Answer: B. thermocouple measures temperatures at the tip and the thermistor at the dimple.
Explanation:
A thermistor is a temperature-sensitive resistor, whilst a thermocouple generates a voltage proportional to the temperature. Thermocouples can work at much higher temperatures than thermistors. They are commonly used for temperature control in heating systems.
Answer:
Rate of Entropy =210.14 J/K-s
Explanation:
given data:
power delivered to input = 350 hp
power delivered to output = 250 hp
temperature of surface = 180°F
rate of entropy is given as
T = 180°F = 82°C = 355 K
Rate of heat = (350 - 250) hp = 100 hp = 74600 W
Rate of Entropy
Answer:
79 kW.
Explanation:
The equation for enthalpy is:
H2 = H1 + Q - L
Enthalpy is defined as:
H = G*(Cv*T + p*v)
This is specific volume.
The gas state equation is:
p*v = R*T (with specific volume)
The specific gas constant for air is:
287 K/(kg*K)
Then:
T1 = 60 + 273 = 333 K
T2 = 200 + 273 = 473 K
p1*v1 = 287 * 333 = 95.6 kJ/kg
p2*v2 = 287 * 473 = 135.7 kJ/kg
The Cv for air is:
Cv = 720 J/(kg*K)
So the enthalpies are:
H1 = 0.8*(0.72 * 333 + 95.6) = 268 kW
H2 = 0.8*(0.72 * 473 + 135.7) = 381 kW
Ang the heat is:
Q = 34 kW
Then:
H2 = H1 + Q - L
381 = 268 + 34 - L
L = 268 + 34 - 381 = -79 kW
This is the work from the point of view of the air, that's why it is negative.
From the point of view of the machine it is positive.
It was determined by the forensic engineers that the historic collapse of the Minneapolis Interstate Bridge was due to design and construction flaws. It led to the collapse of the structure.
<h3>What is the Minneapolis Interstate Bridge?</h3>
The Minneapolis Interstate Bridge is a well-known bridge that was constructed in Minneapolis city in 1967.
It has been demonstrated that the collapse of this bridge constructed in Minneapolis was due to its inadequate load capacity.
In architecture, it is fundamental to measure the amount of weight that a structure (in this case, a bridge) can sustain in a given period of time.
Learn more about the bridge construction here:
brainly.com/question/24686952
Answer:
heat loss per 1-m length of this insulation is 4368.145 W
Explanation:
given data
inside radius r1 = 6 cm
outside radius r2 = 8 cm
thermal conductivity k = 0.5 W/m°C
inside temperature t1 = 430°C
outside temperature t2 = 30°C
to find out
Determine the heat loss per 1-m length of this insulation
solution
we know thermal resistance formula for cylinder that is express as
Rth = 
.................1
here r1 is inside radius and r2 is outside radius L is length and k is thermal conductivity
so
heat loss is change in temperature divide thermal resistance
Q = 
Q = 
Q = 4368.145 W
so heat loss per 1-m length of this insulation is 4368.145 W
