Verizon LTE

1. Technician A says an automotive computer can scan its input and output circuits to detect incorrect voltage problems. Technic

Marta_Voda [28]

B is the correct answer

4 0

3 years ago

Air is pumped from a vacuum chamber until the pressure drops to 3 torr. If the air temperature at the end of the pumping process

malfutka [58]

Answer:

The final pressure is 3.16 torr

Solution:

As per the question:

The reduced pressure after drop in it, P' = 3 torr = $3\times 0.133\ kPa$

At the end of pumping, temperature of air, $T = 5^{\circ}C = 278 K$

After the rise in the air temperature, $T' = 20^{\circ}C = 293 K$

Now, we know the ideal gas eqn:

PV = mRT

So

$P = \frac{m}{V}RT$

$P = \rho_{a}RT$ (1)

where

P = Pressure

V = Volume

$\rho_{a} = air\ density$

R = Rydberg's constant

T = Temperature

Using eqn (1):

$P = \rho_{a}RT$

$\rho_{a} = \frac{P}{RT}$

$\rho_{a} = \frac{3 times 0.133\times 10^{3}}{0.287\times 278} = 0.005 kg/m^{3}$

Now, at constant volume the final pressure, P' is given by:

$\frac{P}{T} = \frac{P'}{T'}$

$P' = \frac{P}{T}\times T'$

$P' = \frac{3}{278}\times 293 = 3.16 torr$

7 0

4 years ago

Wind blows on the side of a fully enclosed hospital located on open flat terrain where V= 120 mi/h. Determine the external press

ss7ja [257]

Answer:

The external pressure is p = -21.9 psf or p = -8.85 psf

Explanation:

Given :

Velocity of wind, v = 120 mi / hr

$k_d = k_c =1$ (wind direction factor)

$k _{zt} = 1$ = topographical factor (for flat terrain)

$q_n$ = velocity pressure at height h

$\therefore q_n = 0.00256 k_z k_{zt} k_d v^2$

$q_n = 0.00256 \times k_z (1)(1)(120)^2$

$= 36.86 k_z$

But for height h = 30 ft, $k_z$ = 0.98 (from table)

$\therefore q_n = 36.86 \times 0.98$

= 36.16

Now, $\frac{L}{B}= \frac{200}{200} =1$ , so $C_p=-0.5$ (from table)

$p = q(G)(C_p)-q_n(GC_{pi})$

where, p = external pressure

G = 0.85 = gust factor (for typical rigid building)

$GC_{pi} = \pm 0.18$ (internal pressure co efficient)

Therefore putting the values,

$p = (36.13)(0.85)(-0.5)-(36.13)(\pm 0.18)$

p = -21.9 psf or p = -8.85 psf

4 0

4 years ago

An isentropic steam turbine processes 2 kg/s of steam at 3 MPa, which is exhausted at50 kPa and 100C. Five percent of this flow

borishaifa [10]

Answer:

2285kw

Explanation:

since it is an isentropic process, we can conclude that it is a reversible adiabatic process. Hence the energy must be conserve i.e the total inflow of energy must be equal to the total outflow of energy.

Mathematically,

$\\ E_{inflow} = E_{outflow}$

Note: from the question we have only one source of inflow and two source of outflow (the exhaust at a pressure of 50kpa and the feedwater at a pressure of 5ookpa). Also the power produce is another source of outgoing energy $\\ E_{inflow} = m_{1} h_{1}$ .

$\\$

$E_{outflow} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$

$\\$

Where $m_{1} h_{1}$ are the mass flow rate and the enthalpies at the inlet at a pressure of 3Mpa $\\$ ,

$m_{2} h_{2}$ are the mass flow rate and the enthalpies at the outlet 2 where we have a pressure of 500kpa respectively. $\\$ ,

and $m_{3} h_{3}$ are the mass flow rate and the enthalpies at the outlet 3 where we have a pressure of 50kpa respectively. $\\$ ,

We can now express write out the required equation by substituting the new expression for the energies $\\$

$m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$ $\\$

from the above equation, the unknown are the enthalpy values and the mass flow rate. $\\$

first let us determine the enthalpy values at the inlet and the out let using the Superheated water table. $\\$

It is more convenient to start from outlet 3 were we have a temperature $100^{0}C$ and pressure value of (50kpa or 0.05Mpa ). using double interpolation method on the superheated water table to determine the enthalpy value with careful calculation we have $\\$

$h_{3}$ = 2682.4 KJ/KG , at this point also from the table the entropy value , $s_{3}$ value is 7.6953 KJ/Kg.K. $\\$

Next we determine the enthalphy value at outlet 2. But in this case, we don't have a temperature value, hence we use the entrophy value since the entropy is constant at all inlet and outlet. $\\$

So, from the superheated water table again, at a pressure of 500kpa (0.5Mpa) and entropy value of 7.6953 KJ/Kg.K with careful interpolation we arrive at a enthalpy value of 3206.5KJ/Kg. $\\$

Finally for inlet one at a pressure of 3Mpa, interpolting with an entropy value of 7.6953KJ/Kg.K we arrive at enthalpy value of 3851.2KJ/Kg. $\\$

Now we determine the mass flow rate at each inlet and outlet. since mass must also be balance, i.e $m_{1} = m_{2} + m_{3}$ $\\$