Answer:
"Biofuels"
Explanation:
I don't know if this counts but I guess it's not one of those.
Answer:
A.) Find the answer in the explanation
B.) Ua = 7.33 m/s , Vb = 7.73 m/s
C.) Impulse = 17.6 Ns
D.) 49%
Explanation:
Let Ua = initial velocity of the rod A
Ub = initial velocity of the rod B
Va = final velocity of the rod A
Vb = final velocity of the rod B
Ma = mass of rod A
Mb = mass of rod B
Given that
Ma = 2kg
Mb = 1kg
Ub = 3 m/s
Va = 0
e = restitution coefficient = 0.65
The general expression for the velocities of the two rods after impact will be achieved by considering the conservation of linear momentum.
Please find the attached files for the solution
Answer:
The answer is 0.727
Explanation:
lemme know if that's right
Answer:
(i) 169.68 volt
(ii) 16.90 volt
(iii) 16.90 volt
(iv) 108.07 volt
(v) 2.161 A
Explanation:
Turn ratio is given as 10:1
We have given that input voltage ![v_p=120volt](https://tex.z-dn.net/?f=v_p%3D120volt)
(i) We know that peak voltage is give by ![v_{peak}=\sqrt{2}v_p=\sqrt{2}\times 120=169.68volt](https://tex.z-dn.net/?f=v_%7Bpeak%7D%3D%5Csqrt%7B2%7Dv_p%3D%5Csqrt%7B2%7D%5Ctimes%20120%3D169.68volt)
(ii) We know that for transformer ![\frac{v_p}{v_s}=\frac{n_p}{n_s}](https://tex.z-dn.net/?f=%5Cfrac%7Bv_p%7D%7Bv_s%7D%3D%5Cfrac%7Bn_p%7D%7Bn_s%7D)
So ![\frac{169.08}{v_s}=\frac{10}{1}](https://tex.z-dn.net/?f=%5Cfrac%7B169.08%7D%7Bv_s%7D%3D%5Cfrac%7B10%7D%7B1%7D)
![v_s=16.90volt](https://tex.z-dn.net/?f=v_s%3D16.90volt)
So peak voltage in secondary will be 16.90 volt
(iii) Peak voltage of the rectifier will be equal to the peak voltage of the secondary
So peak voltage of the rectifier will be 16.90 volt
(iv) Dc voltage of the rectifier is given by ![v_{dc}=\frac{2v_m}{\pi }=\frac{2\times 1.414\times 120}{3.14}=108.07volt](https://tex.z-dn.net/?f=v_%7Bdc%7D%3D%5Cfrac%7B2v_m%7D%7B%5Cpi%20%7D%3D%5Cfrac%7B2%5Ctimes%201.414%5Ctimes%20120%7D%7B3.14%7D%3D108.07volt)
(v) Now dc current is given by ![i_{dc}=\frac{v_{dc}}{R}=\frac{108.07}{50}=2.1614A](https://tex.z-dn.net/?f=i_%7Bdc%7D%3D%5Cfrac%7Bv_%7Bdc%7D%7D%7BR%7D%3D%5Cfrac%7B108.07%7D%7B50%7D%3D2.1614A)
Answer:
COP = 3.828
W' = 39.18 Kw
Explanation:
From the table A-11 i attached, we can find the entropy for the state 1 at -20°C.
h1 = 238.43 KJ/Kg
s1 = 0.94575 KJ/Kg.K
From table A-12 attached we can do the same for states 3 and 4 but just enthalpy at 800 KPa.
h3 = h4 = hf = 95.47 KJ/Kg
For state 2, we can calculate the enthalpy from table A-13 attached using interpolation at 800 KPa and the condition s2 = s1. We have;
h2 = 275.75 KJ/Kg
The power would be determined from the energy balance in state 1-2 where the mass flow rate will be expressed through the energy balance in state 4-1.
W' = m'(h2 - h1)
W' = Q'_L((h2 - h1)/(h1 - h4))
Where Q'_L = 150 kW
Plugging in the relevant values, we have;
W' = 150((275.75 - 238.43)/(238.43 - 95.47))
W' = 39.18 Kw
Formula foe COP is;
COP = Q'_L/W'
COP = 150/39.18
COP = 3.828