What musical form do you hear in Candle in the Wind?

Find the total amount of heat in Q lost through a wall 10' by 18' , with R value from q. 1. Inside temperature is 70 degrees F w

marissa [1.9K]

Answer:

Just think

Explanation:

6 0

4 years ago

Read 2 more answers

A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu

Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

Negligible heat loss through the insulation
Steady state system
One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

$k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}$

During steady state

$\frac{dT}{dt}$ = 0 which gives $k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0$

From which we have $\frac{d^{2}T }{dx^{2} } = -\frac{q'_{G}}{k}$

Considering the boundary condition at x =0 where there is no heat loss

$\frac{dT}{dt}$ = 0 also at the other end of the plane wall we have

$-k\frac{dT }{dx } =$ hc (T - T∞) at point x = L

Integrating the equation we have

$\frac{dT }{dx } = \frac{q'_{G}}{k} x+ C_{1}$ from which C₁ is evaluated from the first boundary condition thus

0 = $\frac{q'_{G}}{k} (0)+ C_{1}$ from which C₁ = 0

From the second integration we have

$T = -\frac{q'_{G}}{2k} x^{2} + C_{2}$

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

$-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k} + C_{2}-T∞)$ → C₂ = $q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$

T(x) = $\frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$ and T(x) = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )$

∴ Tmax → when x = 0 = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))$

Substituting the values we get

T = 167 ° C

4 0

4 years ago

According to the rules of dimension how should written notes appear

Arturiano [62]

Explanation:

Leaders for notes should be straight, not curved, and point to the center of circular views of holes wherever possible. Leaders should slope at 45°, 30° or 60° with horizontal but may be made at any convenient angle except vertical or horizontal.

7 0

3 years ago

A student lab group is brainstorming the design of an experiment that uses an ammeter (measures current) and different resistors

givi [52]

Answer:

Put a 10.0-ohm resistor in the circuit. Measure the current in the circuit. Replace the 10.0-ohm resistor with a 20.0-ohm resistor. Measure the new current. Continue replacing the resistor with a different resistor of known resistance. Measure the current for each resistor. Record all data.

Explanation:

The only design that has resistance varying with everything else remaining the same is the first design. That would be what you'd want to do if you're exploring the effect of resistance on current.

3 0

3 years ago

A metal rod is 0.600 m in length at a temperature of 15.0∘C. When you raise its temperature to 37.0∘C, its length increases by 0

Katyanochek1 [597]

Answer:

The coefficient of linear expansion of the metal is ∝ = 2.91 x 10⁻⁵ °C⁻¹.

Explanation:

We know that Linear thermal expansion is represented by the following equation

Δ L = L x ∝ x Δ T ---- (1)

where Δ L is the change in length, L is for length, ∝ is the coefficient of linear expression and Δ T is the change in temperature.

Given that:

L = 0.6 m

T₁ = 15° C

T₂ = 37° C

Δ L = 0.28 mm

∝ = ?

Solution:

We know that Δ T = T₂ ₋ T₁

Putting the values of T₁ and T₂ in above equation, we get

Δ T = 37 - 15

Δ T = 22 °C

Also Δ L = 0.28 mm

Converting the mm to m

Δ L = 0.00028 m

Putting the values of Δ T, Δ L, L in equation 1, we get

0.00028 = 0.6 x ∝ x 22

Rearranging the equation, we get

∝ = 0.00028 / (0.6 x 16)

∝ = 0.00028 / 13.2

∝ = 2.12 x 10⁻⁵ °C⁻¹

4 0

4 years ago