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3 years ago

. Which of the following formula is used for calculating current:

Engineering

1 answer:

ElenaW [278]3 years ago

8 0

Answer:

I think non of the above because

Explanation:

Current is usually denoted by the symbol I. Ohm's law relates the current flowing through a conductor to the voltage V and resistance R; that is, V = IR. An alternative statement of Ohm's law is I = V/R.

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Find the differential and evaluate for the given x and dx: y=sin2xx,x=π,dx=0.25

Sedaia [141]

By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

<h3>How to determine the differential of a one-variable function</h3>

Differentials represent the instantaneous change of a variable. As the given function has only one variable, the differential can be found by using ordinary derivatives. It follows:

dy = y'(x) · dx (1)

If we know that y = (1/x) · sin 2x, x = π and dx = 0.25, then the differential to be evaluated is:

$y' = -\frac{1}{x^{2}}\cdot \sin 2x + \frac{2}{x}\cdot \cos 2x$

$y' = \frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}}$

$dy = \left(\frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}} \right)\cdot dx$

$dy = \left(\frac{2\pi \cdot \cos 2\pi -\sin 2\pi}{\pi^{2}} \right)\cdot (0.25)$

$dy = \frac{1}{2\pi}$

By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

To learn more on differentials: brainly.com/question/24062595

#SPJ1

4 0

2 years ago

What is the value of the work interaction in this process?

Cloud [144]

Answer:

The answer is " $-121\ \frac{KJ}{Kg}$ ".

Explanation:

Please find the correct question in the attachment file.

using formula:

$\to W=-P_1V_1+P_2V_2 \\\\When \\\\\to W= \frac{P_1V_1-P_2V_2}{n-1}\ \ or \ \ \frac{RT_1 -RT_2}{n-1}\\\\$

$W =\frac{R(T_1 -T_2)}{n-1}\\\\$

$=\frac{0.287(25 -237)}{1.5-1}\\\\=\frac{0.287(-212)}{0.5}\\\\=\frac{-60.844}{0.5}\\\\=-121.688 \frac{KJ}{Kg}\\\\=-121 \frac{KJ}{Kg}\\\\$

7 0

3 years ago

A heat pump with an ideal compressor operates between 0.2 MPa and 1 MPa. Refrigerant R134a flows through the system at a rate of

solmaris [256]

Answer:

The mass flow rate of refrigerant is 0.352 kg/s

Explanation:

Considering the cycle of an ideal heat pump, provided in the attachment, we first find enthalpy at state B and D. For that purpose, we use property tables of refrigerant R134a:

At State A:

From table, we see the enthalpy and entropy value of saturated vapor at 0.2 MPa. Therefore:

ha = 244.5 KJ/kg

Sa = 0.93788 KJ/kg.k

At State B:

Since, the process from state A to B is isentropic. Therefore,

Sb = Sa = 0.93788 KJ/Kg

From table, we see the enthalpy value of super heated vapor at 1 MPa and Sb. Therefore:

hb = 256.85 KJ/kg (By interpolation)

At State C:

From table, we see the enthalpy and entropy value of saturated liquid at 1 MPa. Therefore:

hc = 107.34 KJ/kg

Now, from the diagram it is very clear that:

Heat Loss = m(hb = hc)

m = (Heat Loss)/(hb - hc)

where,

m = mass flow rate = ?

Heat Loss = (180,000 Btu/hr)(1.05506 KJ/1 Btu)(1 hr/3600 sec)

Heat Loss = 52.753 KW

Therefore,

m = (52.753 KJ/s)/(256.85 KJ/kg - 107.34 KJ/kg)

m = 0.352 kg/s

5 0

3 years ago

Hello , how are yall:))))

SVEN [57.7K]

Answer:

eh I'm good hbu?????????

6 0

2 years ago

Read 2 more answers

Steam at 20 bars is in the saturated vapor state (call this state 1) and contained in a pistoncylinderdevice with a volume of 0.

saul85 [17]

Answer:

Explanation:

Given that:

At state 1:

Pressure P₁ = 20 bar

Volume V₁ = 0.03 $\mathbf{m^{3}}$

From the tables at saturated vapour;

Temperature T₁ = 212.4⁰ C ; $v_1 = vg_1$ = 0.0996 $\mathbf{m^{3}}$ / kg

The mass inside the cylinder is m = 0.3 kg, which is constant.

The specific internal energy u₁ = ug₁ = 2599.2 kJ/kg

At state 2:

Temperature T₂ = 200⁰ C

Since the 1 - 2 occurs in an isochoric process v₂ = v₁ = 0.099 $\mathbf{m^{3}}$ / kg

From temperature T₂ = 200⁰ C

$v_f_2 = 0.0016 \ m^3/kg$

$vg_2 = 0.127 \ m^3/kg$

Since $vf_2 < v_2$ , the saturated pressure at state 2 i.e. P₂ = 15.5 bar

Mixture quality $x_2 = \dfrac{v_2-vf_2}{vg_2 -vf_2}$

$x_2 = \dfrac{(0.099-0.0016)m^3/kg}{(0.127 -0.0016) m^3/kg}$

$x_2 = \dfrac{(0.0974)m^3/kg}{(0.1254) m^3/kg}$

$\mathsf{x_2 =0.78}$

At temperature T₂, the specific internal energy $u_f_2 = 850.6 \ kJ/kg$ , also $ug_2 = 2594.3 \ kJ/kg$

Thus,

$u_2 = uf_2 + x_2 (ug_2 -uf_2)$

$u_2 =850.6 +0.78 (2594.3 -850.6)$

$u_2 =850.6 +1360.086$

$u_2 =2210.686 \ kJ/kg$

At state 3:

Temperature $T_3=T_2 = 200 ^0 C ,$

$V_3 = 2V_1 = 0.06 \ m^3$

Specific volume $v_3 = 0.2 \ m^3/kg$

Thus; $vg_3 =vg_2 = 0.127 \ m^3/kg$ ,

SInce $v_3 > vg_3$ , therefore, the phase is in a superheated vapour state.

From the tables of superheated vapour tables; at $v_3 = 0.2 \ m^3/kg$ and T₃ = 200⁰ C

The pressure = 10 bar and v =0.206 $\ m^3/kg$

The specific internal energy $u_3$ at the pressure of 10 bar = 2622.3 kJ/kg

The changes in the specific internal energy is:

$u_2-u_1$

= (2210.686 - 2599.2) kJ/kg

= -388.514 kJ/kg

≅ - 389 kJ/kg

$u_3-u_2$

= (2622.3 - 2210.686) kJ/kg

= 411.614 kJ/kg

≅ 410 kJ/kg

We can see the correct sketches of the T-v plot showing the diagrammatic expression in the image attached below.

3 0

3 years ago