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X 4 y X +6 2 (Y +6) by Substitution

The document contains 4 problems related to mechanical engineering concepts. Problem 1 asks for the age of Harry given information about his and John's ages now and in 6 years. Problem 2 asks for the horizontal pull needed to hold a girder at an angle. Problem 3 asks for the work output of a steam turbine given input and output steam conditions. Problem 4 asks for the thermal conductivity of a material given test panel dimensions and heat transferred over time and temperature difference.

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Eugene Perez
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1K views6 pages

X 4 y X +6 2 (Y +6) by Substitution

The document contains 4 problems related to mechanical engineering concepts. Problem 1 asks for the age of Harry given information about his and John's ages now and in 6 years. Problem 2 asks for the horizontal pull needed to hold a girder at an angle. Problem 3 asks for the work output of a steam turbine given input and output steam conditions. Problem 4 asks for the thermal conductivity of a material given test panel dimensions and heat transferred over time and temperature difference.

Uploaded by

Eugene Perez
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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Laserna, Ray Joseph D.

ME - 5202

Problem No. 7

John is four times as old as Harry. In six years, John will be twice as old as

Harry. What is the age of Harry now? (April 2004)

A. 2 B. 3 C. 4 D.5

Required: Harry’s age

Solution:

Name Present In 6 years

John x x+6

Harry y y+6

x=4 y

x +6=2( y +6)

by substitution,

4 y +6=2 ( y +6 )

y=3

Laserna, Ray Joseph D.


ME – 5202

Problem No. 66

A girder weighing 18,000 lb is suspended by a cable 100ft. long. What horizontal

pull is necessary to hold it 8 ft from the vertical position?

A. 703.23 lb B. 1444.63 lb C. 901.67 lb D. 1082.34 lb

Given: w = 18000 lb

L = 100 ft

y = 8 ft

Required: H =? (Horizontal Pull)

Solution:

92
sin ɵ=
100

ɵ=66.92608193 °

Σ F x =0=H −Tcos ɵ

Σ F y =0=T sin ɵ−W

W =Tsinɵ

18 000 lb=T sin 66.92608193 °

T =19565.21739 lb
Solving for H,

H=Tcos 66.92608193 °

H=7667.96789 lb

Laserna, Ray Joseph D.

ME – 5202
Problem no. 105

A steam turbine with 90% stage efficiency receives steam of 5 MPa and 500 °C

and exhausts at 50 kPa. Determine the turbine work:

At 5 MPa and 500 °C h1 = 3660.3 s1 = 7.1218

At 50 kPa hf = 340.49 hfg = 2305.4 sf = 1.091 sfg = 6.5029

A. 117 kJ/kg B. 132 kJ/kg C. 964.60 kJ/kg D. 143 kJ/kg

Given:

At 5 MPa and 500 °C h1 = 3660.3 s1 = 7.1218

At 50 kPa hf = 340.49 hfg = 2305.4 sf = 1.091 sfg = 6.5029

Required: Wt =?

Solution:

W T =( h 1−h2 )∗η

s1=s 2=7.1218

7.1218=1.091+ x∗6.5029

x=0.9274016208

h2 =340.49+0.9274016208∗2305.4

h2 =2478.521697

W T =( h 1−h2 )∗η

W T =( 3660.3−2478.521697 )∗0.9
W T =1063.600473

Laserna, Ray Joseph D.

ME – 5202

Problem No. 144


Determine the thermal conductivity (W/m-°C) of a material that uses a 4 m 2 test

panel, 25 mm thick with a temperature of 20 °F between surfaces. During the 4 hours of

test period, the heat transmitted is 500 kJ.

A. 0.0432 B. 0.0723 C. 0.0321 D. 0.0195

Given:

A=4 m2

t=25 mm

ΔT = 20° (°F or °C)

Q = 500 kJ @ t= 4 hrs

Required: Thermal Conductivity (k)

Solution:

Q∗t
k=
A∗ΔT

500 kJ∗0.025 m
k=
hr∗3600 s
4 m2∗20 °∗4
hr

W
k =1.085069444 x 10−5
m−° C

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