Ask question

Ask question

All categories

3 years ago

7

The structure of a house is such that it loses heat ata rate of 3800 kJ/h per8C difference between the indoors andoutdoors. A he

at pump that requires a power input of 4 kWis used to maintain this house at 248C. Determine the lowestoutdoor temperature for which the heat pump can meet theheating requirements of this house.

1 answer:

marysya [2.9K]3 years ago

4 0

Answer:

The lowest outdoor temperature for which the heat pump can meet the requirement of the house is -9.6°C

Explanation:

The coefficient of performance of a heat pump, COP, is given by the relation;

$COP_{hp} = \dfrac{T_2}{T_2 - T_1} = \dfrac{Q_1\times (T_2 - T_1)}{\Sigma W}$

The rate of heat loss $\dot Q_1$ = 3800 kJ/h = 3800 ×1/60 × 1/60 kJ/s = 1.06 kJ/s

The power input = 4 kW = 4 kJ/s

The temperature indoors = 24 + 273.15 = 297.15 K

Therefore, we have;

$COP_{hp} = \dfrac{297.15}{297.15- T_1} = \dfrac{1.06 \times (297.15- T_1)}{4}$

4×297.15= (297.15- T₁) × 1.06×(297.15- T₁)

Which gives;

18/19·T₁²-627.32·T₁+92014.97=0

(T₁ -263.59)·(T₁ - 330.71) = 0

The solutions are;

T₁ = 263.59 K or T₁ = 330.71 K

The lowest temperature = 263.59 K which is -9.6°C

The lowest temperature = -9.6°C.

You might be interested in

What is the tool used to take off a wheel nut

pishuonlain [190]

Answer:

A socket wrench

Explanation:

My dad is a mechanic

8 0

4 years ago

Read 2 more answers

What’s the number of gold atoms in a nanogram? a picogram?

zvonat [6]

Answer :

The number of gold atoms in nanogram is, $3.057\times 10^{12}$

The number of gold atoms in picogram is, $3.057\times 10^{9}$

Explanation :

As we know that the molar mass of gold is, 196.97 g/mole. That means, 1 mole of gold has 196.97 grams of mass of gold.

As we know that,

1 mole contains $6.022\times 10^{23}$ number of atoms.

First we have to determine the number of gold atoms in a nanogram.

As, 196.97 grams of gold contains $6.022\times 10^{23}$ number of gold atoms

And, 1 grams of gold contains $\frac{1g}{196.97g}\times (6.022\times 10^{23})$ number of gold atoms

So, $10^{-9}$ nanograms of gold contains $\frac{1g}{196.97g}\times (10^{-9})\times (6.022\times 10^{23})=3.057\times 10^{12}$ number of gold atoms

The number of gold atoms in nanogram is, $3.057\times 10^{12}$

Now we have to determine the number of gold atoms in a picogram.

As, 196.97 grams of gold contains $6.022\times 10^{23}$ number of gold atoms

And, 1 grams of gold contains $\frac{1g}{196.97g}\times (6.022\times 10^{23})$ number of gold atoms

So, $10^{-12}$ picograms of gold contains $\frac{1g}{196.97g}\times (10^{-12})\times (6.022\times 10^{23})=3.057\times 10^{9}$ number of gold atoms

The number of gold atoms in picogram is, $3.057\times 10^{9}$

8 0

3 years ago

A stainless steel (AISI 304) tube used to transport a chilled pharmaceutical has an inner diameter of 36 mm and a wall thickness

Bingel [31]

Answer:

a) heat gain per unit tube length = $\frac{23-6}{1.35} = 12.6W/m$

b) heat gain per unit tube length = $\frac{23-6}{2.20} = 7.7W/m$

Explanation:

Assumptions:

Constant properties
Steady state conditions
Negligible effect of radiation
Negligible constant resistance between tube and insulation
one dimensional radial conduction

a) What is the heat gain per unit tube length

$R_{conv,i}'=\frac{1}{2\pi r_1h_i}$

$d_1=36mm$ Therefore $r_1=\frac{d_1}{2} =36/2=18mm=18*10^{-3}$

$r_2=2mm=2*10^{-3}m$

$k_{st}=14.2W/m.k$

$h_o=6W/m^2$

$h_i=400W/m^2$

$R_{conv,i}'=\frac{1}{2\pi * 1.8*10^{-3}*400}= 0.221m.K/W$

$R_{cond,st}'=\frac{ln(r_2/r_1)}{2\pi k_{st}} =\frac{ln(20/18)}{2\pi *14.2} =1.18*10^{-3}m.K/W$

$R_{conv,o}'=\frac{1}{2\pi r_2h_0}=\frac{1}{2\pi *2*10^{-3}*6}=1.33m.K/W$

$R_{tot}'=R_{conv,i}'+R_{cond,st}'+R_{conv,o}'=0.221+(1.18*10^{-3})+1.33=1.35m.K/W$

heat gain per unit tube length = $\frac{23-6}{1.35} = 12.6W/m$

b) What is the heat gain per unit length if a 10-mm-thick layer of calcium silicate insulation (k_ins = 0.050 W/m.K) is applied to the tube

$r_3=r_1+r_2+10mm=30mm=0.03m$

$R_{conv,i}'$ and $R_{cond,st}'$ are the same, but $R_{conv,o}'$ changes.

Therefore:

$R_{conv,o}'=\frac{1}{2\pi r_3h_0} = \frac{1}{2\pi *0.03*6}=0.88m.K/W$

$R_{conv,ins}'=\frac{ln(r_3/r_)}{2\pi k_{ins}} =\frac{ln(30/20)}{2\pi *0.05} =1.29m.K/W$

The total resistance $R_{tot}'=R_{conv,i}'+R_{cond,st}'+R_{conv,ins}'+R_{conv,o}'=0.221+(1.18*10^{-3})+1.29+0.88=2.20m.K/W$

heat gain per unit tube length = $\frac{23-6}{2.20} = 7.7W/m$

8 0

3 years ago

Read 2 more answers

Daniel Wiseman, a scientist for Gres-Trans Corp., wants to determine if the flow rate of a particular material changes with diff

kiruha [24]

Answer: C. There is no relationship

Explanation:

At the various temperatures, the flow rates do not seem to show any relation as similar temperatures can yield different flow rates.

The flow rates neither appear to generally increase nor decrease as a result of a decrease or an increase in temperature which means that there is a lack of a positive, negative and curvilinear relationship.

There is simply no relationship.

4 0

3 years ago

What is a²+b²=c²????<br> We just learned about it

Furkat [3]

Answer: the answer to you’re problem is Pythagorean Theorem

a²+b²=c² Is the Pythagorean Theorem

Explanation:

5 0

4 years ago