Which of the following steps is performed last in the process of creating a mind map

An electric water heater held at 120° F is kept in a 60°F room. When purchased, its insulation is equivalent to R-5. An owner pu

Trava [24]

Answer:

Total saving would be of 36.917 $\yr

Explanation:

Given Data:

$T_{heater} = 120$ Degree F

$T_{room} = 60$ Degree F

A = 30 ft^2

$\eta = 100$ %

Heat loss before previous final value $= \frac{A \Delta T}{R}$

$=\frac{30\times *(120-60)}{5}$

= 360 Btu/hr

Heat loss after new value $= \frac{30\times \times (120-60)}{15} = 120 Btu/hr$

saving would be $= 360 - 120 Btu/hr \times kw hr/ 3412 Btu\times 24 hr/day \times 365 day/year$

= 616.1782 kw hr/yr

$cost = 616.1782 \times 0.06$ $

= 36.917 $\yr

6 0

4 years ago

(Laminar flow) A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length

EastWind [94]

This question is incomplete, the complete question is;

(Laminar flow) A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length 2 $l$ . The flow is laminar and fully developed. The pressure drop for the first pipe is 1.657 times greater than it is for the second pipe. If the diameter of the first pipe is D, determine the diameter of the second pipe.

D₃ = _____D.

{ the tolerance is +/-3% }

Answer:

the diameter of the second pipe D₃ is 1.13D

Explanation:

Given the data in the question;

Length = 2 $l$

pressure drop in the first pipe is 1.657 times greater than it is for the second pipe.

Now, we know that for Laminar Flow;

V' = πD⁴ΔP / 128μL

where V'₁ = V'₂ and ΔP₁₋₂ = 1.657 ΔP₂₋₃

Hence,

V'₁ = πD⁴ΔP₁₋₂ / 128μL = V'₃ = πD₃⁴ΔP₂₋₃ / 128μL

so

D₃ = D $($ ΔP₁₋₂ / ΔP₂₋₃ $)^{\frac{1}{4}$

we substitute

D₃ = D $($ 1.657 $)^{\frac{1}{4}$

D₃ = D( 1.134568 )

D₃ = 1.13D

Therefore, the diameter of the second pipe D₃ is 1.13D

8 0

3 years ago

A large steel tower is to be supported by a series of steel wires; it is estimated that the load on each wire will be 19,000N. D

zhuklara [117]

Answer:

11.6 mm

Explanation:

With a factor of safety of 5 and a yield strength of 900 MPa the admissible stress is:

σadm = strength / fos

σadm = 900 / 5 = 180 MPa

The stress is the load divided by the section:

σ = P / A

σ = 4*P / (π*d^2)

Rearranging:

d^2 = 4*P / (π*σ)

$d = \sqrt{4*P / (\pi*\sigma)}$

$d = \sqrt{4*19000 / (\pi*180*10^6)} = 0.0116 m = 11.6 mm$

7 0

3 years ago

A steam power plant operating on a simple ideal Rankine cycle maintains the boiler at 6000 kPa, the turbine inlet at 600 °C, and

ohaa [14]

Answer:

ηa=0.349

ηb=0.345

Explanation:

The enthalpy and entropy at state 3 are determined from the given pressure and temperature with data from table:

$h_{3}=3658.8kJ/kg\\ s_{3}=7.1693kJ/kg$

The quality at state 4 is determined from the condition $s_{4} =s_{3}$ and the entropies of the components at the condenser pressure taken from table:

$q_{4} =\frac{s_{4}-s_{liq50} }{s_{evap,50} } \\=\frac{7.1693-1.0912}{6.5019}=0.935$

The enthalpy at state 4 then is:

$h_{4} =h_{liq50} +q_{4} h_{evap,50}\\ (340.54+0.935*2304.7)kJ/kg\\=2495.43kJ/kg\\$

Part A

In the case when the water is in a saturated liquid state at the entrance of the pump the enthalpy and specific volume are determined from A-5 for the given pressure:

$h_{1}=340.54kJ/kg\\ \alpha _{1}=0.00103m^3/kg$

The enthalpy at state 2 is determined from an energy balance on the pump:

$h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1} )$

=346.67 kJ/kg

The thermal efficiency is then determined from the heat input and output in the cycle:

$=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.349$

Part B

In the case when the water is at a lower temperature than the saturation temperature at the condenser pressure we look into table and see the water is in a compressed liquid state. Then we take the enthalpy and specific volume for that temperature with data from and the saturated liquid values:

$h_{1}=293.7kJ/kg\\ \alpha _{1}=0.001023m^3/kg$