If an object has more mass then it will have...

Specific volume for steam at P2 = 0.8 bar in the saturated vapour state can be gotten from saturated steam tables(find a sample of the table attached to this answer).

So from the table,

v2 = 2.087 m^(3)/kg

Now, mass flow rate (m) = (AV) /v

Where AV is the volumetric flow rate.

Thus, the mass flow rate at exit could be calculated as;

m = 15/(2.087) = 7.17 kg/s

We also know energy equation could be defined as;

Q-W = m[(h1 - h2) + {(V2(^2) - (V1(^2)} /2)} + g(Z2 - Z1)]

Since the flow is adiabatic, potential energy can be taken to be zero. Therefore, we get;

-W = m[(h2 - h1) + {(V2(^2) - (V1(^2)} /2)}

From, table 2, i attached , at P1 = 40 bar and T1 = 500°C; specific enthalpy was calculated to be h1 = 3445.3 KJ/Kg

Likewise, at P2 = 0.8 bar; from the table, we get specific enthalpy as;

h2 = 2665.8 KJ/Kg

So we now calculate power developed;

W = - 7.17 [(2665.8 - 3445.3) + {(150^(2) - 200^(2))/2000 = 5701KW

Since the sign is not negative but positive, it means that the power is developed from the system.

4 0

3 years ago

A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing throu

Shtirlitz [24]

Answer:

n = 1.56 x 10¹⁷ electrons

Explanation:

First of all, we will calculate the current passing through wire:

$I = \frac{q}{t}$

where,

I = current = ?

q = charge = 9 mC = 0.009 C

t = time = 3.6 s

Therefore,

$I = \frac{0.009\ C}{3.6\ s}\\\\I = 0.0025\ A = 2.5\ mA$

Now, for the same current in 10 s time the charge will be:

q = It = (0.0025 A)(10 s)

q = 0.025 C

Now, the number of electrons can be given as:

$q = ne\\\\n = \frac{q}{e}\\\\$

where,

n = no. of electrons = ?

q = charge = 0.025 C

e = charge on single electron = 1.6 x 10⁻¹⁹ C

Therefore,

$n = \frac{0.025\ C}{1.6\ x\ 10^{-19}\ C}$

<u>n = 1.56 x 10¹⁷ electrons</u>

8 0

3 years ago

Four copper wires of equal length are connected in series. Thecross sectional areas are 1 cm2, 2 cm2, 3cm2, and 5 cm2. A voltage

vampirchik [111]

Answer:

Voltage across 2 cm² wire = 29.5 V

Explanation:

We equation for resistance,

$R=\frac{\rho L}{A}$

Where ρ is resistivity, L is length and A is area.

Here for the four wires ρ and L is same, only area is different.

So we have

$R_1:R_2:R_3:R_4=\frac{\rho L}{A_1}:\frac{\rho L}{A_2}:\frac{\rho L}{A_3}:\frac{\rho L}{A_3}\\\\R_1:R_2:R_3:R_4=\frac{1}{1}:\frac{1}{2}:\frac{1}{3}:\frac{1}{5}$

Here total voltage is given as 120 V,

In series connection voltage divides in the ratio of resistances

That is

$V_1:V_2:V_3:V_4=\frac{1}{1}:\frac{1}{2}:\frac{1}{3}:\frac{1}{5}$

$\frac{1}{1}\times x+\frac{1}{2}\times x+\frac{1}{3}\times x+\frac{1}{5}\times x=120Vx=59.02V$

Voltage across 2 cm² wire $=\frac{59.02}{2}=29.5V$

Voltage across 2 cm² wire = 29.5 V

3 0

3 years ago