B)

2. A F-22 Raptor has just climbed through an altitude of 9,874 m at 1,567 kph when a disk

BabaBlast [244]

The pressure difference across the sensor housing will be "95 kPa".

According to the question, the values are:

Altitude,

9874

Speed,

1567 kph

Pressure,

122 kPa

The temperature will be:

→ $T = 15.04-[0.00649(9874)]$

→ $= 15.04-64.082$

→ $= -49.042^{\circ} C$

now,

→ $P_o = 101.29[\frac{(-49.042+273.1)}{288.08} ]^{(5.256)}$

→ $= 27.074$

hence,

→ The pressure differential will be:

= $122-27$

= $95 \ kPa$

Thus the above solution is correct.

Learn more about pressure difference here:

brainly.com/question/15732832

3 0

2 years ago

You have a motor such that if you give it 12 Volt, it will eventually reach a steady state speed of 200 rad/s. If it starts from

Aleksandr [31]

Answer:

a) $\frac{Ws}{Es} = \frac{200}{1+1.2s}$

b) attached below

c) type zero system

d) k > $\frac{g}{200}$

e) The gain K increases above % error as the steady state speed increases

Explanation:

Given data:

Motor voltage = 12 v

steady state speed = 200 rad/s

time taken to reach 63.2% = 1.2 seconds

<u>a) The transfer function of the motor from voltage to speed</u>

let ; $\frac{K1}{1+St}$ be the transfer function of a motor

when i/p = 12v then steady state speed ( k1 ) = 200 rad/s , St ( time constant ) = 1.2 sec

hence the transfer function of the motor from voltage to speed

= $\frac{Ws}{Es} = \frac{200}{1+1.2s}$

<u>b) draw the block diagram of the system with plant controller and the feedback path </u>

attached below is the remaining part of the detailed solution

c) The system is a type-zero system because the pole at the origin is zero

d) ) k > $\frac{g}{200}$

7 0

3 years ago

A gas turbine operates with a regenerator and two stages of reheating and intercooling. Air enters this engine at 14 psia and 60

Rzqust [24]

Answer:

flow(m) = 7.941 lbm/s

Q_in = 90.5184 Btu/lbm

Q_out = 56.01856 Btu/lbm

Explanation:

Given:

- T_1 = 60 F = 520 R

- T_6 = 940 = 1400 R

- Heat ratio for air k = 1.4

- Compression ratio r = 3

- W_net,out = 1000 hp

Find:

mass flow rate of the air

rates of heat addition and rejection

Solution:

- Using ideal gas relation compute T_2, T_4, T_10:

T_2 = T_1 * r^(k-1/k)

T_2 = T_4 = T_10 = 520*3^(.4/1.4) = 711.744 R

- Using ideal gas relation compute T_7, T_5, T_9:

T_7 = T_6 * r^(-k-1/k)

T_7 = T_5 = T_9 = 1400*3^(-.4/1.4) = 1022.84 R

- The mass flow rate is obtained by:

flow(m) = W_net,out / 2*c_p*(1400-1022.84-711.744+520)

flow(m) = 1000*.7068 / 2*0.24*(1400-1022.84-711.744+520)

flow(m) = 7.941 lbm/s

- The heat input is as follows:

Q_in = c_p*(T_6 - T_5)

Q_in = 0.24*(1400 - 1022.84)

Q_in = 90.5184 Btu/lbm

- The heat output is as follows:

Q_out = c_p*(T_10 - T_1)

Q_out = 0.24*(711.744 - 520)

Q_out = 56.01856 Btu/lbm

5 0

3 years ago

Problem 4.079 SI A rigid tank whose volume is 3 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large

salantis [7]

Answer:

$Q_{cv} = -1007.86kJ$

Explanation:

Our values are,

State 1

$V=3m^3\\P_1=1bar\\T_1 = 295K$

We know moreover for the tables A-15 that

$u_1 = 210.49kJ/kg\\h_i = 295.17kJkg$

State 2

$P_2 =6bar\\T_2 = 296K\\T_f = 320K$

For tables we know at T=320K

$u_2 = 228.42kJ/kg$

We need to use the ideal gas equation to estimate the mass, so

$m_1 = \frac{p_1V}{RT_1}$

$m_1 = \frac{1bar*100kPa/1bar(3m^3)}{0.287kJ/kg.K(295k)}$

$m_1 = 3.54kg$

Using now for the final mass:

$m_2 = \frac{p_2V}{RT_2}$

$m_2 = \frac{1bar*100kPa/6bar(3m^3)}{0.287kJ/kg.K(320k)}$

$m_2 = 19.59kg$

We only need to apply a energy balance equation:

$Q_{cv}+m_ih_i = m_2u_2-m_1u_1$

$Q_{cv}=m_2u_2-m1_u_1-(m_2-m_1)h_i$

$Q_{cv} = (19.59)(228.42)-(3.54)(210.49)-(19.59-3.54)(295.17)$

$Q_{cv} = -1007.86kJ$

The negative value indidicates heat ransfer from the system

7 0

3 years ago

True or False: Galvanized steel pipe can withstand moderate freezing.

yaroslaw [1]

Answer:

True

Explanation:

Older galvanized steel pipes, which have a tendency to freeze, are a bit more forgiving and will likely not burst. They can withstand extreme cold and warm temperatures.

5 0

3 years ago