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Arlecino [84]
2 years ago
7

A power company desires to use groundwater from a hot spring to power a heat engine. If the groundwater is at 95 deg C and the a

tmosphere is at 20 deg C, what is the maximum power output for a mass flow of 0.2 kg/s? Assume the water is cooled to atmospheric temperature.
Engineering
1 answer:
prisoha [69]2 years ago
3 0

Answer:

W  = 12.8 KW

Explanation:

given data:

mass flow rate = 0.2 kg/s

Engine recieve heat from ground water at 95 degree ( 368 K)  and reject that heat to atmosphere  at 20 degree (293K)

we know that maximum possible efficiency is given as

\eta = 1- \frac{T_L}{T_H}

\eta = 1 - \frac{ 293}{368}

\eta = 0.2038

rate of heat transfer is given as

Q_H = \dot m C_p \Delta T

Q_H = 0.2 * 4.18 8(95 - 20)

Q_H = 62.7 kW

Maximuim power is given as

W = \eta Q_H

W = 0.2038 * 62.7

W  = 12.8 KW

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A large retirement village has a total retail employment of 120. All 1600 of the households in this village consist of two nonwo
polet [3.4K]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

5 0
3 years ago
A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa.
Galina-37 [17]

Answer:

elongation of the brass rod is 0.01956 mm

Explanation:

given data

length = 5 cm = 50 mm

diameter = 4.50 mm

Young's modulus = 98.0 GPa

load = 610 N

to find out

what will be the elongation of the brass rod in mm

solution

we know here change in length formula that is express as

δ = \frac{PL}{AE}    ................1

here δ is change in length and P is applied load  and A id cross section area and E is Young's modulus and L is length

so all value in equation 1

δ = \frac{PL}{AE}  

δ = \frac{610*50}{\frac{\pi}{4} * 4.50^2 * 98*10^3}  

δ = 0.01956 mm

so elongation of the brass rod is 0.01956 mm

7 0
3 years ago
Air at p=1 atm enters a long tube of length 2.5 m and diameter of 12 mm at an inlet temperature of Tm,i=100oC and mass flowrate
Annette [7]

Answer:

The heat transfer q = 18.32W

Explanation:

In this question, we are asked to calculate the heat entering the tube and also evaluate properties at T =400K

Please check attachment for complete solution and step by step explanation

6 0
3 years ago
The vertical and horizontal poles at the traffic-light assembly are erected first. Determine the additional force and moment rea
konstantin123 [22]

Answer:

1. Az=258 lb

2. My=3440 ft lb

3. ∑Mz= 0

Explanation:

∑ Fx= Ax=0

∑ Fy= Ay=0

∑ Fz= Az-86-86-86

Az=258 lb

∑Mx=86 X 31 +Mx=0

Mx=2666 ft lb

∑My=86 X 40 +My=0

My=3440 ft lb

∑Mz= 0

6 0
3 years ago
One method that is used to grow nanowires (nanotubes with solid cores) is to initially deposit a small droplet of a liquid catal
7nadin3 [17]

Answer: maximum length of the nanowire is 510 nm

Explanation:

 

From the table of 'Thermo physical properties of selected nonmetallic solids at At T = 1500 K

Thermal conductivity of silicon carbide k = 30 W/m.K

Diameter of silicon carbide nanowire, D = 15 x 10⁻⁹ m  

lets consider the equation for the value of m

m = ( (hP/kAc)^1/2 )  = ( (4h/kD)^1/2 )  

m =  ( ((4 × 10⁵)/(30×15×10⁻⁹ ))^1/2 ) = 942809.04    

now lets find the value of h/mk    

h/mk = 10⁵ / ( 942809.04 × 30) =  0.00353

lets consider the value θ/θb by using the equation

θ/θb = (T - T∞) / (T - T∞)

θ/θb =  (3000 - 8000) / (2400 - 8000)

= 0.893

the temperature distribution at steady-state is expressed as;

θ/θb = [ cosh m(L - x) + ( h/mk) sinh m (L - x)]   / [cosh mL+  (h/mk) sinh mL]

θ/θb = [ cosh m(L - L) + ( h/mk) sinh m (L - L)]   / [cosh mL+  (h/mk) sinh mL]

θ/θb = [ 1 ]  / [cosh mL+  (h/mk) sinh mL]

so we substitute

0.893 =  [ 1 ]  / [cosh (942809.04 × L) +  (0.00353) sinh (942809.04 × L)]

L = 510 × 10⁻⁹m

L = 510 nm

therefore maximum length of the nanowire is 510 nm

4 0
3 years ago
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