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Nata [24]
3 years ago
12

A house is losing heat at a rate of 1700 kJ/h per °C temperature difference between the indoor and the outdoor temperatures. Exp

ress the rate of heat loss from this house per K, °F, and R difference between the indoor and the outdoor temperature. The rate of heat loss from this house per K difference is kJ/h. The rate of heat loss from this house per °F difference is kJ/h. The rate of heat loss from this house per R difference is kJ/h.
Engineering
1 answer:
Furkat [3]3 years ago
5 0

Answer:

1700kJ/h.K

944.4kJ/h.R

944.4kJ/h.°F

Explanation:

Conversions for different temperature units are below:

1K = 1°C + 273K

1R = T(K) * 1.8

= (1°C + 273) * 1.8

1°F = (1°C * 1.8) + 32

Q/delta T = 1700kJ/h.°C

T (K) = 1700kJ/h.°C

= 1700kJ/K

T (R) = 1700kJ/h.°C

= 1700kJ/h.°C * 1°C/1.8R

= 944.4kJ/h.R

T (°F) = 1700kJ/h.°C

= 1700kJ/h.°C * 1°C/1.8°F

= 944.4kJ/h.°F

Note that arithmetic operations like subtraction and addition of values do not change or affect the value of a change in temperature (delta T) hence, the arithmetic operations are not reflected in the conversion. Illustration: 5°C - 3°C

= 2°C

(273+5) - (273+3)

= 2 K

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A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into t
Nataly [62]

Answer:

A) A(t) = 10(100 - t) + c(100 - t)²

B) Tank will be empty after 100 minutes.

Explanation:

A) The differential equation of this problem is;

dA/dt = R_in - R_out

Where;

R_in is the rate at which salt enters

R_out is the rate at which salt exits

R_in = (concentration of salt in inflow) × (input rate of brine)

We are given;

Concentration of salt in inflow = 2 lb/gal

Input rate of brine = 5 gal/min

Thus;

R_in = 2 × 5 = 10 lb/min

Due to the fact that the solution is pumped out at a faster rate, thus it is reducing at the rate of (5 - 10)gal/min = -5 gal/min

So, after t minutes, there will be (500 - 5t) gallons in the tank

Therefore;

R_out = (concentration of salt in outflow) × (output rate of brine)

R_out = [A(t)/(500 - 5t)]lb/gal × 10 gal/min

R_out = 10A(t)/(500 - 5t) lb/min

So, we substitute the values of R_in and R_out into the Differential equation to get;

dA/dt = 10 - 10A(t)/(500 - 5t)

This simplifies to;

dA/dt = 10 - 2A(t)/(100 - t)

Rearranging, we have;

dA/dt + 2A(t)/(100 - t) = 10

This is a linear differential equation in standard form.

Thus, the integrating factor is;

e^(∫2/(100 - t)) = e^(In(100 - t)^(-2)) = 1/(100 - t)²

Now, let's multiply the differential equation by the integrating factor 1/(100 - t)².

We have;

So, we ;

(1/(100 - t)²)(dA/dt) + 2A(t)/(100 - t)³ = 10/(100 - t)²

Integrating this, we now have;

A(t)/(100 - t)² = ∫10/(100 - t)²

This gives;

A(t)/(100 - t)² = (10/(100 - t)) + c

Multiplying through by (100 - t)²,we have;

A(t) = 10(100 - t) + c(100 - t)²

B) At initial condition, A(0) = 0.

So,0 = 10(100 - 0) + c(100 - 0)²

1000 + 10000c = 0

10000c = -1000

c = -1000/10000

c = -0.1

Thus;

A(t) = 10(100 - t) + -0.1(100 - t)²

A(t) = 1000 - 10t - 0.1(10000 - 200t + t²)

A(t) = 1000 - 10t - 1000 + 20t - 0.1t²

A(t) = 10t - 0.1t²

Tank will be empty when A(t) = 0

So, 0 = 10t - 0.1t²

0.1t² = 10t

Divide both sides by 0.1t to give;

t = 10/0.1

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Answer:

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