Sinks must be used for the correct intended purpose to prevent

A petrol engine produces 20 hp using 35 kW of heat transfer from burning fuel. What is its thermal efficiency, and how much powe

hoa [83]

Answer:

efficiency =42.62%

AMOUNT OF POWER REJECTED IS 20.080 kW

Explanation:

given data:

power 20 hp

heat energy = 35kW

power production = 20 hp = 20* 746 W = 14920 Watt [1 hp =746 watt]

$efficiency = \frac{power}{heat\ required}$

$efficiency = \frac{14920}{35*10^3}$

$= 0.4262*10^100$

=42.62%

b) $heat\ rejected = heat\ required - amount\ of\ power\ generated$

$= 35*10^3 - 14920$

= 20.080 kW

AMOUNT OF POWER REJECTED IS 20.080 kW

5 0

3 years ago

The European Space Agency launched a probe called Rosetta in March 2004. In August 2014, Rosetta reached its destination: a co

Art [367]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The of the lander would be 75.66 pounds

Explanation:

From the question we are given that the

Weight of the Philae on Earth is = 220 pound-mass

gravity of Mars is = 3.71 m/ $s^{2}$

Weight of the Philae on Mars is = ?

Now Mass is quantity of matter in an object so it is always constant for a particular object

Generally Weight = Mass × Gravity :

$\frac{Weight Of Earth}{Gravity Of Earth} = \frac{Weight Of Mars}{Gravity Of Mars}$

Hence

Weight of the Philae on Mars is = $\frac{Weight Of Earth *Gravity Of Mars}{Gravity Of Earth}$

$=\frac{200 *3.71}{9.81}$

$= 75.66 pounds$

6 0

2 years ago

A poundal is the force required to accelerate a mass of 1 lbm at a rate of 1 ft/(s^2). Determine the acceleration of an object o

storchak [24]

Answer:

See it in the pic

Explanation:

See it in the pic

5 0

2 years ago

Calculate the availability of a system where the mean time between failures is 900 hours and the mean time to repair is 100 hour

Debora [2.8K]

Answer:

The availability of system will be 0.9

Explanation:

We have given mean time of failure = 900 hours

Mean time [to repair = 100 hour

We have to find availability of system

Availability of system is given by $\frac{mean\time\ of\ failure}{mean\ time\ of\ failure+mean\ time\ to\ repair}$

So availability of system $=\frac{900}{900+100}=\frac{900}{1000}=0.9$

So the availability of system will be 0.9

7 0

2 years ago

A weighted, frictionless piston-cylinder device initially contains 5.25 kg of R134a as saturated vapor at 500 kPa. The container

kykrilka [37]

Answer:

-6.326 KJ/K

Explanation:

A) the entropy change is defined as:

$delta S_{12}=\int\limits^2_1 \, \frac{dQ}{T}$

In an isobaric process heat (Q) is defined as:

$Q= m*Cp*dT$

Replacing in the equation for entropy

$delta S_{12}=\int\limits^2_1 \frac{m*Cp*dT}{T}$

m is the mass and Cp is the specific heat of R134a. We can considerer these values as constants so the expression for entropy would be:

$delta S_{12}= m*Cp*\int\limits^2_1 \frac{ dT }{T}$

Solving the integral we get the expression to estimate the entropy change in the system

$delta S_{12}= m*Cp *ln(\frac{T_{2}}{T_{1}})$

The mass is 5.25 Kg and Cp for R134a vapor can be consulted in tables, this value is $0.85\frac{kJ}{Kg*K}$

We can get the temperature at the beginning knowing that is saturated vapor at 500 KPa. Consulting the thermodynamic tables, we get that temperature of saturation at this pressure is: 288.86 K

The temperature in the final state we can get it from the heat expression, since we know how much heat was lost in the process (-976.71 kJ). By convention when heat is released by the system a negative sign is used to express it.

$Q= m*Cp*dT$

With $dt=T_{2}-T_{1}$ clearing for T2 we get:

$T_{2}=\frac{Q}{m*Cp}+T1= \frac{-976.71kJ}{5.25Kg*0.85\frac{kJ}{Kg*K}}+288.86 K =69.98 K$

Now we can estimate the entropy change in the system

$delta S_{12}= m*Cp*ln(\frac{T_{2}}{T_{1}})= 5.25Kg*0.85\frac{kJ}{Kg*K}*ln(\frac{69.98}{288.861})= -6.326\frac{kJ}{K}$

The entropy change in the system is negative because we are going from a state with a lot of disorder (high temperature) to one more organize (less temperature. This was done increasing the entropy of the surroundings.

b) see picture.

3 0

3 years ago