All categories

3 years ago

14. A large car fire presents the possibility of

Engineering

2 answers:

dexar [7]3 years ago

5 0

Answer:

Both of the above

Explanation:

Wittaler [7]3 years ago

4 0

Answer:

the answer would possibly be A. if not then C.

You might be interested in

if both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will

egoroff_w [7]

If both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will: a) increase during a climb.

<h3>What is a ram air input?</h3>

A ram air input can be defined as an air intake system which is designed and developed to use the dynamic air pressure that is created due to vehicular motion, or ram pressure, in order to increase the static air pressure within the intake manifold of an internal combustion engine of an automobile.

This ultimately implies that, a ram air input allows a greater mass-flow of air through the engine of an automobile, thereby, increasing the engine's power.

In conclusion, indicated airspeed will increase during a climb when both the ram air input and drain hole of the pitot system become blocked.

Read more on pilots here: brainly.com/question/10381526

#SPJ1

Complete Question:

If both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will

a) increase during a climb

b) decrease during a climb

c) remain constant regardless of altitude change

6 0

2 years ago

Steam enters a turbine steadily at 7 MPa and 600°C with a velocity of 60 m/s and leaves at 25 kPa with a quality of 95 percent.

Rufina [12.5K]

Answer:

a) $\dot m = 16.168\,\frac{kg}{s}$ , b) $v_{out} = 680.590\,\frac{m}{s}$ , c) $\dot W_{out} = 18276.307\,kW$

Explanation:

A turbine is a steady-state devices which transforms fluid energy into mechanical energy and is modelled after the Principle of Mass Conservation and First Law of Thermodynamics, whose expressions are described hereafter:

Mass Balance

$\frac{v_{in}\cdot A_{in}}{\nu_{in}} - \frac{v_{out}\cdot A_{out}}{\nu_{out}} = 0$

Energy Balance

$-q_{loss} - w_{out} + h_{in} - h_{out} = 0$

Specific volumes and enthalpies are obtained from property tables for steam:

Inlet (Superheated Steam)

$\nu_{in} = 0.055665\,\frac{m^{3}}{kg}$

$h_{in} = 3650.6\,\frac{kJ}{kg}$

Outlet (Liquid-Vapor Mix)

$\nu_{out} = 5.89328\,\frac{m^{3}}{kg}$

$h_{out} = 2500.2\,\frac{kJ}{kg}$

a) The mass flow rate of the steam is:

$\dot m = \frac{v_{in}\cdot A_{in}}{\nu_{in}}$

$\dot m = \frac{\left(60\,\frac{m}{s} \right)\cdot (0.015\,m^{2})}{0.055665\,\frac{m^{3}}{kg} }$

$\dot m = 16.168\,\frac{kg}{s}$

b) The exit velocity of steam is:

$\dot m = \frac{v_{out}\cdot A_{out}}{\nu_{out}}$

$v_{out} = \frac{\dot m \cdot \nu_{out}}{A_{out}}$

$v_{out} = \frac{\left(16.168\,\frac{kg}{s} \right)\cdot \left(5.89328\,\frac{m^{3}}{kg} \right)}{0.14\,m^{2}}$

$v_{out} = 680.590\,\frac{m}{s}$

c) The power output of the steam turbine is:

$\dot W_{out} = \dot m \cdot (-q_{loss} + h_{in}-h_{out})$

$\dot W_{out} = \left(16.168\,\frac{kg}{s} \right)\cdot \left(-20\,\frac{kJ}{kg} + 3650.6\,\frac{kJ}{kg} - 2500.2\,\frac{kJ}{kg}\right)$

$\dot W_{out} = 18276.307\,kW$

6 0

3 years ago

Water is pumped steadily through a 0.10-m diameter pipe from one closed pressurized tank to another tank. The pump adds 4.0 kW o

jekas [21]

Complete Question

Complete Question is attached below.

Answer:

$V'=5m/s$

Explanation:

From the question we are told that:

Diameter $d=0.10m$

Power $P=4.0kW$

Head loss $\mu=10m$

$\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+H_m=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+\mu$

$\frac{300*10^3}{\rho g}+35+Hm=\frac{500*10^3}{\rho g}+15+10$

$H_m=(\frac{200*10^3}{1000*9.8}-10)$

$H_m=10.39m$

Generally the equation for Power is mathematically given by

$P=\rho gQH_m$

Therefore

$Q=\frac{P}{\rho g H_m}$

$Q=\frac{4*10^4}{1000*9.81*10.9}$

$Q=0.03935m^3/sec$

Since

$Q=AV'$

Where

$A=\pi r^2\\A=3.142 (0.05)^2$

$A=7.85*10^{-3}$

Therefore

$V'=\frac{0.03935m^3/sec}{7.85*10^{-3}}$

$V'=5m/s$

5 0

3 years ago

A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 82.4 M P a m. If the plate is expos

bekas [8.4K]

Answer:

Answer for the question is given in the attachment.

Explanation:

3 0

3 years ago

Read 2 more answers

In a reversing 2-high mill, a series of cold rolling process is used to reduce the thickness of a plate from 45mm down to 20mm.

podryga [215]

Answer:

Explanation:

Given

Initial Thickness=45 mm

Final thickness=20 mm

Roll diameter=600 mm

Radius(R)=300 mm

coefficient of friction between rolls and strip ( $\nu$ )=0.15

maximum draft $(d_{max})=\nu ^2R$

$=0.15^2\times 300=6.75 mm$

Minimum no of passes $=\frac{45-20}{6.75}=3.70\approx 4$

(b)draft per each pass

$d=\frac{Initial\ Thickness-Final\ Thickness}{min.\ no.\ of\ passes}$

$d=\frac{45-20}{4}=6.25 mm$

5 0

3 years ago