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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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The type of current that flows from the electrode across the arc to the work is called what?
Scrat [10]

Answer:

Direct current.

Explanation:

5 0
3 years ago
An article gave a scatter plot along with the least squares line of x = rainfall volume (m3) and y = runoff volume (m3) for a pa
kompoz [17]

Answer:

y = 0.834X - 1.58015

Slope = 0.8340 ; Intercept = - 1.5802

y = 40.9539

19.93

0.9765

Explanation:

X: Rainfall volume

6

12

14

16

23

30

40

52

55

67

72

81

96

112

127

Y : Runoff

4

10

13

14

15

25

27

48

38

46

53

72

82

99

100

The scatterplot shows a reasonable linear trend between the Rainfall volume and run off.

The estimated regression equation obtained using a linear regression calculator is :

y = 0.834X - 1.58015

y = Runoff ; x = Rainfall volume

Slope = 0.8340 ; Intercept = - 1.5802

Point estimate for Runoff, when, x = 51

y = 0.834X - 1.58015

y = 0.834(51) - 1.58015

y = 40.95385

y = 40.9539

d.)

Point estimate for standard deviation :

s = 5.145

σ = s * √n

σ = √15 * 5.145

= 19.93

e.)

r² = Coefficient of determination gives the proportion of explained variance in Runoff due to the regression line. From the model output, the r² value = 0.9765. Which means That about 97.65% Runoff is due to Rainfall volume.

7 0
3 years ago
A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 a
IrinaVladis [17]

Answer:

The original length of the specimen l_{o} = 104.7 mm

Explanation:

Original diameter d_{o} = 30 mm

Final diameter d_{1} = 30.04 mm

Change in diameter Δd = 0.04 mm

Final length l_{1} = 105.20 mm

Elastic modulus E = 65.5 G pa = 65.5 × 10^{3} M pa

Shear modulus G = 25.4 G pa = 25.4 × 10^{3} M pa

We know that the relation between the shear modulus & elastic modulus is given by

G = \frac{E}{2(1 + \mu)}

25.5 = \frac{65.5}{2 (1 + \mu)}

\mu = 0.28

This is the value of possion's ratio.

We know that the possion's ratio is given by

\mu = \frac{\frac{0.04}{30} }{\frac{change \ in \ length}{l_{o} } }

{\frac{change \ in \ length}{l_{o} } = \frac{\frac{0.04}{30} }{0.28}

{\frac{change \ in \ length}{l_{o} } = 0.00476

\frac{l_{1} - l_{o}  }{l_{o}  } = 0.00476

\frac{l_{1} }{l_{o} } = 1.00476

Final length l_{o} = 105.2 m

Original length

l_{o} = \frac{105.2}{1.00476}

l_{o} = 104.7 mm

This is the original length of the specimen.

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3 years ago
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Do you have a picture of the question?
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Describe the engineering design process in your own words.
MrMuchimi

Answer:

Explanation:

8 0
3 years ago
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