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Sonja [21]
3 years ago
9

What gage pressure does a skin diver experience when they dive to 35 ft in the ocean with a water temperature of 55 °F? Report y

our answer in lbf/in^2.
Engineering
1 answer:
TEA [102]3 years ago
4 0

Answer:

p=15.097lbf/in^2

Explanation:

the manometric pressure is that in which the atmospheric pressure is not taken into account, so for this case only the pressure exerted by the water on the bus is calculated using the following equation.

P=ρgh

where

ρ=density of water at 55°F=999.4kg/m^3

g=9.81m/s^2

h=35ft=10.668m

Solving

P=(994.4)(9.81)(10.668)=104067.02Pa = 15.097lbf/in^2

the gage pressure does a skin diver experience when they dive to 35 ft is 15.097lbf/in^2

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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

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2.  Kinetic and potential energy changes are negligible.

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

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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\\

c. The thermal efficiency of the cycle  n_{th}  =\frac{W^{.} _{net} }{Q^{._{in} } } \\\\= \frac{20,166}{162,656} \\=0.124\\=12.4%

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