!

Engineering

1 answer:

lorasvet [3.4K]3 years ago

6 0

Answer:

Yes! Here's Why

Explanation:

Igneous rocks are often used as setts in areas of heavy wear, because they withstand both weathering and erosion better than most sedimentary rocks, or even concrete.

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List the thermal conductivities of five

ICE Princess25 [194]

Answer:

Sorry I don't understand

Explanation:

8 0

3 years ago

Can i join three 12 volts batteriesto give me 24 volts output

bulgar [2K]

Answer:

YES

Explanation:

If we connect batteries in series then the output voltage is the sum of the individual voltage of each battery i.e if you connect three 12 volts batteries in series then their output voltage will be 12+12+12=36 volts, but the current rating of the batteries in series will be same of the individual battery rating in 'mah'.

But when we connect the batteries in parallel their voltage is not added but their current rating in mah is addition of their individual rating.

So, If you want 24 volts from three 12 volts battery then you can connect two of them in series and the other one in parallel with them this will give 24 volts and the current will be addition of the two series batteries and the third which is in parallel with them. You can use this configuration if current value is not a big factor.

8 0

3 years ago

A binary geothermal power plant uses geothermal water at 160°C as the heat source. The cycle operates on the simple Rankine cycl

bogdanovich [222]

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 ( $c_{geo}$ [) 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 $P_{3} = 3.25mPa = (3.25*1000) kPa\\= 3250kPa\\from the EES TABLE\\h_{3} = 761.54 kJ/kg\\s_{3} = 2.5457 kJ/kg\\P_{4} = 410kPa\\\\s_{4} = s_{3} \\h_{4s} = 470.40kJ/kg\\\\T_{4} = 179.5^{0} C\\\\h_{4} = 689.74 kJ/KG\\\\ The isentropic efficiency, n_{T} = \frac{h_{3}-h_{4} }{h_{3}- h_{4s} }$

= $=\frac{761.54-689.74}{761.54-670.40} \\=\frac{71.8}{91.14} \\=0.788$

b. Pump

$h_{1} = h_{f} @ 410kPa = 273.01kJ/kg\\v_{1} = v_{f} @ 410kPa = 0.001842 m^{3}/kgw_{p,in} = \frac{v_{1}(P_{2}-P_{1}) }{n_{p} } \\\\= \frac{0.01842(3250-410)}{0.9} \\\\ =5.81kJ/kg\\h_{2} =h_{1} + w_{p,in}\\ = 273.01+5.81\\ = 278.82 kJ/kg\\\\w_{T,out} = m^{.} (h_{3} -h_{4} )\\=(305.6)(761.54-689.74)\\=305.6(71.8)\\=21,942kW\\\\$

$W^{.} _ {P,in} = m^{.} (h_{2} -h_{1}) \\=m^{.} w_{p,in \\=305.6(5.81)\\\\=1,777kW\\W^{.} _{net} = W^{.} _{T, out} - W^{.} _{P,in} \\= 21,942-1,777\\=20,166 kW\\\\HEAT EXCHANGER\\\\Q_{in} = m^{.} _{geo} c_{geo} (T_{in-T_{out} } )\\=555.9(4.18)(160-90)\\=162.656kW\\$