Answer:
<em><u>Door knob</u></em>
Explanation:
When the knob on one side of the door is turned, the shaft retracts the spring-loaded latch that holds the door closed. Without the knob in place, more force would be needed to turn the shaft and retract the latch.
We can find the change in the enthalpy through the tables A5 for Saturated water, pressure table.
For 1bar=1000kPa:




Replacing,



With the specific volume we know can calculate the mass flow, that is


Then the heat required in input is,



With the same value required of 15000m^3/h, we can calculate the velocity of the water, that is given by,



Finally we can apply the steady flow energy equation, that is

Re-arrange for Q,




We can note that consider the Kinetic Energy will decrease the heat input.
Answer: u slap birds
Explanation: u order cheeseburgers with n cheese
Answer: the thermal conductivity of the sample is 22.4 W/m . °C
Explanation:
We already know that the thermal conductivity of a material is to be determined by ensuring one-dimensional heat conduction, and by measuring temperatures when steady operating conditions are reached.
ASSUMPTIONS
1. Steady operating conditions exist since the temperature readings do not change with time.
2. Heat losses through lateral surfaces are well insulated, and thus the entire heat generated by the heater is conducted through the samples.
3. The apparatus possess thermal symmetry
ANALYSIS
The electrical power consumed by resistance heater and converted to heat is:
Wₐ = V<em>I</em> = ( 110 V ) ( 0.4 A ) = 44 W
Q = 1/2Wₐ = 1/2 ( 44 A )
Now since only half of the heat generated flows through each samples because of symmetry. Reading the same temperature difference across the same distance in each sample also confirms that the apparatus possess thermal symmetry. The heat transfer area is the area normal to the direction of heat transfer. which is the cross- sectional area of the cylinder in this case; so
A = 1/4πD² = 1/3 × π × ( 0.05 m )² = 0.001963 m²
Now Note that, the temperature drops by 15 degree Celsius within 3 cm in the direction of heat flow, the thermal conductivity of the sample will be
Q = kA ( ΔT/L ) → k = QL / AΔT
k = ( 22 W × 0.03 m ) / (0.001963 m² × 15°C )
k = 22.4 W/m . °C