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vovikov84 [41]
3 years ago
11

A PV battery system has an end-to-end efficiency of 77%. The system is used to run an all-AC load that is run only at night. The

charge controller efficiency is 96% and the inverter efficiency is 85%. How much energy will need to be gathered by the PV array if the load is 120 watts running for 4 hours per night
Engineering
1 answer:
Gre4nikov [31]3 years ago
3 0

Answer:

763.94Wh

Explanation:

If the load is 120 watts running for 4 hours per night, the energy consumed is

120*4 =480 Wh

So, the energy gathered by the PV array is

E=480/(77%*96%*85%)=763.94Wh

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3.
Andreyy89

Answer:

7

Explanation:

5 + 2 = 7

4 0
3 years ago
Water vapor at 10 MPa, 600°C enters a turbine operating at steady state with a volumetric flow rate of 0.36 m3/s and exits at 0.
elena-s [515]

Answer:

Hook's law holds good up to. A elastic limit. B. plastic limit. C.yield point. D.Breaking point

8 0
3 years ago
A prototype boat is 30 meters long and is designed to cruise at 9 m/s. Its drag is to be simulated by a 0.5-meter-long model pul
Ghella [55]

Answer:

a) 1.16 m/s

b)  1/216000

c)  (√15)/6480000

Explanation:

The parameters given are;

Length of boat prototype, lp = 30 m

Speed of boat prototype = 9 m/s

Length of boat model, lm= 0.5 m

a) lm/lp = 0.5/30 = 1/60 = ∝

(vm/vp) = ∝^(1/2) = √∝ = (1/60)^(1/2)

vm = 9 × (1/60)^(1/2) = 1.16 m/s

b) The ratio of the model to prototype drag, Fm/Fp, is given as follows;

Fm/Fp = (vm/vp)²×(lm/lp)² = ∝³

Fm/Fp = (1/60)³ = 1/216000

c) The ratio of the model to prototype power  pm/p_p = (Fm/Fp) × (vm/vp) = ∝³×√∝

The ratio of the model to prototype power  pm/p_p = √(1/60) × (1/60)³

pm/p_p = √(1/60) × (1/60)³ = (√15)/6480000

6 0
3 years ago
A particle moving on a straight line has acceleration a = 5-3t, and its velocity is 7 at time t = 2. If s(t) is the distance fro
Vikki [24]

Given acceleration a = 5-3t, and its velocity is 7 at time t = 2, the value of s2 - s1 = 7

<h3>How to solve for the value of s2 - s1</h3>

We have

= \frac{dv}{dt} =v't = 5-3t\\\\\int\limits^a_b {v'(t)} \, dt

= \int\limits^a_b {(5-3t)} \, dt

5t - \frac{3t^2}{2} +c

v2 = 5x2 -  3x2 + c

= 10-6+c

= 4+c

s(t) = \frac{5t^2}{2} -\frac{t^3}{2} +3t + c

S2 - S1

=(5*\frac{4}{2} -\frac{8}{2} +3*2*c)-(\frac{5}{2} *1^2-\frac{1^2}{2} +3*1*c)

= 6 + 6+c - 2+3+c

12+c-5+c = 0

7 = c

Read more on acceleration here: brainly.com/question/605631

5 0
2 years ago
A house that was heated by electric resistance heaters consumed 1500 kWh of electric
gladu [14]

Answer:

2.5=1500/Whp=> Whp=600 kWh

delWgain=1500-600=900 kWh

Money saved= 900* 6tk*=5400 tk

5 0
3 years ago
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